KSEEB Solutions For Class 8 Maths Chapter 15 Comparing Quantities Points To Remember
Ratio comparing by division is called ratio. Quantities written in ratio have the same unit. Ratio has no unit. Equality of two ratios is called proportion.
Product of extremes = Product of means.
Percentage: Percentage means for every hundred. The result of any division in which the divisor is 100 is a percentage. The divisor is denoted by a special symbol % read as percent.
Profit and Loss.
1)Cost price (P): The amount for which an article is bought.
2)Selling price (SP): The amount for which an article is sold.
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Additional expenses made after buying an article are included in the cost price and are known as overhead expenses. These may include expenses like amount spent on repairs, labor charges, transportation, etc.
Discount is a reduction given on marked price. Discount = marked price – sale price.
Discount can be calculated when discount % is given. Discount = Discount % of marked price.
Additional expenses made after buying an article are included in the cost price and are known as overhead expenses.
CP = Buying price + Overhead expenses.
Sales tax is charged on the sale of an item by the government and is added to the Bill amount. Sales tax = Tax % of Bill amount.
KSEEB Solutions for Class 8 Maths Chapter 15 Comparing Quantities PDF
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Simple interest: If the principle remains the same for the entire loan period, then the interest paid is called simple interest.
\(\mathrm{SI}=\frac{P \times R \times T}{100}\)Compound interest is the interest calculated on the previous year’s amount(A=P+I)
1) Amount when interest is compounded annually
=\(P\left(1+\frac{R}{100}\right)^n\)
P=principle, R=Rate of interest, n=time period.
2) Amount when interest is compounded half yearly =\(P\left(1+\frac{R}{100}\right)^{2 n}\)
R/2 is half-yearly rate and 2n is = Number of ‘half-years’.
Karnataka Board Class 8 Maths Chapter 15 Solutions
Class 8 Maths KSEEB Chapter 15 Solutions Comparing Quantities Exercise 15.1
1 .Find the ratio of the following.
a)Speed of a cycle 15km per hour to the speed of scooter 30 km per hour.
b)5km to 10km
c) 50 paise to ₹5.
Solution: a) Ratio of the speed of cycle to the speed of scooter =\(\frac{15}{30}=1: 2\)
b) Since 1 km = 1000m
Required ratio
\(=\frac{5 m}{10 \mathrm{~km}}=\frac{5 m}{10 \times 1000 \mathrm{~m}}=1: 2000\)c)Since ₹1 = 100 paise.
\(\text { Required ratio }=\frac{50 \text { paise }}{₹ 5}\)=\(\frac{5 \text { paise }}{500 \text { paise }}=1: 10\)
2. Convert the following ratios to percentages
a) 3:4
b) 2:3
Solution: a) \(3: 4=\frac{3}{4}=\frac{3}{4} \times \frac{100}{100}=\frac{3}{4} \times 100 \%=75 \%\)
b) \(2: 3=\frac{2}{3}=\frac{2}{3} \times \frac{100}{100}=\frac{2}{3} \times 100 \%=\frac{200}{3} \%\)
=\(\left(\frac{66 \times 3+2}{3}\right) \%=66 \frac{2}{3} \%\)
3. 72% of 25 students are good in mathematics. How many are not good in mathematics?
Solution: It is given that 72% of 25 students are good in mathematics.
∴ Percentage of students who are not good in mathematics = (100-72)% =28%
∴ Number of students who are not good in mathematics = \(\frac{28}{100} \times 25=7\)
4. A football team won 10 matches out of the total number of matches they played. If their win percentage was 40, then how many matches did they play in all?
Solution: Let the total number of matches played by the team be x.
It is given that the team won 10 matches and the winning percentage of the team was 40%
∴ \(\frac{40}{100} \times x=10\)
\(x=\frac{10 \times 100}{40}=25\)Thus, the team played 25 matches.
Comparing Quantities Class 8 KSEEB Solutions with Answers
5. If chameli had ₹600 left after spending 75% of her money, how much did she have in the beginning?
Solution: Let the amount of money which chameli had, in the beginning, be x.
It is given that after spending 75% of ₹x, she was left with ₹600.
∴ (100-75)% of x = ₹600
or, 25% of x = ₹600
\(\frac{25}{100} \times x=₹ 600\) \(x=\left(600 \times \frac{100}{25}\right)=₹ 2400\)thus, she had ₹2400 in the beginning.
6. If 60% people in city like cricket, 30% like football, and the remaining like other games, then what % of the people like other games? If the total number of people are 50 lahks,
Solution: % of people who like other games
= (100-60-30)%
= 10%
Total number of people = 50 lakh
∴ number of people who like cricket
=\(\left(\frac{600}{100} \times 50\right) \text { lakh }\)
Number of people who like football
=\(\left(\frac{30}{100} \times 50\right) \text { lakh }=15 \text { lakh }\)
Number of people who like other games
= \(\left(\frac{10}{100} \times 50\right) l a k h=5 l a k h\)
Comparing Quantities KSEEB Class Exercise 15.2
1. A man got a 10% increase in his salary. If his new salary is ₹1,54,000, find his original salary.
Solution: Let the original salary be x
it is given that the new salary is ₹1,54,000,
Original salary + Increment = New salary.
