NCERT solutions for class 8 Maths Chapter 8 Comparing Quantities

Solution for Exercise 8.1

1.Find the ratio of the following:
(a) speed of a cycle 15 km per hour to the speed of scooter 30 km per hour.

(b) 5 m to 10 km

(c) 50 paise to ₹ 5


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Answer :

(a)Given, Speed of cycle : Speed of Scooter = 15 km per hour : 30 km per hour

\(\Rightarrow\) 15 :30

\(\Rightarrow\) 1:2

Hence we get the ratio as 1 : 2

(b)Given, 5 m to 10 km

\(\Rightarrow\) 5 m : 10 \(\times\) 1000 m [∵ 1 km = 1000 m]

\(\Rightarrow\) 5 m : 10000 m

\(\Rightarrow\) 1 : 2000

Hence, the ratio = 1 : 2000

(c)Given, 50 paise to ₹ 5

\(\Rightarrow\) 50 paise : 5 \(\times\) 100 paise [∵ ₹ 1 = 100 paise]

\(\Rightarrow\) 50 paise : 500 paise

Hence we get the ratio as 1 : 10

2.Convert the following ratios to percentages:
(a) 3 : 4

(b) 2 : 3


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Answer :

(a) We have, 3:4
\(\Rightarrow \frac{3}4 \times 100=74 \)%

(b) We have, 2:3
\(\Rightarrow \frac{2}3 \times 100=\frac{200}3 =66\frac{2}3 \)%

3.72% of 25 students are good in mathematics. How many are not good in mathematics?


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Answer :

Students who are good in mathematics = 72% of 25

i.e., \(\frac{72}{100}\times25 =18\).

So, the number of students who are not good in mathematics = 25 – 18 = 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?


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Answer :

Win percentage is 40 i.e. 40 matches are won by the team out of 100 matches
So, the team will win 1 match out of \(\frac { 100 }{ 40 }\) matches

And similarly, the team will win 10 matches out of \(\frac { 100 }{ 40 } \times 10 = 25\) matches

Hence, the total number of matches played by the team = 25

5. If Chameli had ₹ 600 left after spending 75% of her money, how much did she have in the beginning?


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Answer :

Let the money with Chameli be ₹ 100

So, the money spent by her = 75% of 100

\(\qquad = \frac { 75 }{ 100 } \times 100 = ₹ 75\)

The money left with her = ₹ 100 – ₹75 = ₹ 25

₹ 25 are left with her out of ₹ 100

₹ 1 is left with her out of ₹ \(\frac { 100 }{ 25 }\)

Similarly, ₹ 600 will be left out of ₹ \(\frac { 100 }{ 25 } \times 600 = ₹ 2400\)

Thus, she had ₹ 2400 in beginning.

6.If 60% of people in a city like a cricket, 30% like football and the remaining like other games, then what per cent of the people like other games? If the total number of people are 50 lakh, find the exact number who like each type of game.


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Answer :

Total number of people = 50,00,000
We have the number of people who like cricket = 60% of 50,00,000

\(\qquad = \frac { 60 }{ 100 } \times 50,00,000\)

\(\qquad = 30,00,000\)

Similarly we have the number of people who like football = 30% of 50,00,000

\(\qquad = \frac { 30 }{ 100 } \times 50,00,000\)

\(\qquad = 15,00,000\)

Hence, the number of people who like other games = 50,00,000 – (30,00,000 + 15,00,000)

\(\qquad\)= 50,00,000 – 45,00,000

\(\qquad\)= 5,00,000

So,the number of people who likes other games=5,00,000 and the percentage of the people who like other games = \(\frac { 500000 }{ 5000000 } \times 100 = 10\)%

Thus, 10% of people like other game.

Solution for Exercise 8.2

1.A man got a 10% increase in his salary. If his new salary is ₹ 1,54,000, find his original salary.


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Answer :

New salary of the man= ₹ 1,54,000
Given, the increase in salary = 10%

New salary of the man = Original salary \(\times\) (1 + \(\frac { Increase }{ 100 }\))

\(\Rightarrow\) 154000 = Original salary \(\times\) (1 + \(\frac{10}{100}\))

\(\Rightarrow\) 154000 = Original salary \(\times\) (\(\frac{110}{100}\))

\(\Rightarrow\) Original salary=\(\frac{154000\times 100}{110}\)

\(\Rightarrow\) Original salary= ₹ 1,40,000.

