SELINA Solution Class 9 Compound interest using formula) Chapter 3 Exercise 3A

Question 1

Find the amount and the compound interest on Rs. 12,000 in 3 years at 5% compounded annually.

Sol:

Given : P= Rs. 12,000; n = 3 years and r = 5%

Amount = ${\displaystyle \text{P}{(1+\frac{\text{r}}{100})}^n}$

              = ${\displaystyle 12000{(1+\frac{5}{100})}^3}$

              = ${\displaystyle 12000{\left(\frac{21}{20}\right)}^3}$
              = Rs. 13,891.50

C.I. = Rs. 13,891.50 - Rs. 12,000 = Rs. 1,891.50.

Quesiton 2

Calculate the amount of Rs. 15,000 is lent at compound interest for 2 years and the rates for the successive years are 8% and 10% respectively.

Sol:

Given : P = Rs. 15,000; n = 2 years;  r₁ = 8 % and r₂ = 10%

Amount = ${\displaystyle \text{P}(1+\frac{r_1}{100})(1+\frac{r_2}{100})}$

              = ${\displaystyle 15000(1+\frac{8}{100})(1+\frac{10}{100})}$

              = ${\displaystyle 15,000\left(\frac{27}{25}\right)\left(\frac{11}{10}\right)}$

              = Rs. 17,820.

Question 3

Calculate the compound interest accrued on Rs. 6,000 in 3 years, compounded yearly, if the rates for the successive years are 5%, 8% and 10% respectively.

Sol:

Given : P = Rs. 6,000; n = 3 years; r₁ = 5%; r₂ = 8% and r₃ = 10%

Amount = ${\displaystyle \text{P}(1+\frac{r_1}{100})(1+\frac{r_2}{100})(1+\frac{r_3}{100})}$

               = ${\displaystyle 6000(1+\frac{5}{100})(1+\frac{8}{100})(1+\frac{10}{100})}$

               = ${\displaystyle 6000\left(\frac{21}{20}\right)\left(\frac{27}{25}\right)\left(\frac{11}{10}\right)}$

               = Rs. 7,484.40
∴ C.I. = Rs. 7,484.40 - Rs. 6,000 = Rs. 1,484.40.

Question 4

What sum of money will amount to Rs. 5,445 in 2 years at 10% per annum compound interest ?

SOl:

Given : P= Rs. 5,445 ; n = 2 years and r = 10%

Amount = ${\displaystyle \text{P}{(1+\frac{\text{r}}{100})}^n}$

⇒ 5,445 = ${\displaystyle \text{P}{(1+\frac{10}{100})}^2}$

⇒ 5,445 = ${\displaystyle \text{P}{\left(\frac{11}{10}\right)}^2}$

⇒ P = ${\displaystyle 5,445{\left(\frac{10}{11}\right)}^2}$

⇒ Rs. 4,500

Quesiton 5

On what sum of money will the compound interest for 2 years at 5% per annum amount to Rs. 768.75?

Sol:

Given : C.I.= Rs. 768.75; n= 2 years and r = 5%

∴ A = ${\displaystyle \text{P}{(1+\frac{r}{100})}^n}$

⇒ A = ${\displaystyle \text{P}{(1+\frac{5}{100})}^2}$

⇒ A = ${\displaystyle P{\left(\frac{21}{20}\right)}^2=\frac{441}{400}}$P

∴ A - P = C.I.
⇒ ${\displaystyle \frac{441}{400}}$P - P = Rs. 768.75

⇒ ${\displaystyle \frac{41}{400}}$P - P = Rs. 768.75

⇒  P = Rs. ${\displaystyle \frac{768.75\times 400}{41}}$ = Rs. 7,500.

Question 6

Find the sum on which the compound interest for 3 years at 10% per annum amounts to Rs. 1,655.

Sol:

Given : C.I. = Rs. 1,655; n = 3 years and r = 10%

∴ A = ${\displaystyle \text{P}{(1+\frac{r}{100})}^n}$

⇒ A = ${\displaystyle \text{P}{(1+\frac{10}{100})}^3}$

⇒ A = ${\displaystyle \text{P}{\left(\frac{11}{10}\right)}^3=\frac{1,331}{1,000}}$P

∴ A - P = C.I.