However, it is given that the increment is 10% of the original salary.
∴ \(x+\frac{10}{100} \times x=154000\)
\(\frac{110 x}{100}=154000\) \(x=154000 \times \frac{100}{110}\)x=140000
Thus, the original salary was ₹1,40,000
2. on Sunday 845 people went to the zoo. On Monday only 169 people went. What is the percent decrease in the people visiting the zoo on Monday?
Solution: It is given that on Sunday, 845 people went to the zoo and on Monday, 169 people went.
Decrease in the number of people
= 845-169 = 676
% decrease = \(\left(\frac{\text { Decrease in the number of people } \times 100}{\text { Number of people who went to zoo on }}\right) \%\)
= \(\frac{676}{845} \times 10\)
=80%
KSEEB Class 8 Maths Comparing Quantities Exercise 15.1 Solutions
3. A shopkeeper buys 80 articles for ₹2400 and sells them for a profit of 16%. Find the selling price of one article.
Solution: It is given that the shopkeeper buys 80 articles for ₹2400.
Cost of one article = \(₹ \frac{2400}{80}=₹ 30\)
profit% = 16
\(\text { Profit } \%=\frac{\text { profit }}{\text { C.P. }} \times 100\) \(16=\frac{\text { profit }}{₹ 30} \times 100\) \( \text { Profit }=₹\left(\frac{16 \times 30}{100}\right)=₹ 4.80\)Selling price of one article = CP + Profit
= 30 + 4.80 = ₹34.80
4. The cost of an article was ₹15,500, and ₹450 were spent on its repairs. If it is sold for a profit of 15%. Find the selling price of the article.
Solution: Total cost of an article = cost + overhead expenses
= ₹1500+₹450
= ₹15950
\(\text { Profit }=₹\left(\frac{15950 \times 15}{100}\right)=₹ 2392.50\)selling price of the article
= c.p. + profit = ₹(15950+2392.50)
=₹18342.50
5. A VCR and TV were bought for ₹8000 each. The shopkeeper made a loss of 4% on the VCR and a profit of 8% on the TV. Find the gain or loss percent on the whole transaction.
Solution: C.P. of VCR = ₹8000
The shopkeeper made a loss of 4% on VCR.
This means if C.P. is ₹100, then S.P. is ₹96.
When C.P. is ₹8000, S.P. = \( ₹\left(\frac{96}{100} \times 8000\right)\) = ₹7680
C.P. of a TV = ₹8000
The shopkeeper made a profit of 8% on TV
This means that if C.P. is ₹100, then S.P. is ₹108.
When C.P. is ₹8000, S.P. = \(\left(\frac{108}{100} \times 8000\right)\) = ₹8640
Total S.P. = ₹7680+₹8640=₹16320
Total C.P. = ₹8000+₹8000=₹16000
Since total S.P.>total C.P. There was a profit.
Profit = ₹16320-₹16000 = ₹320
\(\text { Profit } \%=\frac{\text { Profit }}{\text { C.P. }} \times 100=\frac{320}{16000} \times 100=2 \%\)∴ The shopkeeper had a gain of 2% on the whole transaction.
6. During a sale, a shop offered a discount of 10% on the marked prices of all the items. What would a customer have to pay for a pair of jeans marked at ₹1450 and two shirts marked at ₹850 each?
Solution: Total marked price =₹(1450+2×850)
= ₹3150
Given that, discount% = 10%
Discount = \(₹\left(\frac{10}{100} \times 3150\right)=₹ 315\)
also, Discount=marked price-sale price
₹315=₹3150-sale price
∴ sale price = ₹(3150-315) =₹2835
Thus, the customer will have to pay ₹2835
7. A milkman sold two of his buffaloes for ₹20,000 each on one he made a gain of 5% and on the other a loss of 10%. Find his over all gain or loss.
Solution: S.P. of each buffalo =₹20000
The milkman made a gain of 5% while selling one buffalo.
This mean if C.P. is ₹100, then S.P. is ₹105.
C.P. of one buffalo = \(₹\left(20000 \times \frac{100}{105}\right)\)
= ₹19,0467.62
Also, the second buffalo was sold at a loss of 10%.
This means if C.P. is ₹100, then S.P. is ₹90.
∴ C.P. of other buffalo = \(₹\left(20000 \times \frac{100}{90}\right)\)
= ₹22222.22
Total C.P. =₹19047.62+₹22222.22
= ₹41269.84
Total S.P. = ₹20000+₹20000=₹400000
loss=₹41269.84-₹400000=₹1269.84
Thus, the overall loss of milkman was ₹1269.84.
8. The price of a TV is ₹13,000. The sales tax charged on it is at the rate of 12%. Find the amount that Vinod will have to pay if he buys it.
Solution: On ₹100, the tax to be paid = ₹12
On ₹13000, the tax to be paid
= \(₹\left(\frac{12}{100} \times 13000\right)\)
Required amount = cost+salestax
= ₹13000+₹1560=₹14560
Thus, Vinod will have to pay ₹14560 for the TV.
9. Arun bought a pair of skates at a sale where the discount given was 20%. If the amount he pays is ₹1600, find the marked price.