Thus, the original salary = ₹ 1,40,000

2.On Sunday 845 people went to the Zoo. On Monday only 169 people went. What is the per cent decrease in the people visiting the Zoo on Monday?


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Answer :

Given, Number of people went to Zoo on Sunday = 845
Number of people went to Zoo on Monday = 169

So the decrease in number of people visiting the Zoo = 845 – 169 = 676

Hence we have the decrease in percent:

\(=\frac{decrease \;in \;the \;number \;of\; people\; visiting \;the \;zoo}{total \;number\; of \;people\; visiting \;the \;zoo \;on\; sunday}\times 100\)

\(= \frac{676\times100}{845}=80\)%

Hence we have the decrease in per cent of visiting people= 80%

3.A shopkeeper buys 80 articles for ₹ 2,400 and sells them for a profit of 16%. Find the selling price of one article.


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Answer :

We have Cost price of 80 articles = ₹ 2,400
So, Cost of 1 article = ₹ \(\frac { 2400 }{ 80 } = ₹ 30\)

Given Profit = 16%

So, we know \(SP =CP \times (1+\frac{profit}{100})\)

\(\qquad =30 \times (1+\frac{16}{100})\)

\(\qquad =30 \times (\frac{116}{100})=₹ 34.80\)

Hence, the selling price of one article =₹ 34.80

4.The cost of an article was ₹ 15,500. ₹ 450 were spent on its repairs. If it is sold for a profit of 15%, find the selling price of the article.


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Answer :

Given, CP of the article = ₹ 15,500
The money spent on repairs = ₹ 450

So we have the net CP = ₹ 15,500 +₹ 450 = ₹ 15,950

Given that Profit = 15%

So, we know \(SP =CP \times (1+\frac{profit}{100})\)

\(\qquad =15,950 \times (1+\frac{15}{100})\)

\(\qquad =15,950 \times (\frac{115}{100})=₹ 18342.50\)

Hence, the selling price of article = ₹ 18342.50

5.A VCR and TV were bought for ₹ 8,000 each. The shopkeeper made a loss of 4% on the VCR and a profit of 8% on the TV. Find the gain or loss per cent on the whole transaction.


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Answer :

Given, Cost price of VCR=₹ 8,000.
And Loss=4%

∴ \(SP=CP\; (1-\frac{loss}{100}\))

\(\Rightarrow SP=8000\; (1-\frac{4}{100}\))

\(\Rightarrow SP=8000\;\times (\frac{96}{100}\))

\(\Rightarrow SP=₹\; 7,680\)

Also given that, Cost price of TV=₹ 8,000

And profit=8%

∴ \(SP=CP\; (1+\frac{Profit}{100}\))

\(\Rightarrow SP=8000\; (1+\frac{8}{100}\))

\(\Rightarrow SP=8000\;\times (\frac{108}{100}\))

\(\Rightarrow SP=₹\; 8,640\)

So we have total SP=₹ 8,640 +₹ 7,680=₹ 16,320

So we have total SP=₹ 8000 + ₹ 8000 =₹ 16,000

Therefore, Profit=SP-CP

=₹ 16,320- ₹ 16,000

=₹ 320

∴ Profit % on the whole transaction=\(\frac{Profit}{Total\; CP}\times100\)

\(\qquad = \frac{320}{16000}\times 100 =2\)%

Hence, the shopkeeper gained 2% profit 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?


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Answer :

Marked Price(MP) of Jeans = ₹ 1,450
Marked Price of two shirts = ₹ 850 \(\times\) 2 = ₹ 1,700

Total Marked Price = ₹ 1,450 + ? 1,700 = ₹ 3,150

Given that we have discount = 10%

So, we know \(SP =MP \times (1-\frac{discount}{100})\)

\(\qquad =3,150 \times (1-\frac{10}{100})\)

\(\qquad =3,150 \times (\frac{90}{100})=₹ 2,835\)

Thus, the customer will have to pay ₹ 2,835.

