NCERT Solutions for Class 8 Maths Chapter 8 Comparing Quantities

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

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?

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.

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.

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.

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?

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]

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.

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.

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.

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.

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.

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).

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?

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?

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?

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.

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.

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?

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.

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?

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.

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.

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.