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KSEEB Class 8 Maths Solutions For Chapter 6 Algebraic Expressions And Identities Points To Remember

Expressions are formed from variables and constants

Constant: A symbol having a fixed numerical 2 value. Example: 2, 2/3, 2,1, etc.

Variable: A symbol that takes various numerical values.

Example: x, y, z, etc.

Read and Learn More KSEEB Solutions for Class 8 Maths

Algebraic expression: A combination of constants and variables connected by the sign +, —, x, and ÷ is called Algebraic expression.

Terms are added to form expressions. Terms themselves are formed as product of factors.

Expressions that contain exactly one, two and three terms are called monomials, binomials, and trinomials respectively.

In general, any expression containing one or more terms with non-zero coefficients (and with variables having non-negative exponents) is called a polynomial.

Like terms are formed from the same variables and the powers of these variables are the same, too.
Co-efficient of like terms need not be the same.

While adding (or subtracting) polynomials, first look for like terms and add (or subtract) them, then handle the unlike terms.

There are number of situations in which we need to multiply algebraic expressions for example, in finding the area of a rectangle, the sides of which are given as expressions.

Monomial: An expression containing only one term Example: -3, 4x, 3xy, etc.

Binomial: An expression containing two terms
Example: 2x – 3, 4x + 3y, xy – 4 etc.

Trinomial: An expression containing three terms.
Example: \(2x^2\) + 3xy + 9, 3x + 2y + 5z etc.

Polynomial: In general, any expression containing one or more terms with non-zero coefficients (and with variables having non-negative exponents). A polynomial may contain any number of terms, one or more than one.

A monomial multiplied by a monomial always gives a monomial.

While multiplying a polynomial by a monomial, we multiply every term in the polynomial by the monomial.

In carrying out the multiplication of a polynomial by a binomial (or trinomial), we multiply term by term. i.e. every term of the polynomial is multiplied by every term in the binomial or (trinomial). Note that in such multiplication, we may get terms in the product which are like and have to be combined.

An identity is an equality, which is true for all values of the variables in the equality. On the other hand, an equation is true only for certain values of its variables. An equation is not an identity.

The following are the standard identities.

• \((a + b)^2 = a^2 + 2ab + b^2\)
• \((a – b)^2 = a^2 – 2ab + b^2\)
• \((a + b)(a – b) = a^2 – b^2\)
• \((x + a)(x + b) = x^2 +(a + b)x + ab\)

The above four identities are useful in carrying out squares and products of algebraic expressions, they also allow easy alternative methods to calculate products of numbers and so on.

Coefficients: In the term of an expression any of the factors with the sign of the term is called the coefficient of the product of the other factors.

Terms: Various parts of an algebraic expression that are separated by + and – signs.

Example : The expression 4x + 5 has two terms 4X and 5.

1) Constant term: A term of expression having no lateral factor.

2) Like term: The term having the same literal factors.

Example : 2xy and -4xy are like terms.

3) Unlike terms: The terms having different literal factors.

Example: \(4x^2\) and 3xy are unlike terms.

Factors: Each term in an algebraic expression is a product of one or more number (S) and / or literals. These numbers (S) and/or literal (S) are known as the factor of that term. A constant factor is called numerical factor, while a variable factor is known as a literal factor. The term 4x is the product of its factors 4 and x.

Algebraic Expressions And Identities Solutions KSEEB Class 8 Maths Exercise 6.1

1. Identify the terms, their co-efficients for each of the following expressions.

1) \(5xyz^2 —3zy\)

Solution:

Terms : \(5xyz^2\) ,—3zy

Co – efficients : 5, -3

2) \(1 + x + x^2\)

Solution:

Terms : 1, x, \( x^2\)

Co-efficients: 1,1,1

3) \( 4x^2y^2 – 4x^2 y^2 z^2 + z^2\)

Solution:

Terms : \(4x^2y^2\) , \(-4x^2 y^2 z^2\) , \(z^2\)

Co – efficients : 4, -4, 1

4) 3 – pq + qr – rp

Solution:

Terms : 3 — pq, qr, rp

Co – efficients : 3, -1, 1,-1

5)\( \frac{x}{2}+\frac{y}{2}-x y\)

Solution:

Terms: x/2, y/2, -xy

Coefficients: 1/2, 1/2, -1

6) 0.3a -0.6ab + 0.5/)

Solution:

Terms : 0.3a, — 0.6ab, 0.5b

Coefficient : 0.3, — 0.6, 0.5

2. Classify the following polynomials as monomials, binomials, or trinomials, which polynomials do not fit in any of these three categories?
x+y, 1000, \(x+x^2+x^3+x^4\) , 7+y+5x, \( 2y-3y^2+4y^3\) ,5x-4y+3xy, \( 4z-15z^2\) , ab+bc+cd+da, pqr, \( p^2q+pq^2\) , 2p+2q

Solution:

MonomiaLs: 1000, QPR

Binomials: x + y, \(2y-3y^2\), \(4z-15z^2\),\(p^2q+pq^2\), 2P + 2q

Trinomials : 7 + y + 5x, \(2y-3y^2+4y^3\), 5x – 4y + 3xy

Polynomials that do not fit in any of these categories are: \(x + x^2+x^3+x^4\), ab + bc + cd + da

3. Add the following

1) ab – bc, bc – ca, ca – ab

2) a – b + ab, b – c + bc, c – a + ac

Solution:
1)
2) a -b + ab, b -c + bc,c – a + ac

Solution:
\(a-\not b+a b+b-c+b c+\not c-a c+a c\)
= ab+bc+ac

3) \(2p^2q^2 -3pq + 4,5 + 7pq -3p^2q^2\)
Solution: \(2p^2q^2 — 3pq + 4 + 5+ 7pq— 3p^2q^2\)
= \(-p^2q^2 +4pq + 9\)

4)\(l^2+m^2\), \(m^2+n^2\), 2lm+2mn+2nl
Solution: \(\begin{aligned}
& \ell^2+m^2+m^2+n^2+n^2+\ell^2 \\
&+2 \ell m+2 m n+2 n \ell
\end{aligned}\)
= \(2 \ell^2+2 m^2+2 n^2+2 \ell m+2 m n+2 n \ell\)

4. 1) Subtract 4a-7ab+3b+12 from 12a-9ab+5b-3

Solution:
\(\begin{aligned}
& 12 a-9 a b+5 b-3 \\
& (-) 4 a-7 a b+3 b+12 \\
& 8 a-2 a b+2 b-15 \\
&
\end{aligned}\)

b) Subtract 3xy + 5yz — 7zx from 5 xy — 2 yz — 2 zx +10 xyz

Solution:

c) Subtract \(4 p^2 q-3 p q+5 p q^2-8 p+7 q-10\) from \(18-3 p-11 q+5 p q-2 p q^2+5 p^2 q\)

Solution:

KSEEB Class 8 Maths Chapter 6 Algebraic Expressions And Identities Exercise 6.2

1. Find the product of the following pairs of monomials.

1) 4,7p

2) -4p,7p

3) -4p,7pq

4) \(4 p^3\), -3p

5) 4p,0

Solution:

1) \(4 \times 7 \times p=28 p\)

2) \(-4 p \times 7 p=-28 p^2\)

3)\(-4 p \times 7 p q=-28 p^2 q\)

4) \(4 p^3 \times-3 p=-12 p^4\)

5)\(4 p \times 0=0\)

2. Find the areas of rectangles with the following pairs of monomials as their lengths and breadths respectively.

1)(p,q)

2) (10m, 5n)

3) \(\left(20 x^2, 5 y^2\right)\)

4) \(\left(4 x, 3 x^2\right)\)

5) (3mn,4np)

Solution:

1) (p, q)

Area of rectangle = length x breadth

= pxq = pq

2) (10m, 5n)

Area of rectangle = l x b = 10m x 5n

= 50 mn

3) \(\left(20 x^2, 5 y^2\right)\)

Area of reactangle = l x b = \(20x^2 x 5y^2\)

= \(100x^2y^2\)

4) \(\left(4 x, 3 x^2\right)\)

Solution: Area of rectangle = l x b = \(4x x 3x^2\)

=\(12x^3\)

v) 3mn x 4np

Solution: Area ofrectangle = l x b = 3mn x 4np

=\(12mn^2p\)

3. Complete the table of products

Solution:

4. Obtain the volume of rectangular boxes with the following length, breadth and height respectively.

1) \(5 a, 3 a^2, 7 a^4\)

2) \(2 p, 4 q, 8 r\)

3) \(x y, 2 x^2 y, 2 x y^2\)

4) \(a, 2 b, 3 c\)

Solution: Volume = Length x Breadth x Height

1) \(\vartheta=5 a \times 3 a^2 \times 7 a^4=5 \times 3 \times 7 \times a \times a^2 \times a^4=105 a^7\)

2) \(\vartheta=2 p \times 4 q \times 8 r=2 \times 4 \times 8 \times p \times q \times r=64 p q r\)

3) \(\vartheta=x y \times 2 x^2 y \times 2 x y^2=2 \times 2 \times x \times x^2 \times x \times y \times y \times y^2=4 x^4 y^4\)

4)\(\vartheta=a \times 2 b \times 3 c=2 \times 3 \times a \times b \times c=6 a b c\)

5) Obtain the product of

1) xy, yz, zx

Solution: \(x \times y \times y \times z \times z \times x==x^2 y^2 z^2\)

2) \(a,-a^2, a^3\)

Solution: \(a \times\left(-a^2\right) \times a^3=-a^6\)

3) \( 2,4 y, 8 y^2, 16 y^3\)

Solution: \(2 \times 4 y \times 8 y^2 \times 16 y^3=1024 y^6\)

4) \(a \times 2 b \times 3 c \times 6 a b c=36 a^2 b^2 c^2\)

Solution: \(a \times 2 b \times 3 c \times 6 a b c=36 a^2 b^2 c^2\)

5) \(m,-m n, m n p\)

Solution: \(m \times(-m n) \times(m n p)=-m^3 n^2 p\)

KSEEB Maths Class 8 Algebraic Expressions And Identities Exercise 6.3

1. Carry out the multiplication of the expression in each of the following pairs.

1) \(4 p, q+r\)

2) \(a b, a-b\)

3) \(a+b, 7 a^2 b^2\)

4) \(a^2-9,4 a\)

5) \(p q+q r+r p, 0\)

Solution:

1) \((4 p) \times(q+r)\)

=\((4 p \times q)+(4 p \times r)\)

= \(4 p q+4 p r\)

2) \((a b) \times(a-b)\)

=\(a b \times a-a b(b)\)

=\(a^2 b-a b^2\)

3)\((a+b)\left(7 a^2 b^2\right)\)

=\(a \times 7 a^2 b^2+b \times 7 a^2 b^2\)

=\(7 a^3 b^2+7 a^2 b^3\)

4) \( \left(a^2-9\right)(4 a)\)

=\(a^2 \times 4 a-9 \times 4 a\)

=\(4 a^3-36 a\)

5) \((p q+q r+r p)(0)\)

=\((p q \times 0)+(q r \times 0)+(r p \times 0)\)

=\(0+0+0=0\)

2. Complete the table

Solution:

3. Find the product

1) \(\left(a^2\right) \times\left(2 a^{22}\right) \times\left(4 a^{26}\right)\)

Solution: \(2 \times 4 \times a^2 \times a^{22} \times a^{26}\)

=\(8 a^{50}\)

2) \(\left(\frac{2}{3} x y\right) \times\left(\frac{-9}{10} x^2 y^2\right)\)

Solution: \(\left(\frac{2}{3}\right) \times\left(\frac{-9}{10}\right) \times x \times y \times x^2 \times y^2\)

=\(\frac{-3}{5} x^3 y^3\)

3) \(\left(\frac{-10}{3} p q^3\right) \times\left(\frac{6}{5} p^3 q\right)\)

Solution: \(\frac{-10}{3} \times \frac{6}{5} \times p \times q^3 \times p^3 \times q\)

=\(-4 p^4 q^4\)

4) \(x \times x^2 \times x^3 \times x^4\)

Solution: \(x^{10}\)

4. a) Simplify 3x(4x-5)+3 and find its values for

1) x=3

2) \(x=\frac{1}{2}\)

Solution: \(3 x(4 x-5)+3\)

=\(

= [latex]12 x^2-15 x+3\)

1) For x=3,\(12 x^2-15 x+3\)

=\(12(3)^2-15(3)+3\)

=\(12 \times 9-15 \times 3+3\)

=\(108-45+3\)

= 66

2) For \(x=\frac{1}{2}\)

\(12 x^2-15 x+3\)

= \(12 \times\left(\frac{1}{2}\right)^2-15 \times \frac{1}{2}+3\)=\(12^3\times\frac{1}{4}-\frac{15}{2}+3\)

= \(3-\frac{15}{2}+3\)

= \(6-\frac{15}{2}=\frac{12-15}{2}\)

= \(\frac{-3}{2}\)

b) Simplify \(a\left(a^2+a+1\right)+5\) and find its value for

1) a=0

2) a=1

3) a= -1

Solution:

\( a\left(a^2+a+1\right)+5=a^3+a^2+a+5\)

For a=1, \( a^3+a^2+a+5\)

⇒ \( 1^3+1^2+1+5=1+1+1+5=8\)

For a= -1, \( a^3+a^2+a+5\)

⇒ \( (-1)^3+(-1)^2+(-1)+5\)

= \( -1+1-1+5=4\)

5. a) Add: p(p-q), q(q-r) and r(r-p)

Solution: p(p-q)+q(q-r)+r(r-p)

= \(p^2-p q+q^2-q r+r^2-r p\)

= \(p^2+q^2+r^2-p q-q r-r p\)

b) Add: 2x(z-x-y) and 2y(z-y-x)

Solution: 2x(z-x-y)+2y(z-y-x)

= \(2 x z-2 x^2-2 x y+2 y z-2 y^2-2 x y\)

= \(-2 x^2-2 y^2-4 x y+2 y z+2 z x\)

c) 3l(l-4m+5n) from 4l(10n-3m+5n)

Solution: 4l(10n-3m+2l)-3l(l-4m+5n)

= \(40ln-12ml+8l^2-3l^2+12ml-15ln\)

= \(25ln+5l^2\)

d) Subtract: 3a(a+b+c)-2b(a-b+c) from 4c(-a+b+c)

Solution: \(4 c(a+b+c)-3 a(a+b+c)+2 b(a-b+c)\)

= \(-4 a c+4 b c+4 c^2-3 a^2-3 a b-3 a c+2 a b-2 b^2+2 b c\)

= \(-4 a c-3 a c+4 b c+2 b c-3 a b+2 a b+4 c^2-3 a^2-2 b^2\)

= \(-7 a c+6 b c-a b+4 c^2-3 a^2-2 b^2\)

Chapter 6 Algebraic Expressions And Identities In KSEEB Maths Exercise 6.4

1. Multiply the binomials.

1) (2x+5) and (4x-3)

Solution: (2x+5)x(4x-3)

=2 x(4 x-3)+5(4 x-3)

=\(2 x \times 4 x-2 x \times 3+5 \times 4 x-5 \times 3\)

=\(8 x^2-6 x+20 x-15\)

=\(8 x^2+14 x-15\)

2) (y-8) and (3y-4)

Solution: (y-8)x(3y-4)

=\(y \times(3 y-4)-8(3 y-4)\)

=\(y \times 3 y-y \times 4-8 \times 3 y-8 \times(-4)\)

=\(3 y^2-4 y-24 y+32\)

=\(3 y^2-28 y+32\)

3) (2.5l-0.5m) and (2.5l+0.5m)

Solution: (2.5l-0.5m)x(2.5l+0.5m)

=2.5l(2.5l+0.5m)-0.5m(2.5l+0.5m)

=2.5l x 2.5l + 2.5l x 0.5m -0.5m x 2.5

= \(6.25l^2+1.25lm-1.25lm-0.25m^2\)

=\(6.25l^2-0.25m^2\)

4) (a+3b) and (x+5)

Solution: \((a+3 b) \times(x+5)\)

=\(a(x+5)+3 b(x+5)\)

=\(a x+a 5+3 b x+15 b\)

=\(a x+5 a+3 b x+15 b\)

5) \(\left(2 p q+3 q^2\right) \text { and }\left(3 p q-2 q^2\right)\)

Solution: \(\left(2 p q+3 q^2\right) \text { and }\left(3 p q-2 q^2\right)\)

= \(2 p q \times 3 p q-2 p q \times 2 q^2+3 q^2\times 3 p q-3 q^2 \times 2 q^2\)

= \(6 p^2 q^2-4 p q^3+9 p q^3-6 q^4\)

6) \(\left(\frac{3}{4} a^2+3 b^2\right) \text { and } 4\left(a^2-\frac{2}{3} b^2\right)\)

Solution:  \(\left(2 p q+3 q^2\right) \times\left(3 p q-2 q^2\right)\)

= \(\left(\frac{3}{4} a^2+3 b^2\right) \times\left(4 a^2-\frac{8}{3} b^2\right)\)

= \(\frac{3}{4} a^2\left(4 a^2-\frac{8 b^2}{3}\right)+3 b^2\left(4 a^2-\frac{8 b^2}{3}\right)\)

= \(\frac{3}{4} a^2 \times 4 a^2-\frac{\not \beta}{4} a^2 \times \frac{8^2}{\not 8} b^2+3 b^2\times 4 a^2-\frac{8}{3} b^2 \times \not 3 b^2\)

= \(3 a^4-2 a^2 b^2+12 a^2 b^2-8 b^4\)

= \(3 a^4+10 a^2 b^2-8 b^4\)

2. Find the Product

1) (5-2x)(3+x)

Solution: (5-2x)(3+x)

= 5(3+x)-2x(3+x)

= \(15+5 x-6 x-2 x^2\)

= \(15-x-2 x^2\)

2) (x+7y)(7x-y)

Solution: (x+7y)x(7x-y)

= x(7 x-y)+7 y(7 x-y)

= \(7 x^2-x y+49 x y-7 y^2\)

= \(7 x^2+48 x y-7 y^2\)

3) \(\left(a^2+b\right)\left(a+b^2\right)\)

Solution: \(\left(a^2+b\right)\left(a+b^2\right)\)

= \(a^2\left(a+b^2\right)+b\left(a+b^2\right)\)

= \(a^3+a^2 b^2+a b+b^3\)

4) \(\left(p^2-q^2\right)(2 p+q)\)

Solution: \(\left(p^2-q^2\right) \times(2 p+q)\)

= \(p^2(2 p+q)-q^2(2 p+q)\)

= \(2 p^3+p^2 q-2 p q^2-q^3\)

3. Simplify

1) \(\left(x^2-5\right)(x+5)+25\)

Solution: \(\left(x^2-5\right)(x+5)+25\)

= \(x^2(x+5)-5(x+5)+25\)

= \(x^3+5 x^2-5 x-25+25\)

= \(x^3+5 x^2-5 x\)

2) \(\left(a^2+5\right)\left(b^3+3\right)+5\)

Solution: \(\left(a^2+5\right)\left(b^3+3\right)+5\)

=\(a^2\left(b^3+3\right)+5\left(b^3+3\right)+5\)

= \(a^2 b^3+3 a^2+5 b^3+15+5\)

= \(a^2 b^3+3 a^2+5 b^3+20\)

3) \(\left(t+s^2\right)\left(t^2-s\right)\)

Solution: \(\left(t+s^2\right) \times\left(t^2-s\right)\)

= \(t\left(t^2-s\right)+s^2\left(t^2-s\right)\)

= \(t^3-s t+s^2 t^2-s^3\)

4) \((a+b)(c-d)+(a-b)(c+d)+2(a c+b d)\)

Solution: \((a+b) \times(c-d)+(a-b) \times(c+d)+2(a c+b d)\)

= \(a(c-d)+b(c-d)+a(c+d)-b(c+d)+2(a c+b d)\)

= \(a c-a a+b c-b a+a c+a d-b c-b a+2 a c+2 b a\)

= 4ac

5) (x+y)(2 x+y)+(x+2 y)(x-y)

Solution: \((x+y) \times(2 x+y)+(x+2 y) \times(x-y)\)

= \(x(2 x+y)+y(2 x+y)+x(x-y)+2 y(x-y)\)

= \(2 x^2+y y+2 x y+y^2+x^2-y y+2 x y-2 y^2\)

= \(3 x^2-y^2+4 x y\)

6) \((x+y)\left(x^2-x y+y^2\right)\)

Solution: \((x+y)\left(x^2-x y+y^2\right)\)

= \(x\left(x^2-x y+y^2\right)+y\left(x^2-x y+y^2\right)\)

= \(x^3-y^2 y+y y^2+y^2 y-y y^2+y^3\)

= \(x^3+y^3\)

7) (1.5 x-4 y)(1.5 x+4 y+3)-4.5 x+12 y

Solution: (1.5 x-4 y)(1.5 x+4 y+3)-4.5 x+12 y

= 1.5 x(1.5 x+4 y+3)-4 y(1.5 x+4 y+3)-4.5 x+12 y

= \(2.25 x^2+6 x y+4.5 x-6 x y-16 y^2-12 y-4.5 x+12 y\)

= \(2.25 x^2-16 y^2\)

8) (a+b+c)(a+b-c)

Solution:(a+b+c)x(a+b-c)

= a(a+b-c)+b(a+b-c)+c(a+b-c)

= \(a^2+a b-\not c c+a b+b^2-b c+\not c c+b c-c^2\)

= \(a^2+b^2-c^2+2 a b\)

Practice For Algebraic Expressions and Identities KSEEB Maths Exercise 6.5

1. Use a suitable identity to get each of the following products.

1) (x+3)(x+3)

Solution: (x+3)(x+3)

\((x+3)^2\)

a=x, b=3

\((a+b)^2=a^2+2 a b+b^2\) \((x+3)^2=x^2+2 x \times 3+b^2\) \((x+3)^2=x^2+6 x+b^2\)

2) (2y+5)(2y+5)

Solution: (2y+5)(2y+5)

= \((2y+5)^2\)

\((a+b)^2=a^2+2 a b+b^2\) \((2 y+5)^2=(2 y)^2+2 \times 2 y \times 5+5^2\)

= \(4 y^2+20 y+25\)

3) (2a-7)(2a-7)

Solution: (2a-7)(2a-7)

\((2 a-7)^2\) \((a-b)^2=a^2-2 a b+b^2\) \((2 a-7)^2=(2 a)^2-2 \times 2 a \times 7+7^2\)

=\(4 a^2-28 a+49\)

4) \(\left(3 a-\frac{1}{2}\right)\left(3 a-\frac{1}{2}\right)\)

Solution: \(\left(3 a-\frac{1}{2}\right)\left(3 a-\frac{1}{2}\right)\)

\(\left(3 a-\frac{1}{2}\right)^2\) \((a-b)^2=a^2-2 a b+b^2\) \(\left(3 a-\frac{1}{2}\right)^2=(3 a)^2-2 \times 3 a \times \frac{1}{2}+\left(\frac{1}{2}\right)^2\)

=\(9 a^2-3 a+\frac{1}{4}\)

5) (1.1 m-0.4)(1.1 m+0.4)

solution: (1.1 m-0.4)(1.1 m+0.4)

\((a+b)(a-b)=a^2-b^2\)

(1.1 m-0.4)(1.1 m+0.4)

= \((1.1 m)^2-(0.4)^2\)

= \(1.21 m^2-0.16\)

6) \(\left(a^2+b^2\right)\left(-a^2+b^2\right)\)
Solution: \(\left(a^2+b^2\right)\left(-a^2+b^2\right)\)
= \(\left(a^2+b^2\right)\left(b^2-a^2\right)\)
= \(\left(b^2+a^2\right)\left(b^2-a^2\right)\)
=\(\left(b^2\right)^2-\left(a^2\right)^2\left\{(a+b)(a-b)=a^2-b^2\right\}\)
= \(b^4-a^4\)

7) (6x-7)(6x+7)
Solution: (6x-7)(6x+7)
= \((6 x)^2-7^2 \quad\left\{(a+b)(a-b)=a^2-b^2\right\}\)
= \(36 x^2-49\)

8) (-a+c)(-a+c)
Solution: (-a+c)(-a+c)
= (c-a)(c-a)
= \((c-a)^2 \quad\left\{(a-b)^2=a^2-2 a b+b^2\right\}\)
= \(c^2-2 \times c \times a+a^2\)
= \(c^2-2 a c+a^2\)

9) \(\left(\frac{x}{2}+\frac{3 y}{4}\right)\left(\frac{x}{2}+\frac{3 y}{4}\right)\)
Solution: \(\left(\frac{x}{2}+\frac{3 y}{4}\right)\left(\frac{x}{2}+\frac{3 y}{4}\right)\)
= \(\left(\frac{x}{2}+\frac{3 y}{4}\right)^2 \quad\left\{(a+b)^2=a^2+2 a b+b^2\right\}\)
= \(\left(\frac{x}{2}\right)^2+\not 2\left(\frac{x}{\not 2}\right)\left(\frac{3 y}{4}\right)+\left(\frac{3 y}{4}\right)^2\)
= \(\frac{x^2}{4}+\frac{3 x y}{4}+\frac{9 y^2}{16}\)

10) (7a-9b)(7a-9b)
Solution: (7a-9b)(7a-9b)
\((7 a-9 b)^2 \quad\left\{(a-b)^2=a^2-2 a b+b^2\right\}\)
= \((7 a)^2-2 \times 7 a \times 9 b+(9 b)^2\)
= \(49 a^2-126 a b+81 b^2\)

2. Use the identity \((x+a)(x+b) = x^2+(a+b) x+a b\) to find the following products.
1) (x+3)(x+7)
Solution: (x+3)(x+7)
\((x+a)(x+b)=x^2+(a+b) x+a b\)
a=3 b=7
\((x+3)(x+7)=x^2+(3+7) x+3 \times 7\)
= \(x^2+10 x+21\)