⇒ ${\displaystyle \frac{1,331}{1,000}}$P - P = Rs. 1,655

⇒ ${\displaystyle \frac{331}{1,000}}$P = Rs. 1,665

⇒  P = Rs. ${\displaystyle \frac{1,655\times 1,000}{331}}$ = Rs. 5,000.

Question 7

What principal will amount to Rs. 9,856 in two years, if the rates of interest for successive years are 10% and 12% respectively ?

Sol:

Given : A = Rs. 9,856 ; n = 2 years ;  r₁ = 10 % and r₂ = 12%

Amount = ${\displaystyle \text{P}(1+\frac{r_1}{100})(1+\frac{r_2}{100})}$

⇒ 9,856 = ${\displaystyle \text{P}(1+\frac{10}{100})(1+\frac{12}{100})}$

⇒ 9,856 = ${\displaystyle P\left(\frac{11}{10}\right)\left(\frac{28}{25}\right)}$

⇒ P = Rs. ${\displaystyle \frac{9,856\times 10\times 25}{11\times 28}}$ = Rs. 8,000

Question 8

On a certain sum, the compound interest in 2 years amounts to Rs. 4,240. If the rate of interest for the successive years is 10% and 15% respectively, find the sum.

Sol:

Amount = ${\displaystyle \text{P}(1+\frac{r_1}{100})(1+\frac{r_2}{100})}$

⇒ ( P + 4240 ) = ${\displaystyle \text{P}(1+\frac{10}{100})(1+\frac{15}{100})}$

⇒ ( P + 4240 ) = ${\displaystyle P\left(1.265\right)}$

⇒ P = Rs. 16000

The sum is Rs.16,000.

Question 9

At what per cent per annum will Rs.6,000 amount to Rs.6,615 in 2 years when interest is compounded annually?

Sol:

Amount = ${\displaystyle \text{P}{(1+\frac{r}{100})}^n}$

⇒ 6,615 = ${\displaystyle 6,000{(1+\frac{r}{100})}^2}$

⇒ ${\displaystyle {(1+\frac{r}{100})}^2=\frac{6,615}{6,000}}$

⇒  1 + ${\displaystyle \frac{r}{100}=\frac{21}{20}}$

= r = 5%

At 5% per annum the sum of Rs. 6,000 amounts to Rs. 6,615 in 2 years when the interest is compounded annually.

Question 10

At what rate per cent compound interest, does a sum of money become 1.44 times of itself in 2 years ?

Sol:

Let Principal = Rs. y
Then Amount= Rs 1.44y
n= 2 years

∴ Amount = ${\displaystyle \text{P}{(1+\frac{\text{r}}{100})}^n}$

⇒ 1.44y  = ${\displaystyle y{(1+\frac{r}{100})}^2}$

⇒ ${\displaystyle \frac{1.44y}{y}={(1+\frac{r}{100})}^2}$

⇒ ${\displaystyle \frac{36}{25}={(1+\frac{r}{100})}^2}$

⇒ ${\displaystyle {\left(\frac{6}{5}\right)}^2={(1+\frac{r}{100})}^2}$

On comparing,
${\displaystyle \frac{6}{5}=1+\frac{r}{100}}$

On solving, we get
 r = 20 % 

Quesiton 11

At what rate per cent will a sum of Rs. 4,000 yield Rs.1,324 as compound interest in 3 years ?