Solution: Let the marked price be x.
\(\text { Discount } \%=\frac{\text { Discount }}{\text { Marked price }} \times 100\) \(20=\frac{\text { Discount }}{x} \times 100\) \(\text { Discount }=\frac{20}{100} \times x=\frac{1}{5} x\)Also, Discount = marked price – sale price
\(\frac{1}{5} x=x-₹ 1600\) \(x-\frac{1}{5} x=₹ 1600\) \(\frac{4}{5} x=₹ 1600\) \(x=₹\left(1600 \times \frac{5}{4}\right)=₹ 2000\)Thus, the marked price was ₹2000.
10. I purchased a hair dryer for ₹5400 including 8% VAT. Find the price before VAT was added.
Solution: The price includes VAT
Thus, 8% VAT means that if the price without VAT is ₹ 100. The price including VAT will be 108. When price including VAT is ₹108,
Original price = ₹100
When price including VAT is ₹5400,
Original price = \(₹ \frac{100}{108} \times 5400=₹ 5000\)
Thus, the price of the hair dryer before the addition of VAT was ₹5000
Karnataka Board 8th Maths Chapter 15 Important Questions and Answers
11. An article was purchased for ₹1239 including a GST of 18%. Find the price of the article before GST was added.
Solution: Given, GST = 18%
Cost with GST included =₹1239
Cost without GST = ₹x
Cost before GST + GST = cost with GST
\(x+\left(\frac{18}{100} \times x\right)=1239\) \(x+\frac{9 x}{50}=1239\)x = 1050
Price before GST = 1050 rupees.
KSEEB Class 8 Maths Comparing Quantities Exercise 15.3
1. Calculate the amount and compound interest on
a) ₹10,800 for 3 years at 12 1/3% per annum compounded annually.
Solution: Principle (P) = ₹10800, Number of years (n)= 3
Rate (R) = \(12 \frac{1}{2} \%=\frac{25}{2} \%=\text { (annual) }\)
\(\text { Amount, } A=P\left(1+\frac{R}{100}\right)^n=10800\left(1+\frac{25}{200}\right)^3\)=\(10800\left(\frac{225}{200}\right)^3=10800 \times \frac{225}{200} \times \frac{225}{200} \times \frac{225}{200}\)
= 15377.34375
= ₹15377.34(approximately)
CI=A-P = 15377.34-10800 = ₹4577.34
b) ₹18,000 for \(2 \frac{1}{2}\) years at 10% per annum compounded annually.
Solution: P = ₹18000, R= 10%, n= 2 \(\frac{1}{2}\)years.
The amount for 2 years and 6 months can be calculated by 1 st calculating the amount for 2 years using the compound interest formula, and then calculating the simple interest for 6 months on the amount obtained at the end of 2 years.
Firstly, the amount for 2 years has to be calculated
\(A=1800\left(1+\frac{1}{10}\right)^2=18000\left(\frac{11}{10}\right)^2\)=\(18000 \times \frac{11}{10} \times \frac{11}{10}=₹ 21780\)
By taking ₹21780 as principal, the S.I. for the next \(\frac{1}{2}\) year will be calculated.
\(\mathrm{SI}=₹\left(\frac{21780 \times \frac{1}{2} \times 10}{100}\right)=₹ 1089\)∴ interest for the 1st 2years = ₹(21789-18000) = ₹3780
and interest for the next \(\frac{1}{2}\) year = ₹1089.
Total C.I. = ₹3780 + ₹1089 = ₹4869
A = P + C.I. = ₹18000 + ₹4869 = ₹22869
c) ₹62,500 for \(1 \frac{1}{2}\) years at 8% per annum compounded half yearly
Solution: P = ₹62500, R = 8% per annum or 4% per half-year
Number of years = \(1 \frac{1}{2}\)
There will be 3 half years in \(1 \frac{1}{2}\)
\(\mathrm{A}=\mathrm{P}\left(1+\frac{\mathrm{R}}{100}\right)^n=62500\left(1+\frac{4}{100}\right)^3\)=\(62500 \times \frac{26}{25} \times \frac{26}{25} \times \frac{26}{25}\) = ₹70304
C.I. = A-P = ₹70304-₹62500 = ₹7,804
d) ₹8,000 for 1 year at 9% per annum compounded half yearly.
Solution: P = ₹8000, R = 9% per annum or \(\frac{9}{2}\) % per half year, n=1 year
There will be 2 half years in 1 year.
\(\mathrm{A}=\mathrm{P}\left(1+\frac{\mathrm{R}}{100}\right)^n\)=\(8000\left(1+\frac{9}{200}\right)^2=8000 \times\left(\frac{209}{200}\right)^2\)
C.I. = A-P = ₹8736.20-₹8000
= ₹736.20
e) ₹10000 for 1 year at 8% per annum compounded half yearly.
Solution: P = ₹10,000, R = 8% per annum or 4% per half year, n = 1 year. There are 2 half years in 1 year.
