SChand CLASS 9 Chapter 2 Compound Interest Exercise 2(C)

 Exercise 2(C)

Question 1

What amount of money should Mohan invest in a bank in order to get ₹ 1323 in 2 years at 5% compounded annually?

Sol :

After 2 years amount =1323

Rate=5%

Time= 2 years

Amount $P\left[1+\frac{R}{100}\right]^{n}$

$1323=P\left(1+\frac{5}{100}\right)^{2}$

$1323=P\left[1+\frac{1}{20}\right\rfloor^{2}$

$1323=P\left[\frac{21}{20}\right\rfloor^{2}$

$P=1323 \times\left(\frac{20}{21}\right)^{2}$

$P \Rightarrow 1323 \times \frac{20}{21} \times \frac{20}{21}$

P=1200

Question 2

Find the sum which amounts to ₹ 1352 in 2 years at 4% compound interest.

Sol :

After 2 years amount=1352

Rate=4%

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

$1352=P\left[1+\frac{4}{100}\right]$

$1352= P\left[1+\frac{1}{25}\right\rfloor^{2}$

$1352=P\left[\frac{26}{25}\right]^{2}$

$P=1352 \times\left(\frac{25}{26}\right)^{2}$

$P=1352 x \frac{25}{26} \times \frac{25}{26}$

P=1250

Question 3

What principal will amount to ₹ 9768 in two years. If the rates of interest for the successive years are 10% p.a. and 11% p.a. respectively.

Sol :

After 2 year amount=9768

First year rate of interest (r₁)=10%

Second year rate of interest (r₂)=11%

Amount $\Rightarrow P\left[1+\frac{r_{1}}{100}\right]\left[1+\frac{r_{2}}{100}\right]$

$\begin{aligned} 9768 &=P\left[1+\frac{10}{100}\right)\left(1+\frac{11}{100}\right) \\ 9768 &=P\left(\frac{11}{10}\right)^{10}\left[\frac{111}{100}\right\rfloor \\ P & \Rightarrow 9768 \times \frac{10}{11} \times \frac{100}{111} \end{aligned}$

P=8000

Question 4

On what sum of money does the difference between the simple interest and compound interest in 2 years at 5% per annum is Rs. 15?

Sol :

Let Principal =100

Rate=5%

Time (n)=2 years

Difference between S.I. and C.I=15

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

$\frac{100 \times 5 \times R}{100}$

S.I.=10

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

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

$=100\left[\frac{21}{20}\right]^2$

$=100\left[\frac{21}{20} \times \frac{21}{20}\right]$

$=\frac{441}{4}$

C.I.=A-P

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

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

Difference between S.I. and C.I.

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

$=\frac{41-40}{4}=\frac{1}{4}$

If difference is $\frac{1}{4}$ then principal is 100

If difference is 1 then principal is equal to $\frac{100 \times 4}{1} \Rightarrow 400$

According tot he question if difference is 15 then principal 400×15=6000

Question 5

The difference between simple and compound interest on the same sum of money at $6\frac{2}{3}/%$ for 3 years is ₹ 184. Determine the sum.

Sol :

Difference between C.I and S.I=184

Let principal=100

Rate$=6\frac{2}{3}\%=\frac{20}{3}\%$

Time(n) = 3 years

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

$\frac{100 \times 20 \times 3}{100 \times 3}$

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

$100\left(1+\frac{20}{3 \times 100}\right)^{3}$

$100\left(1+\frac{2}{30}\right)^{3}$

$100\left(1+\frac{1}{15}\right)^{3}$

$100 \times\left(\frac{16}{15}\right)^{3}$

$100 \times \frac{16}{15} \times \frac{16}{15} \times \frac{16}{15}$

$\frac{16384}{135}$

C.I.=A-P

$\frac{16384}{135}-100$

$\frac{16384-13500}{135}$

$C.I.=\frac{2804}{135}$

Difference between C.I. and S.I.

$\frac{2884}{135}-20$

$\frac{2884-2400}{135}=\frac{184}{135}$

If difference is $\frac{184}{135}$ then principal is 100

If difference is 1 then principal

$=100 \times \frac{135}{184}$

According to the question difference is 184 then principal 

$=100 \times \frac{135}{184} \times 184$

P=13500

Question 6

On what sum of money will the difference between the simple interest and the compound interest for 2 years at 5% per annum be equal to ₹ 50.