Sol:

Given : P = Rs. 4,000, C.I. = Rs. 1,324 and n = 3 years
Now, A = P + I
⇒ A = Rs. ( 4,000 + 1,324 ) = Rs. 5,324

${\displaystyle A=P{(1+\frac{r}{100})}^3}$

⇒ ${\displaystyle 5324=4000{(1+\frac{r}{100})}^3}$

⇒ ${\displaystyle \frac{5324}{4000}={(1+\frac{r}{100})}^3}$

⇒ ${\displaystyle \frac{1331}{1000}={(1+\frac{r}{100})}^3}$

⇒ ${\displaystyle {(1+\frac{R}{100})}^3=\frac{1331}{1000}={\left(\frac{11}{10}\right)}^3}$

⇒ 1 + ${\displaystyle \frac{r}{100}=\frac{11}{10}}$

⇒ ${\displaystyle \frac{r}{100}=\frac{11}{10}-1=\frac{1}{10}}$

⇒ ${\displaystyle r=\frac{100}{100}=10\%}$

Thus, the rate of interest is 10%.

Quesiton 12

A person invests Rs5,000 for three years at a certain rate of interest compounded annually. At the end of two years this sum amounts to Rs6,272. Calculate :
(i) the rate of interest per annum.
(ii) the amount at the end of the third year.

Sol:

Given: P = Rs. 5,000; A = Rs. 6,272 and n = 2 years.

(i) ∴ ${\displaystyle \text{A}=\text{P}{(1+\frac{r}{100})}^n}$

⇒ ${\displaystyle 6,272=5,000{(1+\frac{r}{100})}^2}$

⇒ ${\displaystyle 6,\frac{272}{5},000={(1+\frac{r}{100})}^2}$

⇒ ${\displaystyle \frac{784}{625}={(1+\frac{r}{100})}^2}$

⇒ ${\displaystyle {\left(\frac{28}{25}\right)}^2={(1+\frac{r}{100})}^2}$

On comparing,

${\displaystyle \frac{28}{25}=1+\frac{r}{100}}$

On solving, we get
r = 12%

(ii) Amount at the third year

= ${\displaystyle 5,000{(1+\frac{12}{100})}^3}$

= ${\displaystyle 5000{\left(\frac{28}{25}\right)}^3}$

= Rs. 7,024.64

Question 13

In how many years will Rs. 7,000 amount to Rs. 9,317 at 10% per annum compound interest ?

Sol:

Given : P = Rs. 7,000; A = Rs. 9,317 and r = 10%.

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

⇒ ${\displaystyle 9,317=7,000{(1+\frac{10}{100})}^n}$

⇒ ${\displaystyle \frac{9,317}{7,000}={\left(\frac{11}{10}\right)}^n}$

⇒ ${\displaystyle \frac{1,331}{7,000}={\left(\frac{11}{10}\right)}^n}$

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

On comparing,
n = 3 years

Quesitoin 14

Find the time, in years, in which Rs. 4,000 will produce Rs. 630.50 as compound interest at 5% compounded annually.

Sol:

Given : P= Rs. 4,000; C.I.= Rs. 630.50 and r = 5%

∴ C.I. = ${\displaystyle P[{(1+\frac{r}{100})}^n-1]}$

⇒ 630.50 = ${\displaystyle 4,000[{(1+\frac{5}{100})}^n-1]}$

⇒ ${\displaystyle \frac{630.50}{4,000}=[{\left(\frac{21}{20}\right)}^n-1]}$

⇒ ${\displaystyle \frac{1,261}{8,000}=[{\left(\frac{21}{20}\right)}^n-1]}$

⇒ ${\displaystyle \frac{1,261}{8,000}+1=[{\left(\frac{21}{20}\right)}^n-1]}$

⇒ ${\displaystyle \frac{9,261}{8,000}={\left(\frac{21}{20}\right)}^n}$

⇒ ${\displaystyle {\left(\frac{21}{20}\right)}^3={\left(\frac{21}{20}\right)}^n}$

On comparing,

n = 3 years

Question 15

Divide Rs. 28,730 between A and B so that when their shares are lent out at 10% compound interest compounded per year, the amount that A receives in 3 years is the same as what B receives in 5 years.