\(\mathrm{A}=\mathrm{P}\left(1+\frac{\mathrm{R}}{100}\right)^n=10000\left(1+\frac{4}{100}\right)^2\)=\(10000\left(1+\frac{1}{25}\right)^2\)
=\(10000 \times \frac{26}{25} \times \frac{26}{25}=₹ 10,816\)
CI = A-P = ₹10816-₹10000 = ₹816
2. Kamala borrowed ₹26400 from a bank to buy a scooter at a rate of 15% p.a. compounded yearly. What amount will she pay at the end of 2 years and 4 months to clear the loan?
(Hint: Find A for 2 years with interest compounded yearly and then find SI on the 2nd year amount for \(\frac{4}{12}\) years)
Solution: P = ₹26400, R = 15% per annum, n = \(2 \frac{4}{12}\) years. The amount for 2 years and 4 months can be calculated by first calculating the amount for 2 years using the compounded interest formula, and then calculating the simple interest for 4 months on the amount obtained at the end of the years. Firstly, the amount for 2 years has to be calculated.
\(A=26400\left(1+\frac{15}{100}\right)^2=26400\left(1+\frac{3}{20}\right)^2\)=\(26400 \times \frac{23}{20} \times \frac{23}{20}=₹ 34914\)
By taking ₹34916 as principal, the S.I. for the next \(\frac{1}{3}\) years will be calculated.
\(\text { S.I. }=₹\left(\frac{34914 \times \frac{1}{3} \times 15}{100}\right)=₹ 1745.70\)Interest for the 1 st two years = 34914 – 26400 =₹8514
and interest for the next \(\frac{1}{3}\) years =₹1,745.70
Total C.I. = 8514+ 1745.70 = ₹10,259.70
Amount = P + C.I. = ₹26400 + ₹10259.70 = ₹36659.70
3. Fabina borrows ₹12500 at 12% per annum for 3 years at simple interest and Radha borrows the same amount for the same time period at 10% per annum, compounded annually, who pays more interest, and by how much?
Solution: Interest paid by Fabina = \(\frac{\text { PTR }}{100}\)
=\(\frac{12500 \times 3 \times 12}{100}=₹ 4500\)
Amount paid by Radha at the end of 3 years
\(\mathrm{A}=\mathrm{P}\left(1+\frac{\mathrm{R}}{100}\right)^n=12500\left(1+\frac{10}{100}\right)^3\)=\(12500 \times \frac{110}{100} \times \frac{110}{100} \times \frac{110}{100}=₹ 16,637.50\)
C.I. — A – P= 16637.50 – 12500 = 4137.50.
The interest paid by Fabina is ₹4500 and by Radha is ₹4137.50.
Thus, Fabina pays more interest.
₹4500 -4137.50 = ₹362.50
Hence, Fabina will have to pay ₹362.50 more.
Download KSEEB Solutions For 8th Maths Chapter 15
4. I borrowed ₹12000 from Jamshed at 6% per annum simple interest for 2 years. Had I borrowed this sum at 6% per annum compound interest. What extra amount would I have to pay?
Solution: P = ₹12000, R = 6% per annum, T = 2 years
\(\text { S.I. }=\frac{\text { PTR }}{100}=\frac{12000 \times 2 \times 6}{100}=₹ 1440\)To find the C.I., the amount (A) has to be calculated.
\(\mathrm{A}=\mathrm{P}\left(1+\frac{\mathrm{R}}{100}\right)^n=12000\left(1+\frac{6}{100}\right)^2\)=\(12000\left(1+\frac{3}{50}\right)^2=12000 \times\left(\frac{53}{50}\right) \times\left(\frac{53}{50}\right)\)
= ₹13483.20
∴ CI = A-P = 13483.20-12000 = ₹1483.20
CI – SI = ₹1483.20 – ₹1400 = ₹43.20
Thus the extra amount to be paid is ₹43.20.
KSEEB Class 8 Maths Chapter 15 Textbook Solutions
5. Vasudevan invested ₹60000 at an interest rate of 12% per annum compounded half yearly. What amount would he get?
1) after 6 months?
2) after 1 year?
Solution:1) P = ₹60,000, R = 12% p.a. = 6% per half year. n = 6 months = 1 half year.
\(\mathrm{A}=\mathrm{P}\left(1+\frac{\mathrm{R}}{100}\right)^n\)=\(60000\left(1+\frac{6}{100}\right)^1=60000 \times \frac{106}{100}\)
= ₹63,600.
2) There are 2 half years in 1 year.
n=2
= ₹67,416
6. Arif took a loan of ₹80,000 from a bank. If the rate of interest is 10% p.a. Find the difference in amounts he would be paying after 1 \(\frac{1}{2}\) years, if the interest is
1)compounded annually
2)compounded half-yearly.
Solution: 1) P = ₹80000, R= 10% p.a., n = 1 1/2years.
The amount for 1 year and 6 months can be calculated by first calculating the amount for 1 year using the compound interest formula and then calculating the simple interest for 6 months on the amount obtained at the end of 1 year.
Firstly, the amount for 1 year has to be calculated.
\(A=80000\left(1+\frac{10}{100}\right)^1=80000 \times \frac{11}{10}=₹ 88,000\)By taking ₹88000 as principal. The SI for the next 1/2 year will be calculated
\(\mathrm{SI}=\frac{\mathrm{PTR}}{100}=\frac{88000 \times 10 \times 1 / 2}{100}=₹ 4,400\)Total C.I. = ₹8000 + ₹4,400 = ₹12,400
A = P + C.I. = 80000 + 12400 = ₹92400
2) The interest is compounded half-yearly. R = 10% p.a. = 5% per half year.
