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

Question 1

If the interest is compounded half-yearly, calculate the amount when principal is Rs. 7,400; the rate of interest is 5% per annum and the duration is one year.

Sol:

Given: P = Rs. 7,400; r = 5% p.a. and n = 1 year

Since the interest is compounded half-yearly,

Then, A = ${\displaystyle \text{P}{(1+\frac{r}{2\times 100})}^{n\times 2}}$ 

= ${\displaystyle 7,400{(1+\frac{5}{2\times 100})}^{1\times 2}}$

= ${\displaystyle 7,400{\left(\frac{41}{40}\right)}^2}$

= Rs. 7,774.63

Question 2

Find the difference between the compound interest compounded yearly and half-yearly on Rs. 10,000 for 18 months at 10% per annum.

Sol:

(i) When interest is compounded yearly :
Given : P = Rs. 10,000 ; n = 18 months = ${\displaystyle 1\frac{1}{2}}$ year and r = 10% p.a.
For 1 year

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

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

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

= Rs. 11,000

For ${\displaystyle \frac{1}{2}}$ year
P = Rs. 11,000 ; n = ${\displaystyle \frac{1}{2}}$ year and r = 10 %

A = ${\displaystyle \text{P}{[1+\frac{r}{2\times 100}]}^{n\times 2}}$

= ${\displaystyle 11,000{(1+\frac{10}{2\times 100})}^{\frac{1}{2}\times 2}}$

= ${\displaystyle 11,000{\left(\frac{21}{20}\right)}^1}$

= Rs. 11,550.

C.I.= Rs. 11,550 - Rs. 10,000 = Rs. 1,550

(ii) When interest is compounded half-yearly :

P = Rs. 10,000 ; n = ${\displaystyle 1\frac{1}{2}}$ year and r = 10% p.a.

A = ${\displaystyle \text{P}{(1+\frac{r}{2\times 100})}^{n\times 2}}$

= ${\displaystyle 10,000{[1+\frac{10}{2\times 100}]}^{\frac{3}{2}\times 2}}$

= ${\displaystyle 10,000{\left(\frac{21}{20}\right)}^3}$

= Rs. 11,576.25

 C.I.= Rs.11,576.25 - Rs.10,000 = Rs. 1,576.25

Difference between both C.I. = Rs. 1,576.25 - Rs. 1,550 = Rs. 26.25

Question 3

A man borrowed Rs.16,000 for 3 years under the following terms:
20% simple interest for the first 2 years.
20% C.I. for the remaining one year on the amount due after 2 years, the interest being compounded half-yearly.
Find the total amount to be paid at the end of the three years.

Sol:

For the first 2 years

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

⇒ S.I. = ${\displaystyle \frac{16,000\times 2\times 20}{100}}$ ⇒ S.I. = 6,400

Amount = S.I. + P 
⇒  Amount = 6,400 + 16,000 = Rs. 22,400

Amount in the account at the end of the two years is Rs.22,400.
For the remaining one year

A = ${\displaystyle P{(1+\frac{r}{2\times 100})}^{n\times 2}}$

⇒ A = ${\displaystyle 22,400{(1+\frac{20}{200})}^2}$

⇒ A = ${\displaystyle 22,400{\left(\frac{11}{10}\right)}^2}$

⇒ A = 27,104

The total amount to be paid at the end of the three years is Rs. 27,104.

Question 4

What sum of money will amount to Rs. 27,783 in one and a half years at 10% per annum compounded half yearly ?

Sol:

A = ${\displaystyle P{(1+\frac{r}{2\times 100})}^{n\times 2}}$

⇒ ${\displaystyle 27,783= \text{P}{(1+\frac{10}{200})}^{\frac{3}{2}\times 2}}$

⇒ ${\displaystyle 27,783= \text{P}{\left(\frac{21}{20}\right)}^3}$

⇒ P = ${\displaystyle 27,783{\left(\frac{20}{21}\right)}^3}$

⇒ P = 24,000

The sum of Rs. 24,000 amount Rs. 27,783 in one and a half years at 10% per annum compounded half yearly.

Question 5

Ashok invests a certain sum of money at 20% per annum, compounded yearly. Geeta invests an equal amount of money at the same rate of interest per annum compounded half-yearly. If Geeta gets Rs. 33 more than Ashok in 18 months, calculate the money invested.

Sol:

(i) For Ashok (interest is compounded yearly) : 

Let P = Rs. y; n = 18 months = ${\displaystyle 1\frac{1}{2}}$ year and r = 20% p.a.

For 1 year
A = ${\displaystyle \text{P}{(1+\frac{r}{100})}^n=y{(1+\frac{20}{100})}^1=\left(\frac{6}{5}\right)y}$

For ${\displaystyle \frac{1}{2}}$ year
P = Rs. ${\displaystyle \left(\frac{6}{5}\right)y; n=\frac{1}{2}}$ year and r = 20%

A = ${\displaystyle \text{P}{(1+\frac{r}{2\times 100})}^{n\times 2}=Rs.\left(\frac{6}{5}\right)y{(1+\frac{20}{2\times 100})}^{\frac{1}{2}\times 2}=Rs.\left(\frac{66}{50}\right)y}$

(ii) For Geeta ( interest is compounded half-yearly )

 P = Rs. y ;  n = ${\displaystyle 1\frac{1}{2}}$ year and r = 20% p.a.

${\displaystyle \text{A}=\text{P}{(1+\frac{r}{2\times 100})}^{n\times 2}=y{(1+\frac{20}{2\times 100})}^{\frac{3}{2}\times 2}=y{\left(\frac{11}{10}\right)}^3}$

= Rs. ${\displaystyle \frac{1,331}{1,000}y}$

According to question

∴ ${\displaystyle \frac{1,331}{1,000}y-\left(\frac{66}{50}\right)y=Rs.33}$

= ${\displaystyle \left(\frac{11}{1,000}\right)y=Rs.33}$

= y = Rs. ${\displaystyle \frac{33\times 1,000}{11}=Rs.3,000}$

Money invested by each person=Rs. 3,000.