Sol:

Let share of A = Rs. y
share of B = Rs (28,730 - y)
rate of interest= 10%

According to question,
Amount of A in 3 years= Amount of B in 5 years
⇒ ${\displaystyle y{(1+\frac{10}{100})}^3=(28,730-y){(1+\frac{10}{100})}^5}$

⇒ ${\displaystyle y=(28,730-y){(1+\frac{10}{100})}^2}$

⇒ ${\displaystyle y=(28,730-y)\left(\frac{121}{100}\right)}$

⇒ 100y = 121( 28,730 - y )

⇒ 100y + 121y = 121 x 28,730

⇒ 221y = 121 x 28,730

⇒ y = ${\displaystyle \frac{121\times 28,730}{221}}$ = Rs. 15,730

Therefore share of A = Rs. 15,730

Share of B = Rs. 28,730 - Rs.15,730 = Rs. 13,000

Quetion 16

A sum of Rs 44,200 is divided between John and Smith, 12 years and 14 years old respectively, in such a way that if their portions be invested at 10% per annum compound interest, they will receive equal amounts on reaching 16 years of age.
(i) What is the share of each out of Rs44,200 ?
(ii) What will each receive, when 16years old ?

Sol:

(i) Let share of John = Rs y
share of Smith = Rs (44,200 - y)
rate of interest= 10%

According to question,
Amount of John in 4 years = Amount of Smith in 2 years

⇒ ${\displaystyle y{(1+\frac{10}{100})}^4=(44,200-y){(1+\frac{10}{100})}^2}$

⇒ ${\displaystyle y{(1+\frac{10}{100})}^2=(44,200-y)}$

⇒ ${\displaystyle y{\left(\frac{11}{10}\right)}^2=(44,200-y)}$

⇒ ${\displaystyle 121y=100(44,200-y)}$

⇒ ${\displaystyle 121y=100\times 44,200-100y}$

⇒ ${\displaystyle 121y+100y=100\times 44,200}$

⇒ ${\displaystyle 221y=100\times 44,200}$

⇒ ${\displaystyle y=100\times 44,\frac{200}{221}}$ = Rs. 20,000.

Therefore share of John =Rs. 20,000

Share of Smith = Rs. 44,200 - Rs. 20,000 = Rs. 24,200

(ii) Amount that each will receive

= ${\displaystyle 20,000{(1+\frac{10}{100})}^4}$

= ${\displaystyle 20,000{\left(\frac{11}{10}\right)}^4}$

= Rs. 29,282

Question 17

The simple interest on a certain sum of money and at 10% per annum is Rs. 6,000 in 2 years, Find:

1. the sum.
2. the amount due to the end of 3 years and at the same rate of interest compounded annually.
3. the compound interest earned in 3 years. 

Sol:

(i) I = Rs. 6,000, T = 2 years and R = 10%
∴ P = ${\displaystyle \frac{\text{I}\times 100}{\text{R}\times \text{T}}=\frac{6,000\times 100}{10\times 2}}$ = Rs.30,000

(ii) P = Rs. 30,000, n = 3 years and r = 10%

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

= ${\displaystyle 30,000{(1+\frac{10}{100})}^3}$

= ${\displaystyle 30,000{\left(\frac{11}{10}\right)}^3}$

= ${\displaystyle 30,000\times \frac{11}{10}\times \frac{11}{10}\times \frac{11}{10}}$

= Rs. 39,930

(iii) C.I. earned in 3 years = A - P = Rs. (39,930 - 30,000) = Rs. 9,930.

Question 18

Find the difference between compound interest and simple interest on Rs. 8,000 in 2 years and at 5% per annum.

Sol:

Given : P = Rs. 8,000, R = 5%, T = 2 years
For simple interest,

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

= ${\displaystyle \frac{8,000\times 5\times 2}{100}}$

= Rs. 800

For compound interest,

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

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

= ${\displaystyle 8,000\times \frac{21}{20}\times \frac{21}{20}}$

= Rs. 8,820

C.I. = A - P
      = Rs. (8,820 - 8,000)
      = Rs. 820

Now, C.I. - S.I. = Rs. ( 820 - 800 ) = Rs. 20.
Thus, the difference between the compound interest and the simple interest is Rs. 20.

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