There will be three half year in 1 \frac{1}{2} years
\(A=80000\left(1+\frac{5}{100}\right)^3=80000\left(1+\frac{1}{20}\right)^3\)=\(80000 \times \frac{21}{20} \times \frac{21}{20} \times \frac{21}{20}=₹ 92,610\)
Difference between the amounts = ₹92,610-₹92400 = ₹210
KSEEB Maths 8th Standard Chapter 15 Comparing Quantities guide
7. Maria invested ₹8000 in a business, and she would be paid interest at 5% per annum compounded annually. Find
1)The amount credited against her name at the end of the second year.
2)The interest for the 3rd year.
Solution: a) 1) P = ₹8000, R = 5% p.a. n = 2 years.
\(A=8000\left(1+\frac{5}{100}\right)^2=8000\left(1+\frac{1}{20}\right)^2\)=\(80000\left(\frac{21}{20}\right)^2=8000 \times \frac{21}{20} \times \frac{21}{20}=₹ 8820\)
2) The interest for the next one year.
i.e., The third year has to be calculated. By taking ₹8820 as principal. The S.I. for the next year will be calculated.
\(\text { S.I. }=\frac{8820 \times 1 \times 5}{100}=₹ 441\)8. Find the amount and the compound interest on ₹10,000 for 1 1/2 years at 10% per annum compounded half yearly. Would this interest be more than the interest he would get if it was compounded annually?
Solution: p = ₹10000, R=10%p.a. = 5% per half year \(n=1 1 / 2 \text { years. }\)
There will be 3 half years in 1 1/2 years.
\(A=10000\left(1+\frac{5}{100}\right)^3=10000\left(1+\frac{1}{20}\right)^3\)=\(10000 \times \frac{21}{20} \times \frac{21}{20} \times \frac{21}{20}=₹ 11576.25\)
C.I. = A-P= 11576.25 – 10000= 1576.25
The amount for 1 year and 6 months can be calculated by first calculating the amount for 1 year using the compound interest formula, and then calculating. The simple interest for 6 months on the amount obtained at the end of 1 year.
The amount for the first year has to be calculated first.
\(A=10000\left(1+\frac{10}{100}\right)^1=10000\left(1+\frac{1}{10}\right)\)=\(10000 \times \frac{11}{10}=₹ 11000\)
By taking ₹11000 as the principal, the S.I. for the next 1/2 year will be calculated.
\(\text { S.I. }=₹\left(\frac{11000 \times 10 \times \frac{1}{2}}{100}\right)=₹ 550\)∴ Interest for the first year = ₹11000=₹10000
= ₹1000+₹550 = ₹1550
∴ The interest would be when compounded half yearly than the interest when compounded annually.
9. Find the amount which Ram will get on ₹4096, he gave it for 18 months at 12 1/2%p.a., interest being compounded half-yearly.
Solution: P = ₹4096, R= 12 1/2% p.a. = 25/4% per half year. n=18 months.
There will be 3 half years in 18 months.
∴ \(A=P\left(R+\frac{1}{100}\right)^n=4096\left(1+\frac{25}{400}\right)^3\)
=\(4096\left(1+\frac{1}{16}\right)^3\)
=\(4096\left(\frac{17}{16}\right)^3=4096 \times \frac{17}{16} \times \frac{17}{16} \times \frac{17}{16}=4913\)
Thus, the required amount is ₹4913.
10. The population of a place increased to 54,000 in 2003 at a rate of 5% per annum.
1)Find the population in 2001
2)What would be its population in 2005?
Solution: 1) it is given that, the population in the year 2003 = 54,000
∴ \(54000=(\text { Population in 2001 })\left(1+\frac{5}{100}\right)^2\) population in 2001
=\(\frac{54000}{\left(1+\frac{5}{100}\right)^2}=\frac{54000}{\left(1+\frac{1}{20}\right)^2}=\frac{54000}{\left(\frac{21}{20}\right)^2}\)|
=\(54000 \times\left(\frac{20}{21}\right)^2=54000 \times \frac{20}{21} \times \frac{20}{21}\)
= 48979.59
Thus, the population in the year 2001 was approximately 48980.
2)\(\text { Population in } 2005=54000 \times\left(1+\frac{5}{100}\right)^2\)
=\(5400 \times\left(1+\frac{1}{20}\right)^2=54000 \times \frac{20}{21} \times \frac{20}{21}=59535\)
Thus the population in the year 2005 would be 59535.
11. In a laboratory, the count of bacteria in a certain experiment was increasing at the rate of 2.5%. per hour. Find the bacteria at the end of 2 hours if the count was initially 506000.
Solution: The initial count of bacteria is given as 506000.
Bacteria at the end of 2 hours
=\(506000\left(1+\frac{1}{40}\right)^2=506000 \times \frac{41}{40} \times \frac{41}{40}\)
= 531616.25 = 531616(approx.)
Thus, the count of bacteria at the end of 2 hours will be 531616 approx.