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 overall gain or loss. [Hint: Find CP of each]


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Answer :

Given, SP of a buffalo=₹ 20,000.
And Gain=5%

∴ \(SP=CP\; (1+\frac{Gain}{100}\))

\(\Rightarrow 20000=CP\; (1+\frac{5}{100}\))

\(\Rightarrow 20000=CP\;\times (\frac{105}{100}\))

\(\Rightarrow CP=\frac{20000\times100}{105}\)

\(\Rightarrow CP=₹\;\frac{4,00,000}{21}\)

SP of another buffalo=₹ 20,000

And given loss=10%

∴ \(SP=CP\; (1-\frac{loss}{21}\))

\(\Rightarrow 20000=CP\; (1-\frac{10}{100}\))

\(\Rightarrow 20000=CP\;\times (\frac{90}{100}\))

\(\Rightarrow CP=\frac{20000\times100}{90}\)

\(\Rightarrow CP=₹\;\frac{2,00,000}{9}\)

So we have total CP=\(₹\;\frac{4,00,000}{21}\) +\(₹\;\frac{2,00,000}{9}\)

\(=₹\;\frac{12,00,000+14,00,000}{63}\)

\(=₹\;\frac{26,00,000}{63}\)

And we have total SP=₹ 20,000 + ₹ 20,000 =₹ 40,000

Here we see that,SP < CP i.e., it was a loss.

∴ Loss=CP-SP

\(=₹\;\frac{26,00,000}{63}-₹ 40,000\)

\(=₹\;\frac{26,00,000-25,20,000}{63}\)

\(=₹\;\frac{80,000}{63}=₹\;1,269.84\)

Therefore, %Loss=\(\frac{Loss}{CP}\)

=\(\frac{\frac{80,000}{63}}{\frac{26,00,000}{63}}\times 100\)

=\(\frac{80,000}{63}\times\frac{26,00,000}{63}\times100\)

=\(\frac{80}{26}\)%=\(\frac{40}{13}\)%=\(3\frac{1}{13}\)%

Hence, the complete percent loss=\(3\frac{1}{13}\)%.

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.


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Answer :

Marked price of TV = ₹ 13000
Given that the Rate of sales tax = 12%

∴ Sales tax = 12% of ₹ 13000

\(\qquad= ₹ \left( \frac { 12 }{ 100 } \times 13000 \right) = ₹ 1560\)

So, the total amount which Vinod had to pay for purchasing TV = ₹ (13000 +1560) = ₹ 14560.

Hence, The required amount that Vinod has to pay = ₹ 14,560

9. Arun bought a pair of skates at a sale where the discount is given was 20%. If the amount he pays is ₹ 1,600, find the marked price.


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Answer :

Let the marked price of the skates be ₹ 100
We are given discount = ₹ 20% of 100 = ₹ 20

So the Sale price =₹ 100 – ₹ 20 = ₹ 80

If SP is ₹ 80 then marked price =₹ 100

so,if SP is ₹ 1 then marked price = ₹ \(\frac { 100 }{ 80 }\)

Similarly, if SP is ₹ 1,600 then marked price = ₹ \(\frac { 100 }{ 80 } \times 1600 = ₹ 2,000\)

Hence, we got the marked price = ₹ 2000.

10. I purchased a hair-dryer for ₹ 5,400 including 8% VAT. Find the price before VAT was added.


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Answer :

Let the original price of the hair-dryer be ₹ 100
Given, VAT = 8% of 100 = ₹ 8

So, the Sale price = ₹ 100 + ₹ 8 = ₹ 108

If SP is ₹ 108 then original price =₹ 100

So, if SP is ₹ 1 then the original price = ₹ \(\frac { 100 }{ 108 }\)

Similarly, if SP is ₹ 5,400 then the original price = ₹ \(\frac { 100 }{ 108 } \times 5,400 = ₹ 5,000\)

Thus, the price of hair-dryer before the addition of VAT = ₹ 5000.

Solution for Exercise 8.3

1. Calculate the amount and compound interest on
(a)₹ 10,800 for 3 years at \(12\frac { 1 }{ 2 }\) % per annum compounded annually.

(b)₹ 18,000 for \(2\frac { 1 }{ 2 }\) years at 10% per annum compounded annually.

(c)₹ 62,500 for \(1\frac { 1 }{ 2 }\) years at 8% per annum compounded half yearly.