2) (4x+5)(4x+1)
Solution: (4x+5)(4x+1)
x=4x, a=5, b=1
\((x+a)(x+b)=x^2+(a+b) x+a b\)
\((4 x+5)(4 x+1)=(4 x)^2+(5+1)(4 x)+5 \times 1\)
= \(16 x^2+24 x+5\)

3) (4x-5)(4x-1)
Solution: (4x-5)(4x-1)
= \((4 x)^2+(5-1)(4 x)+5(-1)\)
= \(16 x^2+16 x-5\)

4)(4x+5)(4x-1)
Solution: (4x+5)(4x-1)
= \((4 x)^2+(5-1)(4 x)+5(-1)\)
= \(16 x^2+16 x-5\)

5) (2x+5y)(2x+3y)
Solution: \((2x+5y)(2x+3y)\)
= \(2 x)^2+(5 y+3 y)(2 x)+(5 y)(3 y)\)
= \(4 x^2+(8 y)(2 x)+15 y^2\)
= \(4 x^2+16 x y+15 y^2\)

6) \(\left(2 a^2+9\right)\left(2 a^2+5\right)\)
Solution: \(\left(2 a^2+9\right)\left(2 a^2+5\right)\)
= \(\left(2 a^2\right)^2+(9+5)\left(2 a^2\right)+9 \times 5\)
= \(4 a^4+28 a^2+45\)

7) (xyz-4)(xyz-2)
Solution: (xyz-4)(xyz-2)
= \((x y z)^2+(-4-2)(x y z)+(-4)(-2)\)
= \(x^2 y^2 z^2-6 x y z+8\)

3. Find the following squares by using the identities.
1) \((b-7)^2\)
Solution: \((b-7)^2\)
= \(b^2-2 \times b \times 7+7^2\)
\((a-b)^2=a^2-2 a b+b^2\)
= \(b^2-14 b+49\)

2) \((x y+3 z)^2\)
Solution: \((x y+3 z)^2\)
\(\left\{(a+b)^2=a^2+2 a b+b^2\right\}\)
= \((x y)^2+2(x y)(3 z)+(3 z)^2\)
= \(x^2 y^2 z^2+6 x y z+9 z^2\)

3)\(\left(6 x^2-5 y\right)^2\)
Solution: \(\left(6 x^2-5 y\right)^2\)
\((a-b)^2=a^2-2 a b+b^2\)
= \(\left(6 x^2\right)^2-2\left(6 x^2\right)(5 y)+(5 y)^2\)
= \(36 x^4-60 x^2 y+25 y^2\)

4) \(\left(\frac{2}{3} m+\frac{3}{2} n\right)^2\)
Solution: \(\left(\frac{2}{3} m+\frac{3}{2} n\right)^2\)
\(\left\{(a+b)^2=a^2+2 a b+b^2\right\}\)
= \(\frac{4}{9} m^2+2 m n+\frac{9}{4} n^2\)

5) \((0.4 p-0.5 q)^2\)
Solution: \((0.4 p-0.5 q)^2\)
\((a-b)^2=a^2-2 a b+b^2\)
= \((0.4 p)^2-2 \times 0.4 p \times 0.5 q+(0.5 q)^2\)
= \(0.16 p^2-0.4 p q+0.25 q^2\)

6) \((2 x y+5 y)^2\)
Solution: \((2 x y+5 y)^2\)
\(\left\{(a+b)^2=a^2+2 a b+b^2\right\}\)
= \((2 x y)^2+2(2 x y)(5 y)+(5 y)^2\)
= \(4 x^2 y^2+20 x y^2+25 y^2\)

4. Simplify
1) \(\left(a^2-b^2\right)^2\)
Solution: \(\left(a^2-b^2\right)^2\)
\((a-b)^2=a^2-2b+b^2\)
= \(\left(a^2\right)^2-2 \times a^2 \times b^2+\left(b^2\right)^2\)
= \(a^4-2 a^2 b^2+b^4\)

2) \((2 x+5)^2-(2 x-5)^2\)
Solution: \((2 x+5)^2-(2 x-5)^2\)
\(\begin{aligned}
& (a+b)^2=a^2+2 a b+b^2 \\
& (a-b)^2=a^2-2 a b+b^2
\end{aligned}\)
\(\left\{(2 x)^2+2 \times 2 x \times 5+5^2\right\}-\left\{(2 x)^2-2 \times 2 x \times 5+5^2\right\}\)
= \(\left(4 x^2+20 x+25\right)-\left(4 x^2-20 x+25\right)\)
= \(4 x^2+20 x+25-4 x^2+20 x-25\)
= 40x

3) \((7 m-8 n)^2+(7 m+8 n)^2\)
Solution: \((7 m-8 n)^2+(7 m+8 n)^2\)
\(\left\{\begin{array}{l}
(a+b)^2=a^2+2 a b+b^2 \\
(a-b)^2=a^2-2 a b+b^2
\end{array}\right\}\)
\(\left\{(7 m)^2-2 \times 7 m \times 8 n+(8 n)^2\right\}+\left\{(7 m)^2+2(7 m)(8 n)+(8 n)^2\right\}\)
= \(49 m^2-112 m n+64 n^2+49 m^2+112 m n+64 n^2\)
= \(98 m^2+128 n^2\)

4) \((4 m+5 n)^2+(5 m+4 n)^2\)
Solution: \((4 m+5 n)^2+(5 m+4 n)^2\)
=\(\left\{(4 m)^2+2 \times 4 m \times 5 n+(5 n)^2\right\}+\left\{(5 m)^2+2(5 m)(4 n)+(4 n)^2\right\}\)
= \(16 m^2+40 m n+25 n^2+25 m^2+40 m n+16 n^2\)
= \(41 m^2+80 m n+41 n^2\)

5) \((2.5 p-1.5 q)^2-(1.5 p-2.5 q)^2\)
Solution: \((2.5 p-1.5 q)^2-(1.5 p-2.5 q)^2\)
\(\left\{(a-b)^2=a^2-2 a b+b^2\right\}\)
= \(\left\{(2.5) p^2-2(2.5 p)(1.5 q)+(1.5 q)^2\right\}-\left\{(1.5 p)^2-2(1.5 p)(2.5 q)+(2.5 q)^2\right\}\)
= \(\left(6.25 p^2-7.5 p q+2.25 q^2\right)-\left(2.25 p^2-7.5 p q+6.25 q^2\right)\)
= \(6.25 p^2-7.5 p q+2.25 q^2-2.25 p^2+75 p q-6.25 q^2\)
= \(4 p^2-4 q^2\)

6) \((a b+b c)^2-2 a b^2 c\)
Solution: \((a b+b c)^2-2 a b^2 c\)
\(\left\{\begin{array}{l}
(a+b)^2 \\
=a^2+2 a b+b^2
\end{array}\right\}\)
= \((a b)^2+2(a b)(b c)+(b c)^2-2 a b^2 c\)
= \(a^2 b^2+2 a b^2 c+b^2 c^2-2 a b^2 c\)
= \(a^2 b^2+b^2 c^2\)

7) \(\left(m^2-n^2 m\right)^2+2 m^3 n^2\)
Solution: \(\left(m^2-n^2 m\right)^2+2 m^3 n^2\)
\(\left\{(a-b)^2=a^2-2 a b+b^2\right\}\)
=\(\left(m^2\right)^2-2 \times m^2 \times n^2 m+\left(n^2 m\right)^2+2 m^3 n^2\)
= \(m^4-2 m^3 n^2+n^4 m^2+2 m^3 n^2\)
= \(m^4+n^4 m^2\)

5. Show that
1) \((3 x+7)^2-84 x=(3 x-7)^2\)
Solution: LHS = \((3 x+7)^2-84 x\)
= \((3 x)^2+2 \times 3 x \times 7+7^2-84 x\)
= \(9 x^2+42 x+49-84 x\)
= \(9 x^2-42 x+49\)
RHS = \((3 x-7)^2\)
= \((3 x)^2-2 \times(3 x)(7)+7^2\)
= \(9 x^2-42 x+49\)
∴ LHS=RHS

2) \((9 p-5 q)^2+180 p q=(9 p+5 q)^2\)
Solution: LHS = \((9 p-5 q)^2+180 p q\)
= \((9 p)^2-2 \times 9 p(5 q)+(5 q)^2+180 p q\)
= \(81 p^2-90 p q+25 q^2+180 p q\)
= \(81 p^2+90 p q+25 q^2\)
RHS = \((9 p+5 q)^2\)
= \((9 p)^2+2 \times 9 p \times 5 q+(5 q)^2\)
= \(81 p^2+90 p q+25 q^2\)
∴LHS=RHS

3) \(\left(\frac{4}{3} m-\frac{3}{4} n\right)^2+2 m n=\frac{16}{9} m^2+\frac{9}{16} n^2\)
Solution: LHS = \(\left(\frac{4}{3} m-\frac{3}{4} n\right)^2+2 m n\)
= \(\frac{16}{9} m^2-2 m n+\frac{9 n^2}{16}+2 m n\)
= \(\frac{16}{9} m^2+\frac{9 n^2}{16}\)
RHS = \(\frac{16}{9} m^2+\frac{9}{16} n^2\)
∴ LHS = RHS

4) \((4 p q+3 q)^2-(4 p q-3 q)^2=48 p q^2\)
Solution: LHS =\((4 p q+3 q)^2-(4 p q-3 q)^2\)
= \(\left\{(4 p q)^2+2(4 p q)(3 q)+(3 q)^2\right\}-\left\{(4 p q)^2-2(4 p q)(3 q)+(3 q)^2\right\}\)
= \(16 p^2 q^2+24 p q^2+9 q^2-16 p^2 q^2+24 p q^2-9 q^2\)
= \(48 p q^2\)
RHS = \(48 p q^2\)
∴ LHS = RHS

5) (a-b)(a+b)+(b-c)(b+c)+(c-a)(c+a)=0
Solution: LHS = (a-b)(a+b)+(b-c)(b+c)+(c-a)(c+a)=0
= \(\left(a^2-b^2\right)+\left(b^2-c^2\right)+\left(c^2-a^2\right)\)
= \(a^2-b^2+b^2-c^2+c^2-a^2\)
=0
=RHS =0
∴ LHS = RHS

6. Using identities, evaluate
1) \(71^2\)
Solution: \(71^2=(70+1)^2\)
=\(\left\{\begin{array}{l}
(a+b)^2 \\
=a^2+2 a b+b^2
\end{array}\right\}\)
= \((70)^2+2 \times 70 \times 1+1^2\)
= 4900+140+1
= 5041

2) \(99^2\)
Solution: \((99)^2=(100-1)^2\)
=\(\left\{\begin{array}{l}
(a+b)^2 \\
=a^2+2 a b+b^2
\end{array}\right\}\)
= \((100)^2-2 \times 100 \times 1+1^2\)
= 10000-200+1
= 9801

3)\(102^2\)
Solution: \((102)^2=(100+2)^2\)
=\(\left\{\begin{array}{l}
(a+b)^2 \\
=a^2+2 a b+b^2
\end{array}\right\}\)
= \((100)^2+2 \times 100 \times 2+2^2\)
= 10000+400+4
= 10404

4) \(998^2\)
Solution: \((998)^2=(1000-2)^2\)
=\(\left\{\begin{array}{l}
(a+b)^2 \\
=a^2+2 a b+b^2
\end{array}\right\}\)
= \((1000)^2-2 \times 1000 \times 2+2^2\)
= 1000000-4000+4
= 996004

5) \(5.2^2\)
Solution: \((5.2)^2=(5.0+0.2)^2\)
=\(\left\{\begin{array}{l}
(a+b)^2 \\
=a^2+2 a b+b^2
\end{array}\right\}\)
= \((5.0)^2+2 \times 5.0 \times 0.2+(0.2)^2\)
= 25+2+0.04
= 27.04

6) 297 x 303
Solution: 297 x 303
= (300-3)(300+3)
= \((300)^2-(3)^2\)
=\(\left\{\begin{array}{l}
(a+b)^2 \\
=a^2+2 a b+b^2
\end{array}\right\}\)
= 90000-9
= 89991

7) 78 x 82
Solution: 78 x 82
= (80-2)(80+2)
=\(\left\{\begin{array}{l}
(a+b)^2 \\
=a^2+2 a b+b^2
\end{array}\right\}\)
= \((80)^2-2^2\)
= 6400-4
= 6396

8) \(8.9^2\)
Solution: \((8.9)^2=(9.0-0.1)^2\)
= \((9.0)^2-2 \times(9.0)(0.1)+(0.1)^2\)
= 81-1.8+0.01
= 79.21

9) 10.5 x 9.5
Solution: (10.5)(9.5)
= (10+0.5)(10-0.5)
= \((10)^2-(0.5)^2\)
= 100-0.25
= 99.75

7. Using \(a^2-b^2=(a+b)(a-b)\) find
1) \(51^2-49^2\)
Solution: \((51)^2-(49)^2\)
= (51+49)(51-49)
= 100 x 2
= 200

2) \((1.02)^2-(0.98)^2\)
Solution: \((1.02)^2-(0.98)^2\)
= (1.02+0.98)(1.02-0.98)
= 2 x 0.04
= 0.08

3) \(153^2-147^2\)
Solution: (153+147)(153-147)
= 300 x 6
= 1800

4) \((12.1)^2-(7.9)^2\)
Solution: \((12.1)^2-(7.9)^2\)
= (12.1+7.9)(12.1-7.9)
= 20 x 4.2
= 84

8. Using \((x+a)(x+b)=x^2+(a+b) x+a b\), find
1) 103 x 104
Solution: 103 x 104
= (100+3)(100+4)
= \(100^2+(3+4)(100)+3 \times 4\)
= 10000+700+12
= 10712

2) 5.1 x 5.2
Solution: \((5.1) \times(5.2)=(5+0.1)(5+0.2)\)
= \(5^2+(0.1+0.2) 5+(0.1 \times 0.2)\)
= 25+1.5+0.02
= 26.52

3)103 x 98
Solution: \((103) \times(98)\)
= (100+3)(100-2)
= \(100^2+(3-2) \times 100+(3)(-2)\)
= 10000+100-6
= 10094

4) 9.7 x 9.8
Solution: 9.7 x 9.8
= (10-0.3)(10-0.2)
= \(10^2+(-0.3-0.2) 10+(-0.3)(-0.2)\)
= 100 – (0.5 x 10) + 0.06
= 100 – 5 + 0.06
= 95.06

Algebraic Expressions And Identities Questions And Answers KSEEB Maths Additional Problems

1. Subtract \(6 x^2-4 x y+5 y^2\) from \(8 y^2+6 x y-3 x^2\)
Solution:

2. Using suitable identity, evaluate
\((69.3)^2-(30.7)^2\)
Solution: \((69.3)^2-(30.7)^2\)
= (69.3+30.7)(69.3-30.7)
\(\left\{\begin{array}{l}
a^2-b^2 \\
=(a+b)(a-b)
\end{array}\right\}\)
= 100 x 38.6
= 3860

3. Add: 9ax, 3by-cz and -5by+ax+3cz
Solution: 9ax+3by-cz-5by7+ax+3cz
= 10ax-2by+2cz

4. Expand the following, using suitable identities
1) \(\left(\frac{4 x}{5}+\frac{y}{4}\right)\left(\frac{4 x}{5}+\frac{3 y}{4}\right)\)
Solution: \((x+a)(x+b)=x^2+(a+b) x+a b\)
\(\left(\frac{4 x}{5}+\frac{y}{4}\right)\left(\frac{4 x}{5}+\frac{3 y}{4}\right)\)
= \(\left(\frac{4 x}{5}\right)^2+\left(\frac{y}{4}+\frac{3 y}{4}\right)\left(\frac{4 x}{5}\right)+\frac{y}{4} \times \frac{3 y}{4}\)
= \(\frac{16 x^2}{25}+\frac{16^4 x y}{4 \times 5}+\frac{3 y^2}{16}\)
= \(\frac{16 x^2}{25}+\frac{4 x y}{5}+\frac{5}{16} y^2\)

2) \((0.9 p-0.5 q)^2\)
Solution: \((0.9 p)^2-2 \times 0.9 p \times 0.5 q+(0.5 q)^2\)
= \(0.81 p^2-0.90 p q+0.25 q^2\)

5. Evaluate \(\left(\frac{a}{b}+\frac{b}{a}\right)^2\) using identity
Solution: \(\left(\frac{a}{b}+\frac{b}{a}\right)^2=\left(\frac{a}{b}\right)^2+2 \times \frac{a}{b} \times \frac{b}{a}+\left(\frac{b}{a}\right)^2\)
\(\left\{\begin{array}{l}
(a+b)^2 \\
=a^2+2 a b+b^2
\end{array}\right\}\)
=\(\frac{a^2}{b^2}+2+\frac{b^2}{a^2}\)

6. Evaluate: \((\sqrt{5} a+\sqrt{3} b)^2\)
Solution: =\((\sqrt{5} a)^2+(\sqrt{3} b)^2+2 \times \sqrt{5} \times a \times \sqrt{3} \times b\)
= \(5 a^2+3 b^2+2 \sqrt{15} a b\)
= \(5 a^2+3 b^2+2 \sqrt{15} a b\)
\((\sqrt{5} a+\sqrt{3} b)^2=5 a^2+3 b^2+2 \sqrt{15} a b\)

7. Subtract the sum of \( 3l-4m-7n^2 and 2l+3m-4n^2 from the sum of 9l+2m-3n^2 and -3l+m+4n^2\)
Solution:

Subtracting \(5l-m-11n^2 from 6l+3m+n^2 we get\)

8. Subtract 7p(3q+7p) from 8p(2p-7q)
Solution: 8p(2p-7p)-7p(3q+7p)
= \(16 p^2-56 p q-21 p q-49 p^2\)
= \(-33 p^2-77 p q\)

9. Simplify \((ab-c)^2\)
Solution: \((ab-c)^2+2abc\)
= \((a b)^2-2 \times a b \times c+c^2+2 a b c\)
= \(a^2 b^2-2 a b c+c^2+2 a b c\)
= \(a^2 b^2+c^2\)

10. Expand (2x+9)(2x-7) using identity
Solution: (2x+9)(2x-7)
= \((2 x)^2+(9-7)(2 x)+(9)(-7)\)
= \(4 x^2+4 x-63\)

11. Expand: \((49)^2\)
Solution: \((49)^2=(50-1)^2\)
\(\left\{\begin{array}{l}
(a-b)^2 \\
=a^2-2 a b+b^2
\end{array}\right\}\)
= \((50)^2-2 \times 50 \times 1+1^2\)
= 2500-100+1
= 2400+1
= 2401

12. Write the greatest common factor in each of the following.

1) \(2 x y,-y^2, 2 x^2 y\)

Solution: \(2 x y,-y^2, 2 x^2 y\)

\(2 x y=2 \times x \times y\) \(-y^2 = -y x y\) \(2 x^2 y=2 \times x \times x \times y\)

G.C.F = y-16

2) \(3 x^3 y^2 z,-6 x y^3 z^2, 12 x^2 y z^3\)

Solution: \(3 x^3 y^2 z,-6 x y^3 z^2, 12 x^2 y z^3\)

\(3 x^3 y^2 z=3 \times x \times x \times x \times y \times y \times z\) \(-6 x y^3 z^2=-3 \times 2 \times x \times y \times y \times y \times z \times z\) \(12 x^2 y z^3=2 \times 2 \times 3 \times x \times x \times y \times z \times z \times z\)

G.C.F = 3xyz

13. Multiply the binomials

a) (2x+5)(4x-3)

Solution: (2x+5)(4x-3)

= 2x(4x-3)+5(4x-3)

= 2x x 4x +2x(-3)+5 x 4x + 5(-3)

= \(8 x^2-6 x+20 x-15\)

= \(8 x^2+14 x-15\)

14. Find the areas of rectangles with the following pairs of monomials as their lenghts and breadths respectively.

Solution: W.K.T. Area of rectangle = Length x Breadth

Area of 1st rectangle = \(3mn x 4np = 12mn^2p\)

Area of 2nd reactangle =\(4 x \times 3 x^2=12 x^3\)

15. If \(a+\frac{1}{a}=7, find a^2+\frac{1}{a^2}\)

Solution: Consider \(a+\frac{1}{a}=7\) square bothside, we get

\(\left(a+\frac{1}{a}\right)^2=7^2\) \(a^2+2 \times a+\frac{1}{a}+\left(\frac{1}{a}\right)^2=49\) \(a^2+2+\frac{1}{a^2}=49\) \(a^2+\frac{1}{a^2}=49-2=47\) \(\text { ie } a^2+\frac{1}{a^2}=47\)

16. Multiply \(x^2+2 y \text { by } x^3-2 x y+y^3 and find the value of the product for x=1 and y= -1\)

Solution: \(\left(x^2+2 y\right)\left(x^3-2 x y+y^3\right)\)

= \(x^2\left(x^3-2 x y+y^3\right)+2 y\left(x^3-2 x y+y^3\right)\)

= \(x^5-2 x^3 y+x^2 y^3+2 x^3 y-4 x y^2+2 y^4\)

= \(x^5+x^2 y^3-4 x y^2+2 y^4\)

put x=1 & y= -1

= \((1)^5+(1)^2(-1)^3-4(1)(-1)^2+2(-1)^4\)

= 1+(1)(-1)-4(1)(1)+2(1)

= 1-1-4+2 = -2

17. If \(x^2+\frac{1}{x^2}=38\), find the values of

1) \(x-\frac{1}{x}\)

2) \(x^4+\frac{1}{x^4}\)

Solution:

1)\(\left(x-\frac{1}{x}\right)^2=x^2+\frac{1}{x^2}-2 \times x \times \frac{1}{x}\)

= \(x^2+\frac{1}{x^2}-2=38-2=36\)

∴ \(x-\frac{1}{x}=\sqrt{36}=6\)

2)\(\left(x^2+\frac{1}{x^2}\right)^2=x^4+\frac{1}{x^4}+2 \times x^2 \times \frac{1}{x^2}\)

\((38)^2=x^4+\frac{1}{x^4}+2\)

= \(1444-2=x^4+\frac{1}{x^4}\)

∴ \(x^4+\frac{1}{x^4}=1442\)

18. Find the area of the reactangle whose length and breadths are \(3 x^2 y m and 5 x y^2 m\) respectively.

Solution: Length = \(3 x^2 y m\), breadth = \(5 x y^2m\)

Area of rectangle = length x breadth

= \(3 x^2 y \times 5 x y^2 \text { sq m }\)

= \(15 x^3 y^3 \text { sq } \mathrm{m}\)

19. Find the value of the expression \(\left(81 x^2+16 y^2-72 x y\right) \text {, when } x=\frac{2}{3} \& y=\frac{3}{4}\)

Solution: \(\left(81 x^2+16 y^2-72 x y\right)\)

= \((9 x)^2+(4 y)^2-2 \times 9 x+4 y\)

= \((9 x-4 y)^2\)

= \(\left(9 \times \frac{2}{3}-4 \times \frac{3}{4}\right)^2 \text {, when } x=\frac{2}{3} \& y=\frac{3}{4}\)

= \((6-3)^2=3^2=9\)

KSEEB Class 8 Maths Solutions For Chapter 5 Squares And Square Roots Points To Remember

Square: Number obtained when a number is multiplied by itself. It is the number raised to the power 2, \(2^2=2 \times 2=4\) (square of 2 is 4)

If a natural number m can be expressed as \(\mathrm{n}^2\), where n is also a natural number, then m is a square number.

All square numbers end with 0, 1,4, 5, 6, or 9 at unit’s place.

Square numbers can only have even number of zero’s at the end.

Square root is the inverse operation of square.

There are two integral square roots of a perfect square number.

Positive square root of a number is denoted by the symbol √. For example \(3^2=9\) gives √9=3

Read and Learn More KSEEB Solutions for Class 8 Maths

Perfect square or square number: It is the square of some natural number.If\(\mathrm{m}=\mathrm{n}^2\) , then m is a perfect square number where m and n are natural numbers.

Example: \(1=1 \times 1=1^2\)

\( 4=2 \times 2=2^2\)

Properties of square number

(1) A number ending in 2, 3, 7 or 8 is never a perfect square. Ex: 152, 1028 etc.,

(2) A number ending in 0, 1,4, 5, 6 or 9 may not necessarily be a square number Example: 20, 31, 24 etc.,

(3) Square of even numbers are even

Example:\(2^2=4,4^2=16\) etc.

(4) Square of odd numbers are odd

Example: \(5^2 = 25, 9^2 = 81\) etc.,

(5) A number ending in an odd number of zeroes cannot be a perfect square

Example: 10,1000, 900000etc.,

(6) The difference of squares of two consecutive natural number is equal to their sum.

\((n +1)^2 — n^2 = n +1 + n\)

Example: \(4^2 – 3^2 = 4 + 3 = 7\)

\(12^2 – 11^2 = 12 + 11 =23\) etc.,

(7) A triplet (m, n, p) of three natural numbers m, n and p is called a Pythagorean triplet.
If \( m^2 + n^2 = p^2\)

⇒ \( 3^2 + 4^2 = 25 = 5^2\)

 

 

Squares And Square Roots Solutions KSEEB Class 8 Maths Exercise 5.1

1. What will be the unit digit of the squares of the following numbers?
1)81
2)272
3)799
4)3853
5)1234
6)26387
7)52698
8)99880
9)12796
10)55555

Solution: wkt, if a number has its unit’s place digit as a then its square, will end with the unit digit of the multiplication a x a.

1) 81 since the given number has its unit’s pace digit as 1, its square will end with the unit digit of the multiplication (1 x 1 = 1) i.e., 1

2) 272 Since the given number has its unit’s place digit as 2 its square will end with the unit digit of the multiplication (2×2 = 4) i.e., 4.