12. A scooter was bought of ₹42000. Its value depreciated at the rate of 8% p.a. Find its value after one year.
Solution: Principal = cost price of the scooter = ₹42000
Depriciation=8% of ₹42000 per year
=\(\left(\frac{42000 \times 8 \times 1}{100}\right)=₹ 3360\).
Value after 1 year = ₹(42000-3360) = ₹38640.
Class 8 Maths Chapter 15 KSEEB Comparing Quantities Additional Problems
1. A jacket was sold for ₹4120 after allowing a discount of 20%. Find marked price of the Jacket.
Solution: Let the marked price of the jacket be ₹x.
Discount % on marked price = 20%
Selling price of Jacket = ₹1120.
Then \(1120=x-x \times \frac{20}{100}\)
\(1120=x-\frac{x}{5}\) \(1120=\frac{5 x-x}{5}=\frac{4 x}{5}\) \(\Rightarrow x=\frac{1120 \times 5}{4}=280 \times 5=₹ 1400\)So, marked price of the jacket is ₹1400.
2. Anubha’s income is 20% less than that of Shubra. How much percent is Shubra’s income more than Anubha’s?
Solution: I Method: Suppose Shubra’s income = ₹100.
\(20 \% \text { of } 100=\frac{20}{100} \times 100=₹ 20\) .
∴ Anubha’s income = ₹(100 – 20) = ₹80
If Anubha’s income is ₹80.
Then Shubhra’s income = ₹100
If Anubha’s income is ₹1
Then Shubhra’s income = \( ₹ \frac{100}{80}\)
If Anubha’s income is ₹100
Then Shubhra’s income = \( ₹ \frac{100}{80} \times 100=₹ 125\)
∴ Shubhra’s income is ₹(125 – 100) or ₹25 i.e. 25% more than Anubha’s
(or)
II Method
Let Shubhra’s income = ₹100 then Anubha’s income = ₹(100 – 20) = ?80
Then Shubhra’s income is ₹20 more than Anubha’s.
Shubhra’s income = \(\frac{20}{80} \times 100\) = 25% more than 80 Anubha’s.
KSEEB 8th Standard Maths Chapter 15 Notes and Solutions
3. The price of sugar increased by 20%. By how much percent should Divesh reduce her consumption of sugar? So that her expenditure of sugar does not increase?
Solution: Suppose original consumption of sugar = 100kg
Suppose price of 100kg of sugar = ₹100
Increase in the price of sugar = 20% = ₹20.
Price of sugar after increase=₹100 + 20 = ₹120.
But expenditure has to remain ₹100 only.
So for ₹120, Divesh gets 100kg sugar
For ₹1 Divesh gets = \(\frac{100}{120}\)kg sugar
For ₹100 Divesh gets =\(\frac{100}{120} \times 100=\frac{10000}{120} \mathrm{~kg}\)
=\(83 \frac{1}{3} \mathrm{~kg} \text { sugar }\)
∴ Reduction in consumption =\(\left(100-83 \frac{1}{3}\right) \%\)
=\(\left(100-\frac{250}{3}\right) \mathrm{kg}=\frac{50}{3} \%=16 \frac{2}{3} \%\)
4. Tarun purchased a calculator for ₹500 and sold it to his friend at ₹560. Find his gain percent.
Solution: Here C.P. ₹500 and S.P. ₹560
Since S.P > C.P So, there is a profit.
Profit = S.P. – C.P. = ₹560 – ₹500 = ₹60
Profit percent
=\(\frac{\text { Profit }}{\text { C.P. }} \times 100=\frac{60}{500} \times 100=12 \%\)
5. If the selling price of 16 water bottles is equal to the cost price of 17 water bottles, find the gain percent earned by the dealer.
Solution: Let the cost price of each water bottle be ₹1
So the cost price of 17 water bottles will be ₹17.
C.P. of 16 water bottles = ₹16
S.P. of 16 water bottles = ₹17
∴ S.P. > C.P.
∴ There is gain and
Gain = S.P. – C.P. = ₹17- ₹l6 = ₹1
Gain% =
\(\frac{\text { Gain }}{\text { C.P. }} \times 100=\frac{1}{16} \times 100=\frac{25}{4}=6 \frac{1}{4} \%\)6. A dealer marks his goods at 35% above the cost price and allows a discount of 20% on the marked price. Find his gain or loss percentage.
Solution: Suppose C.P. =₹100
Marked price = ₹100 + 35% of 100
= ₹100 + ₹35 = ₹135
Discount = \(₹ \frac{20}{100} \times 135=₹ 27\)
S.P.= ₹135-₹27 = ₹108
∴ Gain = ₹108 – ₹100 = ₹8
∴ Gain % = \(\frac{8}{100} \times 100=8 \%\)
7. Meena bought a pen for ₹23.75, after getting 5% discount on it. What is its list price? Ans:
Solution: Let the list price be = ₹x.
Discount % = 5%
Discount = 5% of x
Net price = ₹23.75
Net price = \(x-\frac{x}{20}=\frac{19 x}{20}\)
\(\Rightarrow \frac{19 x}{20}=23.75\) \(\text { or } \frac{₹ 23.75 \times 20}{19}=\frac{2375 \times 20}{19 \times 100}=₹ 25 \text {. }\)8. Calculate the time in which ₹8000 would become ₹9500 at an interest rate of 7 1/2% p.a.