(d) ₹ 8,000 for 1 year at 9% per annum compounded half yearly. (You could use the year by year calculation using SI formula to verify).

(e)₹ 10,000 for 1 year at 8% per annum compounded half yearly.


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Answer :

(a) Given, P=₹ 10,800,n=3 years.
We have, R=\(12\frac{1}2\)% = \(\frac{25}2\)%p.a.

So, we know that \(Amount=P(1+\frac{R}{100})^n\)

\(=10,800(1+\frac{25}{100})^3\)

\(=10,800(\frac{9}{8})^3\)

\(=10,800\times(\frac{9}{8})\times(\frac{9}{8})\times(\frac{9}{8})\)

\(=\frac{4,92,075}{32}=₹ \; 15,377.34\)
Thus, we have Compound interest=A - P

=₹ 15,377.34-₹ 10,800 = ₹ 4,577.35

Hence we have, amount = ₹ 15,377.34 and CI = ₹ 4,577.34

(b)We have: P = ₹ 18,000, n = \(2\frac { 1 }{ 2 }\) years and R = 10% p.a.
The amount for \(2\frac { 1 }{ 2 }\) years, can be calculated by first calculating the amount to 2 years using CI formula and then calculating the simple interest by using SI formula.

The amount for 2 years:

Amount = \(18000(1+\frac{10}{100})^2\)

= \(18000\times(\frac{110}{100})^2\)

= \(18000\times(\frac{11}{10})\times(\frac{11}{10})=₹ \; 21,780\)

∴ Interest after 2 years = Amount - P

=21,780-18,000=₹ 3,780

So, now taking principal amount as ₹ 21,780, the SI for the next \(\frac{1}2\)years will be as:

SI=\(\frac{P\times R\times n}{100}\)

=\(\frac{21,780\times10\times1}{100\times2}\)

=\(₹\; 1,089\)

Therefore,Total CI = ₹ 3780 +₹ 1,089= ₹ 4,869

Amount = P + I = ₹ 21,780 + ₹ 1,089 =₹ 22,869

Hence, the amount = ₹ 22,869 and CI = ₹ 4,869
(c)Given, P=₹ 62,500,n=\(1\frac{1}2=\frac{3}2\) years per annum compounded half yearly i.e., \(\frac{3}2\times 2\) years=3 half years.
We have, R=\(8\)%=\(\frac{8}2\)% = \(4\)%half yearly

So, we know that \(Amount=P(1+\frac{R}{100})^n\)

\(=62,500(1+\frac{4}{100})^3\)

\(=62,500(\frac{26}{25})^3\)

\(=62,500\times(\frac{26}{25})\times(\frac{26}{25})\times(\frac{26}{25})\)

\(=4 \times 26\times 26\times 26=₹ \; 70,304\)
Thus, we have Compound interest=A - P

=₹ 70,304-₹ 62,500 = ₹ 7,804

Hence we have, amount = ₹ 70,304 and CI = ₹ 7804
(d)Given, P=₹ 8,000,n=1 years and R = 9% per annum compounded half yearly
Since, the interest is compounded half yearly n =\(1 \times 2 = 2\) half years.So, R=\(\frac{9}2\)% per half years.

So, we know that \(Amount=P(1+\frac{R}{100})^n\)

\(=8,000(1+\frac{9}{2\times100})^2\)

\(=8,000(\frac{209}{200})^2\)

\(=8,000\times(\frac{209}{200})\times(\frac{209}{200})\)

\(=₹ \;(\frac{8,7362}{10})\)

\(=₹ \; 8,736.20\)

Thus, we have Compound interest=A - P

=₹ 8,736.20-₹ 8,000 = ₹ 736.20

Hence we have, amount = ₹ 8736.20 and CI = ₹ 736.20
(e)Given, P=₹ 10,000,n=1 years and R = 8% per annum compounded half yearly
Since, the interest is compounded half yearly n =\(1 \times 2 = 2\) half years.So, R=\(\frac{8}2=4\)% per half years.