3) 799 Since the given number has its unit’s place digit as 9 its square will end with the unit digit of the multiplication (9 x 9 = 81) i.e., 1

4)3853 Since the given number has its unit’s place digit as 3 its square will end with the unit digit of the mul¬tiplication (3 x 3 = 9) i.e., 9

5)1234 Since the given number has its unit’s place digit as 4 its square will end with the unit digit of the multiplication (4 x 4 = 16) i.e., 16

6)26387 Since the given number has its unit’s place digit as 7 its square will end with the unit digit of the multiplication (7 x 7 = 49) i.e., 9

7)52698 Since the given number has its unit’s place digit as 8 its square will end with the unit digit of the multiplication (8 x 8 = 64) i.e., 4

8)99880 Since the given number has its unit’s place digit as 0 its square will have two zeroes at the end

∴ The unit digit of the square of the given number is 0.

9)12796 Since the given number has its unit’s place digit as 8, its square will end with the unit digit of the multiplication (6 x 6 = 36) i.e., 6

10)55555 Since the given number has its unit’s place digit as 5 its square will end with the unit digit of the multiplication (5×5= 25) i.e., 5

2. The following numbers are obviously not perfect squares. Give reason.
1)1057
2)23453
3)7928
4)222222
5)64000
6)89722
7)222000
8)505050

Solution: The square of numbers may end with any one of the digits 0, 1,5, 6 or 9. Also, a perfect square has even number of zeroes at the end of it.

1) 1057 has its unit place digit as 7

∴ it cannot be a perfect square.

2) 23453 has its unit place digit as 3

∴ it cannot be a perfect square.

3) 7928 has its unit place digit as 8

∴ it cannot be a perfect square.

4) 222222 has its unit place digit as 2

∴ it cannot be a perfect square.

5) 64000 has three zeros at the end ofit. However, since a perfect square cannot end with odd number of zeroes, it is not a perfect square.

6) 89722 has its unit place digit as 2

∴ it cannot be a perfect square.

7) 222000 has three zeros at the end ofit. However, since a perfect square cannot end with odd number of zeroes, it is not a perfect square.

8) 505050 has one zero at the end of it, However since a perfect square cannot end with odd number of zeroes, it is not a perfect square.

3. The squares of which of the following would be odd numbers?
1)431
2)2826
3)7779
4)82004

Solution: The square of an odd number is odd and the square of an even number is even. Here 431 and 7779 are odd numbers.

Thus, the square of 431 and 7779 will be an odd number.

4. Observe the following pattern and find the missing digits. \( 11^2 = 121\)
\( 101^2 = 10201\)
\( 1001^2 = 1002001\)
\( 100001^2 = 1—-2—–1\)
\( 10000001^2= —–\)

Solution: In the given pattern, it can be observed that the squares of the given numbers have the same number of zeroes before and after the digit 2 as it was in the original number.

∴ \( 100001^2 = 10000200001\)

\( 10000001^2= 100000020000001\)

5. Observe the following pattern and supply the
missing number.
\( 11^2 = 121\)
\( 101^2 = 10201\)
\( 10101^2 = 102030201\)
\( 1010101^2 = \)
\( ^2 = 10203040504030201\)

Solution: By following the given pattern, we obtain

\( 1010101^2= 1020304030201\) \( 101010101^2 =10203040504030201\)

6. Using the given pattern, find the missing numbers
\(l^2 + 2^2 + 2^2 = 3^2\)
\(2^2 + 3^2 + 6^2 = 7^2\)
\( 3^2 + 4^2 + 12^2 = 13^2\)
\( 4^2 + 5^2 + ^2 = 21^2\)
\( 5^2 + ^2 + 30^2 = 31^2\)
\( 6^2 + 7^2 + ^2 = ^2\)
Solution: From the given pattern, it can be observed that

1) The third number is the product of the first two numbers.

2) The fourth number can be obtained by adding

1 to the third number. Thus, the missing numbers in the pattern will be as follows.

\( 4^2+ 5^2+ 20^2 = 21^2\) \( 5^2+_6^2 + 30^2 = 31^2\) \( 6^2+ 7^2+ 42^2 = 43^2\)

7. Without adding find the sum.
1)1+ 3 + 5 + 7 + 9
2) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19
3) 1+3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19+ 21+ 23

Solution: wkt The sum of first ‘n’ odd natural number is n^2.

1) Here, we have to find the sum of first five odd natural numbers.

∴ \(1 + 3 + 5 + 7 + 9 = 5^2 = 25\)

ii) Here, we have to find the sum of first ten odd natural numbers.

∴\(1+3 + 5 + 7 + 9+11 + 13 + 15+17+19 = 10^2 = 100\)

iii) Here, we have to find the sum of first twelve odd natural numbers.

∴ \(1+3 + 5 + 7 + 9+11 + 13 + 15+17+19+21+23 = 12^2=144\)

8. 1) Express 49 as the sum of 7 odd numbers.

Solution: wkt The sum of first ‘n’ odd natural numbers is \(n^2\).

\(49 = 7^2\)

∴ 49 is the sum of first 7 odd natural numbers

49=1+3 + 5 + 7 + 9+11 + 13

2) Express 121 as the sum of 11 odd numbers.

Solution: \(121 = ll^2\)

∴ 121 is the sum of first 11 odd natural numbers.

121 = 1+ 3 + 5 + 7 + 9+11 + 13 + 15+17 + 19 + 21

KSEEB Maths Class 8 Squares and Square Roots Notes

9. How many numbers lie between squares of the following numbers?
1) 12 & 13
2) 25 & 26
3) 99 and 100.

Solution: wkt There will be 2n numbers in between the squares of the numbers n and (n + 1)

1)  \(12^2  13^2\), there will be 2 x 12 = 24 numbers.

2)  \(25^2  26^2\), there will be 2 x 25 = 50 numbers.

3)  \(99^2  100^2\), there will be 2 x 99 =198 numbers.

KSEEB Class 8 Maths Chapter 5 Square And Square Roots Exercise 5.2

1. Find the square of the following numbers.

1) 32

Solution: \(32 = (30 + 2)^2 = 30(30 + 2) + 2 (30+2)\)

= \(30^2 + 30 x 2 + 2 x 30 + 2^2\)

= 900 + 60 + 60 + 4

= 1024

2)35

Solution: 35 = 3 (3 + 1) hundreds + 25

= (3×4) hundreds + 25

= 1200 + 25

= 1225

3) 86

Solution: \(86^2 = (80 + 6)^2\)

= 80 (80 + 6) + 6 (80 + 6)

= \(80^2 + 80 x 6 + 6 x 80 + 6^2\)

= 6400 + 480 + 480 + 36

= 7396

4) 93

Solution: \( (93)^2 = (90 + 3)^2\)

= 90 (90 + 3) + 3 (90 + 3)

= \( 90^2 + 90 x 3 + 3 x 90 + 3^2 \)

= 8100 + 270 + 270 + 9

= 8649

5) 71

Solution: \( (71)^2 = (70 + l)^2\)

= 70(70+ 1)+ 1 (70+ 1)

= \( 70^2 + 70 x 1 + 1 x 70 + 1\)

= 4900 + 70 + 70+1

= 5041

6)46

Solution: \( 46^2 = (40 + 6)^2\)

= 40 (40 + 6) + 6 (40 + 6)

= \( 40^2 + 40 x 6 + 6 x 40 + 6^2\)

= 1600 + 240 + 240 + 36

= 2116

2. Write a Pythagorean triplet whose one member is
1) 6
2) 14
3) 16
4) 18

Solution: For any natural number m > 1, 2m,\( m^2 – 1, m^2 + 1\) forms a pythagorean triplet.

1) if we take \( m^2 + 1 = 6\) , then \( m^2 = 5\) The value of m will not be an integer.

If we take m2 – 1 = 6, then m2 = 7 Again the value of m is not an integer.

Let 2m = 6

m = 3

∴ The pythagorean triplets are \(2 x 3, 3^2 – 1, 3^2 + 1 or 6, 8 and 10\)

2) 14

If we take \(m^2 + 1 = 14, then m^2 = 13\) The value of m will not be an integer.

If we take \(m^2 – 1 = 14, then m^2 = 15\)

Again the value of m is not an integer.

Let 2m =14⇒ m = 7

Thus, \(m^2 – 1 = 49 – 1 = 48\) and

\(m^2 + 1 = 49 + 1 = 50\)

3) 16

If we take \(m^2 + 1 = 16, then m^2 = 15\) The value of m will not be an integer.

If we take \(m^2 – 1 = 16, then m^2 = 17\)

Again the value of m is not an integer.

Let 2m =16⇒ m = 8

Thus, \(m^2 – 1 = 64 – 1 = 63\) and

\(m^2 + 1 = 64 + 1 = 65\)

∴ The Pythagorean triplet is 16,63 and 65

4) 18

If we take \( m^2 + 1 = 18, then m^2 = 17\)

The value of m will not be an integer.

If we take \( m^2 – 1 = 18, then m^2 = 19\)

Again the value of m is not an integer.

Let 2m =18 ⇒ m = 9

Thus, \( m^2-1=81-1=80\) and

\( m^2+1=81+1=82\)

∴ The Pythagorean triplet is 18, 80 and 82

Solutions For Squares And Square Roots KSEEB Maths Exercise 5.3

1. What could be the possible one’s digits of the square root of each of the following numbers?

1) 9801

Solution: If the number ends with 1, then the one’s digit of the square root of that number may be 1 or 9.

∴ one’s digit of the square root of 9801 is either 1 or 9.

2) 99856

If the number ends with 6, then the one’s digit of the square root of th at number may be 4 or 6.

∴ one’s digit of that square root of 99856 is either 4 or 6.

3) 998001

If the number ends with 6, then the one’s digit of the square root of that number may be 1 or 9.

∴ one’s digit of the square root of 998001 is either 1 or 9

4) 657666025

If the number ends with 5, then the one’s digit of the square root of that number will be 5.

∴ The one’s digit of the square root of 657666025 is 5.

2. Without doing any calculation, find the numbers which are surely not perfect squares
1) 153
2)257
3) 408
4) 441

Solution: The perfect squares of a number can end with any of the digits 0, 1,4, 5, 6 or 9 at units place. Also, a perfect square will end with even number of zeroes, if any.

i) since the number 153 has its unit’s place digit as 3 it is not a perfect square

ii) since the number 257 has its unit’s place digit as 7 it is not a perfect square.

iii) since the number 408 has its unit’s place digit as 8. it is not a perfect square.

iv) since the number 441 has its unit’s place digit as 1. It is a perfect square.

Squares And Square Roots Problems In KSEEB Maths

3. Find the square roots of 100 and 169 by the method of repeated subtraction.

Solution: w.k.t. the sum of the first n odd natural numbers is \( n^2\) .

consider √100

1)100 – 1 = 99

2)99 – 3 = 96

3)96-5 = 91

4)91 -7 = 84

5)84- 9 = 75

6)75 – 11 = 64

7)64-13 = 51

8)51 – 15 = 36

9)36- 17= 19

10)19- 19 = 0

We have subtracted successive odd numbers starting from 1 to 100 and obtained 0 at 10th step.

∴ √100=10

The square root of 169 can be obtained by the method of repeated subtractions as follows.

1) 169- 1 = 168

2) 168-3 = 165

3) 165 -5 = 160

4) 160-7 = 153

5) 153-9 = 144

6) 144-11 = 133

7) 133 – 13 = 120

8) 120 – 15 = 105

9) 105 – 17 = 88

10) 88 – 19 = 69

11) 69-21 =48

12) 48 – 23 = 25

13) 25 – 25 = 0

We have subtracted successive odd numbers starting from 1 to 169 and obtained 0 at 13th step.

∴ √169=13

4. Find the square roots of the following numbers by the prime Factorisation method.

1) 729

Solution: 729= 3x3x3x3x3x3
√729= 3x3x3
=27

2)400


Solution: 400= 2x2x2x2x5x5
√400= 2x2x5
=20

3)1764


Solution: 1764= 2x2x3x3x7x7
√1764= 2x3x7
∴√1764= 42

4)4096

Solution: 4096= 2x2x2x2x2x2x2x2x2x2x2x2
∴√4096= 2x2x2x2x2x2
= 64

5)7744


Solution: 7744= 2x2x2x2x2x2x11x11
∴√7744= 2x2x2x11
= 88

6) 9064

Solution: 9064= 2x2x7x7x7x7
∴√9064= 2x7x7
= 98

7) 5929

Solution: 5929= 7x7x11x11
∴√5929= 7×11
= 77

8) 9216

Solution: 9216=2x2x2x2x2x2x2x2x2x2x3x3
=2x2x2x2x2x3
√9216=96

9) 529

Solution: √529=23

10) 8100

Solution: 8100= 2x2x3x3x3x3x5x5
√8100= 2x3x3x5
=90

KSEEB Class 8 Maths Key Concepts Of Squares And Square Roots

5. For each of the following numbers, find the smallest whole number by which it should be multiplied so as to get a perfect square number. Also, find the square root of the square number so obtained.
1)252
2)180
3)1008
4)2028
5)1458
6)768

Solution:
1)252

252 = 2 x 2 x 3 x 3 x7

Here, prime factor 7 does not have its pair.

If 7 gets a pair, then the number will become a perfect square.

∴ 252 has to be multiplied with 7 to obtained a perfect square

252 x7 = 2x2x3x3 x7 x7

∴252 x 7 = 1764 is a perfect square

∴ √1764 =2x3x7

= 42

2) 180

180= 2x2x3x3x5

Here, prime factor 5 does not have its pair, if 5 gets a pair, then the number will become a perfect square.

∴ 180 has to be multiplied with 5 to obtain a perfect square.

180 x 5 = 900 = 2 x 2 x 3 x 3 x 5×5

∴ 180 x 5 = 900 is a perfect square.

∴ √900 = 2X3X5 = 30

3) 1008

1008= 2x2x2x2x3x3x7

Here, prime factor 7 does not have its pair, If 7 gets a pair, then the number will become a perfect square.

∴ 1008 can be multiplied with 7 to obtained a perfect square.

1008×7 = 7056 = 2x2x2x2x3x3x7x7

∴ 1008×7 = 7056 is a perfect square.

∴ √7056 = 2x2x3x7 =84

4) 2028

2028= 2x2x3x13x13

Here prime factor 3 does not have its pair, If 3 gets a pair, then the number will become a perfect square. 1

∴ 2028 has to be multiplied with 3 to obtain a perfect square.

∴ 2028 x 3 = 6084 is a perfect square

2028 x 3 = 6084= 2x2x3x3x13x13

∴ √6084 = 2x3x13 = 78

5) 1458

1458= 2x3x3x3x3x3x3

Here, prime factor 2 does not have its pair, If 2 gets a pair, then the number will become a perfect square.

∴ 1458 has to be multiplied with 2 to obtained a perfect square.

∴ 1458 x 2 = 2916 is a perfect square

1458 x 2 = 2916= 2x2x3x3x3x3x3x3

∴ √2916 = 2x3x3x3 = 54

6) 768

768= 2x2x2x2x2x2x2x2x3x2

Here, prime factor 3 does not have its pair, if 3 gets a pair, then the number will become a perfect square.

∴ 768×3 = 2304 is a perfect square

768 x 3 = 2304

= 2x2x2x2x2x2x2x2x3x3

∴ √2304 = 2x2x2x2x3 = 48

Squares And Square Roots Questions and Answers KSEEB Maths

6. For each of the following numbers, find the smallest whole number by which it should be divided so as to get a perfect square number. Also find the square root of the square number so obtained.
1) 252
2) 2925
3) 396
4) 2645
5) 2800
6) 1620

Solution:

1) 252

252 = 2x2x3x3x7

Here, prime factor 7 does not have its pair.

If we divide this number by 7, then the number will become a perfect square.

∴ 252 has to be divided by 7 to obtain a perfect square

252 ÷ 7 = 36 is a perfect square

36 = 2 x 2 x 3 x 3,

∴ √36 = 2×3 = 6

2) 2925

Here, prime factor 13 does not have its pair. If we divide this number by 13, then the number will become a perfect square.

∴ 2925 has to be divided by 13 to obtain a perfect square.

2925 ÷ 13 = 225 is a perfect square

225= 3x3x5x5

∴ √225= 3×5=15

3) 396

396 = 2 x 2 x 3 x 3 x 11

Here, prime factor 11 does not have its pair.

If we divide this number by 11, then the number will become a perfect square.

∴ 396 has to be divided by 11 to obtain a perfect square.

396 ÷ 11 = 36 is a perfect square

36 = 2x2x3x3

∴ √36 = 2×3 = 6

4) 2645

If we divide this number by 5, then the number will become a perfect square.

∴ 2645 has to be divided by 5 to obtain a perfect square.

2645 ÷ 5 = 529 is a perfect square

529 = 23 x 23

∴ √529 = 23

5) 2800

Here, prime factor 7 does not have its pair. If we divide this number by 7, then the number will become a perfect square.

∴ 2800 has to be divided by 7 to obtain a perfect square.

2800 ÷ 7 = 400 is a perfect square

400= 2x2x2x2x5x5

∴ √400 = 2x2x5 = 20

6) 1620

Here, prime factor 5 does not have its pair.

If we divide this number by 5, then theh number will become a perfect square.

∴ 1620 has to be divided by 5 to obtain a perfect square

1620÷5 = 324 is a perfect square.

324= 2x2x3x3x3x3

∴ √324 = 2x2x3 = 18

7. The student of class 8 of a school donated Rs. 2401 in all, for prime minister’s National Relief Fund. Each student donated as many rupees as the number of students in the class. Find the number of students in the class.

Solution: It is given that each student donated as many rupees as the number of students of the class. Number of students in the class will be the square root of the amount donated by the students of the class.

The total amount of donation is ₹2401.

Number of students in the class = √2401

2401 = 7x7x7x7

√2401 = 7×7 = 49

Hence, the number of students in the class is 49.

KSEEB Class 8 Maths Chapter 5 Squares and Square Roots 

8. 2025 plants are to be planted in a garden in such a way that each row contains as many plants as the number of rows. Find the num-ber of rows and the number of plants in each row.

Solution: It is given that in the garden, each row contains as many plants as the number of rows.

Hence, number of rows = Number of plants in each row

Total number of plants = Number of rows x

Number of plants in each row.

Number of rows x Number of plants in each row = 2025

\((Number of rows)^2\) = 2025

Number of rows = √2025

2025 = 5x5x3x3x3x3

∴ √2025 = 5x3x3= 45

Thus, the number of rows and the number of plants in each row is 45.

9. Find the smallest square number that is divisible by each of the numbers 4,9 and 10.

Solution: The number that will be perfectly divisible by each one of 4, 9, 10 is their LCM.

The LCM of these numbers is as follows

LCM of4, 9, 10= 2x2x3x3x5 = 180

Here, prime factor 5 does not have its pair,

∴ 180 is not a perfect square, If we multiply 180 with 5, then the number will become a perfect square.

∴ 180 should be multiplied with 5 to obtain a perfect square.

Hence, the required square number is

180 x 5 = 900

10. Find the smallest square number that is divisible by each of the numbers 8,15 and 20.

Solution: The number that is perfectly divisible by each of the numbers 8,15 and 20 is their LCM.

LCM of 8, 15 and 20 = 2 x 2 x 2 x 3 x 5 = 120

Here, prime factors 2,3 and 5 do not have their respective pairs,

∴ 120 is not a perfect square.

∴ 120 should be multiplied by 2 x 3 x 5 ie 30, to obtain a perfect square.

Hence, the required square number is 120 x 2 x 3 x 5 = 3600

KSEEB Class 8 Maths Chapter 5 Solved Problems Square And Square Roots Excercise 5.4

1. Find the square root each of the following numbers by Division Method.

1)2304

∴ √2304=48

2) 4489

∴ √4489=67

3) 3481

∴ √3481=59

4) 529

∴ √529=23

5) 3249

∴ √3249=57

6) 1369

∴ √1369=37

7) 5776

∴ √5776=76

8) 7921

∴ √7921=89

9) 576

∴ √576=34

10) 1024

 

∴ √1024=32

11) 3136

∴ √3136=56

12) 900

∴ √900=30

2. Find the number of digits in the square root of the following numbers (without any calculation)
1) 64
2) 144
3) 4489
4) 27225
5) 390625

Solution:

1) By placing bars, we obtain

\(64=\overline{64}\)

Since there is only one bar, the square root of 64 will have only one digit in it.

2) 144

= \(\overline{144}\)

Since there are two bars, the square root of 144 will have 2 digits in it.

3) 4489

= \(\overline{4489}\)

Since there are two bars, the square root of 4489 will have 2 digits in it.

4) 27225

= \(\overline{27225}\)

Since there are three bars, the square root of 27225 will have three digits in it.

5) 390625

= \(\overline{390625}\)

Since there are three bars, the square root of 390625 will have three digits in it.

3. Find the square root of the following decimal numbers.
1) 2.56
2) 7.29
3) 51.84
4) 42.25
5) 31.36

Solution:

1)2.56

∴ √2.56=1.6

2)7.29

∴ √7.29=2.7

3) 51.84

∴ √51.84=7.2

4)42.25

∴ √42.25=6.5

5)31.36

∴ √31.36=5.6

4. Find the least number which must be subtracted from each of the following numbers so as to get a perfect square so obtained.
1) 402
2) 1989
3) 3250
4) 825
5) 4000

Solution:

1) 402

The remainder is 2, it represents that the square of 20 is less than 402 by 2

∴ a perfect square will be obtained by subtracting 2 from the given number 402.

∴ required perfect square = 402 – 2 = 400

∴ √400 = 20

ii) 1989

The remainder is 53, it represents that the square of 44 is less than 1989 by 53.

∴ a perfect square will be obtained by subtracting 53 from the given number 1989

∴ required perfect square = 1989 – 53

= 1936

∴ √l936 = 44

3) 3250

The remainder is 1, it represents that the square of 57 is less than 3250 by 1.

∴ a perfect square can be obtained by subtracting 1 from the given number 3250.

∴ required perfect square = 3250 – 1 = 3249

∴ √3249 =57

4) 825

The remainder is 41, it represents that the square of 28 is less than 825 by 41.

∴ a perfect square can be calculated by subtracting 41 from the given number 825.

∴ required perfect square = 825 – 41 = 784 and √784 = 28

5) 4000

The remainder is 31, it represents that the square of 63 is less than 4000 by 31.

∴ a perfect square can be obtained by subtracting 31 from the given number 4000.

∴ required perfect square = 4000 -31 = 3969 and √3969 = 63

KSEEB Maths Class 8 Squares and Square Roots Notes

5. Find the least number which must be added to each of the following numbers so as to get a perfect square. Also, find the square root of the perfect square so obtained
1) 525
2) 1750
3) 252
4) 1825
5) 6412

Solution:

1) 525

The remainder is 41.

It represents that the square of 22 is less than 525. Next number is 23 and \(23^2 = 529\)

Hence, number to be added to 525

= \(23^2 – 525 = 529 – 525 = 4\)

The required perfect square is 529 and

√529 = 23

2) 1750

The remainder is 69.

It represents that the square of41 is less than 1750, The next number is 42 and \(42^2 = 1764\)

Hence, number to be added to 1750

= \(42^2 – 17250\)

= 1764- 1750= 14

∴ The required perfect square is 1764 and √l764 = 42

3) 252

The remainder is 27. It represents that the square of 15 is less than 252.

The next number is 16 and \(16^2 = 256\)

Hence, number to be added \(252 = 16^2- 252

= 256 – 252 = 4\)

The required perfect square is 256 and

√256 = 16

4) 1825

The remainder is 61.

It represents that the square of42 is less than 1825

The next number is 43 and \(43^2 = 1849\)

Hence, number to be added to \(1825 = 43^2 – 1825\)

= 1849 – 1825 = 24

The required perfect square is 1849 and

√l849 = 43

5) 6412

The remainder is 12.

It represents that the square of 80 is less than 6412. The next number is 81 and \(81^2 = 6561\)

Hence, number to be added to \(6412 = 81^2- 6412 = 6561 – 6412= 149\)

The required perfect square is 6561 and

√6561 = 81

KSEEB Class 8 Maths Key Concepts Of Squares And Square Roots

6. Find the length of the side of a square whose area is \(441m^2\).

Solution: Let the length of the side of the square be x m.

Area of square = \(x^2 = 44lm^2\)

x= √441

∴ x=21m.

Hence, the length of the side of the square is 21m

7. In a right triangle ABC, [B = 90° a) If AB = 6cm, BC = 8cm, find AC b) If AC = 13cm, BC = 5cm, find AB

 Solution: a) △ABC is right angled at B.

∴ by applying Pythagoras theorem, we obtain

\(AC^2 = AB^2 + BC^2\) \(AC^2 = (6cm)^2 + (8cm)^2\) \(AC^2 = (36 + 64)^2 = 100cm^2\)

AC = √100 = 10 cm

b) △ABC is right angled at B.

∴by applying Pythagoras theorem, we obtain

\(AC^2 = AB^2 + BC^2\) \((13)^2 = (AB)^2 + (5)^2\) \(AB^2 = 13^2 – 5^2 = 169 – 25 = 144 cm^2\)

AB = √l44 cm= 12cm

AB = 12cm

8. A gardener has 1000 plants. He wants to plant these in such away that the number of rows and the number of columns remain same. Find the minimum number of plants he needs more for this.

Solution: It is given that the gardener has 1000 plants. The number of rows and the number of columns is the same. We have to find the number of more plants that should be there, so that when the gardner plants them, the number of rows and columns are same. That is, the number which should be added to 1000 to make it a perfect square has to be calculated.

The remainder is 39, it represents that the square of 31 is less than 1000.

The next number is 32 and 322 = 1024

Hence, number to be added to 1000 to make it a perfect square = \(32^2-1000=1024-1000=24\)

Thus, the required number of plants is 24.

9. These are 500 children in a school. For a P.T. drill they have to stand in such a manner that the number of rows is equal to number of columns. How many children would be left out in this arrangement?

Solution: It is given that there are 500 children in the school. They have to stand for a P.T. drill such that the number of rows is equal to the number of columns. The number of children who will be left out in this arrangement has to be calculated. That is, the number which should be subtracted from 500 to make it a perfect square has to be calculated.

The remainder is 16. It shows that the square of 22 is less than 500 by 16.

∴ If we subtract 16 from 500, we will obtain a perfect square.

