SChand CLASS 9 Chapter 2 Compound Interest Test

 Test

Question 1

Nikita invests ₹ 6,000 for two years at a certain rate of interest compounded annually. At end of first year it amounts to ₹ 6,720. Calculate:

(i) the rate of interest

(ii) the amount at the end of the second year.

Sol :

Principal=6000

Time=2 year

and after 1 year she go back 720

(i) Rate of Interest

$=\frac{72 \times 100}{6000}$

=12%

(ii) end of the second year interest

$=\frac{6720 \times 12 \times 1}{100}$

=806.40

Amount at the end of second year

=6720+806.40

=7526.40

Question 2

Rohit borrows ₹ 86,000 from Arun for two years at 5% per annum simple interest. He immediately lends out this money to Akshay at 5% compound interest compounded annually for the same period. Calculate Rohit’s profit in the transaction at the end of two years.

Sol :

Rohit borrow (P)=86000

Time=2 year

Rate=5%

∴Interest$=\frac{86000 \times 2 \times 5}{100}$

=8600

In 2 year S.I Rohit get 8600

Rohit lends money to Akshay at 5% compounded annually for 2 year

Amount he get back after two year

$A=P\left(1+\frac{r}{100}\right)^n$

$=86000 \left(1+\frac{5}{100} \right)$

$=8600\left(1+\frac{1}{20}\right)^{2}$

$=86000 \times \left(\frac{21}{20}\right)^{2}$

$=86000 \times \frac{21}{20} \times \frac{21}{20}$

A=215×21×21=94815

CI=A-P

=94815-86000

=8815

Profit=CI-Interest

=8845-8600

=215

Question 3

Mr. Kumar borrowed ₹ 15,000 for two years. The rate of interest for the two successive years are 8% and 10% respectively. If he repays ₹ 6,200 at the end of the first year, find the outstanding amount at the end of the second year.

Sol :

For 1st year :

$\text{S.I.}=\frac{15000 \times 8 \times 1}{100}$

= ₹ 1200

Amount = ₹ (15000 + 1200) = ₹ 16200

Remaining amount after repayment

= ₹ (16200 – 6200) = ₹ 10000

For 2nd year :

P = ₹ 10000

$\text{S.I.}=\frac{10000 \times 10 \times 1}{100}$

 = ₹ 1000

Amount at the end of 2nd year = ₹ 10000 + ₹ 1000 = ₹ 11000

Question 4

In what period of time will ₹ 12,000 yield ₹ 3,972 as compound interest at 10% per annum, if compounded on an yearly basis?

Sol :

P=12000

A=12000+3972

=15972

R=10%

Time (n)=?

$A \Rightarrow P\left(1+\frac{R}{100}\right)^{n}$

$15972=12000\left(1+\frac{10}{100}\right)^{n}$

$\frac{15972}{12000}=\left(1+\frac{1}{10}\right)^{n}$

$\frac{1331}{1000}=\left(\frac{11}{10}\right)^{n}$

$\left(\frac{11}{10}\right)^{3}=\left(\frac{11}{10}\right)^{n}$

n=3

Question 5

On what sum of money will the difference between the compound interest and simple interest for 2 years be equal to ₹ 25 if the rate of interest charged for both is 5%?

Sol :

Difference between CI and SI=25

Let principal=100

Rate=5%

Time (n)=2 year

$S.I.=\frac{P\times R \times T}{100}$

$=\frac{100 \times 5 \times 2}{100}$

=10

$A \Rightarrow P\left(1+\frac{R}{100}\right)^{n}$

$\Rightarrow 100\left(1+\frac{5}{100}\right)^2$

$\Rightarrow 100 \times\left(1+\frac{1}{20}\right)^{2}$

$\Rightarrow 100 \times\left(\frac{21}{20}\right)^{2}$

$\Rightarrow 100 \times \frac{21}{20} \times \frac{21}{20}$

$\Rightarrow \frac{21 \times 21}{4}$

$\Rightarrow \frac{441}{4}$

CI=A-P

$=\frac{441}{4}-100$

$=\frac{441-400}{4}$

$=\frac{41}{4}$

Difference between CI and S.I

$=\frac{41}{4}-10$

$=\frac{41-41}{4}$

$=\frac{1}{4}$

When Difference $\frac{1}{4}$ then principal is 100 and difference is 1 then principal$=\frac{100 \times 4}{1}$

=400

and difference is 25 then principal is 

=25×400=10000

Question 6

A sum of ₹ 12,000 deposited at compound interest becomes double after 5 years. After 20 years, it will become

(a) ₹ 48,000

(b) ₹ 96,000

(c) ₹ 1,90,000

(d) ₹ 1,92,000

Sol :