So, we know that \(Amount=P(1+\frac{R}{100})^n\)

\(=10,000(1+\frac{4}{100})^2\)

\(=10,000(\frac{26}{25})^2\)

\(=10,000\times(\frac{26}{25})\times(\frac{26}{25})\)

\(=16\times 26\times 26\)

\(=₹ \; 10,816\)

Thus, we have Compound interest=A - P

=₹ 10,816 -₹ 10,000 = ₹ 816

Hence we have, amount = ₹ 10,816 and CI = ₹ 816

2.Kamala borrowed ₹ 26,400 from a Bank to buy a scooter at a rate of 15% per annum compounded yearly. What amount will she pay at the end of 2 years and 4 months to clear the loan? (Hint: Find amount for 2 years with interest is compounded yearly and then find SI on the 2nd year amount for \(\frac { 4 }{ 12 }\) years).


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Answer :

We have:
P = ₹ 26,400

R = 15% p.a. compounded yearly

n = 2 years and 4 months

So, amount for 2 years=\(P(1+\frac{R}{100})^n\)

=\(26,400(1+\frac{15}{100})^2\)

=\(26,400(\frac{23}{20})^2\)

=\(26,400\times\frac{23}{20}\times\frac{23}{20}\)

=\(66\times529=₹\; 34,914\)

Principal for 2 years=₹ 34,914

So, SI for 4 months =\(\frac{P\times R\times n}{100\times 12}\)

=\(\frac{34,914\times 15 \times 4}{100\times 12}\)

=\(₹ 1745.70 \)

Thus, the mount after 2 years and 4 months = ₹ 34,914 + ₹ 1745.70 = ₹ 36,659.70

Hence, the total amount to be paid by Kamla = ₹ 36,659.70

3. Fabina borrows Rs 12,500 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?


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Answer :

We have given data for Fabina as: P = ₹ 12,500, R = 12% p.a. and n = 3 years
So, SI for 4 months =\(\frac{P \times R \times n}{100\times 12}\)

=\(\frac{12,500\times 12 \times 3}{100}\)

=\(₹ 4500 \)

Therefore we have, CI=A - P

=\(P(1+\frac{R}{100})^n\)-P

=\(12,500(1+\frac{10}{100})^3\)-12,500

=\(12,500(\frac{11}{10})^3\)-12,500

=\(12,500\times (\frac{10}{100}\times \frac{10}{100}\times \frac{10}{100})\)-12,500

=\(12,500\times (\frac{1331}{1000})\)-12,500

=\(12,500 \times (\frac{1331}{1000}-1)\)

=\(12,500 \times (\frac{1331-1000}{1000})\)

=\(12,500 \times (\frac{331}{1000})\)

=\(12.5\times 331 = ₹ 4137.50 \)

So, the difference between the two interests = ₹ 4500 – ₹ 4137.50 = ₹ 362.50

Hence, Fabina pays more interest by ₹ 362.50.

I borrowed ₹ 12,000 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?


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Answer :

We have given data : P = ₹ 12,000, R = 6% p.a. and n = 2 years
So, SI for 4 months =\(\frac{P \times R \times n}{100\times 12}\)

=\(\frac{12,000\times 6 \times 2}{100}\)

=\(₹ 1440 \)

Therefore we have, CI=A - P

=\(P(1+\frac{R}{100})^n\)-P

=\(12,000(1+\frac{6}{100})^2\)-12,000

=\(12,000(\frac{53}{50})^3\)-12,000

=\(12,000\times (\frac{53}{50}\times \frac{53}{50})\)-12,000

=\(12,000\times (\frac{2809}{2500})\)-12,000

=\(12,000 \times (\frac{2809}{2500}-1)\)

=\(12,000 \times (\frac{2809-2500}{1000})\)

=\(12,000 \times (\frac{309}{2500})\)

=\(\frac{7416}5 = ₹ 1483.20 \)

So, the difference between the two interests = ₹ 1483.20 – ₹ 1440 = ₹ 43.20

Hence, Fabina pays more interest by ₹ 43.20.

5.Vasudevan invested ₹ 60,000 at an interest rate of 12% per annum compounded half yearly. What amount would he get
(i) after 6 months?

(ii) after 1 year?


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Answer :

(i) Given: P = ₹ 60,000, R = 12% p.a. compounded half yearly and n=6 months=\(\frac{6}{12}=\frac{1}2 \;year\)
∴ Simple Interest=\(\frac{P \times R \times n}{100\times 12}\)

=\(\frac{60,000\times 12 }{100}\times \frac{1}2\)

=\(₹ \;3600 \)

Therefore , we have Amount=P+SI

=₹ 60,000 + ₹ 3600

=₹ 63600

So the required amount =₹ 63600
(ii)We have given with :P = ₹ 60,000, R = \(\frac{12}2=6\)% and n=\(1\times 2=2\)half years.