Sol :

Let P=100

Rate=5%

Time(n)=2 years

∴$\text{S.I}=\frac{PRT}{100}$

$=\frac{100\times 5 \times 2}{100}=10$

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

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

$=\left.10211+\frac{1}{20}\right)^{2}$

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

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

$=\frac{441}{4}$

C.I=A-P

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

$=\frac{41-40}{4}$

$=\frac{1}{4}$

If difference $\frac{1}{4}$ then principal is 100 

If difference is 1 then principal 

$\frac{100 \times 4}{1} \Rightarrow 400$

According to the question , difference is 50 then 

P=400×50=20000

Question 7

Find the rate per cent per annum, if compounded yearly

(i) Principal = ₹ 196, Amount = ₹ 225, time = 2 years

(ii) Principal = ₹ 3136, Compound interest = ₹ 345, Time = 2 years

Sol :

(i) Principal=196

Amount=225

Time=2 years

We know that

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

$\frac{225}{196}=\left(1+\frac{R}{100}\right)^{2}$

$\left(\frac{15}{14}\right)^{2}=\left(1+\frac{R}{100}\right)^{2}$

$\frac{15}{14}=1+\frac{R}{10}$

$\frac{R}{100}=\frac{15}{14}-1$

$\frac{R}{100}=\frac{15-14}{14}$

$\frac{R}{100}=\frac{1}{14}$

$R=\frac{100}{14}=\frac{50}{7}=7\frac{1}{7}\%$

(ii) Principa=3136

C.I=345

∴Amount=P+C.I.

=3136+345

=34841

Time=2 years

We know that

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

$\frac{A}{P} \Rightarrow\left[1+\frac{R}{1000}\right]^{n}$

$\frac{3481}{3136}=\left(1+\frac{R}{100}\right)^{2}$

$\left( \frac{59}{56}\right)^{2}=\left(1+\frac{R}{100}\right)^{2}$

$\frac{59}{56}=1+\frac{R}{100}$

$\frac{R}{100}=\frac{59}{56}-1$

$\frac{R}{100}=\frac{59-56}{56}$

$\frac{R}{100}=\frac{3}{56}$

$R \Rightarrow \frac{3 \times 100}{56}$

$R=\frac{75}{14}$

∴Rate$=\frac{75}{14}\%$

$=5\frac{5}{14}\%$

Question 8

Hari purchased Relief Bonds for 1000, a sum which will fetch him ₹ 2000 after 5 years. Find the rate of interest if the interest is compounded half-yearly.

(Given that $\sqrt[10]{2}$ = 1.072)

Sol :

Principal=1000

Amount=2000

Time (n)= 5 years, 

We know that

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

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

$\frac{2000}{1000}=\left(1+\frac{R}{100}\right)^{10}$

$\frac{2}{1}=\left( 1+\frac{R}{100}\right)^{10}$

$1+\frac{R}{100}=10 \sqrt{2}$  (∵ Given that 10√2=1.072)

$1+\frac{R}{100}=1.072$

$\frac{R}{100}=1.072-1.000$

$\frac{R}{10}=0.072$

R=7.2

Rate half yearly=7.2

Annually=7.2×2=14.4%

Question 9

₹ 8000 became ₹ 9261 in a certain interval of time at the rate of 5% per annum C.l. Find the time.

Sol :

Principal=8000

Amount=9961

Rate=5%

We know that

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

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

$\frac{9261}{8000}=\left(1+\frac{5}{200}\right)^{n}$

$\frac{9261}{8000}=\left(1+\frac{1}{20}\right)^{n}$

$\frac{9261}{8000}=\left(\frac{21}{20}\right)^n$

$\left(\frac{21}{20}\right)^{30}=\left(\frac{21}{20}\right)^{n}$

n=3

∴Time = 3 years

Question 10

In how many years will a sum of ₹ 3000 at 20% per annum compounded semi-annually become ₹ 3993.

Sol :

Principal=3000

Amount=3993

Rate=20% ,10% half yearly

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

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

$\frac{3993}{3000}=\left(1+\frac{10}{1000}\right)^{n}$

$\frac{3993}{3000}=\left(1+\frac{1}{10}\right)^{n}$

$\frac{3993}{3000}=\left(\frac{11}{10}\right)^{n}$

$\frac{3}{3}\left(\frac{1331}{1000}\right)=\left(\frac{11}{10}\right)^{n}$

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

n=3

Time = 3 years or $1\frac{1}{2}$ half years

Question 11

A sum of money put out at compound interest amounts in 2 years to ₹ 578.40 and in 3 years to ₹ 614.55. Find the rate of interest.

Sol :

Amount of 3 years=614.55

Amount of 2 years=578.40

Subtracting 614.55-578.40

=36.15 (Interest for 1 year)

∴36.15 is interest on 578.40 for 1 years

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

$=\frac{36.15 \times 100}{572.40 \times 1}$

$=\frac{3615 \times 100}{57040}$

$=\frac{25}{4}$

Rate$=6\frac{1}{4}\%$

Question 12

A sum compounded annually becomes $\frac{25}{16}$ times of itself in 2 years. Determine the rate of interest per annum?

Sol :

Let principal=1 Rs

Amount$=\frac{25}{16}$ Rs

Time (n) =2 years

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

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

$\frac{25}{16 \times 1}=\left(1+\frac{R}{100}\right)^{2}$

$\frac{25}{16}=\left(1+\frac{R}{100}\right)^{2}$

$\left(\frac{5}{4}\right)^{2}=\left(1+\frac{R}{100}\right)^{2}$

$\frac{R}{100}=\frac{5}{4}-1$

$R \Rightarrow \frac{1}{4} \times 100$

Rate=25%