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
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2) y + 3 = 10
solution: Given y + 3 = 10
By subtracting 3 from both sides we get
y + 3 – 3 = 10 -3
y = 7
KSEEB Solutions for Class 8 Maths Chapter 3 Linear Equations in One Variable
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
\(\frac{14 y}{14}=\frac{21}{14}\) \(y=3 / 2\)Kseeb Solutions For 8th Class Maths Chapter 3 Pdf
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
Karnataka Board 8th Class Maths Chapter 3 Linear Equations Solutions
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.
Karnataka Board Class 8 Maths Chapter 3 Important Questions
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.
Linear Equations In One Variable Kseeb 8th Class Notes
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
⇒ \(y=\frac{-24}{3}=-8\)
How To Solve Linear Equations In One Variable Class 8 Kseeb
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
Kseeb 8th Maths Chapter 3 Exercise Solutions Step By Step
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
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