So, we have amount=\(P(1+\frac{R}{100})^n\)

=\(60,000(1+\frac{6}{100})^2\)

=\(60,000(\frac{53}{50})^2\)

=\(60,000\times\frac{53}{50}\times\frac{53}{50}\)

=\(24\times2809=₹\; 67,416\)

Hence the required amount= ₹ 67,416

6. Arif took a loan of ₹ 80,000 from a bank. If the rate of interest is 10% per annum, find the difference in amounts he would be paying after \(1\frac { 1 }{ 2 }\) years if the interest is
(i) compounded annually.

(ii) compounded half yearly.


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Answer :

(i) We are given: P = ₹ 80,000 , R = 10% p.a. and n = \(1\frac { 1 }{ 2 }\) years Since the interest is compounded annually so we have

Simple Interest=\(\frac{P \times R \times n}{100}\)

=\(\frac{80,000\times 10\times 1}{100}\)

=\(₹ \;8,000 \)

Principal for the second year=P+SI

=₹ 80,000 + ₹ 8,000 = ₹ 88,000

Now for \(\frac{1}2\) year we have interest==\(\frac{88,000\times 10\times 1}{100\times 2}=₹ \;4,400\)

Thus, amount=₹ 88,000 + ₹ 4,400=₹ 92,400

(ii)So for compounded interest half yearly we have:

R=\(\frac{10}2=5\)% half yearly.

n=\(1\frac{1}2 \)years=\(\frac{3}2\times 2=3\) half years

∴ Amount=\(P(1+\frac{R}{100})^n\)

=\(80,000(1+\frac{5}{100})^3\)

=\(80,000(\frac{21}{20})^3\)

=\(80,000\times(\frac{21}{20}\times \frac{21}{20}\times \frac{21}{20})\)

=\(10\times 9261\)

=\(₹ \; 92610\)

Difference between the amounts = ₹ 92,610 – ₹ 92,400 = ₹ 210

7.Maria invested ₹ 8,000 in a business. She would be paid interest at 5% per annum compounded annually. Find
(i) The amount credited against her name at the end of the second year.

(ii) The interest for the third year.


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Answer :

(i) Given: P = ₹ 8,000, R = 5% p.a. and n = 2 years

Amount=\(P(1+\frac{R}{100})^n\)

=\(8,000(1+\frac{5}{100})^2\)

=\(8,000(\frac{21}{20})^2\)

=\(8,000\times(\frac{21}{20}\times \frac{21}{20})\)

=\(20\times 441\)

=\(₹ \; 8,820\)

Thus, the amount credited at the end of 2 years=₹ 8,820

(ii) Interest for the third year=Amount after 3 years - Amount after 2 years

=\(P(1+\frac{R}{100})^n\) - ₹ 8,820

=\(8,000(1+\frac{5}{100})^3\) - ₹ 8,820

=\(8,000(\frac{21}{20})^3\) - ₹ 8,820

=\(8,000\times(\frac{21}{20}\times \frac{21}{20}\times \frac{21}{20})\) - ₹ 8,820

= ₹ 9,261 - ₹ 8,820

= ₹ 441

Hence we got, interest for the third year = ₹ 441

8. Find the amount and the compound interest on ₹ 10,000 for \(1\frac { 1 }{ 2 }\) years at 10% per annum, compounded half yearly. Would this interest be more than the interest he would get if it was compounded annually?