Required perfect square = 500 – 16 = 484 Thus, the number of children who will be left out is 16.

Chapter 5 Squares And Square Roots Solutions KSEEB Maths Additional Problems

1. Which of the following are perfect squares
a) 16
b) 54

Solution:

a) 16

Prime factors of 16 are = 2x2x2x2

Prime factors occur in pairs of like factors, so 16 is a perfect square.

b) 54

Prime factors of 54 are

= 2x3x3x3

= 2 x (3 x 3) x 3

We observe here that there are three 3’s, two form a pair but one is left. 2 also does not have a like factor to form a pair with, thus 54 is not a perfect square.

2. Find the greatest number of 3 – digit which is a perfect square.

Solution: w.k.t. the greatest number of three digit is 999. But 999 is not a perfect square so we must find the smallest number which must be subtracted from 999 to make it a perfect square.

This shows that if we subtract 38 from 999, the resulting number will be a perfect square.

Thus 999 – 3 = 961 is the greatest number of 3 digits which is a perfect square.

3. Find the square root of the following decimals.
a) 1043.29
b) 0.000676

Solution:

a) 1043.29

 

∴ √1043.29 = 32.3

b) 0.000676

∴ √0.000676=0.026
4. Write a Pythagorean triplet whose one number is 6.

Solution: For any natural number m > 1, 2m, \(m^2 – 1 and m^2+ 1\) forms a pythagorean triplet.

If we take \(m^2 – 1 = 6\)

then, \(m^2 = 6 + 1 = 7\)

the value of m will not be an integer.

If we take \(m^2 + 1 = 6\)

then \(m^2 = 6 – 1 = 5\)

Again the value of m will not be an integer So, let 2m = 6

m= 3

thus, \(m^2 – 1 = 3^2 – 1 = 8\)

\(m^2 + 1 = 3^2 + 1 = 10\)

So, the required triplet is 6,8,10

KSEEB Class 8 Maths Chapter 5 Squares And Square Roots

5. Rahul walks 12 m North from his house and turns west to walk 35m to reach his friend’s house. While returning, he walks diagonally from his friend’s house to reach back to his house. What distance did he walk while returning?

Solution: Applying Pythagoras in △ABC we have

\( AC^2 = AB^2 + BC^2\) \( AC^2 = 35^2 + 12^2\) \( AC^2 = 1225 + 144\) \( AC^2 = 1369\)

AC= √1369 = 37m

6. Find three numbers in the ratio 2:3:5, the sum of whose squares is 608.

Solution: Let three numbers be 2x, 3x and 5x According to question,

\( (2x)^2 + (3x)^2 + (5x)^2 = 608\) \( 4x^2 + 9x^2 + 25x^2 = 608\) \( 38x^2 = 608\) \( x^2 = 608 /38\) \( x^2= 16\)

X= √16

x = √2x2x2x2 = 4 Numbers are 8, 12 and 20

7. The area of a square plot is 101 1/400 \( m^2\) . Find the length of one side of the plot.

Solution: Let the length of one side be x m Then area of square = \( x^2\)

∴ \(x^2 = 101 1/400\)

\(x^2=40401/400\)

x= √40401/400

x= 201/20

x=101/20

So, length of each side = 10 1/20m

8. The area of a rectangular field whose length is twice its breadth is \(2450m^2\). Find the perimeter of the field.

Solution: Let the breadth of the field be x metres, then length of the field is 2x metres.

∴ area of the rectangular field = length x breadth

= \(2x x X = (2x^2)m^2\)

Given that area is \( 2450m^2\)

\( 2x^2 = 2450\) \( x^2=2450/2\)

⇒ X = √l225 or x = 35 m

Hence breadth = 35m and length = 35 x 2 = 70m

Perimeter of the field =2 (l + b)

= 2(70 + 35)m

= 2 x 105m = 210m

9. 13 and 31 is a strange pair of numbers such that their squares 169 and 961 are also mirror images of each other. Find two other such pairs?

Solution: As per the question, 13 and 31 are palindrome and llly its square will also be a palindrome. Now if I take 12 and 21 then there square will be 144 and 441

Hence, 12 and 21 are such pair.

10. During a mass drill exercise 6,250 students of different schools are arranged in rows such that the number of students in each row is equal to the number of rows. In doing so, the instructor finds out that 9 children are left out. Find the number of children in each row of the square.

Solution: Total number of students = 6250 Number of students forming a square = 6250-9 = 6241

Thus, 6241 students form a big square which has number of rows equal to the number of students in each row. Let the number of students in each row be x, then the number of rows

=  x

∴ X x X = 6241

\( x^2 = 6241\)

⇒ x= √6241

x=79

Hence, there are 79 students in each row of the square formed.

KSEEB Maths Class 8 Squares and Square Roots

11. Find the perfect square numbers between 40 and 50.

Solution: Perfect square numbers between 40 and 50 = 49

12. Which of the following \( 24^2,49^2, 77^2, 131^2 or 189^2\) end with digit 1?

Solution: Only \( 49^2, 131^2and 189^2\) end with digit 1.

13. Check whether (6, 8, 10) is a Pythagorean triplet.

Solution: 2m, \( m^2 -1 and m^2 + 1\) represent the Pythagorean triplet.

Let 2m = 6 ⇒ m = 3

\(m^2 – 1 = 3^2 – 1 = 9 – 1 = 8\) and

\(m^2 + 1 = 3^2+ 1 = 9+ 1 = 10\)

Hence (6,8,10) is a pythagorean triplet.

14. Find the square root of the following using successive subtraction of odd numbers starting from 1
a) 81
b) 25

Solution:

a) 81

81 – 1 = 80,

80 – 3 = 77,

77 – 5 = 72,

12 – 1 = 65,

65 – 9 = 56,

56 – 11 = 45,

45 – 13 = 32,

32- 15= 17,

17-17 = 0

We have subtracted 9 times to get 0.

∴ √81 = 9

b) 25

25-1 =24,

24-3 = 21,

21 – 5 = 16,

16-7 = 9,

9-9 = 0

We have subtracted 5 times to get 0

∴ √25 = 5

15. Find the cost erecting a fence around a square field whose area is 9 hectares if fencing costs ₹35 per metre.

Solution: Area of the square field =\( (9 x 10000)m^2\)

= \(90000m^2\)

length of each side of the field = √90000 m

= 300m

Perimeter of the field = (4 x 300)m = 1200m Cost of fencing = ? (1200 x35) = ₹42000

16. In an auditorium, the number of rows is equal to the number of chairs in each row, If the capacity of the auditorium is 2025, find the number of chairs in each row.

Solution: Let the number of chairs in each row be x

Then, the number of rows =x

Total number of chairs in the auditorium

= \((x x x) = x^2\)

but, the capacity ofthe auditorium = 2025 (given)

∴ x^2 =2025 = 5x5x3x3x3x3

x = 5x3x3 = 45

Hence, the number of chairs in each row = 45.

17. A ladder 10m long rests against a vertical wall. If the foot of the 1 adder is 6m away from the wall and the ladder just reaches the top of the wall, how high is the wall?

Solution: Let AC be the ladder

∴ AC = 10m,

Let BC be the distance between the foot of the ladder and the wall,

∴ BC = 6cm

△ABC forms a right-angled triangle, right-angled at B

By Pythagoras theorem,

\(AC^2 + AB^2 + BC^2\) \(10^2 = AB^2 + 6^2\) \(AB^2 = 10^2 – 6^2\)

= 100 – 36 = 64

⇒AB = √64 = 8m

Hence, the wall is 8m high.

KSEEB Class 8 Maths Solutions For Chapter 4 Understanding Quadrilaterals Points To Remember

Polygons: A simple closed curve made up of only line segments.

A line segment connecting two non-consecutive vertices of a polygon is called a diagonal.

Convex Polygon: The measure of each angle is less than 180°.

Concave Polygon: The measure of at least one angle is more than 180°.

Read and Learn More KSEEB Solutions for Class 8 Maths

For a regular polygon of n sides (n≥3)

Each exterior angle=[360/n]°

Each interior angle = [2n-4/n x 90]°

Interior angle+exterior angle = 180°.

For a convex polygon of n sides

Sum of all exterior angles=360°

Sum of all interior angles =(2n-4)x90°

Number of diagonals of n sides (n>3)
= n(n-3)/2
A triangle has no diagonals

Quadrilaterals: Polygon having four sides

Element of Quadrilateral:

Sides: Lined segments joining the points.

Vertices: Point of intersection of two consecutive sides.

Opposite sides: Two sides of a quadrilateral having no common end point.

Opposite angles: Two angles of a quadrilateral not having a common area.

Diagonals: Line segment obtained by joining the opposite vertices.

Adjacent angles: Two angles of a quadrilateral having a common arm

Adjacent sides: Two sides of a quadrilateral having a common end point.

Parallelogram: A quadrilateral with each pair of opposite sides parallel.
1) Opposite sides are equal
2) Opposite angles are equal
3) Diagonals bisect one another

Rhombus: A parallelogram with sides of equal length

1) All the properties of a parallelogram
2) Diaganols are perpendicular to each other.

Rectangle: a parallelogram with a right angle

1) All the properties of a parallelogram.
2) Each of the angles is a right angle.
3) Diagonals are equal.

Square: A rectangle with sides of equal length,
1) All the properties of a parallelogram, rhombus and a rectangle.

Kite: A quadrilateral with exactly two pairs of equal consecutive sides.

1) The diagonals are perpendicular to one antoher

2) One of the diagonals bisects the other

3) In the figure m∠R=m∠P

but m∠Q = = m∠S

KSEEB Class 8 Maths Chapter 4 Understanding Quadrilaterals Solutions 

Trapezium: A quadrilateral having exactly one pair of parallel sides.

Understanding Quadrilaterals Solutions KSEEB Class 8 Maths Exercise 4.1

1) Given here are some figures

Classify each of them on the basis of the
following.
a) Simple curve
b) Simple closed curve
c) polygon
d) Convex polygon
e) Concave polygon
Solution:

a) Simple curve – 1, 2, 5, 6, 7
b) Simple closed curve – 1, 2, 5, 6, 7
c) Polygon – 1,2
d) Convex polygon-2
e) Concave polygon-1

2) How many diagonals does each of the following have?
a) Convex quadrilateral
Solution: There are 2 diagonals in a convex quadrilateral
b) A regular hexagon
Solution: There are 9 diagonals in a regular hexagon
c) A triangle
Solution: A triangle does not have any diagonal in it.|

3)What is the sum of the measures of the angles of a convex quadrilateral? Will this property hold if the quadrilateral is not convex? (Make a non-convex quadrilateral and try!)

Solution: The sum of the measure of the angles of a convex quadrilateral is 360° as a convex quadrilateral is made of two triangles.

Here ABCD is a convex quadrilateral made of two triangles △ABD and △BCD

∴ The sum of all the interior angles of this quadrilateral will be same as the sum of all the interior angles of these two triangles ie 180° +180° 360°

Yes, this property also holds true for a quadrilateral which is not convex, this is because any quadrilateral can be divided into two triangles.

Here, again, ABCD is a concave quadrilateral, made of two triangles

△ABD and △BCD

∴ sum of all the interior angles of this quadrilateral will also be 180° +180°= 360°

4. Examine the table. (Each figure is divided into triangles and the sum of the angles deduced from that)

What can you say about the angle sum of a convex polygon with number of sides?
a) 7
b) 8
c) 10
d) n

Solution: From the table, it can be observed that the angle sum of a convex polygon of n sides is (n-2)x180°. Hence, the angle sum of the convex polygons having number of sides as
above will be as follows:

a) (7-2)x180°=900°

b) (8-2)x180°=1080°

c) (10-2)x180°=1440°

d) (n-2)x180°

5) What is a regular polygon? State the name of a regular polygon of
1080°
1) 3 sides
2) 4 sides
3) 6 sides

Solution: A polygon with equal sides and equal angles is called a regular polygon.

i) Equilateral triangle

ii) Square

 

iii)Regular Hexagon

6) Find the angle measure x in the following figures.
a)

Sum of the measures of all interior angles of a quadrilateral is 360°.

∴ In the given quadrilateral

50°+130°+120°+x=360°

300°+x=360°

x=360°-300=60°
b) 

From the figure, it can be concluded that,

90° +a 180° (linear pair)

a=180°-90=90°

Sum of the measures of all interior angles of a quadrilateral is 360°.

∴ In the given quadrilateral

60° +70°+x+90°= 360°

220°+x=360°

x=360°-220=140°
c)
Solution: From the figure, it can be concluded that

70°+a=180° (linear pair)

a=110°

60° +b=180° (linear pair)

b=120°

Sum of the measures of all interior angles of a pentagon is 540°

∴In the given pentagon

120°+110°+30°+x+x=540°

260°+2x = 540°

2x=540°-260°-280°

x=280/2=140°
d) 
sum of the measure of all interior angles of a pentagon is 540°

5x=540°

x=108°

7) Find x+y+z
a)
Solution:

x+90°= 180° (linear pair)

x=90°

z+30=180° (linear pair)

z=150°

y=90°+30° (Exterior angle theorem)

y =120°

∴ x+y+z 90+150+120=360°

b)Find x+y+z+w

Solution: Sum of the measures of all interior angles of a quadrilateral is 360°

∴In the given quadrilateral

a+60° +80°+120°=360°

a=360-260=100°

x+120° 180° (linear pair)

x = 60°

y+80°=180°

∴ y=100°

z+60°=180°

∴ z=120°

w+100°=180°

∴ w=80°

sum of the measures of all interior angles

= x+y+z+w

=60°+100+120°+80°

x+y+z+w = 360°

KSEEB Class 8 Maths Chapter 4 Understanding Quadrilaterals Exercise 4.2

1) Find x in the following figures.
a)

Solution: W.K.T. the sum of all exterior angles of any polygon is 360°

125°+125°+x = 360°

x+250°= 360°

⇒ x=360°-250°=110°

⇒ x=110°

b)

Solution:

60°+90°+70°+x+90° = 360°

310°+x = 360°

x=360-310 = 50°

KSEEB Class 8 Maths Chapter 4 Solved Problems

2) Find the measure of each exterior angle of a regular polygon of
1) 9 sides
2) 15 sides

Solution:

i) Sum of all exterior angles of the given polygon = 360°

Each exterior angle of a regular polygon has the same measure.

Thus, measure of each exterior angle of a regular polygon of 9 sides = 360°/9 =40°

ii) Sum of all exterior angles of the given polygon=360°

Each exterior angle of a regular polygon has the same measure.

Thus, measure of each exterior angle of a regular polygon of 15 sides = 360°/15 = 24°

3)How many sides does a regular polygon have if the measure of an exterior angle is 24°?

Solution: Sum of all exterior angles of the given polygon=360º.

Measure of each exterior angle = 24°

Thus, number of sides of the regular polygon

= 360°/24 = 15°

4) How many sides does a regular polygon have if each of its interior angles is 165°?

Solution: Measure of each interior angle = 165°

Measure of each exterior angles = 180° – 165° = 15°

The sum of all exterior angles of any polygon is 360º.

Thus, number of sides of the polygon

= 360º/15 = 24

5) a) Is it possible to have a regular polygon with measure of each exterior angle is 22º?
b) Can it be an interior angle of a regular polygon? Why?

Solution:

The sum of all exterior angles of all polygon is 360°. Also in a regular polygon each exterior angle is of the same measure. Hence, if 360° is a perfect multiple of the given exterior angle, then the given polygon will be possible.

a) Exterior angle = 22°

360° is not a perfect multiple of 22º. Hence such polygon is not possible.

b) Interior angle = 22°

Exterior angle = 180-22=158°

Such a polygon is not possible as 360° is not a perfect multiple of 158°.

6) a) What is the minimum interior angle possible for a regular polygon?
b) What is the maximum exterior angle possible for a regular polygon?

Solution: Consider a regular polygon having the lowest possible number of sides (ie, an equilateral triangle). The exterior angle of this triangle will be the maximum exterior angle possible for any regular polygon.

Exterior angle of an equilateral triangle = 360°/3 = 120°

Hence maximum possible measure of exterior

angle for any polygon is 120°. Also, we know

that an exterior angle and an interior angle are always in a linear pair. Hence, minimum interior angle.

= 180°-120° = 60°

Understanding Quadrilaterals KSEEB Maths Exercise 4.3

1. Given a parallelogram ABCD, complete each statement along with the definition or property used
1) AD=—–
2)∠DCB=——
3) OC= —-
4) m∠DAB+M∠CDA

Solution:

i) In a parallelogram, opposite sides are equal in length.

∴ AD = BC

ii) In a parallelogram, opposite angles are equal in measure

∴ ∠DCB = ∠DAB

iii) In a parallelogram, diagonals bisect each other

Hence OC = OA

iv) In a parallelogram, adjacent angles are supplementary to each other

Hence m∠DAB+m∠CDA = 180º.

2)Consider the following parallelogram. Find the values of the unknown .x, y, z

1. 
Solution:

x+100° 180° (adjacent angles are supplementary)

x=80°

z = x = 80° (opposite angles are equal)

y = 100° (opposite angles are equal)

2. 50 + y —180° (Adjacent angles are supplementary)

Solution:

⇒ y = 130°

x = y = 130° (opposite angles are equal)

z = x = 130° (corresponding angles)

x = y = z = 130°

3.
Solution:

x = 90° (vertically opposite angles)

x + y + 30°=180° (angle sum property of triangle)

90 + y + 30 = 180°

⇒ y = 180 – 120 = 60°

z = y -60° (Alternate interior angles)

4.
Solution:

z = 80° (corresponding angles)

y = 80° (opposite angles are equal)

x + y = 180° (adjacent angles are supplementary)

x = 180-y = 180°-80°= 100°

∴ x = 100, y = z = 80°

5.

Solution:

y = 112° (opposite angles are equal)

x + y + 40° = 180° (Angle sum property of triangles)

x + 112 + 40°= 180°

x = 180 —152°= 28°

x = z = 28° (Alternate interior angles)

x = 28°,y = 112°, z = 28°

KSEEB Maths Class 8 Understanding Quadrilaterals Notes

3) Can a quadrilateral ABCD be a parallelogram if
1) ∠D + ∠B = 180°
2) AB = DC = 8cm, AD = 4cm and BC = 4.4cm
3) ∠A = 70° and ∠C = 65° ?

Solution: 1) For ∠D + ∠B = 180° , quadrilateral ABCD may or may not be a parallelogram. Along with this condition, the following conditions should also be fulfilled.

The sum of the measures of adjacent angles should be 180°.

Opposite angles should also be of same measures

2) No. opposite sides AD and BC are of different lengths.

3) No. opposite angles A and C have different measures.

4. Draw a rough figure of a quadrilateral that is not a parallelogram but has exactly two opposite angles of equal measure.

Solution: Here, quadrilateral ABCD (kite) has two of its interior angles

∠B and ∠D of same measures.

However, still the quadrilateral

ABCD is not a parallelogram as the measures of the remaining pair of opposite angles ∠A and ∠C are not equal.

5.The measures of two adjacent angles of a parallelogram are in the ratio 3 :2. Find the measure of each of the angles of the parallelogram.

Solution: Let the measures of two adjacent angles [A and [B of parallelogram ABCD are in the ratio of3 : 2. Let ∠A = 3X and ∠B = 2x

w.k.t The sum of the measures of adjacent angles is 180° for a parallelogram

A+B= 180°

3x+2x=180°

5x=180°

x=180°/5 = 36°

∠A = ∠C = 3x = 108° (opposite angles)

∠B = ∠D = 2x = 72° (opposite angles)

Thus, the measures of the angles of the parallelogram are 108°, 72°, 108° and 72°.

6. Two adjacent angles of a parallelogram have equal measure. Find the measure of each of the angles of the parallelogram.

Solution: Sum of adjacent angles = 180°

∠A + ∠B = 180°

2 ∠A = 180° (∴ ∠A = ∠B)

∠A = 90°

∠B = [A = 90°

∠C = ∠A = 90° (opposite angles)

∠D = ∠B = 90° (opposite angles)

Thus, each angle of the parallelogram measures 90°.

7. The adjacent figure HOPE is a parallelogram. Find the angle measures x, y & z, State the properties you use to find them.

Solution: y = 40° (alternate interior angles)

70° = z + 40° (corresponding angles)

z = 70°-40°

z = 30°

x + (z + 40°) = 180

(Adjacent pair of angles)

x+70° = 180°

x= 180°-70°

x=110°

8) The following figures GUNS and RUNS are parallelogram. Find x and y (lengths are in cm)

Solution: 1) WKT The length of opposite sides of a parallelogram are equal to each other

GU = SN SG = NU

3y – 1 = 26 3x=18

3y = 26 + 1 = 27 x=18/3=6

y = 9

Hence, the measure of x and y are 6cm and 9cm respectively.

2) WKT the diagonals of a parallelogram bisect each other

y + 7 = 20

y = 13

x+y=16

x+13=16

∴ x=3

Hence, the measures of x and y are 3cm and 13cm respectively

9)

Solution:

In the above figure both RISK and CLUE are parallelograms. Find the value of x.

Adjacent angles of a parallelogram are supplementary

In parallelogram RISK,

∠20°+ ∠ISK = 180°

∠ISK=180°-120°=60°

Also, opposite angles of a parallelogram are equal

In parallelogram CLUE,∠ULE=∠CEU=70°

The sum of the measures of all interior angles of a traingle is 180°.

x+60°+70° = 180°

x = 50°

10) Explain how this figure is a trapezium which of its two sides are parallel?

Solution: If a transversal line is intersecting two given lines such that the sum of the measures of the angles on the same side of transversal is 180°, then each other.

Here, LNML+LMLK = 180°

Hence, NM ∥ LK

As quadrilateral KLMN has a pair of parrallel lines, ∴ it is a trapezium

11) Find c in the following figure if AB∥DC

Solution: Given that, Given that, \(\overline{\mathrm{AB}} \| \overline{\mathrm{DC}}\)

∠B + ∠C = 180° (Angles on the same side of transversal)

120°+∠C = 180°

∠C = 60°

KSEEB Class 8 Maths key concepts of Understanding Quadrilaterals

12) Find the measure of p and s, if \(\overline{\mathrm{SP}} \| \overline{\mathrm{RQ}}\) in the following figure, (If you find m R is there more than one method to find m P?)

Solution: \(\left\lfloor\mathrm{P}+\left\lfloor\mathrm{Q}=180^{\circ}\right.\right.\)(angles on the same side4 of transversal)
\(\left\lfloor\mathrm{P}+130^{\circ}=180^{\circ}\right.\)

\(\left\lfloor\underline{P}=50^{\circ}\right.\) \(\left\lfloor\underline{R}+\left\lfloor\underline{S}=180^{\circ}\right.\right.\)

(angles on the same side of transversal)

\(90^{\circ}+\left\lfloor\mathrm{S}=180^{\circ}\right.\) \(\mathrm{S}=180^{\circ}-90^{\circ}=90^{\circ}\)

Yes, there is one more method to find the measure of \(\mathrm{m} \mid \mathrm{P}\)

m R and m Q are given, after finding m S the angle sum property of a quadrilateral can be applied to find m P.

KSEEB Maths Class 8 Chapter 4 Understanding Quadrilaterals Exercise 4.4

1. state whether True or False.

1) All rectangles are squares

Solution: False, All squares are rectangles but all rectangles are not square.

2) All rhombuses are parallelograms

Solution: True, opposite sides of a rhombus are equal and parallel to each other.

3) All squares are rhombuses and also rectangles.

Solution: True, all squares are rhombuses as all sides of a square are of equal lengths.
All squares are also rectangles as each internal angle measure 90°

4) all squares are not parallelograms.

Solution: False, All squares are parallelograms as opposite sides are equal and parallel.

5) All kites are rhombuses.

Solution: False, A kite does not have all sides of the same length.

6) All rhombuses are kites

Solution: True, A rhombus also has two distinct consecutive pairs of sides of equal length.

7) All parallelograms are trapeziums

Solution: True, All parallelograms have a pair of parallel sides.

8) All squares are trapezium.

Solution: True, All squares have a pair of parallel sides

2. Identify all the quadrilaterals that have

1) Four sides of equal length.

Solution: Rhombus and square are the quadrilaterals that have 4 sides of equal length.

2) Four right angles

Solution: Square and rectangle are the quadrilaterals that have 4 right angles.

3) Explain how a square is

1) a quadrilateral

Solution: A square is a quadrilateral since it has four sides

2) a parallelogram

Solution: A square is a parallelogram since its opposite sides are parallel to each other

3) a rhombus

Solution: A square is a rhombus since its four sides are of the same length.

4) a rectangle

Solution: A square is a rectangle since each interior angle measures 90°

4. Name the quadrilaterals whose diagonals

1) bisect each other

Solution: The diagonals of a parallelogram, rhombus, square and rectangle bisect each other.

2) are perpendicular bisectors of each other

Solution: The diagonals of a rhombus and square act as perpendicular bisectors

3) are equal

Solution: The diagonals of a rectangle and square are equal.

5. Explain why a rectangle is a convex quadrilateral

Solution: In a rectangle, there are two diagonals both lying in the interior of the rectangle. Hence it is a convex quadrilateral.

6. ABC is a right angled triangle and O is the mid-point of the side opposite to the right angle. Explain why O is equidistant from A, B and C (The dotted lines are drawn additionally to help you)

Solution: Draw lines AD and DC such that AD ∥ BC

AB ∥ DC

AD = BC, AB = DC

ABCD is a rectangle as opposite sides are equal and parallel to each other and all the interior angles are of 90°.

In a rectangle diagonals are of equal length and also these bisect each other.

Hence AO = OC = BO = OD

Thus, O is equidistant from A, B and C.

KSEEB Class 8 Maths Key Concepts of Understanding Quadrilaterals Additional Problems

1. Two adjacent angles of a parallelogram are in the ratio 1: 3. Find its angles

Solution: Let the angles be x and 3x

then x + 3x = 180°

4x= 180°

⇒ x = 45° (so angles are 45° and 135°)

2. A photo frame is in the shape of a quadrilateral with one diagonal longer than the other. Is it a rectangle? Why or Why not?