Question 6

At what rate of interest per annum will a sum of Rs. 62,500 earn a compound interest of Rs. 5,100 in one year? The interest is to be compounded half yearly.

Sol:

C.I. = ${\displaystyle P[{(1+\frac{r}{2\times 100})}^{2\times n}-1]}$

⇒ 5,100 = 62,500${\displaystyle [{(1+\frac{r}{200})}^2-1]}$

⇒ ${\displaystyle {(1+\frac{r}{200})}^2=\frac{67,600}{62,500}}$

⇒ ${\displaystyle 1+\frac{r}{200}=\frac{260}{250}}$

⇒ r = 8

The rate of interest is 8%.

Question 7

In what time will Rs. 1,500 yield Rs. 496.50 as compound interest at 20% per year compounded half-yearly ?

Sol:

Given: P=Rs. 1,500; C.I.= Rs. 496.50 and r = 20%
Since interest is compounded semi-annually

Then, C.I. = ${\displaystyle \text{P}[{(1+\frac{r}{2\times 100})}^{n\times 2}-1]}$

⇒ 496.50 = 1,500${\displaystyle [{(1+\frac{20}{2\times 100})}^{n\times 2}-1]}$

⇒ ${\displaystyle \frac{496.50}{1500}={\left(\frac{11}{10}\right)}^{2n}-1}$

⇒ ${\displaystyle \frac{331}{1000}+1={\left(\frac{11}{10}\right)}^{2n}}$

⇒ ${\displaystyle \frac{1331}{1000}={\left(\frac{11}{10}\right)}^{2n}}$

⇒ ${\displaystyle {\left(\frac{11}{10}\right)}^3={\left(\frac{11}{10}\right)}^{2n}}$

On comparing, we get,

2n = 3 ⇒ n = ${\displaystyle 1\frac{1}{2}}$ years

Question 8

Calculate the C.I. on Rs. 3,500 at 6% per annum for 3 years, the interest being compounded half-yearly.
Do not use mathematical tables. Use the necessary information from the following:
(1.06)³ =1.191016; (1.03)³ = 1.092727
(1.06)⁶ =1.418519; (1.03)⁶ = 1.194052

Sol:

Given : P = Rs. 3,500; r = 6% and n = 3 years

Since interest is being compounded half-yearly
Then, ${\displaystyle \text{C.I.}=\text{P}[{(1+\frac{r}{2\times 100})}^{n\times 2}-100]}$

= 3,500${\displaystyle [{(1+\frac{6}{2\times 100})}^{n\times 2}-1]}$

= 3,500${\displaystyle [{\left(\frac{103}{100}\right)}^6-1]}$

= 3,500[(1.03)⁶ - 1 ]

= 3,500[ 1.194052 - 1 ]

= 3,500 x 0.194052

= Rs. 679.18

Question 9

Find the difference between compound interest and simple interest on Rs.12,000 and in  ${\displaystyle 1\frac{1}{2}}$ years at 10% compounded yearly.

Sol:

Given: P = Rs. 12,000;  n =  ${\displaystyle 1\frac{1}{2}}$ years and r = 10%

S.I. = ${\displaystyle \frac{P\times R\times T}{100}=\frac{12,000\times 10\times \frac{3}{2}}{100}=Rs.1,8000}$

To Calculate C.I.
For 1 Year
P = Rs. 12,000; n = 1 year and r = 10%

A = ${\displaystyle \text{P}{[1+\frac{r}{100}]}^n =12,000{(1+\frac{10}{100})}^1 =Rs.13,200}$

For next ${\displaystyle \frac{1}{2}}$ year
P = Rs. 13,200; n = ${\displaystyle \frac{1}{2}}$ year and r = 10%

A = ${\displaystyle \text{P}{(1+\frac{r}{2\times 100})}^{n\times 2}}$

= ${\displaystyle 13,200{(1+\frac{10}{2\times 100})}^{\frac{1}{2}\times 2}}$

= ${\displaystyle 13,200{\left(\frac{21}{20}\right)}^1}$

= Rs. 13,860

∴ C.I. = Rs. 13,860 - Rs. 12,000 = Rs. 1,860
∴Difference between C.I. and S.I.
= Rs. 1,860 - Rs. 1,800 = Rs. 60.

Question 10

Find the difference between compound interest and simple interest on Rs. 12,000 and in ${\displaystyle 1\frac{1}{2}}$ years at 10% compounded half-yearly.

Sol:

Given : P = Rs. 12,000 ; n =  ${\displaystyle 1\frac{1}{2}}$ years and r = 10%

S.I. = ${\displaystyle \frac{P\times R\times T}{100} =\frac{12,000\times 10\times \frac{3}{2}}{100} =Rs.1,800}$

To calculate C.I. ( Compound half - yearly ) :
P = Rs. 12,000; n = ${\displaystyle 1\frac{1}{2}}$ years  and r = 10%

A = ${\displaystyle \text{P}{(1+\frac{r}{2\times 100})}^{n\times 2}}$ 

A = ${\displaystyle 12,000{(1+\frac{10}{2\times 100})}^{\frac{3}{2}\times 2}}$ 

A = ${\displaystyle 12,000{\left(\frac{21}{20}\right)}^3}$ 

A = Rs. 13,891.50

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

∴ Difference between C.I. and S.I = Rs. 1,891.50 - Rs. 1,800 = Rs. 91.50.