Solution: P = ₹8000, A= ?9500, R = \(7 \frac{1}{2} \%=\frac{15}{2} \%\)
I =A-P = 9500 – 8000 = 1500
\(\Rightarrow \mathrm{I}=\frac{\mathrm{PTR}}{100}=\mathrm{T}=\frac{\mathrm{I} \times 100}{\mathrm{PR}}=\frac{1500 \times 100}{8000 \times \frac{15}{2}}\)=\(\frac{150^{1 \phi^5} \emptyset \times 100 \times 2^1}{4 28 \emptyset \emptyset 0 \times 15}=\frac{5}{2}=2 \frac{1}{2} \text { years. }\)
9. What principal will earn an interest of ₹576 at 6% per annum in 3 years?
Solution: I = ₹576, R = 6% p.a, T = 3 y ears.
\(\mathrm{I}=\frac{\mathrm{PTR}}{100} \Rightarrow \mathrm{P}=\frac{\mathrm{I} \times100}{\mathrm{TR}}=\frac{576 \times 100}{3 \times 6}\)=\(\frac{576 \times 100}{18}=₹ 3200\)
∴ p = ₹3200
Step-by-step Solutions for Comparing Quantities Class 8 Karnataka Board
10. Reema borrowed ₹5000. At the end of 5 years. Reema had to pay back ₹6225. What was the rate of interest?
Solution: P = ₹5000, A = ₹6225, T = 5 years.
Since A = P + I
∴ I = A – P = 6225 – 5000 = ₹1225
\(\mathrm{I}=\frac{\mathrm{PTR}}{100}\) \(\mathrm{R}=\frac{100 \times \mathrm{I}}{\mathrm{PT}}=\frac{100 \times 1225}{5000 \times 5}=\frac{1225}{250}\)= 4.9% p.a.
R = 4.9% p.a.
11. If P = ₹3000, R = 8%, T = 13 weeks, find the total amount to be paid.
Solution: P = ₹3000, R = 8%, and T= 13 weeks
∴ \(\text { S.I. }=\frac{\text { PTR }}{100}=\frac{₹ 3000 \times 8 \times 13}{100 \times 52}=₹ 60\)
12. Heema deposited a sum of ₹500 at the rate of 5% in the bank on August 5. On October 17, she withdraw the sum deposited along with the interest. Find the amount she got.
Solution: P=₹500, T=Aug 5 to Oct = 17, R = 5% p.a.
Aug 5 to Aug 31 26 days
Sept 30 days
Oct 17 days
Total \(\frac{73}{365} \text { years }\)
\(I=\frac{P T R}{100}=\frac{₹ 500 \times 5 \times 73}{100 \times 365}=₹ 5\)Manavi got amount = P + I = ₹500+₹5
= ₹505
13. In what time will ₹1600 amount to ₹2025 and 12 1/2% per annum compounded annually.
Solution: Let time be n conversion periods.
Amount = ₹2025, principal = 1600 and Rate = \(12 \frac{1}{2} \% \text { p.a. }=\frac{25}{2} \% \text { p.a. }\)
\(\mathrm{A}=\mathrm{P}\left(1+\frac{\mathrm{R}}{100}\right)^n\) \(2025=1600\left(1+\frac{\frac{25}{2}}{100}\right)^n\) \(2025=1600\left(1+\frac{25}{2} \times \frac{1}{100}\right)^n\) \(2025=1600\left(1+\frac{1}{8}\right)^n\) \(\frac{2025}{1600}=\left(\frac{9}{8}\right)^n\) \(\frac{81}{64}=\left(\frac{9}{8}\right)^n\) \(\left(\frac{9}{8}\right)^2=\left(\frac{9}{8}\right)^n\)∴ n=2
Hence, time = 2 years.
14. Find the amount on compound interest a sum of ₹5000 for 2 years at the rate of 2% per annum.
Solution: P1 = 5000, R = 2% p.a. T = 1 year.
\(\mathrm{I}_1=\frac{5000 \times 2 \times 1}{100}=₹ 100\) \(A_1=P_1+I_1=₹ 5000+₹ 100=₹ 5100=P_2\) \(\mathrm{P}_2=₹ 5100, \mathrm{R}=2 \% \text { p.a., } \mathrm{T}=1 \text { year. }\) \(I_2=\frac{₹ 5100 \times 2 \times 1}{100}=₹ 102\) \(\mathrm{A}_2=\mathrm{P}_2+\mathrm{I}_2=₹ 5100+₹ 102=₹ 5202\)Thus, amount at the end of the second year = ₹5202, and compound interest
= A – P =₹(5202-5000) = ₹202.
15. On selling a chair of ₹736, a shopkeeper suffers a loss of 8%. At what price should he sell it, so as to gain 8%?
Solution: Let the cost price of a chair = ₹x
Selling price of a chair = ₹736.
At selling a shopkeeper suffers 8% loss.