Principal=12000

Time (n)=5 year

Amount=12000×2=24000

$A\Rightarrow P\left(1+\frac{R}{100}\right)^{n}$

$\frac{A}{P}=\left(1+\frac{R}{100}\right)^{n}$

$\frac{24000}{12000}=\left(1+\frac{R}{100}\right)^{5}$

$2=\left(1+\frac{R}{100}\right)^{5}$

∴$\left(1+\frac{R}{100}\right)^{20} \Rightarrow 2^{4} \Rightarrow 16$

Now after 20 year amount

$P\left(1+\frac{R}{100}\right)^{20}$

$=12000 \left(1+\frac{R}{100}\right)^{20}$

=12000×(2)⁴

=12000×16=192000

Question 7

The difference between C.I. (compound interest) and S.I. (simple interest) on a sum of ₹ 4,000 for 2 years at 5% p.a. payable yearly is

(a) ₹ 20

(b) ₹ 10

(c) ₹ 50

(d) ₹ 60

Sol :

Principal=4000

Rate=5%

Time (n)=2 years

∴$\text{S.I}=\frac{PRT}{100}$

$=\frac{4000 \times 5 \times 2}{100}$

=400

$A=P\left(1+\frac{R}{100}\right)^n$

$=4000\left(1+\frac{5}{100}\right)^{2}$

$=4000\left(\frac{21}{20}\right)^{2}$

$=4000 \times \frac{21}{20} \times \frac{21}{20}$

=4410

CI=A-P

=4410-4000

=410

Difference Between CI and SI

=410-400=10

Question 8

If the difference between simple interest and compound interest on a certain sum of money for 3 years at 10% per annum is ₹ 31, the sum is

(a) ₹ 500

(b) ₹ 50

(c) ₹ 1000

(d) ₹ 1250

Sol :

Difference in CI and SI=31

Let principal=100

Rate=10%

Time (n)=3 year

∴SI $\Rightarrow \frac{P R T}{100}$

$=\frac{100 \times 10 \times 3}{100}$

=30

$A=P\left(1+\frac{R}{100}\right)^n$

$A=100\left(1+\frac{10}{100}\right)^3$

$=100\times \frac{11}{10}\times \frac{11}{10}\times \frac{11}{10}$

$=\frac{1331}{10}$

∴C.I.=A-P

$=\frac{1331}{10}-100$

$=\frac{1331-1000}{10}=\frac{331}{10}$

Difference between CI and SI

$=\frac{331}{10}-30$

$=\frac{331-300}{70}=\frac{31}{10}$

If difference $\frac{31}{10}$ then principal 100 

If difference 1 the principal

$=\fac{100 \times 10}{31}$

and difference is 31 the principal

$=\frac{100 \times 10 \times 31}{31}$

=1000

Question 9

On what sum of money will compound interest for 2 years at 5% per annum amount to ₹164?

(a) ₹ 1600

(b) ₹ 1500

(c) ₹ 1400

(d) ₹ 1700

Sol :

CI=164

Rate=5%

Let principal=100

$A=P\left(1+\frac{R}{100}\right)^n$

$=100 \left(1+\frac{5}{100}\right)$

$100 \times\left(1+\frac{1}{20}\right)^{2}$

$ 100\left(\frac{21}{20}\right)^{2}$

$=100 \times \frac{21}{20} \times \frac{21}{20}$

$=\frac{441}{4}$

CI=A-P

$=\frac{441}{4}-100$

$=\frac{441-400}{4}=\frac{41}{4}$

If C.I. $\frac{41}{4}$ then principal is 100

If C.I. is 1 then principal $=\frac{100 \times 4}{41}$

and C.I is 164 then principal

$\frac{100 \times 4 \times 164}{41}$

=1600

Question 10

A sum of money becomes eight times in 3 years. If the rate is compounded annually, in how much time will the same amount at the same compound rate becomes sixteen times?

(a) 6 years

(b) 4 years

(c) 8 years

(d) 5 years

Sol :

Let principal=100

Amount=100×8=800

Time (n)= 3 year

$A=P\left(1+\frac{R}{100}\right)^{n}$

$\frac{A}{P}=\left(1+\frac{R}{100}\right)^{n}$

$\frac{800}{100}=\left(1+\frac{R}{100}\right)^{n}$

$8=\left(1+\frac{R}{100}\right)^{3}$

$\left(1+\frac{R}{100}\right)=2$...(i)

Second Case

P=100

A=100×16=1600

$A=P\left(1+\frac{R}{100}\right)^{n}$

$\frac{A}{P}=\left(1+\frac{R}{100}\right)^{n}$

$\frac{1600}{100}=\left(1+\frac{R}{100}\right)^{n}$

$16=\left(1+\frac{R}{100}\right)^{n}$

$16=2^n$ from equation (i)

$(2)^4=2^n$

n=4

∴ Period = 4 years