Solution: No, it is not a rectangle. In a rectangle diagonals are equal.

3. A playground in the town is in the form of a kite. The perimeter is 106 metres. If one of its sides is 23 metres what are the lengths of other three sides?

Solution: A kite has two pairs of equal consecutive sides.

∴ one side = 23m

other side = 23m

Sum of rest two sides = 106- (23 + 23)

Length of other two = 60m

Sides are 30m and 30 m

Hence, the length of other three sides are 23m, 30m and 30m.

4. Find the measure of an exterior angle of a regular pentagon and an exterior angle of a regular decagon. What is the ratio between these two angles?

Solution: The measure of each exterior angle of n sided regular polygon= (360/n)°

Measure of exterior angle of a regular pentagon (360/5)° = 72°

Measure of exterior angle of a regular decagon =(360/10)° = 36°.

Ratio = 72°/36° =2/1

5. In rhombus BEAM, find ∠AME and ∠AEM

Solution: Since diagonals of rhombus bisect each other at right angles.

∴ ∠AOM = 90°

∠AOM+∠OAM+∠AMO=180°

90°+70°+∠AMO=180°

160°+∠AMO=180°

∠AMO=180°-160° ⇒ ∠AMO=20°

∠AME=∠AMO=20°

In △AEM

AE=AM (sides of rhombus are equal)

∠AME=∠AEM (Angles opposite to equal sides)

∴ ∠AEM =20°

6. In the given parallelogram YOUR, ∠RUO=120° and OY is extended to point S such that ∠SRY=50°, Find ∠YSR.

Solution: ∠RYO=∠RUO=120° (opposite angles are equal)

\(\left\lfloor\mathrm{RYS}+\left\lfloor\mathrm{RYO}=180^{\circ}\right. \text { (linear pair) }\right.\) \(\left\lfloor\text { RYS }+120^{\circ}=180^{\circ}\right.\)

⇒ ∠RYS=60°

In △RSY

∠YSR+∠RYS+∠SRY=180°

∠YSR+60°+50°=180°

∠YSR=180°-110°=70°

7. Two angles of a quadrilateral measures 75° each and the other two angles are equal. What is the measure of these two angles? Name the possible figure so formed

Solution: Let equal angles be x.

∴\(x+x+75^{\circ}+75^{\circ}=360^{\circ}\)

2x+150°=360°

2x=360°-150°=210°

⇒ x=105°

Each equal angle is 105°, the possible figure so formed is parallelogram.

8. ABCD is a parallelogram. The bisector of angle of A intersects CD at x and bisector of angle C intersects AB at Y. Is AXCY a parallelogram? Give reason

Solution:

∠A=∠C (opposite angles of parallelogram)

or \(\frac{\lfloor\mathrm{A}}{2}=\frac{\lfloor\mathrm{C}}{2} \text { or }\lfloor 1=\lfloor 2\)

but 2=3 (alternate angles)

∴ ∠1=∠3

But they are a pair of corresponding angles

∴ AX ∥ YC…….(1)

AY ∥ XC…….(2) (∵ AB ∥ DC)

from (1)and(2)

AXCY is a ∥gm

9. A diagonal of an parallelogram bisects an angle.
(1)Will it also bisect the other angle?
(2)It is a rhombus? Give reason.

Solution: Let ABCD is a ∥gm 1=2

1) ∠1=∠4 (alternate angles)

& ∠2=∠3 (alternate angles)

but ∠3=∠4

so, the diagonal will bisect the other angle.

2) ∠1=∠2 (Given)

∠2=∠3 (Alternate angles)

∴ ∠1=∠3

Hence CD=DA (sides opposite to equal angles are equal)

∴ ABCD is a rhombus.

10. In rectangle PQRS, ∠OPQ=48°, find ∠RSQ

Solution:

∠QPS = 90°

(As each angle of a rectangle is a right angle)

∠1 + 48°= 90°

⇒∠1 = 90°-48°

∠1 = 42°

Since in a rectangle diagonals are equal and bisect each other.

OS = OP = OQ = OR

⇒∠1 = ∠2 = 42°

Also ∠2 + ∠3 = 90° (each angle of a rectangle is a right angle)

⇒ 42°+∠3 = 90°

∠3 = 90°-42°= 48°

thus ∠RSQ = 48°

11. The angles of a quadrilateral are in the ratio 1 : 2 : 2 : 4. Find the measure of each angle.

Solution: Let the ratio constant hex, thus the measure of the given angles will be (lx)°, (2x)° (2x)° and (4x)°.

w.k.T, The sum of the angles of a quadrilateral is 360°.

1x°+ 2x°+2x° + 4x°= 360°

9x = 360°

x=360°/9 =40°

∴ The angles are

1x = 1×40 = 40°, 2x= 2x 40° = 80°

2x = 2×40 = 80° and 4x = 4×40 = 160°

Hence the angles are 40°, 80°, 80° and 160°.

12. In fig, ABCD is a quadrilateral with ∠B = 56° and ∠D = 104°. If ∠BAD and ∠BCD are in the ratio, 2 : 3, find the values of ∠1 and ∠2.

Solution: wkT, The sum of the angles of a quadrilateral is 360°

∴∠BAD + ∠56° + ∠BCD + 104°= 360°

(or) ∠BAD+ ∠BCD+ 160° = 360°

⇒ ∠BAD + ∠BCD = 200°

∴∠BAD +∠BCD = 2:3

Let the ratio constant be x.

∠BAD = 2x and ∠BCD = 3x

∴ 2x + 3x = 200

⇒ 5x = 200

x = 200/5 = 40°

∠BAD = 2x = 2 x 40 = 80°

∠BCD = 3x = 3 x 40 = 120°

but ∠BAD + 1 = 180°(linear pair)

80°+∠1 = 180°

∠1 = 180°-80°= 100°

and ∠BCD + ∠2 = 180°(linear pair)

120°+∠2 = 180°

∠2 = 180° -120°= 60°

13. In a parallelogram PQRS, the bisectors of ∠P & ∠Q meet at O. Find ∠POQ. P Q

Solution: Since PO and QO are the bisectors of P&[Q respectively.

∴ \(\left\lfloor 1=\frac{1}{2}\left\lfloor\mathrm { P } \& \left\lfloor 2=\frac{1}{2}\lfloor\mathrm{Q}\right.\right.\right.\)

but ∠P + ∠Q = 180°

\(\frac{1}{2}\left\lfloor\mathrm{P}+\frac{1}{2}\left\lfloor\mathrm{Q}=90^{\circ}\right.\right.\)

∠1+ ∠2 = 90°

but ∠1 + ∠2 + ∠POQ = 180°

90°+ ∠POQ = 180°

∠POQ = 180°-90° -90°

⇒ ∠POQ = 90°

14. The diagonals of a rhombus ABCD intersect at O, If AC = 8cm and BD = 6cm, find the length of its sides.

Solution: Since diagonals of a rhombus bisect each other at right angle.

AC=8cm ⇒ OA = OC = 4cm

and BD = 6cm

⇒ OB = 3cm

Also ∠BOC = 90°

Hence △BOC is a right angled triangle.

∴ \(B C^2=O B^2+O C^2\)

(using pythagoras theorem)

⇒\(\mathrm{BC}^2=4^2+3^2=16+9\)

\(\mathrm{BC}^2=25\) \(\mathrm{BC}=5 \mathrm{~cm}\)

Length of each side of the rhombus is 5cm.

15. Both the pairs of opposite angles of a quadrilateral are equal and supplementary.

Solution: Since pairs opposite angles of a quadrilateral are equal and supplementary.

∴ x+x=180°

2x=180°

⇒ x=90°

Also, y+y=180°

2y=180°⇒ y=90°

Each angle =90°

16. Length and breadth of a rectangular wire are 9cm and 7cm respectively. If the wire is bent into a square, find the length of its side.

Solution: Perimeter of the rectangle

= 2 (l + b)

= 2 (9 + 7) = 2 x 16 = 32cm

Now perimeter of the square

= perimeter of rectangle = 32cm

side of the square = 32/4 = 8cm

Hence, the length of the side of square = 8cm.

17. The sides AB & CD of a quadrilateral ABCD are extended to points P & Q respectively. Is ∠ADQ + ∠CBP =∠A + ∠C? Give reason.

Solution: Join AC, then

∠CBP = ∠BCA + ∠BAC &

∠ADQ = ∠ACD + ∠DAC

(Exterior angles of △le )

∠CBP +∠ADQ = ∠BCA + ∠BAC + ∠ACD + ∠DAC

= (∠BCA + ∠ACD) + (∠BAC + ∠DAC)

= ∠C + ∠A

18. ABCD is a parallelogram with ∠A = 80°. The internal bisectors of ∠B & ∠C meet at O. Find the measure of the three angles of △BCO•

Solution: ∠C = ∠A

(opposite angles of a parallelogram are equal)

∠C = 800 (Given ∠C = 80°)

OCB = 1/2 ∠C =1/2×80 = 40°

∠B = 180° – ∠A (sum of interior angles on the same side of the transversal is 180°)

= 180° – 80° = 100°

∠CBO = 1/2B =1/2xl00°

= 50° ∠BOC = 180°- ( ∠OBC + ∠CBO)

(angle sum of a △le )

= 180° – 40° + 50° = 180° – 90° = 90°

∴The three angles of the △le BCO namely ∠OCB, ∠CBO, ∠BCO are 40°, 50° and 90° respectively.

19. The ratio of two sides of a parallelogram is 4:3. If its perimeter is 56cm, find the lengths of its sides.

Solution: Let the lengths of two sides of parallelogram be 4x cm and 3x cm respectively.

Then, its perimeter

= 2(4x + 3x)cm = 14x cm

∴ 14x = 56 ⇒ x = 56/14 = 4

∴ one side = (4 x 4)cm= 16cm & other side

= (3 x 4cm) = 12cm

KSEEB Class 8 Maths Solutions For Chapter 3 Linear Equations In One Variable Points To Remember

An algebraic equation is an equality involving variables. It says that the value of the expression on one side of the equality sign is equal to the value of the expression on the other side.

A linear equation may have for its solution any rational number.

An equation may have linear expressions on both sides.

Just as numbers, variables can also be trSolution-posed from one side of the equation to the other.

Occasionally, the expressions forming equations have to be simplified before we can solve them by usual methods. Some equations may not even be linear to begin with, but they can be brought to a linear form by multiplying both sides of the equation by a suitable expression.

The utility of linear equations is in their diverse applications; different problems on numbers, ages, perimeters combinations of currency notes and so on can be solved using linear equations.

Read and Learn More KSEEB Solutions for Class 8 Maths

Linear Equations In One Variable Solutions KSEEB Class 8 Maths Exercise 3.1

1. Solve the following equations

1) x-2=7

Solution: Given x -2 = 7

By adding 2 to both sides we get

x – 2 + 2 = 7 + 2

x = 9

2) y + 3 = 10

solution: Given y + 3 = 10

By subtracting 3 from both sides we get

y + 3 – 3 = 10 -3

y = 7

3) 6 = z + 2

Solution:

Given 6 = z + 2

By subtracting 2 from both sides, we get

6 -2 = z + 2 -2

4 = z

⇒z = 4

4) \(\frac{3}{7}+x=\frac{17}{7}\)

Solution: Given \(\frac{3}{7}+x=\frac{17}{7}\)

By subtracting 3/7 from both sides.

\(\frac{3}{7}+x-\frac{3}{7}=\frac{17}{7}-\frac{3}{7}\)

⇒ \(x=\frac{17-3}{7}\)

⇒ \(x=14 / 7\)

5) 6x = 12

Solution: Given, 6x = 12

By dividing both sides with 6, we get

6x/6=12/6

⇒ x=12/6

⇒ x=2

6) \(\frac{t}{5}=10\)

Solution: Given, \(\frac{t}{5}=10\)

By multiplying both sides with 5, we get

\(\frac{t}{5} \times 5=10 \times 5\)

⇒ t=10×5

⇒ t=50

7)  \(\frac{2 x}{3}=18\)

Solution: Given,  \(\frac{2 x}{3}=18\)

By multiplying both sides with 3, we get

\(\frac{2 x}{3} \times 3=18 \times 3\)

⇒ 2x=54

By dividing both sides with 2, we get

\(\frac{2 x}{2}=\frac{54}{2}\)

⇒ x=27

KSEEB Class 8 Maths Chapter 3 Linear Equations In One Variable 

8)  \(1.6=\frac{y}{1.5}\)

Solution: Given, \(1.6=\frac{y}{1.5}\)

By multiplying both sides with 1.5 we get

\(1.6 \times 1.5=\frac{y}{1.5} \times 1.5\)

1.6×1.5 = y

2.40 = y

⇒  y =2.40

9) 7x -9 = 16

Solution: Given, 7x -9 = 16

By adding 9 to both sides, we get

7x -9 + 9 = 16 + 9

7x = 25

After dividing both sides by 7, we get

⇒  \(\frac{7 x}{7}=\frac{25}{7}\)

⇒  \(x=\frac{25}{7}\)

10) 14y – 8 = 13

Solution: Given, 14y – 8 = 13

By adding 8 to both sides, we get

14y -8 + 8 = 13 + 8

⇒ 14y = 21

After dividing both sides by 14, we get

\(\frac{14 y}{14}=\frac{21}{14}\) \(y=3 / 2\)

11) 17 + 6p = 9

Solution: Given, 17 + 6p = 9

After subtracting 17 from both sides we get

17 + 6p-17 = 9-17

⇒ 6/7 = -8

After dividing both sides by 6, we get

6p/6 = 8/6

⇒ \(p=\frac{\phi^+}{\phi_3} \)

⇒  p=4/3

12)  \(\frac{x}{3}+1=\frac{7}{15}\)

Solution: Given, \(\frac{x}{3}+1=\frac{7}{15}\)

By subtracting 1 from both sides we get

\(\frac{x}{3}+1-1=\frac{7}{15}-1\) \(\frac{x}{3}=\frac{7-15}{15}\)

⇒ \(\frac{x}{3}=\frac{-8}{15}\)

After multiplying both sides with 3, we get -8

⇒  \(\frac{x}{3} \times 3=\frac{-8}{15_5} \times \beta^1\)

\(x=\frac{-8}{5}\)

KSEEB Class 8 Maths Chapter 3 Solved Problems Linear Equations In One Variable Exercise 3.2

1)If you subtract \(\frac{1}{2}\) from a number and multiply the result by \(\frac{1}{2}\), you get\(\frac{1}{8}\), What is the number?

Solution: Let the number be x.

According to the question

\(\left(x-\frac{1}{2}\right) \times \frac{1}{2}=\frac{1}{8}\)

On multiplying both sides by 2, we obtain

\(\left(x-\frac{1}{2}\right) \times \frac{1}{2} \times 2=\frac{1}{8} \times 2\) \(x-\frac{1}{2}=\frac{1}{4}\)

Adding 1/2 on both sides we get

⇒ x-\(\frac{1}{2}+\frac{1}{2}=\frac{1}{4}+\frac{1}{2}\)

⇒ \(x=\frac{1+2}{4}=\frac{5}{4}\)

∴ The number is 3/4

2)The perimeter of a rectangular swimming pool is 15+m. Its length is 2m more than twice the breadth of the pool?

Solution: Let the breadth be Xm.

The length will be (2x+2)m

Perimeter of swimming pool

=2(l+b)=154m

=2(2x+2+x)=154m

2(3x+2)=154

dividing both sides by 2, we get

\(\frac{2(3 x+2)}{2}=\frac{15+}{2}\)

3x+2=77

subtracting 2 both sides, we get

3x+2-2=77-2

⇒ 3x=75

on dividing both sides by 3, we obtain

⇒ \(\frac{3 x}{3}=\frac{75}{3}\)

⇒ x=25

⇒ 2x+2=2×25+2

=50+2

=52

Hence, the breadth and length of the pool are 25m and 52m respectively.

3)The base of an isosceles triangle is \(\frac{4}{3}\)cm. The perimeter of the triangle is 4\( \frac{2}{15}\)Cm, What is the length of either of the remaining equal sides?

Solution: Let the length of equal sides be ‘x’ cm

Perimeter = x + x+ base = 4\( \frac{2}{15}\) cm

\( 2 x+\frac{4}{3}=\frac{62}{15}\)

on transposing 4/3 to RHS we obtain

\( 2 x=\frac{62}{15}-\frac{4}{3}=\frac{62-4 \times 5}{15}=\frac{62-20}{15}\)

⇒ \( 2 x=\frac{42}{15}\)

On dividing both sides by 2, we get

\(\frac{2 x}{2}=\frac{+2^{21^7}}{15_5} \times \frac{1}{2}\) \(x=\frac{7}{5}=1 \frac{2}{5}\)

∴ The length of equal sides is \(1 \frac{2}{5}\)cm.

KSEEB Class 8 Maths Solutions For Chapter 3 Linear Equations In One Variable

4) Sum of two numbers is 95. If one exceeds the other by 15, find the numbers.

Solution: Let one number be ‘x’

∴ The other number will be x +15

According to the question,

x +x + 15 = 95

2x + 15 = 95

Subtracting 15 on both sides, we get

2x + 15 -15 = 95 -15

2x = 80

On dividing both sides by 2, we get

2x/2=80/2

x=40

∴ x+15=40+15=55

Hence, the number are 40 and 55.

5) Two numbers are in the ratio 5 : 3. If they differ by 18, what are the numbers?

Solution: Let the common ratio between these numbers be ‘x’.

∴ The numbers will be 5x and 3x respectively.

Difference between these numbers =18

5x – 3x = 18

2x = 18

Dividing both sides by 2, we get

\(\frac{2 x}{2}=\frac{18}{2}\)

⇒ x = 9

∴ 1st number 5x = 5 x 9 = 45

2nd number 3x = 3 x 9 = 27

6) Three consecutive integers add up to 51. What are these integers?

Solution: Let three consecutive integers be x, x +1 and x +2.

Sum of these numbers

= x + x + 1 + x + 2 = 51

3x + 3 = 51

On transposing 3 to RHS we get

3x = 51-3

3x = 48

On dividing both sides by 3, we get

\(\frac{3 x}{3}=\frac{48}{3}\)

x = 16

x + 1 = 17

x + 2 = 18

Hence, the consecutive integers are 16,17 and 18.

7) The sum of three consecutive multiples of  8 is 888. find the multiples.

Solution: Let the three consecutive multiples of+8 be 8x, 8(x +1), 8(x + 2)

Sum of these numbers =

8x + 8(x +1) + 8(x + 2) = 888

8(x + x +1 + x + 2) = 888

8(3x + 3) =888

On dividing both sides by 8, we get

\(\frac{8(3 x+3)}{8}=\frac{888}{8}\)

3x + 3 =111

On transposing 3 to RHS, we get

3x = 111 – 3

3x = 108

On dividing both sides by 3, we get

\(\frac{3 x}{3}=\frac{108}{3}\)

1 st Multiple = 8x = 8 x 36 = 288

2nd Multiple = 8 (x +1) = 8 x (36 +1)

= 8 x 37 = 296

3rd Multiple = 8 (x + 2) = 8 (36 + 2)

= 8 x 38 = 304

Hence, the required numbers are 288,296 and 304.

8) Three consecutive integers are such that when they are taken in increasing order and multiplied by 2,3 and 4 respectively. They add up to 74. find these numbers.

Solution: Let three consecutive integers be x, x + 1, x + 2

According to the question,

2x + 3(x +1) + 4x(x + 2) = 74

2x + 3x + 3 + 4x + 8 = 74

9x+11 = 74

On transposing 11 to RHS, we get

9x = 74 -11

9x = 63

On dividing both sides by 9, we get

\(\frac{9 x}{9}=\frac{63}{9}\)

x=7

x + 1 = 7 + 1 = 8

x + 2 = 7 + 2 = 9

Hence, the numbers are 7, 8 and 9.

Linear Equations In One Variable Solutions KSEEB Class 8 Maths

9)The ages of Rahul and Haroon are in the ratio 5:7; four years later the sum of the ages will be 56 years. What are their present ages?

Solution: Let common ratio between Rahul’s age and Haroon’s age be x

∴ Age of Rahul and Haroon will be 5x years and 7x years respectively. After 4 years, the age of Rahul and Haroor will be (5x + 4) years and (7x + 4) years respectively.

According to the given question, after 4 years, the sum of the ages of Rahul and Haroon is 56 years.

∴(5x + 4 + 7x + 4) = 56

12x + 8 = 56

On transposing 8 to RHS we get

12x = 56-8

12x = 48

On dividing both sides by 12, we get

\(\frac{12 x}{12}=\frac{48}{12}\)

Rahul’s age = 5x years = (5 x 4) yrs = 20yrs.

Haroon’s age = 7x years =(7 x 4) yrs = 28 yrs.

10. The number of boys and girls in a class are in the ratio 7 : 5, The number of boys is 8 more than the number of girls. What is the total class strength?

Solution: Let the common ratio between the number of boys and number of girls be x.

Number of boys =7x

Number of girls = 5x

According to the given question

Number of boys = Number of girls + 8

∴7x=5x+8

On transposing 5x to LHS. We get

7x -5x = 8

2x = 8

On dividing both sides by 2 we get \(\frac{2 x}{2}=\frac{8}{2}\)

⇒ x = 4

Number of boys = 7x=74+=28

Number of girls = 5x = 5 x 4 = 20

Hence, total class strength

= 28 + 20 = 48 students

11. Baichung’s father is 26 years younger than Baichung’s grandfather and 29 years older than Baichung. The sum of the ages of all the three is 135 years. What is the age of each one of them?

Solution: Let Baichung’s father’s age be ‘x’ years.

∴The Baichung’s age and Baichung’s grandfather’s age will be (x -29) years and

(x + 26) years respectively. According to the

given question, the sum of the ages of these 3 people is 135 years.

∴x + x-29 + x + 26 = 135

3x-3 = 135

On transposing 3 to RHS we obtain

3x = 135 + 3

3x = 138

On dividing both sides by 3, we get

\(\frac{3 x}{3}=\frac{138}{3}\)

x=46

Baichung’s father’s age = x years = 46 years

Baichung’s age = (x -29) years

= (46-29) years = 17 years

Baichung’s grandfather’s age

= (x + 26) years = (46 + 26)years = 72 years

KSEEB Class 8 Maths Chapter 3 Solved Problems

12. fifteen years from now Ravi’s age will be four times his present age. What is Ravi’s present age?

Solution: Let Ravi’s present age be ‘x’ years  fifteen years late, Ravi’s age

= 4x His present age

x + 15 = 4x

On transposing x to RHS we get

15 = 4x -x

⇒15 = 3x

On dividing both sides by 3, we get

\(\frac{15}{3}=\frac{3 x}{3}\)

5=x

⇒ x=5

Hence, Ravi’s present age = 5 years.

13. A rational number is such that when you multiply it by \(\frac{5}{2}\) and add \(\frac{2}{3}\) to the product, you get \(\frac{7}{12} \), What is the number?

Solution: Let the number be x.

According to the given question,

\(\frac{5}{2} x+\frac{2}{3}=\frac{-7}{17}\)

On transposing \(\frac{2}{3}\) to RHS, we obtain

\(\frac{5}{2} x=\frac{-7}{12}-\frac{2}{3}\)

₹ \(\frac{5}{2} x=\frac{-7-(2 \times 4)}{12}\)

₹ \(\frac{5 x}{2}=\frac{-15}{12}\)

On multiplying both sides by\(\frac{2}{5}\) we get

\( x=\frac{-15}{12} \times \frac{2}{5}=\frac{-1}{2}\)

Hence, the rational number is \( -1 / 2\)

14. Lakshmi is a cashier in a bank, she has currency notes of denominations Rs. 100,₹50and ₹10 respectively. The ratio of the number of these notes is 2 : 3 : 5. The total cash with Lakshmi is ₹4,00,000. How many notes of each denomination does she have?

Solution: Let the common ratio between the numbers of notes of different denominations be x.

∴The numbers of ₹100 notes, ₹50 notes and

₹10 notes will be 2x, 3x and 5x respectively.

Amount of ₹100 notes

= ₹(l 00 x 2x) = ₹200x

Amount of ₹50notes = ₹(5Ox3x) = ₹150x

Amount of ₹10 notes = ₹ (I0x5x) = ₹50x

It is given that total amount is ₹4,00,000

∴ 200x + 150x + 50x = 400000

⇒ 400x = 400000

⇒ on dividing both sides by 400, we get

x = 1000

Number of ₹100 notes = 2x = 2 x 1000 = 2000

Number of ₹50 notes = 3x = 3 x 1000 = 3000

Number of ₹10 notes = 5x = 5 x 1000 = 5000

15. I have a total of ₹300 in coins of denomination ₹1, ₹2 and ₹5. The number of ₹2 coins is 3 times the number of ₹5 coins. The total number of coins is 160. How many coins of each denomination are with me?

Solution: Let the number of ₹5 coins be x.

Number of ₹2 coins

= 3 x Number of ₹ 5 coins = 3x

Number of ₹ 1 coins = 160 – (Number of coins of ₹5 and of₹2)

= 160 – (3x + x) = 160 -4x

Amount of ₹1 coins = ₹ [1 x (160-4x)]

= ₹ (160-4x)

Amount of 2 coins = ₹ (2 x 3x) = ₹ 6x.

Amount of 5 coins = ₹ (5 x x) = ₹ 5x.

It is given that the total amount is ₹ 300

160 – 4x + 6x + 5x = 300

160 + 7x = 300

On transposing 160 to RHS, we get

7x = 300 – 160

7x= 140

On dividing both sides by 7, we get

\( \frac{7 x}{7}=\frac{140}{7}\)

x = 20

Number of ₹1 coins = 160-4x= 160 -4 x 20

= 160 – 80 = 80

Number of ₹ 2 coins = 3x = 3 x 20 = 60

Number of ₹ 5 coins = x = 20

Linear Equations In One Variable KSEEB Maths

16. The organizers of an essay competition decide that a winner in the competition gets a prize of ₹ 100 and a participant who does
not win gets a prize of ₹ 25. The total prize money distributed is ₹ 3000. find the number of winners, if the total number of participants is 63.