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Answer :

We are given: P = ₹ 10,000, n = \(1\frac { 1 }{ 2 }\)years and R = 10% per annum
Since the interest is compounded half yearly:

∴ n=\(1\frac{1}2 \) years

=\(\frac{3}2\times 2 = 3\) half yearly

R=\(\frac{10}2\)%=5% half yearly

Amount=\(P(1+\frac{R}{100})^n\)

=\(10,000(1+\frac{5}{100})^3\)

=\(10,000(\frac{21}{20})^3\)

=\(10,000\times(\frac{21}{20}\times \frac{21}{20}\times \frac{21}{20})\)

=\(\frac{5}4\times 9261 = \frac{46305}4\)

=\(₹ \; 11576.25\)

Thus, we have: Compound interest=A - P

=₹ 11576.25 - ₹ 10,000 = ₹ 1576.25

If the interest is compounded annually, then we have:

n=\(1\frac{1}2\) year and R=10 %

Therefore, we have SI= \(\frac{P \times R \times n}{100}\)

=\(\frac{10,000\times 10 \times 1}{100} = ₹ 1,000\)

Principal for the second year:

=₹ 10,000 + ₹ 1,000

=₹ 11,000

∴ Interest for \(\frac{1}2\) year=

=\(\frac{11,000\times 10 \times 1}{100\times 2}=₹ 550\)

Total interest = ₹ 1,000 + ₹ 550 = ₹ 1,550 Difference between the two interests = ₹ 1,576.25 – ₹ 1,550 = ₹ 26.25

Hence, the interest will be ₹ 26.25 more when compounded half yearly than the interest when compounded annually.

9. Find the amount which Ram will get on ₹ 4,096, if he gave it for 18 months at \(12\frac { 1 }{ 2 }\) per annum, interest being compounded half yearly.


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Answer :

We are given: P = ₹ 4,096, R = \(12\frac { 1 }{ 2 }\) % pa, n = 18 months=\(\frac{18}{12}\)years=\(\frac{18}{12} \times 2=3\) half years.

R=\(12\frac{1}2\)% =\(\frac{25}2\times \frac{1}2=\frac{25}4\)% half yearly

Amount=\(P(1+\frac{R}{100})^n\)

=\(4,096(1+\frac{25}{4\times 100})^3\)

=\(4,096(\frac{17}{16})^3\)

=\(4,096\times(\frac{17}{16}\times \frac{17}{16}\times \frac{17}{16})\)

=\(₹ \; 4,913\)

Thus, the required amount =₹ 4913

10.The population of a place increased to 54,000 in 2003 at a rate of 5% per annum.
(i) Find the population in 2001.

(ii) What would be its population in 2005?


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Answer :

(i) We are given: Population in 2003 = 54,000, Rate = 5% p.a.
Time = 2003 – 2001 = 2 years

\(Population\; in \;2003 = Population\; in \;2001\times(1+\frac{R}{100})^n\)

\(\Rightarrow 54,000=Population\; of \;2001\times(1+\frac{5}{100})^2\)

\(\Rightarrow 54,000=Population \;of \;2001\times(\frac{21}{20})^2\)

\(\Rightarrow 54,000=Population\; of \;2001\times(\frac{441}{400})\)

∴ Population of 2001= \(\frac{54,000\times 400}{441}\)

= \(\frac{21,6,00,000}{441}=48,979.59≈48,980\)

(ii) \(Population \;in \;2005= Population\;in\;2003\times(1+\frac{R}{100})^n\)

\( =54,000\times(1+\frac{5}{100})^2\)

\( =54,000\times(\frac{21}{20})^2\)

\( =54,000\times(\frac{441}{400})^2\)

\(=135\times441=59,535\)

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 5,06,000.


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Answer :

We have: Initial count of bacteria = 5,06,000
Rate = 2.5% per hour

n = 2 hours

So, we have:

num. of bacteria(end of 2 hours)=Num. of count of bacteria initially\(\times(1+\frac{R}{100})^n\)

\( =5,06,000\times(1+\frac{2.5}{100})^2\)

\( =5,06,000\times(\frac{41}{40})^2\)

\( =5,06,000\times(\frac{1681}{1600})^2\)

\( =5,31,616.25\)

Hence we have, the number of bacteria after two hours = 5,31,616
(approx).

12.A scooter was bought at ₹ 42,000. Its value depreciated at the rate of 8% per annum. Find its value after one year.


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Answer :

We have: Cost price of the scooter = ₹ 42,000
Rate of depreciation = 8% p.a.

Time(n)= 1 year

So, Final value of the scooter = Present value\(\times(1-\frac{R}{100})^n\)

\(=42,000\times(1-\frac{8}{100})\)

\(=42,000\times(\frac{23}{25})\)

\(=1,680\times23=38,640\)

Thus, the value of scooter after 1 year =₹ 38,640.



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