According to the question,
\(x-x \times \frac{8}{100}=736\) \(\frac{92 x}{100}=736\) \(x=\frac{736 \times 100}{92}=₹ 800\)To gain 8% profit the price should be
=\(800+\frac{8}{100} \times 800=800+64\)
= ₹864
16. Find the S.P. if M.P. = ₹5450 and discount = 5%
Solution: Discount = 5% of ₹5450= \(\frac{5}{100} \times 5450\)
= ₹272.50
S.P. = M.P. – Discount = ₹5450-₹272.50
= ₹5177.50
17. Find the compound interest by calculating simple interest when principal is ₹3000, rate is 5% per annum and time is 2 years.
Solution: Principal for the 1 st year = ₹3000, R = 5, T = 2 years.
Interest for the 1st year= \(\frac{3000 \times 5 \times 1}{100}=₹ 150\)
Amount at the end of first year = ₹3000+150 = ₹3150.
Principle for the second year = ₹3150.
Interest for the second year = \(\frac{₹ 3150 \times 5 \times 1}{100}\)
= ₹157.50
Amount at the end of second year = ₹3150 + 157.50 = ₹3307.50
Compound interest = ₹3307.50 + ₹3000 = ₹307.50.
18. The difference between SI and CI of a certain sum of money is ₹48 at 20% p.a. for 2 years. Find the principal.
Solution: Let the certain sum be ₹P
\(\mathrm{SI}=\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}=\frac{\mathrm{P} \times 20 \times 2}{100}=\frac{2 P}{5}\) \(\mathrm{CI}=\mathrm{P}\left[\left(1+\frac{R}{100}\right)^n-1\right]=P\left[\left(1+\frac{20}{100}\right)^2-1\right]\)=\(P\left(\frac{36}{25}-1\right)=\frac{11 P}{25}\)
As CI-SI = 48
\(\frac{11 P}{25}-\frac{2 P}{5}=48 \Rightarrow \frac{11 \mathrm{P}-10 \mathrm{P}}{25}=48\)p = 48 x 25 = ₹1200.
19. In a factory, women are 35% of all the workers the rest of the workers being men. The number of men exceeds that of women by 252. Find the total number of workers in the factory.
Solution: Let total number of workers be x.
Women = 35% of \(x = \frac{35}{100} x\)
Men = (100-35)% of x = 65% of \(x = \frac{65}{100} x\)
According to question,
\(\frac{65}{100} x=\frac{35 x}{100}+252=\frac{65 x-35 x}{100}=252\) \(30x = 252 x 100\) \(x=\frac{252 \times 100}{30}\)x = 840
20. Karthik purchased a hair dryer for ₹5450 including 8% VAT. Find the price before VAT was added.
Solution: Let the original price of dryer be ₹100
Price including VAT = ₹108
When price including VAT is ₹108, original price = ₹100
When price including VAT is ₹5400, original price
=\(\frac{100}{108} \times 5400=₹ 5000\)
Thus, the price including VAT was added = ₹5000
21. A student used the proportion w100 = 532 to find 5% of 32. What did the student do wrong?
Solution: 5% of 32 will be calculated as
\(\frac{5}{100} \times 32=\frac{32}{20}=16\)but student finding percent is 5 of 32
22. The human body is made up mostly of water. In fact about 67% of a person total body weight is water. If Jyoti weights 56kg. How much of her weight is water?
Solution: Jyoti’s weight = 56kg
where 67% of a person is total body weight is water.
Water in Jyoti body =
\(\frac{67}{100} \times 56=\frac{3752}{100}=37.52 \mathrm{~kg}\)23. At a toy shop price of all the toys is reduced to 66% of the original price.
a)What is the sale price of a toy that originally costs ₹90?
b)How much money would you save on a toy costing ₹90?
Solution: a) Original cost of the toy = ₹90
If the price, reduced to 66% of the original price, then
price became = \(90-\frac{66}{100} \times 90=90-59.4\) = ₹30.6
b) We have discounted amount = 66% of 90
=\(\frac{66}{100} \times 90=₹ 59.4\)
24. In a certain experiment, the count of bacteria was increasing at the rate of 2.5% per hour. Initially, the count was 512000. Find the bacteria at the end of 2 hours.
Solution: Bacteria at the end of 2 hours
=\(\left\{512000 \times\left(1+\frac{5}{2 \times 100}\right)^2\right\}\)
=\(\left(512000 \times \frac{41}{40} \times \frac{41}{40}\right)=537920\)
Hence, the bacteria at the end of 2 hours = 537920.
25. A motorcycle is bought at ₹160000. Its value depreciates at the rate of 10% per annum. Find its value after
1) 1 year
2) 2 years.
Solution: 1) Value of the motorcycle after 1 year
=\(₹\left\{160000 \times\left(1-\frac{10}{100}\right)\right\}\)
=\(₹\left(160000 \times \frac{9}{10}\right)=₹ 144000\)
∴ value after 1 year = ₹144000
2) Value of the motorcycle after 2 years
=\(₹\left\{160000 \times\left(1-\frac{10}{100}\right)^2\right\}\)
=\(₹\left(160000 \times \frac{9}{10} \times \frac{9}{10}\right)=₹ 129600\)
∴ Value after 2 years = ₹129600