Solution: Let the number of winners be x.

∴ The number of participants who did not win will be 63 – x.

Amount given to the winners = ₹(100 x X)

= ₹ 100x

Amount given to the participants who did not

win = ₹ 25 (63 – x)

= ₹(1575 -25x)

According to the given question.

100 x + 1575 – 25x = 3000

On transposing 1575 to RHS we get

75x = 3000 – 1575

75x = 1425.

On dividing both sides by 75, we get

\( \frac{75 x}{75}=\frac{1425}{75}\)

x = 19, Hence, number of winners =19.

Linear Equations In One Variable KSEEB Maths Exercise 3.3

1. Solve the following equations and check your results.
1) 3x = 2x + 18

Solution: 3x = 2x + 18

On transposing 2x to LHS we get

3x – 2x = 18

x = 18

LHS = 3x = 3 x 18 = 54

RHS = 2x +18 = (2×18)+ 18 = 36+18 = 54

LHS = RHS

Hence, the result obtained above is correct.

2) 5t -3 = 3t – 5

Solution: 5t-3 = 3t-5

On transposing 3t to LHS and -3 to RHS, we obtain

5t-3t= -5-(-3)

2t= -2

On dividing both sides by 2, we obtain

\( \frac{2 t}{2}=\frac{-2}{2}\)

⇒ t= -1

LHS= 5t-3 =5(-1)-3= -5-3= -8

RHS = 3t – 5 = 3(- 1) – 5 = -3 – 5 = – 8

Hence the result obtained above is correct.

3. 5x + 9 = 5 + 3x

Solution: 5x + 9 = 5 + 3x

On transposing 3x to LHS and 9 to RHS we obtain

5x -3x = 5 -9

2x = -4

On dividing both sides by 2, we obtain

x = -2

LHS = 5x + 9 = 5(-2) + 9 = -10 + 9 = -1

RHS = 5 + 3x = 5 + 3(-2) = 5 – 6 = -1

LHS = RHS

Hence, the result is verified.

4) 4z+3 = 6 + 2z

Solution: 4z + 3 = 6 + 2z

On transposing 2z to LHS and 3 to RHS, we obtain

4z -2z = 6 – 3

2z = 3

Dividing both sides by 2, we obtain

\( \frac{2 z}{2}=\frac{3}{2}\) \( z=\frac{3}{2}\)

LHS= \(+ z+3=A^2 \times \frac{3}{\not 2_1}+3=6+3=9 \)

RHS= \(6+2 z=6+\not 2^1 \times \frac{3}{\not 2}=6+3=9\)

LHS = RHS

Hence the result is verified.

5. 2x -1 = 14 -x

Solution: 2x -1 = 14-x

On transposing x to LHS and 1 to RHS we obtain

2x+x=14+1

3x = 15

Dividing both sides by 3, we obtain \(\frac{3 x}{3}=\frac{15}{3}\)

x = 5

LHS = 2x -1 = 2(5) – 1 = 10 – 1 = 9

RHS = 14 -x = 14 -5 = 9

LHS = RHS

Hence, the result is verified.

6. 8x + 4 = 3(x -1) + 7

Solution: 8x + 4 = 3(x -1) + 7

8x + 4 = 3x – 3 + 7

8x + 4 = 3x + 4

transposing 3x to LHS and + to RHS we obtain

8x -3x =4-4

5x = 0

⇒ x = 0

LHS = 8x + 4 = 8(0) + 4 = 4

RHS = 3(x-1) + 7 = 3(0-1)+7 = -3+7=4

= 3(x – 1) + 7 = 3(0 -1) + 7 = – 3 + 7 = 4

LHS = RHS

Hence, the result is verified.

Linear Equations In One Variable KSEEB Maths

7. \(x=\frac{4}{5}(x+10)\)

Solution: \(x=\frac{4}{5}(x+10)\) multiplying both sides by 5, we obtain

5x = 4(x + 10)

5x = 4x + 40

transposing 4x to LHS, we obtain

5x – 4x = 40

x = 40

LHS = x = 40

RHS = \( \frac{4}{5}(x+10)=\frac{4}{5}(40+10)=\frac{4}{5} \times 50\)

= 4 x 10 = 40

LHS = RHS

Hence, the result is verified.

8. \( \frac{2 x}{3}+1=\frac{7 x}{15}+3\)

Solution: \( \frac{2 x}{3}+1=\frac{7 x}{15}+3\)

transposing\( \frac{7 x}{15}\) to LHS and 1 to RHS we obtain

\( \frac{2 x}{3}-\frac{7 x}{15}=3-1\) \( \frac{10 x-7 x}{15}=2 \Rightarrow \frac{3 x}{15}=2\) \(\frac{x}{5}=2\)

Multiplying both sides by 5, we obtain

x = 10

\(\text { LHS }=\frac{2 x}{3}+1=\frac{2}{3} \times 10+1=\frac{20+3}{3}=\frac{23}{3}\) \(\text { RHS }=\frac{7 x}{15}+3=\frac{7 \times 10^2}{15_3}+3=\frac{14}{3}+3\)

= \(\frac{14+9}{3}=\frac{23}{3}\)

LHS = RHS

Hence, the result is verified.

9) \(2 y+\frac{5}{3}=\frac{26}{3}-y\)

Solution: \(2 y+\frac{5}{3}=\frac{26}{3}-y\)

Transposing y to LHS and \(\frac{5}{3}\) to RHS, we obtain

\(2 y-y=\frac{26}{3}-\frac{5}{3}\) \(3 y=\frac{26-5}{3}=\frac{21}{3}=7\)

3y=7

Dividing both sides by 3, we obtain \(y=\frac{7}{3}\)

\(\text { LHS }=2 y+\frac{5}{3}=\frac{2 \times 7}{3}+\frac{5}{3}=\frac{14}{3}+\frac{5}{3}=\frac{19}{3}\) \(\mathrm{RHS}=\frac{26}{3}-y=\frac{26}{3}-\frac{7}{3}=\frac{26-7}{3}=\frac{19}{3}\)

LHS = RHS

Hence the result is verified.

10. \(3 m=5 m-\frac{8}{5}\)

Solution: \(3 m=5 m-\frac{8}{5}\)

transposing 5m to LHS we obtain

\(3 m=5 m-\frac{8}{5}\)

\(-2 m=\frac{-8}{5}\) Dividing both sides by -2,

We obtain \(\frac{-2 m}{-2}=\frac{-8}{5 \times-2}\)

\(m=\frac{4}{5}\) \(\text { LHS }=3 m=\frac{3 \times 4}{5}=\frac{12}{5}\) \(\mathrm{RHS}=5 m-\frac{8}{5}=5 \times \frac{4}{5}-\frac{8}{5}\)

= \(\frac{20}{5}-\frac{8}{5}=\frac{12}{5}\)

LHS = RHS

Hence, the result is verified.

KSEEB Maths Class 8 Linear Equations in One Variable Notes

KSEEB Maths Class 8 Linear Equations In One Variable Exercise 3.4

1. Amina thinks of a number and subtracts 5/2 from it. She multiplies the result by 8. The result now obtained is 3 times the same number she thought of, what is the number?

Solution: Let the number be x.

According to the given question.

\(8\left(x-\frac{5}{2}\right)=3 x\) \(8 x-8 \times \frac{5}{2}=3 x\)

8x-20=3x

transposing 3x to LHS and – 20 to RHS,

we obtain 8x-3x = 20

5x = 20

Dividing both sides by 5, we obtain x =4

Hence the number is 4.

2)A positive number is 5 times another number. If 21 is added to both numbers, then one of the new numbers becomes twice the other new number. What are the numbers?

Solution: Let the numbers be x and 5x According to the question.

21 + 5x = 2(x + 21)

21 + 5x = 2x + 42

Transposing 2x to LHS and 21 to RHS, we obtain

5x – 2x = 42 – 21

3x = 21

dividing both sides by 3, we obtain \(\frac{3 x}{3}=\frac{21}{3}\)

x = 7

5x = 5×7 = 35

Hence, the numbers are 7 and 35 respectively.

3) Sum of the digits of a two digit number is 9, when we interchange the digits, it is found that the resulting new number is greater than the original number by 27, What is the two-digit number?

Solution: Let the digits at tens place and one’s place be x and 9 – x respectively.

∴ Original number = 10x + (9 – x)

= 9x + 9

On interchanging the digits, the digits at ones place and tensplace will be x and 9 -x respectively.

∴ The new number after interchanging the digits = 10 (9 – x) + x

= 90 – 10x + x

= 90 -9x

According to the given question.

New Number = Original number + 27

90 -9x = 9x + 9 + 27

90 -9x = 9x + 36

transposing 9x to RHS and 36 to LHS

We obtain

90 – 36 = 9x + 9x = 18x

5+ = 18x

Dividing both sides by 18, we obtain

3 = x

⇒ x = 3

9-x=9-3=6

Hence, the digits at tens place and ones place of the number are 3 and 6 respectively.

∴ The two-digit number is

9X + 9 = 9X3 + 9 = 27 + 9 = 36

KSEEB Class 8 Maths Key Concepts Of Linear Equations In One Variable

4) One of the two digits of a two-digit number is three times the other digit. If you interchange the digits of this two digit number and add the resulting number to the original number, you get 88. What is the Original number?

Solution: Let the digits at tens place and one place be x and 3x respectively.

∴ original number= 10x+3x=13x

On interchanging the digits, the digits at ones place, and tens place will be x and 3x

Number after interchanging = 10 x 3x + x

= 30x + x = 31x

According to the given question

Original number+New number = 88

13x + 31x = 88

44x = 88

Dividing both sides by 44, we obtain x = 2

Number after interchanging = 10 x 3x + x

= 30x + x = 31x According to the given question Original number+New number = 88 13x + 31x = 88 ++x = 88

Dividing both sides by 44, we obtain x = 2

Number after interchanging = 10 x 3x + x

= 3 Ox + x = 3 1x According to the given question Original number+New number = 88

13x + 31x = 88

44x = 88

Dividing both sides by 44, we obtain x = 2

∴ Original number = 13x = 13×2 = 26

By considering the tens place and ones place as

3x and x respectively, the two digit number obtained is 62.

∴ The two-digit number maybe 26 or 62.

5) Shobo’s mother’s present age is six times Shobo’s present age. Shobo’s age five years from now will be one third of his mother’s present age. What are their present ages?

Solution: Let Shobo’s age be x years,

∴ His mother’s age will be 6x years.

According to the given question.

After 5 years, Shobo’s age

= Shobo’s mother’s present age /3

\(x+5=\frac{6 x}{3}\)

x+5=2x

transposing x to RHS, we obtain

5=2x-x

₹5=x

₹6x=6 x 5=30

∴ The present age of Shobo and Shobo’s mother will be 5 years and 30 years respectively.

6. There is a narrow rectangular plot, reserved for a school, in Mahuli village, the length and breadth of the plot are in the ratio 11: 4. At the rate ₹100 per metre it will cost the village panchayat ₹75000 to fence the plot. What are the dimensions of the plot?

Solution: Let the common ratio between the length and breadth of the rectangular plot be x.

Hence, the length and breadth of the rectangular plot will be 1lx m and +x m respectively.

Perimeter of the plot = 2 (l + b)

= [2(11x + 4x)]m = 30x m

It is given that the cost of fencing the plot at the rate of ₹100 per metre is ₹75,000

∴100 x Perimeter = 75000

100 x 30x = 75000

3000x = 75000

dividing both sides by 3000, we obtain x = 25 length= 11x m = (11×25)w = 215m

Breadth = 4x m = (4 x 25) m = 100m

Hence, the dimensions of the plot are 275m and 100m respectively.

Practice Questions For Linear Equations In One Variable KSEEB Maths

7. Hasan buys two kinds of cloth materials for school uniform, shirt material that costs him ₹50 per metre and trouser material that costs him ₹90 per metre. for every 3 metres of the shirt material he buys 2 metres of the trouser material. He sells the materials at 12% and 10% profit respectively. His total sale is ₹36600. How much trouser material did he buy?

Solution: Let 2 xm of trouser material and 3x m of shirt material be bought by him.

Per metre selling price of trouser material

=₹ \(\left(90+\frac{90 \times 12}{100}\right)\)

=₹ 100.80

Per metre selling price of shirt material

=₹ \(\left(50+\frac{50 \times 10}{100}\right)\)

=₹ 55

Given that, total amount of selling = ₹36660 100.80 x (2x) + 55 x (3x) = 36660

201.60x +165x = 36660

366.60x = 36660

Dividing both sides by 366.60, we obtain

x = 100

Trouser material = 2x = (2 x 100) m = 200 m

8) Half of a herd of deer are grazing in the field and three fourths of the remaining are playing nearby. The rest 9 are drinking water from the pond. find the number of deer in the herd.

Solution: Let the number of deer be x

Number of deer grazing in the field = x/2

Number of deer playing nearby

=3/4 x Number of remaining deer

= \(\frac{3}{4} \times\left(x-\frac{x}{2}\right)=\frac{3}{4} \times \frac{x}{2}=\frac{3 x}{8}\)

Number of deer drinking water from the pond= 9

\(x-\left(\frac{x}{2}+\frac{3 x}{8}\right)=9\) \(x-\left(\frac{4 x+3 x}{8}\right)=9\) \(x-\frac{7 x}{8}=9\) \(\frac{8 x-7 x}{8}=9\) \(\frac{x}{8}=9\)

Multiplying both sides by 8, we obtain x=72

Hence, the total number of deer in the herd is 72.

9. A grandfather is ten times older than his granddaughter. He is also 54 years older than her. find their present ages.

Solution: Let the granddaughter’s age be x years.

grandfather’s age will be 10x years.

According to the question

Grandfather’s age

= Granddaughter’s age + 54 years.

10x = x + 54

Transposing x to LHS we obtain

10x -x = 54

9x = 54

x = 6

Granddaughter’s age = x years = 6 years Grandfather’s age = 10x years

= (10×6) years = 60 years

10. Aman’s age is three times his son’s age. Ten years ago he was five times his son’s age. find their present ages.

Solution: Let Aman’s son’s age be x years.

∴ Aman’s age will be 3x years.

Ten years ago their age was (x – 10) years and

(3x -10) respectively.

According to the question.

10 years ago, Aman’s age = 5 x Aman’s son’s age 10 years ago.

3x -10 = 5(x -10)

3x -10 = 5x – 50

Transposing 3x to RHS and 50 to LHS

we obtain 50 -10 = 5x – 3x

40 = 2x

dividing both sides by 2, we obtain 20=x Aman’s son’s age =x years = 20 years

Aman’s age = 3x years = (3 x 20) years

= 60 years.

Linear Equations In One Variable questions And Answers KSEEB Maths

KSEEB Class 8 Maths Chapter 3 Linear Equations In One Variable Exercise 3.5

1. Solve the following linear equations.

1)  \(\frac{x}{2}-\frac{1}{5}=\frac{x}{3}+\frac{1}{4}\)

Solution: LCM of the denominators, 2, 3,4 and5 is 60.

Multiplying both sides by 60, we obtain

\(60\left(\frac{x}{2}-\frac{1}{5}\right)=60\left(\frac{x}{3}+\frac{1}{4}\right)\)

⇒ 30x – 12 = 20x +15 (opening the bracket)

⇒ 30x -20x = 15 + 12

⇒10x = 27

⇒ \(x=\frac{27}{10}\)

2) \(\frac{n}{2}-\frac{3 n}{4}+\frac{5 n}{6}=21\)

Solution: \(\frac{n}{2}-\frac{3 n}{4}+\frac{5 n}{6}=21\)

LCM of the denominators 2,4 and 6 is 12.

Multiplying both sides by 12, we obtain

6n – 9n + 10n = 252

⇒ 7n = 252

⇒ \(n=\frac{252}{7}\)

⇒ n = 36

3) \(x+7-\frac{8 x}{3}=\frac{17}{6}-\frac{5 x}{2}\)

Solution: \(x+7-\frac{8 x}{3}=\frac{17}{6}-\frac{5 x}{2}\)

LCM of the denominators 2,3 and 6 is 6.

Multiplying both sides by 6, we obtain

⇒ 6x+42-16x=17-15x

⇒ 6x-16x+15x=17-15x

⇒ 5x = -25

⇒ \(x=\frac{-25}{5}\)

⇒ x = -5

 

4) \(\frac{x-5}{3}=\frac{x-3}{5}\)

Solution:  \(\frac{x-5}{3}=\frac{x-3}{5}\)

LCM of the denominators, 3 and 5 is 15

Multiplying both sides by 15, we obtain

5(x-5)=3(x-3)

=> 5x-25=3x-9 (opening the bracket)

=> 5x-3x=25-9

=> 2x=16

=> \(x=16 / 2\)

=> x=8

5) \(\frac{3 t-2}{4}-\frac{2 t+3}{3}=\frac{2}{3}-t\)

Solution: LCM of the denominators,3 and + is 12.

Multiplying both sides by 12, we obtain

3(3t-2)-4(2t+3)=8-12t

⇒ 9t-6-8t-12=8-12t (opening the bracket)

⇒ 9t-8t+12t=8+6+12

⇒ 13t=26

⇒ \(t=26 / 13\)

⇒ t=2

6) \(m-\frac{m-1}{2}=1-\frac{m-2}{3}\)

Solution: \(m-\frac{m-1}{2}=1-\frac{m-2}{3}\)

LCM of the denominators 2 and 3 is 6

Multiplying both sides by 6, we obtain

6m-3(m-1)=6-2(m-2)

⇒ 6m-3m+3=6-2m+4 (opening the bracket)

⇒ 6m-3m+2m=6+4-3

⇒ 5m=7

⇒ \(m=7 / 5\)

KSEEB Class 8 Maths Chapter 3 Linear Equations In One Variable

Simplify and solve the following linear equations

7. 3(t-3) = 5(2t + 1)

Solution: 3(t-3) = 5(2t + 1)

3t -9 = 10t+ 5

⇒- 9 -5 = 10t -3t

⇒-14 = 7t

⇒ \(t=\frac{-14}{7}\)

⇒ t= -2

8. 15(y-4)-2(y-9) + 5(y + 6) = 0

Solution: 15(y-4)-2(y-9) + 5(y + 6) = 0

15y – 60 -2y + 18 + 5y+ 30 = 0

18y -12 = 0

⇒ 18y = 12

⇒ \(y=\frac{12}{18}=\frac{2}{3}\)

9. 3(5z-7)-2(9z-11) = +(8z-13)-17

Solution: 3 (5z – 7) – 2 (9z – 11) = + (8z – 13) – 17

15z -21 -18z + 22 = 32z -52 -17

⇒ -3z + l = 32z -69

⇒ -3z -32z = -69 – 1

⇒ -35z = -70

⇒ \(z=\frac{70}{35}=2\)

z=2

10. 0.25 (4f – 3) = 0.05 (10f – 9)

Solution: 0.25 (4f-3) = 0.05 (10f-9)

1/4(4f-3)=1/20(10f-9)

Multiplying both sides by 20, we obtain

⇒ 5(4f -3) = 1(10f -9)

⇒20f -15 = 10f -9

⇒ 20f -10f = -9 + 15

⇒10f = 6

⇒ \(f=\frac{6}{10}=\frac{3}{5}=0.6\)

f = 0.6

Linear Equations In One Variable Exercise 3.6

1. Solve the following equations.

1) \(\frac{8 x-3}{3 x}=2\)

Solution: \(\frac{8 x-3}{3 x}=2\)

On multiplying both sides by 3x, we obtain

⇒8x-3=6x

⇒ 8x-6x=3

⇒ 2x=3

⇒ \(x=\frac{3}{2}\)

2) \(\frac{9 x}{7-6 x}=15\)

Solution: \(\frac{9 x}{7-6 x}=15\)

On multiplying both sides by 7-6x, we obtain

⇒9x=15(7-6x)

⇒9x=105-90x

⇒9x+90x=105

⇒ 99x=105

⇒ \(x=\frac{105}{99}=\frac{35}{33}\)

3) \(\frac{z}{z+15}=\frac{4}{9}\)

Solution: \(\frac{z}{z+15}=\frac{4}{9}\)

On multiplying both sides by 9(z+15)

we obtain

9z=+(z+15)

⇒ 9z=+z+60

⇒5z=60

⇒ \(z=60 / 5=12\)

⇒ z=12

4) \(\frac{3 y+4}{2-6 y}=\frac{-2}{5}\)

Solution: \(\frac{3 y+4}{2-6 y}=\frac{-2}{5}\)

On multiplying both sides by 5 (2 -6y) We obtain

⇒5(3y+4)= -2(2-6y)

⇒15y+20= -4+12y

⇒15y-12y= -4-20

⇒3y= -24

⇒ \(y=\frac{-24}{3}=-8\)

5) \(\frac{7 y+4}{y+2}=\frac{-4}{3}\)

Solution: \(\frac{7 y+4}{y+2}=\frac{-4}{3}\)

On multiplying both sides by 3(y+2)

we obtain

⇒3(7y+4)= -4(y+2)

⇒ 21y+12= -4y-8

⇒ 21y+4y= -8-12

⇒25y= -20

⇒ \(y=\frac{-20}{25}=\frac{-4}{5}\)

KSEEB Class 8 Maths solutions for Chapter 3 Linear Equations In One Variable

6. The ages of Hari and Harry are in the ratio 5 : 7. Four years from now the ratio of their ages will be 3 : 4, find their present ages.

Solution: Let the common ratio between their ages be x.

∴ Hari’s age and Harry’s age will be 5x years and 7x years respectively and four years later, their ages will be (5x + 4) years and (7x + 4)
years respectively.

According to the situation given in the question.

\(\frac{5 x+4}{7 x+4}=\frac{3}{4}\)

⇒  4(5x + 4) = 3(7x + 4)

⇒  20x + 16 = 21x + 12

⇒  16 -12 = 21x -20x

⇒  4= x

Hari’s age = 5x years = (5x+) years = 20 years Harry’s age = 7x years = (7x+) years

= 28 years

∴ Hari’s age and Harry’s age are 20 years and 28 years respectively.

7. The denominator of a rational number is greater than its numerator by 8, If the numerator is increased by 17 and the denominator is decreased by 1. The number obtained is\(\frac{3}{2}\). find the rational number.

Solution: Let the numerator of the rational number be x.

∴its denominator will be x + 8.

The rational number will be \(\frac{x}{x+8}\)

According to the question

\(\frac{x+17}{x+8-1}=\frac{3}{2}\)

⇒  \(\frac{x+17}{x+7}=\frac{3}{2}\)

⇒  2(x+17)=3(x+7)

⇒  2x+34=3x+21

⇒  34-21=3x-2x

⇒  13=x

Numerator of the rational number = x = 13 Denominator of the rational number

= x+8 = 13+8 = 21

Rational number = \(\frac{13}{21}\)

Linear Equations in One Variable Additional Problems KSEEB Maths

1) Radha takes some flowers in the basket and visits three temples one by one. At each temple, she offers one half of the flowers from the basket. If she is left with 3 flowers at the end, find the number of flowers she had in the beginning.

Solution: Let number of flowers in the beginning =x

flowers offers at 1st temple = \(\frac{x}{2}\)

flowers offered at 2nd temple = \(\frac{1}{2} \text { of } \frac{x}{2}=\frac{x}{4}\)

flowers offered at 3rd temple = \(\frac{1}{2} \text { of } \frac{x}{4}=\frac{x}{8}\)

flowers left = 3

According to the question,

\(\frac{x}{2}+\frac{x}{4}+\frac{x}{8}+3=x\) \(x-\frac{x}{2}-\frac{x}{4}-\frac{x}{8}=3=\frac{8 x-4 x-2 x-x}{8}=3\)

= \(\frac{4 x-3 x}{8}=3\)

=\(\frac{x}{8}=3 \text { (or) } x=24\)

So, the number of flowers she had, in the beginning, is 24.

2) The sum of three consecutive numbers is 156. find the number which is a multiple of 13 out of these numbers.

Solution: Let one number =x

Second number = x +1

Third number=x+2

According to question,

x+x+1+x+2=156

or 3x+3=156,3x=156-3

⇒  3x=153

\(x=\frac{153}{3}=51\)

3) The volume of water in a tank is twice of that in the other. If we draw out 25 litres from the 1st and add it to the other, the volumes of the water in each tank will be the same. find the volume of water in each tank.

Solution: Let the volume of smaller tank = x L

The volume of larger tank = 2x L

According to the question.

2x – 25 = x + 25

2x -x = 50

⇒  x = 50

Volume of smaller tank = 50L

Volume of larger tank = 2×50 = 100L

4) one number is 6 more than another number. Also, 7 times the smaller number is equal to 6 times the larger number. find the two numbers.

Solution: Let the smaller number be x

Then the larger number is x + 6

Now 7 x smaller number = 6 x larger number

∴ 7x = 6(x + 6)

7x = 6x+36

7x – 6x = 36

x = 36

⇒  ∴ Smaller number = 36

Larger number = 36 + 6 = 42

5) Solve the equation, 0.36+0.7=0.85x

Solution: 0.36+0.7=0.85x

0.3=0.85x-0.7x

0.3=0.15x

⇒ \(x=\frac{0.30}{0.15}\)

x = 2

6. Solve the equation and check your solution

\(\frac{1-9 y}{19-3 y}=\frac{5}{8}\)

Solution: \(\frac{1-9 y}{19-3 y}=\frac{5}{8}\)

cross multiplying, we have

8x(1-9y)=5x(19-3y)

8-72y=95-15y

8-95= -15y+72y

-87=57y

\(y=\frac{87}{-57}=\frac{-29}{19}\)

Thus \(y=\frac{-29}{19}\) is the solution of the given equation

LHS = \(\frac{1-9 y}{19-3 y}\)=\(\frac{1-9\left(\frac{-29}{19}\right)}{19-3 \times\left(\frac{-29}{19}\right)}\)

=\(\frac{1+\frac{261}{19}}{19+\frac{87}{19}}=\frac{\frac{19+261}{\not 9}}{\frac{361+87}{19}}\)

LHS=\(\frac{280}{448}\)

=\(\frac{5}{8}\)

=RHS

Hence, LHS=RHS

7) \(\frac{x-4}{7}-x=\frac{5-x}{3}+1\)

Solution: \(\frac{x-4}{7}-x=\frac{5-x}{3}+1\)

\(\frac{x-4-x \times 7}{7}=\frac{5-x+1 \times 3}{3}\) \(\frac{x-4-7 x}{7}=\frac{5-x+3}{3}\) \(\frac{-6 x-4}{7}=\frac{8-x}{3}\)

3(-6x-4)=7(8-x)

-18x-12=56-7x

-18x+7x=56+12

-11x=68

x=\(\frac{-68}{11}\)

8) find the measure of the angle x, marked

Solution: We know that,

The sum of the angles, of a △le is 180°

∴[A + [B + [C = 180°

58°+x + x = 180°

2x = 180 -58°

2x = 122°

\(x=\frac{122}{2}\)

x=61°

9) Sum of two numbers is 40, if one of them is 10 more than the other, find the numbers

Solution: Let one number be x

Then the other number be x + 10

Sum of two numbers 40

(According to question)

x + (x + 10) = 40

2x + 10 = 40

2x = 30 or x = 15

one number is 15 and the other number is x + 10 = 15 + 10 = 25

10) Two numbers differ by 40, when each number is increased by 8, the bigger become thrice the lesser number. If one number is x, then find the other number.

Solution: If one number = x

then other number = x + 40

According to question, x + 40 + 8 = 3 (x + 8)

⇒ x + 48 = 3x + 24

⇒ 48 -24 = 3x-x

⇒ 24 = 2x ⇒  \(x=\frac{24}{2}=12\)

So, the numbers are 12 and (12+40) i.e., 52.

11) If 1/2 is subtracted from a number and the difference is multiplied by 4, the result is 5. What is the number?

Solution: Let the number be x,

According to question

\(4\left(x-\frac{1}{2}\right)=5\)

⇒ 4x-2=5

⇒ 4x=5+2

⇒ \(x=\frac{7}{4}\)

∴ Required number =\(7 / 4\)

12)The perimeter of a rectangle is 240cm. If its length is increased by 10% and its breadth is decreased by 20%, we get the same perimeter. find the length and breadth of the rectangle.

Solution: Let the length ‘l’ be x

2(l+b) = perimeter

2(x+b) = 2+0

x+b=120 ⇒ b=120-x

Now, new length= x+10% of x

= x+10xf100 = x+xf10 = 11xf10

New breadth= (120-x)-20% of (120-x)

= \((120-x)-\frac{20}{100} \times(120-x)\)

= \(\frac{5(120-x)-(120-x)}{5}\)

= \(\frac{600-5 x-120+x}{5}=\frac{480-4 x}{5}\)

According to question,

\(2\left(\frac{11 x}{10}+\frac{480-4 x}{5}\right)=240\) \(\Rightarrow \frac{11 x+960-8 x}{10}=120\)

⇒ 3+960=1200

⇒ 3x=1200-960

3x=240

x=80

so, length = 80cm

breadth=120-x=120-80=40cm

13. If Dennis is \(\frac{1}{3}\) rd the age of his father keith now, and was \(\frac{1}{4}\) th the age of his father 5 years ago, then how old will his father keith be 5 years from now?

Solution: Let Keith’s age now be x years.

Dennis’s age now = \( \frac{x}{3}\) years

Keith’s age 5 years ago = (x – 5) years

Dennis’s age 5 years ago =\(\left(\frac{x}{3}-5\right)\) years

According to question

\(\left(\frac{x}{3}-5\right)=\frac{1}{4}(x-5)\) \(\frac{x-15}{3}=\frac{x-5}{4}\)

⇒ 4(x-15)=3(x-5)

⇒  x= -15+60

⇒ x=45

∴ Keith’s age 5 years from now

= (45 + 5) years = 50 years.

14) Two numbers are in the ratio 5: 8. If the sum of the numbers is 182, find the numbers.

Solution: Let the numbers be 5x and 8x

then, 5x + 8x = 182

13x = 182

\(x=\frac{182}{13}=14\)

∴One number = 5 x 14 = 70

Other number = 8 x 14=112.

15) A steamer goes downstream and covers the distance between two ports in 4 hours while it covers the same distance upstream in 5 hours. If the speed of the stream is 2km/h. find the speed of the steamer in still water.

Solution: Let the speed of the steamer in still water be x km/h,

then the speed downstream = (x + 2) km/h and the speed upstream = (x – 2) km /h

The distance covered in 4 hours while going downstream = 4(x + 2) km

The distance covered in 5 hours while going upstream = 5 (x – 2)km.

but, each of these distances is the distance between the two ports which is the same.

∴ 4(x + 2) = 5(x-2)

=> 4x + 8 = 5x -10

=> x = 18

∴ speed of the steamer in still water = 18km/h

16) Two years ago, Dilip was three times as old as his son and two years hence, twice his age will be equal to five times that of his son. find their present ages.

Solution: Let Dilip’s son’s age 2 years ago be x years then, Dilip’s age 2 years ago = 3x years.

∴ The son’s age 2 years hence = (x + 4) years Dilip’s age 2 years hence = (3x + 4) years

∴ 2(3x + 4) = 5(x + 4)

⇒ 6x + 8 = 5x + 20

⇒ x = 12

∴ Dilip’s son’s age 2 years ago = 12 years and Dilip’s age 2 years ago = 36 years

∴Son’s present age = 1+ years and Dilip’s present age = 38 years

17) The distance between two stations is 425km, two trains start simultaneously from these stations on parallel tracks to cross each other. The speed of one of them is greater than that of the other by 5 km/h. If the distance between the two trains after 3 hours of their start is 20km. find the speed of each train.

Solution: Let the speed of one train be x km/h

then, the speed of the other train = (x+5)km/h

The distance travelled by the 1st train in 3 hours

The distance travelled by the 2nd train=(3x)km in 3 hours = 3(x + 5)km.

∴ 425-(3x + 3(x + 5)) = 20

⇒ 425 – 3x -3x -15 = 20

⇒6x = 390

⇒ x = 65

So, the speed of the 1 st train = 65km/h

The speed of the 2nd train = (65+5) = 70km/h

KSEEB Class 8 Maths Solutions For Chapter 2 Rational Numbers Points To Remember

Rational nos. are closed under the operations of addition, subtraction and multiplication.

The operations addition and multiplication are

1)commutative for rational numbers.

2)associative for rational numbers.

The rational no. ‘0’ is the additive identity for rational nos.

The rational no.’1′ is the multiplicative identity for rational numbers.

The additive inverse of the rational number [a/b is -a/b] and vice- versa.

Read and Learn More KSEEB Solutions for Class 8 Maths

The reciprocal or multiplicative inverse of the rational numbers. a/b is c/d if a/b x c/d =1

Distributivity of rational nos. For all rational numbers a, b and c,
a(b + c) = ab + ac and a(b-c) =ab-ac

Rational numbers can be represented on a number line.

Between any two given rational numbers there are countless rational numbers. The idea of mean helps us to find rational numbers between two rational numbers.

Positive Rationals: Numerator and denominator both are either opposite or negative.
Example: 4/3,-3/-5

Negative Rationals: Numerator and denominator
both are of opposite signs.
Example: -2/11, 4/-9

Additive inverse: Additive inverse (negative)
(a/b)+(-a/b) = (-a/b)+(a/b)=0
-a/b is the additive inverse of -a/b and a/b is the additive inverse of -a/b

Multiplicative inverse (reciprocal):
a/b x c/d=1=c/d x a/b where c/d is the reciprocal of a/b.
0 has no reciprocal
The reciprocal of 1 is 1 and of -1 is -1.

Rational Numbers Solutions KSEEB Class 8 Maths Exercise 2.1

1. Using appropriate properties find.

1. \(\frac{-2}{3} \times \frac{3}{5}+\frac{5}{2}-\frac{3}{5} \times \frac{1}{6}\)

Solution: \( \frac{-2}{3} \times \frac{3}{5}+\frac{5}{2}-\frac{3}{5} \times \frac{1}{6}\)

= \( \frac{-2}{3} \times \frac{3}{5}-\frac{3}{5} \times \frac{1}{6}+\frac{5}{2}\)(Using commutativity of rational numbers)

= \( \left(-\frac{3}{5}\right) \times\left(\frac{2}{3}+\frac{1}{6}\right)+\frac{5}{2}\)(Distributivity)

= \( \left(-\frac{3}{5}\right) \times\left(\frac{2 \times 2+1}{6}\right)+\frac{5}{2}\)

= \(\left(-\frac{3}{5}\right)\left(\frac{5}{6}\right)+\frac{5}{2}=\left(\frac{-3}{6}\right)+\frac{5}{2}\)

=\(\left(\frac{-3+5 \times 3}{6}\right)=\frac{-3+15}{6}=\frac{12}{6}=2\)

2. \(\frac{2}{5} \times\left(\frac{-3}{7}\right)-\frac{1}{6} \times \frac{3}{2}+\frac{1}{14} \times \frac{2}{5}\)

Solution: =\(\frac{2}{5} \times\left(\frac{-3}{7}\right)+\frac{1}{14} \times \frac{2}{5}-\frac{1}{6} \times \frac{3}{2}\)                                                                                             (By

(By commutativity)

= \(\frac{2}{5} \times\left(\frac{-3}{7}+\frac{1}{14}\right)-\frac{1}{4}\) (By distributivity)

= \(\frac{2}{5}\left(\frac{-3 \times 2+1}{14}\right)-\frac{1}{4}\)

= \(\frac{2}{5}\left(\frac{-5}{14}\right)-\frac{1}{4}\)

= \(\frac{-1}{7}-\frac{1}{4}=\frac{-4-7}{28}=\frac{-11}{28}\)

2. Write the additive inverse of each of the following.
1) 2/8 
2)-5/9
3)-6/-5
4)2/-9
5)19/-6

Solution:

1) -2/8
2)5/9
3)-6/5
4)2/9
5)19/6

3. Verify that -(-x)=x for
1) x=11/15
2) x=-13/17

Solution: 1) x=11/15 The additive inverse of x=11/15 is

\(-x=\frac{-11}{15} \text { as } \frac{11}{15}+\left(\frac{-11}{15}\right)=0\)

This equality (11/15)+(-11/15)=0 represents that the additive inverse of -11/15 is 11/15 (or) it can be said that -(-11/15)=11/15

i.e., -(-x)=x

2) x=-13/17

The additive inverse of x=13/17 is

-x=13/17 as -13/17+13/17=0

This equality -13/17+13/17=0 represents that the additive inverse of 13/17 is 13/-17 i.e., -(-x)=x.

4. Find the multiplicative inverse of the following.

1) -13
Solution:  Multiplicative inverse = 1/-13
2) -13/19
Solution:  -19/13
3) 1/5
Solution:  5
4)-5/8 x-3/7
Solution:  -5/8 x -3/7 = 15/56 multiplicative inverse = 56/15
5) -1 x -2/5
Solution:  -1 x -2/5 = 2/5 multiplicative inverse = 5/2
6) -1
Solution:  Multiplicative inverse = -1

5. Name the property under multiplication used in each of the following

1)-4/5×1=1x-4/5=-4/5
Solution: 1 is the multiplicative identity

2)-13/17x-2/7=-2/7x-13/17
Solution: Commutativity

3)-19/29×29/-19=1
Solution: Multiplicative inverse

Rational Numbers Solutions KSEEB Class 8 Maths

6. Multiply 6/13 by the reciprocal of -7/16

Solution: 6/13x[Reciprocal of -7/16]

=6/13x-16/7=-96/91

7. Tell what property allows you to compute

1/3 x (6 x 4/3) as (1/3 x 6)x4/3

Solution: Associativity

8. Is 8/9 the multiplicative inverse of -1 1/8 ? Why or Why not?

Solution: If it is the multiplicative inverse, then the product should be 1. However, here, the product is not 1 as 8/9 x(-1 1/8)=8/9 x (-9/8) = -1≠1

Solution: 3 1/3 = 10/3

\(0.3 \times 3 \frac{1}{3}=0.3 \times \frac{10}{3}=\frac{3}{10} \times \frac{10}{3}=1\)

Here, the product is 1.

Hence, 0.3 is the multiplicative inverse of 3 1/3.

10. Write.

(1) The rational number that does not have a reciprocal.
Solution: 0 is a rational number but its reciprocal is not defined.

2) The rational numbers that are equal to their reciprocals.Solution: 1 and -1 are the rational numbers that are equal to their reciprocals.

3) The rational number that is equal to its negative.
Solution: 0 is the rational number that is equal to its negative.

11. Fill in the blanks:
1) Zero has No reciprocal.
2) The numbers 1 and -1 are their own reciprocals.
3) The reciprocal of-5 is -1/5
4) Reciprocal of 1/x, where x ≠ 0 is x.5) The product of two rational numbers is always
a Rational Number.
6) The reciprocal of a positive rational number is positive rational number.

KSEEB Class 8 Maths Chapter 2 Rational Numbers Exercise 2.2

1. Represent these numbers on the number line.
1) 7/4
2)-5/6

Solution:1) 7/4cann be represented on the number line as follows

 

2) -5/6

Solution:

 

 

2. Represent-2/11, -5/11, -9/11 on the number line.

Solution:

 

3. Write five rational numbers which are smaller than 2.

Solution: 2 can be represented as 14/7

∴ five rational numbers smaller than 2 are 13/7, 12/7, 11/7, 10/7, 9/7

4. Find ten rational numbers between -2/5 and 1/2.

Solution: -2/5 and 1/2 can be represented as -8/20 and 10/20 respectively

∴ Ten rational numbers between-2/5 and 1/2 are

-7/20, -6/20, -5/20, -4/20, -3/20, -1/20, 0, 1/20, 2/20

5. Find five rational numbers between
1) 2/3 and 4/5 2) -3/2 and 5/3
3) 1/4 and 1/2

Solution: l)2/3 and 4/5 can be represented as 30/45 and 30/45 and 36/45 respectively.

∴ 5 rational numbers between 2/3 and 4/5 are

31/45, 32/45, 33/45, 34/45, 35/45

2) -3/2 and 5/3 can be represented as -9/6 and 10/6 respectively.

∴ 5 rational number between -3/2 and 5/3 are

-8/6, -7/6, -1, -5/6, -4/6

3) 1/4 and 1/2 can be represented as 8/32 and 16/32 respectively

∴ 5 rational numbers between 1/4 and 1/2 are

9/32, 10/32, 11/32, 12/32, 13/32

KSEEB Class 8 Maths Chapter 2 Rational Numbers

6. Write five rational numbers greater than -2.

Solution: -2 can be represented as -14/7

∴ 5 rational numbers greater than -2 are

-13/7, -12/7, -11/7, -10/7, -9/7

7. Find ten rational numbers between 3/5 and 3/4

Solution: 3/5 and 3/4 can be represented as 48/80 and 60/80 respectively.

∴ 10 rational numbers between 3/2 and 3/4 are

49/80, 50/80, 51/80, 52/80, 53/80, 54/80, 55/80, 56/80, 57/80, 58/80

KSEEB Maths Class 8 Rational numbers Additional Problems

1. The product of two rational numbers is -14/27.

If one of the numbers be 7/9, find the other.

Solution: Product of two rational numbers =-14/27

one number =7/9

∴ other number = -14/27 ÷ 7/9

= -14/27 x 9/7 = -2/3

2. Find three rational numbers between -2 and 7

Solution: A rational number lying between -2 and 7

= (-2+7)÷2

=5÷2=5/2

∴ -2<5/2<7

Now a rational number lying between -2 and 5/2

-2 and 5/2=(-2+5/2)÷2

= \( \left(\frac{-4+5}{2}\right) \div 2=\frac{1}{2} \times \frac{1}{2}=\frac{1}{4} \)

∴ -2<1/4<5/2<7

A rational number lying between 5/2 and 7

=\(\left(\frac{5}{2}+7\right) \div 2=\left(\frac{5+14}{2}\right) \div=\frac{19}{2} \times \frac{1}{2}=\frac{19}{4}\)

∴ -2<1/4<5/2<19/4<7

Hence 1/4, 5/2 and 19/4 are three rational numbers between -2 and 7.

3. Arrange the following numbers in ascending order 4/5, -2/3, 1/-2, -4/7

Solution: Rewrite each rational number with a positive denominator so 4/5, -2/3, 1/-2, -4/7 is same as 4/5, -2/3, 1/-2, -4/7

LCM of 5,3,2 and 7 =210

Now \(\frac{4}{5}=\frac{4 \times 42}{5 \times 42}=\frac{168}{210}\)

\(\frac{-2}{3} \times \frac{70}{70}=\frac{-140}{210}, \frac{-1 \times 105}{2 \times 105}=\frac{-105}{210} \text { and }\)

\(\frac{-4}{7}=\frac{-4 \times 30}{7 \times 30}=\frac{-120}{210}\)

∵ -140<-120<-105<168

∴ -140/210<-120/210<-105/210<168/210

∴ -2/3<-4/7<-1/2<4/5

4. What should be added to 2/3 so as to get 1/15?

Solution: Let us suppose that the required number to be added be x, then

2/3 + x =1/15

\(x=\frac{1}{15}-\frac{2}{3}=\frac{1-2 \times 5}{15}=\frac{1-10}{15}=\frac{-9}{15}\)

x=-3/5 (reducing to lowest terms)

Hence, the required number is -3/5.

Rational Numbers Questions And Answers KSEEB Maths

5. Using the rearrangement property find the sum of 4/3+3/5+ -2+3+ -11/5

Solution: Using commutative and Associative property, we can rearrange terms in any suitable order, using this rearrangement property, we have

4/3+3/5+ -2+3+ -11/5

=\(\left(\frac{4}{3}+\frac{-2}{3}\right)+\left(\frac{3}{5}+\frac{-11}{5}\right)\)

=\(\left\{\frac{4+(-2)}{3}\right\}+\left\{\frac{3+(-11)}{5}\right\}\)

=\(\left(\frac{4-2}{3}\right)+\left(\frac{3-11}{5}\right)=\frac{2}{3}+\frac{(-8)}{5}\)

= \(\frac{10+(-24)}{15}=\frac{10-24}{15}=\frac{-14}{15}\)

6. Find the rational number between 1/4 and 1/3

Solution: A rational number between 1/4 and 1/3 is

\(\frac{1}{2}\left(\frac{1}{4}+\frac{1}{3}\right)=\frac{1}{2}\left(\frac{3+4}{12}\right)=\frac{1}{2} \times \frac{7}{12}=\frac{7}{24}\)

7. Verify that
a) 2 x (-3/5)=(-3/5) x 2
b) \(\frac{-2}{7} \times\left(\frac{-3}{7} \times \frac{(-9)}{14}\right)=\left(\frac{-2}{7} \times \frac{-3}{7}\right) \times \frac{(-9)}{14}\)

Solution: a) 2 x (-3/5)=(-3/5) x 2

LHS =\( 2 \times\left(\frac{-3}{5}\right)=\frac{2}{1} \times \frac{-3}{5}=\frac{2 \times(-3)}{1 \times 5}=\frac{-6}{5}\)

RHS = \( \left(\frac{-3}{5}\right) \times 2=\frac{-3 \times 2}{5 \times 1}=\frac{-6}{5}\)

LHS=RHS

Hence verified that 2 x (-3/5)=(-3/5) x 2

b) \(\frac{-2}{7} \times\left(\frac{-3}{7} \times \frac{(-9)}{14}\right)=\left(\frac{-2}{7} \times \frac{-3}{7}\right) \times \frac{(-9)}{14}\)

LHS = \(\frac{-2}{7} \times\left(\frac{-3}{7} \times \frac{(-9)}{14}\right)\)

= \(\frac{-2}{7} \times\left\{\frac{(-3) \times(-9)}{7 \times 14}\right\}\)

=\(\frac{-2}{7} \times\left\{\frac{(-3) \times(-9)}{7 \times 14}\right\}\)

RHS = \(\left(\frac{-2}{7} \times \frac{-3}{7}\right) \times\left(\frac{-9}{14}\right)\)

=\(\left(\frac{6}{49}\right) \times \frac{-9}{14}=\frac{-54}{686}=\frac{-27}{343}\)

Hence verified, LHS=RHS

8. A farmer has a field of area 49 4/5 ha. He wants to divide it equally among his one son and two daughters. Find the area of each one’s share.
(ha means hectare: 1 hectare = 10000m2)

Solution:

49 4/5 ha =249/5 ha

Each share = 1/3 x 249/5 ha

= 83/5 ha

= 16 3/5 ha

9. Let O, P and Z represent the numbers 0, 3 and -5 respectively on the number line. Points Q, R and S are between O and P such that OQ = QR = RS = SP. What are the rational numbers represented by the point Q, R and S Next Choose a point T between Z and O so that ZT = TO, which rational number does T represent?

Solution:

 

As OQ=QR=RS=SP

and OQ+QR+RS+SP=OP

∴ Q,R and S divide OP into four equal parts.

so, R is the mid-point of OP.

i.e., R = \(\frac{0+3}{2}=\frac{3}{2}\)

⇒ i.e., Q=\(\frac{1}{2}\left(\frac{0+3}{2}\right)=\frac{3}{4}\) and S is the midpoint of RP.

i.e., s =\(\frac{1}{2}\left(\frac{3}{2}+3\right)=\frac{9}{4}\)

∴ Q=3/4, R=3/2 and S=9/4

also ZT=TO

So T is the mid-point of OZ

i.e., T=\(\frac{0+(-5)}{2}=\frac{-5}{2}\)

10. From a rope 40m long, pieces of equal size are cut, if the length of one piece is 10/3m, find the number of such pieces.

Solution: Total length of rope =40m

Length of one piece =10/3m

Let the number of pieces be x

Then, according to be the question 10x/3=40

⇒ x=\(\frac{40 \times 3}{10}=12\)

Hence, number of pieces cut from the rope are 12.

Practice Questions For Rational Numbers KSEEB Maths

11. Find the sum of 1/2 and 5/-6.

Solution: \(\frac{1}{2}+\left(\frac{5}{-6}\right)=\frac{1}{2}-\frac{5}{6}\)

L.C.M. of 2 and 6 is 6.

\(\frac{1}{2}-\frac{5}{6}=\frac{1 \times 3-5 \times 1}{6}=\frac{3-5}{6}=\frac{-2}{6}=\frac{-1}{3}\)

12. The sum of two rational numbers is -2. If one of the number is -14/5, find the other

Solution: Let the other number be x, then

\(x+\left(\frac{-14}{5}\right)=(-2)\)

⇒  \(x-\frac{14}{5}=-2\)

⇒ \(x=\frac{14}{5}-2\)

⇒ \(x=\frac{14-2 \times 5}{5}\)

\(x=\frac{14-10}{5}\)

=4/5

Hence, the required number is 4/5.

13. If the product of two numbers is -5/8 and one of them is -3/4. Find the other.

Solution: Let the other number be 3 x.

then -3/4 x X=-5/8

⇒ \(x=\frac{-5}{8} \div \frac{-3}{4}=\frac{-5}{8} \times \frac{-4}{3}\)

= 20/24 = 5/6

Hence, the other number is 5/6

14. Find three rational number between -3 and -2.

Solution: A rational numbers between -3 and -2 is

\(\frac{1}{2}\{[-3+(-2)]\}=\frac{1}{2}(-3-2)=\frac{-5}{2}\)

A rational number between -3 and -5/2 is

=\(\frac{1}{2}\left\{-3+\left(\frac{-5}{2}\right)\right\}=\frac{1}{2}\left(-3-\frac{5}{2}\right)=\frac{1}{2}\left(\frac{-11}{2}\right)=\frac{-11}{4}\)

A rational number between -5/2 and -2 is

=\(\frac{1}{2}\left\{\frac{-5}{2}+(-2)\right\}\)

=\(\frac{1}{2}\left\{\frac{-5-4}{2}\right\}=\frac{1}{2}\left(\frac{-9}{2}\right)=\frac{-9}{4}\)

∴ -11/4, -5/2, -9/4 are three rational numbers between -3 and -2.

15. One fruit salad receipe requires 1/2 cup of sugar. Another receipe for the same fruit salad requires 2 tablespoons of sugar. If 1 tablespoon is equivalent to 1/16 cup, how much more sugar does the first receipe require?

Solution: 1 tablespoon =1/16 cup

2 tablespoon= 1/16×2 cup =1/8 cup

The quantity of more sugar the first receipe requires

=\(\frac{1}{2}-\frac{1}{8}=\frac{4-1}{8}=\frac{3}{8} \text { cup. }\)

KSEEB Class 8 Maths key concepts of Rational Numbers

16. Tarun purchased 15 3/4m of cloth from the market and gave 7 1/3m of cloth to his sister. How much cloth is left with him?

Solution: Total cloth purchased =15 3/4m =63/4m

Cloth given to his sister =7 1/3m =22/3m

∴ cloth left with Tarun = \(\frac{63}{4}-\frac{22}{3}=\frac{189-88}{12}\)

101/12m or 8 5/12m

17. Roller coaster at an amusment park is 2/3m high. If a new roller coaster is built that is 3/5 times the height of the existing coaster, what will be the height of the new roller coaster?

Solution: Height of roller coaster =2/3m

New roller coaster =3/5 x2/3m =2/5m

Hence, height of the new roller coaster is 2/5m

18. The product of two rational numbers is -14/27, If one of the numbers be 7/9, find the other.

Solution: Product of two rational numbers = -14/27

one number = 7/9

∴ Other number= \(\frac{-14}{27} \div \frac{7}{9}=\frac{-14}{27} \times \frac{9}{7}=\frac{-2}{3}\)

19. Rajini had a certain amount of money in her purse. She spent ₹10 1/4 in the school canteen, bought a gift worth ₹25 3/4 and gave ₹16 1/2 to her friend. How much she have to begin with?

Solution: Amount give to school canteen= ₹10 1/4

Amount given to buy gift= ₹25 3/4

Amount given to her friend= ₹16 1/2

To begin with Rajni had

=₹\(10 \frac{1}{4}+₹ 25 \frac{3}{4}+₹ 16 \frac{1}{2}=₹\left(\frac{41}{4}+\frac{103}{4}+\frac{33}{2}\right)\)

=₹\(\left(\frac{41+103+66}{4}\right)=₹ \frac{210}{4}=₹ 52 \frac{2}{4}\)

=₹52 1/2.

20. One third of a group of people are men. If the nuimber of women is 200 more than the men, find the total number of people.

Solution: Number of men in the group =1/3 of the group.

Number of women = 1-1/3 = 2/3

Difference between the number of men and women = 2/3-1/3 = 1/3

If difference is 200, then total lnumber of people

= \(200 \div \frac{1}{3}=200 \times 3=600\)

Hence, the total number of people= 600.

KSEEB Class 8 Maths Solutions For Chapter 1 Playing With Numbers Points To Remember

Numbers in general form:

• A number is said to be in a general form if it is expressed as the sum of the products of its digits with their respective place values.

Two-digit numbers:

• A two-digit number  is written as, ab = 10a+b where a and b are whole numbers taking values from 0 to 9 such that , a ≠ 0
  Example:  63 = 10 x 6 + 3

Three-digit numbers:

• A three digit number abc is written as abc = 100a +10b +c  are whole numbers taking values from 0 to 9 such that a ≠0
  Example: 765 = 100 x 7 +10 x 6 +5

 

 

Tests of Divisibility

Read and Learn More KSEEB Solutions for Class 8 Maths

Divisibility by 2 :

• A number is divisible by 2 when its one’s digit is0, 2, 4, 6  or 8.
  Explanation: Given number

        abc = 100a + 10b +c

       100a and 10b are divisible by 2 because 100 and 10 are divisible by only when a= 0, 2, 4, 6 or 8.

Divisibility by 3:

• A number is divisible by 3 when the sum of its digits is divisible by 3.
  Example: Given number = 17658
  Sum of digits = 1+7+6+5+8 =27  which is divisible by 3
  ∴ 17658 is divisible by 3

Divisibility by 4 :

• A number is divisible by 4 when the number formed by its last two digits is divisible by 4.
  Example: 6216,548 etc.

Divisibility by 5:

• A number is divisible by 5 when its ones digit is 0 or 5 .
  Example: 545,640 etc.

Divisibility by 6:

• A number is divisible by 6 when it is divisible by both 2 and 3 .
  Example: 246,7230 etc.

Divisibility by 9:

• A number is divisible by 9 , when the sum of its digits is divisible by 9 .
  Example: Consider a number 215847
  Sum of digits = 2+1+5+8+4+7=27  which is divisible by 9.
  ∴ 215847 is divisible by 9 .

Divisibility by 10:

• A number is divisible by 10 when its one digit is 0 .
  Example: 640,420 etc.

Letters for digits (puzzles)

• Many number puzzles involving different letters for different digits are solved using rules of number operations.
• Two rules :

1. Each letter in the puzzle must stand for just one digit. Each digit must be represented by just one letter.
2. The first digit of a number cannot be zero, thus, we write the number “sixty-three” as 63 and not as 063 , or 0063 .

• Each letter in the puzzle must stand for just one digit. Each digit must be represented by just one letter.
• The first digit of a number cannot be zero, thus, we write the number “sixty-three” as 63 and not as 063 , or 0063 .

KSEEB Maths Class 8 Playing With Numbers Notes

Playing With Numbers Solutions KSEEB Class 8 Maths Exercises 1.1

1 . Find the values of the letters in each of the following and give reasons for the steps involved.

\(\begin{array}{r}
3 \mathrm{~A} \\
+25 \\
\hline \mathrm{B} 2 \\
\hline
\end{array}\)

Solution:

The addition of A and 5 is giving 2 i.e., a number whose ones digit is 2. This is possible only when digit A is 7. In that case, the addition of A(7) and 5 will give 12 and thus, I will be the carry for the next step. In the next step 1 + 3+2+ = 6

∴ The addition is as follows

\(\begin{array}{r}
37 \\
+25 \\
\hline 62 \\
\hline
\end{array}\)

Clearly, B is 6

Hence, A and Bare 7 and 6 respectively.

2. \( \begin{array}{r}
4 \mathrm{~A} \\
+98 \\
\hline \mathrm{CB3} \\
\hline
\end{array} \)

Solution: The addition of A and 8 is giving 3 i.e, a number whose ones digit is 3 . This is possible only when digit A is 5 . In that case, the addition of A and 8 will give 13 and thus, 1 will be the carry for the next step. In the next step.

1+4+9 = 14

∴ The addition is as follows

\(\begin{array}{r}
45 \\
+98 \\
\hline 143 \\
\hline
\end{array}\)

Clearly, B and C are 4 and 1 respectively. Hence, A,B  and C  are 5,4 and 1 respectively.

3.  \(\begin{array}{r}
1 \mathrm{~A} \\
\times \mathrm{A} \\
\hline 9 \mathrm{~A} \\
\hline
\end{array}\)

Solution: The multiplication of A with A itself gives a number whose ones digit is A again. This happens only when A=1,5 or 6.

If A=1, then the multiplication will be 11 x 1 =1. However, here the tens digit is given as 9 .

∴ A=1 is not possible.

Thirdly, if A=5, then the multiplication will be 15 x 5 75, thus A = 5  is also not possible.

If we take A=6  , then 16 x 6 = 96

∴ A should be 6

The multiplication is as follows

\(\begin{array}{r}
16 \\
\times 6 \\
\hline 96 \\
\hline
\end{array}\)

Hence, the value of A is 6.

4. \(\begin{array}{r}
\mathrm{AB} \\
+37 \\
\hline 6 \mathrm{~A} \\
\hline
\end{array}\)

Solution: The addition of A and 3 is giving 6 . There can be two cases.

1. 1st step is not producing a carry
   In that case, A comes to be 3 as  3 + 3 = 6  considering the first step in which the addition of B and 7 is giving A (i.e., 3), B should be a number such that the unit digit of this addition comes to be 3 , it is possible, only when B = 6. In this case,A = 6 + 7= 13.  However, A is a single digit number, hence it is not possible.
2. 2nd step is producing a carry
   In that case, A comes to be 2 as 1 + 2 + 3 = 6, considering the first step in which the addition of B and 7 is giving A  (i.e., 2) B should be a number such that the units digit of this addition comes to be 2 . It is possible only when B = 5 and 5 + 7 = 12.

\(\begin{array}{r}
25 \\
+37 \\
\hline 62 \\
\hline
\end{array}\)

Hence, the values of A and B are 2 and 5 respectively.

5. \(\begin{array}{r}
\text { A B } \\
\times 3 \\
\hline \text { CAB } \\
\hline
\end{array}\)

Solution: The multiplication of 3 and B gives a number whose ones digit is  B again. Hence B  must be 0 or 5 .

Let B  is 5.

Multiplication of 1st step = 3 x 5 = 15

1 will be a carry for the next step.

We have, 3 x A +1 = CA

This is not possible for any value of A. Hence, B must be 0 only. If B = 0, then there will be no carry for the next step we should obtain, 3 x A= CA  . That is, the one’s digit of 3 x A should be A.  This is possible When A=5 or 0.

However, A cannot be 0 as AB is a two digit number.

∴ A must be 5 only, the multiplication is as follows.

\(\begin{array}{r}
50 \\
\times 3 \\
\hline 150 \\
\hline
\end{array}\)

Hence, the values of A, B and C are 5, 0 and 1 respectively.

6. \(\begin{array}{r}
\mathrm{AB} \\
\times 5 \\
\hline \mathrm{CAB} \\
\hline
\end{array}\)

Solution: The multiplication of B and 5 is giving a number whose ones digit is B again this is possible when B=5 or B=0  only. In case of B=5 . The product, B x 5 = 5 x 5 = 25.5 will be a carry for the next step. We have,5 x A = 2 = CA, which is possible for A = 2  or 7

The multiplication is as follows

\(\begin{array}{r}
25 \\
\times 5 \\
\hline 125 \\
\hline
\end{array}\)
\(\begin{array}{r}
75 \\
\times 55 \\
\hline 375 \\
\hline
\end{array}\)

If B=0

B x 5 = B ⇒ 0 x 5 = 0

There will not be any carry in this step.

In the next step, 5 x A = CA

It can happen only when A = 5 orA = 0

However, A cannot be 0 as AB is a two digit number.

Hence, A can be 5 only, the multiplication is as follows.

\(\begin{array}{r}
50 \\
\times 5 \\
\hline 250 \\
\hline
\end{array}\)

Hence, there are 3 possible values of A, B and C.

1) 5,0 , and 2 respectively

2) 2,5 and 1 respectively

3) 7,5 and 3 respectively

7. \(\begin{array}{r}
\text { AB } \\
\times 6 \\
\hline \text { B BB B } \\
\hline
\end{array}\)

Solution: The multiplication of 6 and B gives a number whose one’s digit is B again. It is possible only when B =0,2,4,6 or 8. If B =0, then the product will be 0.

∴This value of B is not possible.

If B=2, then B x 6 = 12 and 1 will be a carry for the next step.

6A +1=BB=22 ⇒ 6A = 21and hence, any integer value of A is not possible.

If B=6 , then B x 6 = 36 and 3 will be a carry for the next step.

6A+3 =B=66 ⇒ 6A=63 and hence, any integer value of  is not possible.

If B=8 , then B x 6 =48 and 4 will be a carry for the next step.

6A +4=BB=88 ⇒ 6A = 84 and hence,A =14 however,  is a single digit number.

∴This value of A is not possible.

If B=4, then B x 6 =24 and 2 will be a carry for the next step.

6A +2=BB=44

⇒ 6A = 42 and hence A = 7 The multiplication is as follows

\(\begin{array}{r}
74 \\
\times 6 \\
\hline 444 \\
\hline
\end{array}\)

Hence, the values of A and B  are 7 and 4 respectively.

8.  \(\begin{array}{r}
\mathrm{A} 1 \\
+\underline{1 \mathrm{~B}} \\
\underline{\mathrm{BO}}
\end{array}\)

Solution: The addition of 1 and B is giving 0 . i.e., a number whose ones digits is 0 . This is possible only when digit B , is 9 . In that case, the addition of 1 and B will give 10 and thus, 1 will be the carry for the next step. In the next step. 1+A+1=B

Clearly, A is 7 as 1+7+1 =9 = B

∴The addition is as follows

\(\begin{array}{r}
71 \\
+19 \\
\hline 90 \\
\hline
\end{array}\)

Hence, the values of A and B are 7 and 9 respectively.

9.  \(\begin{array}{r}
2 \mathrm{AB} \\
+\mathrm{AB} 1 \\
\hline \mathrm{B} 18 \\
\hline
\end{array}\)

Solution: The addition of B and 1 is giving 8 i.e., a number, whose ones digit is 8 . This is possible only when digit B is 7 . In that case, the addition of B and 1 will give 8.

In the next step,

A+B=1

Clearly, A is 4 .

4+7=11 and 1 will be a carry for the next step. In the next step, 1+2+A=B

1+2+4=7

∴ the addition is as follows

\(\begin{array}{r}
247 \\
+471 \\
\hline 718 \\
\hline
\end{array}\)

Hence the values of A and B are 4 and 7 respectively.

10.   \(\begin{array}{r}
12 \mathrm{~A} \\
+6 \mathrm{AB} \\
\hline \mathrm{A} 09 \\
\hline
\end{array}\)

Solution: The addition of A and B is giving 9 i.e., a number whose ones digits is 9 . The sum can be 9 only as the sum of two single digit numbers cannot be 19 .

∴ There will not be any carry in this step.

In the next step, 2+A=0

It is possible only when A=8

2+8=10 and 1 will be the carry for the next step 1+1+6=A

Clearly, A is 8 , we know that the addition of A and B is giving 9. As A is 8 .

is 1

∴The addition is as follows.

\(\begin{array}{r}
128 \\
+681 \\
\hline 809 \\
\hline
\end{array}\)

Hence, the values of A and B are 8 and 1 respectively.

Playing With Numbers Problems In KSEEB Maths

KSEEB Class 8 Maths Chapter 1 Playing With Numbers Exercise 1.2

1. If 21y5 is a multiple of 9 , where y is a digit. What is the value of y ?

Solution: If a number is a multiple of 9 , then the sum of its digits will be divisible by 9 .

Sum of digits of 21y5=2+1+y+5= 8+y

Hence, 8+y should be a multiple of 9 .

This is possible when 8+y  is any one of these numbers 0,9,18,27 and so on.

However, since y is a single digit number. This sum can be 9 only.

∴ y should be 1 only.

2. If 31z5 is a multiple of 9 , where z is a digit, what is the value of z ? You will find that there are two answers for the last problem. Why is this so?

Solution: If a number is a multiple of 9 , then the sum of its digits will be divisible by 9 .

Sum of digits of 31z5 = 3+1+z+5+9 = 9+z.

Hence, 9+z should be a multiple of 9 . This is possible when 9+z is any one of these numbers 0,9,18,27 and so on.

However, since  is a single digit number this sum can be cither 9 or 18 .

∴ z should be either 0 or 9.

3. If 24x is a multiple of 3 , where x is a digit, what is the value of x ?

Solution: Since 24x is a multiple of 3 , the sum of its digits is a multiple of 3 , sum of digits of 24x = 2+4+x=6+x

Hence, 6 is a multiple of 3 .

This is possible when 6+x is any one of these numbers 0,3,6,9 and so on.

Since x is a single digit number, the sum of the digits can be 6 or 9 or 12 or 15 and thus, the value of x comes to 0 or 3 or 6 or 9 respectively.

Thus  can have its value as any of the four different values  or 9 .

4. If 31z5 is a multiple of 3 , where  is a digit, what might be the values of z ?

Solution: Since 31z5 is a multiple of 3 , the sum of its digits will be a multiple of 3 . i.e., 3+1+z+5=9+z  is a multiple of 3. This is possible when 9+z is any one of 0,3,6,9,12,15,18 and so on.

Since z is a single-digit number, the value of 9+z can only be 9 or 12 or 15 or 18 and thus, the value of x comes to 0 or 3 or 6 or 9 respectively.

Thus z can have its value as any one of the four different values 0,3,6 or 9 .

KSEEB Maths Class 8 Playing With Numbers Additional Problems

1. Without actual division find the remainder when 3,79,843 is divided by 3 .

Solution: We can find the remainder by dividing the sum of all the digits of the given number = 3+7+9+8+4+3=34. When 34 divided by 3 , we get 1 as remainder.

Hence, division 3,79,843 of  by 3 leaves remainder of 1.

2. Find the values of letter A and B.

\(\begin{array}{r}
2 \mathrm{AB} \\
+\mathrm{AB} 1 \\
\hline \mathrm{B} 18 \\
\hline
\end{array}\)

Solution: B+1=8 ⇒ B=8-1=7

A+B =1⇒ A+7=1 ( is ones digit)

A+7 can either 11 or 21 .

When A+7=11 ⇒ A=11-7=4

When A+7=21 ⇒ A=21-7=14 (not possible)

Thus A=4

∴ \(\begin{array}{r}
2 \text { A B } \\
+\text { A B 1 } \\
\hline \text { B 18 } \\
\hline
\end{array}\)
is \(\begin{array}{r}
247 \\
+471 \\
\hline 718 \\
\hline
\end{array}\)

∴ A is 4 and B is 7

KSEEB Class 8 Maths Key Concepts Of Playing With Numbers

3. If from a two digit number, we subtract the number formed by reversing its digits then the result so obtained is a perfect cube. How many such numbers are possible? Write all of them.

Solution: Let ab be any two digit number. Then the digit formed by reversing its digits is ba.

Now, \(\begin{aligned}
a b-b a & =(10 a+b)-(10 b+a) \\
& =(10 a-a)+(b-10 b) \\
& =9 a-9 b=9(a-b)
\end{aligned}\)

Further, since  ab – ba is a perfect cube and is a multiple of 9.

∴ The possible value of a-b is 3

i.e., a = b+3

Here, b can take value from 0 to 6

For b = 0, a = 3 i.e., 3 0

For b = 1, a = 4, i.e. 4 1

For b = 2, a = 5, i.e., 5 2

For b = 3, a = 6, i.e., 6 3

Forb = 4, a = 7, i.e., 7 4

For b = 5, a= 8, i.e., 8 5

For b = 6, a= 9, i.e., 9 6

4. Find the value of the letters in each of the following :

1.\(\begin{array}{r}
\mathrm{PQ} \\
\times 6 \\
\hline \mathrm{QQQ} \\
\hline
\end{array}\)

2. \( \begin{array}{r}
2 \text { L M } \\
+\text { L M 1 } \\
\hline \text { M18 } \\
\hline
\end{array}\)

Solution: 1) We have, \( \begin{array}{r}
\mathrm{PQ} \\
\times 6 \\
\hline \mathrm{QQQ} \\
\hline
\end{array}\)

Here, in first column, we see that 6 x QP = Q

∴ The possible values of Q are 2,4,6 and 8

For Q=2, 6 X P +1 cannot be equal to 22 for any value of P.

So, Q=2 is not possible

For Q=4

⇒ 6 x p +2= 44

⇒  6P = 44-2=42

⇒  P=2

Hence P =7 and Q = 4

2. \(\begin{array}{r}
2 \text { L M } \\
+\text { L M 1 } \\
\hline \text { M1 } 8 \\
\hline
\end{array}\)

In first column M+1=8

Clearly, M=7

In second column L+M=1

⇒ L+7=1

∴The value of L can be 4

In third column, 2+L+1=M

⇒ 2+4+1=7

⇒ 7=7, so the third column is satisfied for L=4, M=7

Hence, L=4 andM=7

5. If \(\begin{array}{r}
1 \mathrm{P} \\
\times \mathrm{P} \\
\hline \mathrm{Q} 6 \\
\hline
\end{array}\) where Q-P=3 , then find the values of P  and Q

Solution: We have \(\begin{array}{r}
1 \mathrm{P} \\
\times \mathrm{P} \\
\hline \mathrm{Q} 6 \\
\hline
\end{array}\)and Q-P=3

Here, P x P  is 6 , so the value of P is either 4 or 6 . But, if P=4,Q=5, which does not satisfy the relation Q-P=3

Hence,P=6  and then Q=9 .

6. If A3+8B=150, then find the value of A+B.

Solution: We have A3+8B=150 .

Here 3+B=0, So 3+B is a two-digit number whose unit’s digit is zero.

∴ 3+B=10 ⇒ B=7

Now, considering ten’s column, A+8+1=15

⇒ A+9=15

⇒ A=6

Hence A+B=6+7=13

A3+8B=150

i.e., 63+87=150

150=150

7. If 56 x 32y  is divisible by 18 , find the least value of y .

Solution: It is given that, the number 56 x 32y  is divisible by 18 , then it is also divisible by each factor of 18. Thus, it is divisible by 2 as well as 3 . Now, the number is divisible by 2 its unit digit must be an even number that is 0,2,4,6.

∴The least value of y is 0

Again, the number is divisible by 3 also, sum of its digits is a multiple of 3 , ie., 5+6+x+3+2+y is a multiple of 3.

⇒ 16+x+y+=0,3,6,9,………

⇒ 16+x=18

⇒ x=2, which is the least value of x.

∴ x=2 and y=0

Practice Questions For Playing With Numbers KSEEB Maths

8. Without performing actual division, find the remainder when 430346 is divided by 9.

Solution: Here sum of digits 4+3+0+3+4+6 is 20. When 20 is divide by 9 then the remainder will be 2 .

9. In a two digit number the digit in the one’s place is three times the digit in the ten’s place and the sum of the digits is equal to 8 . What is the number?

Solution: Let ten’s digit be ‘ a ‘ then unit digit be ‘3a  ‘

Number= 10 x a +3a

=10a+3a=13a

According to the question

a+3a=8

4a=8

a=2

∴ Number = 13 x 2 =26

10. If 24x is a multiple of 3 , where x is a digit, what is the value of x? 

Solution: Since 24x is a multiple of 3 its sum of digits 6+x is a multiple of 3 . So 6+x is one of these numbers;0,3,6,9,12,15,18, ………….

But since x is a digit, it can only be that 6+x=6 or 9 or 12 or 15.

∴ x=0 or 3 or 6 or 9.

Thus, x  can have any of four different values.

11. Find the values of the letter?
\(\begin{gathered}
\mathrm{AB} \\
\times 5 \\
\hline \mathrm{CAB}
\end{gathered}\)

Solution: This means that 5 x B is a number whose units digit is B . Clearly, B can take value 5 .

Taking B=5, we have

\(\Rightarrow \overline{\mathrm{A} 5} \times 5=\overline{\mathrm{CA} 5}\) \(\Rightarrow(10 A+5) \times 5=100 \times C+A \times 10+5 \times 1\) \(\Rightarrow 10 A+5=\frac{1}{5}(100 C+10 A+5)\) \(\Rightarrow 10 A+5=20 C+2 A+1\) \(\Rightarrow 8 A+4=20 C \Rightarrow A(2 A+1)=20 C\)

\(\Rightarrow 2 A+1=5 C\)

⇒ 2A + is an odd multiple of 5, hence 2A+1 is odd [∴ o< A≤ 9]

∴ 2A+1=5 or 2A+1=15

A=2, A=7

By putting A=2 in (i) we get C = 1

⇒ A=2, B=5 and C=1

and by putting A=7 in (i) we get C=3

∴ A=7,B=5 and C=3

12.The ratio of tens digit to the units digit of a two digit number is 2:3. If 27 is added to the number, the digits interchange their places, find the number.

Solution: Let the unit digit of the required number be X and the tens digit of the required number be Y

then the required number is  and number obtained by reversing the digits is also \(\frac{y}{x}=\frac{2}{3}\)

\(y=\frac{2}{3} x\) ……..(i)

According to the question

10 y+x+27=10 x+y  ………(ii)

Putting  \(y=\frac{2}{3} x\)  from equation (i) in equation (ii)

\(10 \times \frac{2}{3} x+x+27=10 x+\frac{2}{3} x\) \(\frac{20}{3} x+x+27=\frac{30 x+2 x}{3}\) \(\frac{20 x+3 x}{3}+27=\frac{32 x}{3}\) \(23 x+27 \times 3=32 x\)

32 x-23 x=81

9 x=81

x=81 / 9

\(y=2 / 3 x=\frac{2}{3} \times 9=6\)

Thus the required number is 10 x 6+9=69

Playing With Numbers Questions And Answers KSEEB Maths

13. Find the value of k where 31k2 is divisible by 6 .

Solution: 31k2 is divisible by 6 , then, it is also divisible by 2 and 3 both.

Now, 31k2 is divisible by 3 , sum of its digits is a multiple of 3.

⇒ k+6=0,3,6,9,12……..

⇒ k=0 or 3,6,9.

14. Prove that the difference of the given numbers and the numbers obtained by reversing their digits is divisible by 9.
1. 59
2. 203

Solution: 1. Given number = 59

Number obtained by reversing the digits = 95

Difference = 95-59=36÷9=4

Hence, the required number is 9.

2) Given number = 203.

Number obtained by reversing the digits = 302

Difference = 302-203=99÷9=11

Hence, the required number is 9.

15. If 1AB + CCA = 697 and there is no carry over in addition, find the value of A+B+C.

Solution: We have 1AB

\(\frac{+\mathrm{CAA}}{697}\)

Since there is no carry over – in addition

⇒  1+C=6

C=5

⇒  A+C=9

A+5=9

⇒  A=4

B+A=7

⇒  B=3

Hence, A+B+C=4+3+5=12

16. Observe the following patterns
1×9-1=8
21×9-1=188
321×9-1=2888
4321×9-1=38888
Find the value of 87654321×9-1

Solution: From the pattern, we observe that there are as many eights in the result as the first digit from the right which is to be multiplied by 9 and reduced by 1
87654321×9-1 = 788888888.

17. In a 3 – digit number, the hundreds digit is twice the tens digit while the units digit is thrice the tens digits Also, the sum of its digits is 18 . Find the number.

Solution:  Let the tens digit be x

then, the hundreds digit = 2x and the unit digit =3x

∴ 2x+x+3x=18 ⇒  6x=18 ⇒ x=3

∴ hundreds digit=(2×3)=6,

tens digit = 3 and

units digit=(3×3)=9

Hence, the required number

=(100×6+10×3+9)=639

18. Find all possible values of x for which the 4-digit number 754x si divisible by 3 . Also, find each such number.

Solution:  It is given that the number 754x is divisible by 3.

∴ Sum of its digits = (7+5+4+x) must be.

This happens when x=2 or 5 or 8 divisible by 3 . Since x a digit, it cannot be more than 9.

∴ x=2 or x=5 or x=8 or  are only required values of . Hence, all required numbers are 7542,7545,7548

19. If a,b, c are three digits of a three digit number, prove that abc+  cab+  bca is a multiple of 37.

Solution: We have abc + cab + bca

abc=100a+10b+c

cab=100c+10a+b

bca=100b+10c+a

adding abc+cab+bca = 111a+111b+111c

=111(a+b+c) = 3 x 3(a+b+c)

which is a multiple of 37.

Hence proved.

20. The product of two 2 – digit numbers is 1431 . The product of their tens digits is 10 and the product of their units digits is 21 . Find the numbers.

Solution: Let the required two 2 digit numbers be 10a+b and 10p+q as per the condition.

We have a x p=10 and b x q=21

a =2 and p=5 or a=5 and p=2

111ly b x q=21, b=3 and q = 7 or b =7 and q =3

10p+q=57 or 10p+q =53 and

10a+b =23 or 10a+b =27

Since the units digit of product 1431 is 1 numbers are 57 & 23  or 53 & 27

Now  &  which is given. Hence, the required numbers are 53 & 27.