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

Question 1

The difference between simple interest and compound interest on a certain sum is Rs. 54.40 for 2 years at 8 per cent per annum. Find the sum.

Sol:

Let principal (P) = x
R = 8%
T = 2 years

SI = ${\displaystyle \frac{\text{X}\times 8\times 2}{100}=\frac{4X}{25}}$

CI = A - P = ${\displaystyle \text{X}{(1+\frac{8}{100})}^2-\text{X}}$

                 = ${\displaystyle \text{X}|{[1+\frac{2}{25}]}^2-1|}$

                 = ${\displaystyle \text{X}|{\left[\frac{27}{25}\right]}^2-1|}$

                 = ${\displaystyle \frac{104}{625}X}$

Given, CI = SI = 54.40
${\displaystyle \frac{104X}{625}-\frac{4X}{25}}$ = Rs. 54.40

${\displaystyle X(\frac{104}{625}-\frac{4}{25}\times \frac{25}{25})=54.40}$

${\displaystyle X\left(\frac{4}{625}\right)=54.40}$

X = ${\displaystyle \frac{54.40\times 625}{4}}$

X = Rs. 8500 
Thus, principal sum = Rs. 8500

Question 2

A sum of money, invested at compound interest, amounts to Rs. 19,360 in 2 years and to Rs. 23,425.60 in 4 years. Find the rate per cent and the original sum of money.

Sol:

(for 2 years) A = Rs. 19360
T = 2 years
Let P = X

${\displaystyle \text{X}{(1+\frac{\text{R}}{100})}^2=19,360}$                      ......(1)

A (for 4 years) = Rs. 23425.60

${\displaystyle \text{X}{(1+\frac{\text{R}}{100})}^4=23425.60}$                   ......(2)

2 ÷ 1

${\displaystyle {[1+\frac{\text{R}}{100}]}^2=\frac{23425.60}{19360}}$

${\displaystyle {[1+\frac{\text{R}}{100}]}^2=\frac{2342560}{1936000}}$

${\displaystyle {[1+\frac{\text{R}}{100}]}^2=\frac{14641}{12100}}$

${\displaystyle {[1+\frac{\text{R}}{100}]}^2={\left[\frac{121}{110}\right]}^2}$

${\displaystyle 1+\frac{\text{R}}{100}=\frac{121}{110}}$

${\displaystyle \frac{\text{R}}{100}=\frac{121}{110}-1}$

R = 10%

From (1) x ${\displaystyle {[1+\frac{10}{100}]}^2}$ = 19360

X = ${\displaystyle \frac{19360\times 10\times 10}{11\times 11}}$

X = Rs. 16,000

Thus, sum = Rs. 16000

Question 3

A sum of money lent out at C.I. at a certain rate per annum becomes three times of itself in 8 years. Find in how many years will the money becomes twenty-seven times of itself at the same rate of interest p.a.

Sol:

Let principal = X, A = 3X, T = 8 years, R = ?

Case I,
${\displaystyle \text{A}=\text{P}{[1+\frac{\text{R}}{100}]}^{\text{T}}}$

${\displaystyle \text{3X}=\text{X}{[1+\frac{\text{R}}{100}]}^8}$

${\displaystyle 3^{\frac{1}{8}}=1+\frac{\text{R}}{100}}$                ...(1)

Case II,
P = X, A = 27X, T = ?
${\displaystyle 27\text{X}=\text{X}{[1+\frac{\text{R}}{100}]}^{\text{T}}}$

${\displaystyle {27}^{\frac{1}{T}}=1+\frac{\text{R}}{100}}$             ...(2)

From (1) and (2)
${\displaystyle 3^{\frac{1}{8}}={27}^{\frac{1}{T}}}$

${\displaystyle 3^{\frac{1}{8}}=3^{\frac{1}{8}}=\frac{3}{\text{T}}}$

T = 24

Time = 24 years.

Question 4

On what sum of money will compound interest (payable annually) for 2 years be the same as simple interest on Rs. 9,430 for 10 years, both at the rate of 5 per cent per annum ?

Sol:

P = Rs. 9430
R = 5%
T = 10 years

SI = ${\displaystyle \frac{9430\times 5\times 10}{100}}$ = Rs. 4715

Let sum = X
CI = 4715, T = 2 years, Rs= 5%

CI = AP

${\displaystyle 4715=\text{X}{[1+\frac{\text{R}}{100}]}^{\text{T}}-\text{X}}$

${\displaystyle 4715=\text{X}{[1+\frac{5}{100}]}^2-\text{X}}$

${\displaystyle 4715=\text{X}|{\left(\frac{21}{20}\right)}^2-1|}$

${\displaystyle 4715=\text{X}\left(\frac{441-400}{400}\right)}$

${\displaystyle \text{X}=\frac{4715\times 400}{41}}$ = Rs. 46,000

Thus principal from = Rs. 46,000

Question 5

Kamal and Anand each lent the same sum of money for 2 years at 5% at simple interest and compound interst respectively. Anand recived Rs. 15 more than Kamal. Find the amount of money lent by each and the interest received.

Sol:

Let principal = Rs. 100, R = 5% T = 2 years

For Kamal, SI = ${\displaystyle \frac{100\times 5\times 2}{100}}$ = Rs. 10

For Anand,        ${\displaystyle \text{A}=\text{P}{[1+\frac{\text{R}}{100}]}^{\text{T}}}$ 

= ${\displaystyle 100{[1+\frac{5}{100}]}^2}$

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

= ${\displaystyle \frac{441}{4}}$

CI = ${\displaystyle \frac{441}{4}-100=\frac{41}{4}}$

Difference of CI and SI = ${\displaystyle \frac{41}{4}-10}$

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

= Rs. ${\displaystyle \frac{1}{4}}$

When difference is Rs.${\displaystyle \frac{1}{4}}$ , then principal = Rs. 100.
If difference is 1, principal = 100 x 4
If difference is Rs.15, principal = 100 x 4 x 15 = Rs. 6000

For kamal, interest = ${\displaystyle \frac{6000\times 5\times 2}{100}}$ = Rs. 600.

For Anand, interest = ${\displaystyle 6000{[1+\frac{5}{100}]}^2-6000}$

= ${\displaystyle 6000[{\left(\frac{21}{20}\right)}^2-1]}$

= ${\displaystyle 6000[\frac{441}{400}-1]}$

= ${\displaystyle 6000\times \frac{41}{400}}$

= Rs. 615

Let principal = Rs. 100, R = 5% T = 2 years

For Kamal, SI = ${\displaystyle \frac{100\times 5\times 2}{100}}$ = Rs. 10

For Anand,        ${\displaystyle \text{A}=\text{P}{[1+\frac{\text{R}}{100}]}^{\text{T}}}$ 

= ${\displaystyle 100{[1+\frac{5}{100}]}^2}$

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

= ${\displaystyle \frac{441}{4}}$

CI = ${\displaystyle \frac{441}{4}-100=\frac{41}{4}}$

Difference of CI and SI = ${\displaystyle \frac{41}{4}-10}$

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

= Rs. ${\displaystyle \frac{1}{4}}$

When difference is Rs.${\displaystyle \frac{1}{4}}$ , then principal = Rs. 100.
If difference is 1, principal = 100 x 4
If difference is Rs.15, principal = 100 x 4 x 15 = Rs. 6000

For kamal, interest = ${\displaystyle \frac{6000\times 5\times 2}{100}}$ = Rs. 600.

For Anand, interest = ${\displaystyle 6000{[1+\frac{5}{100}]}^2-6000}$

= ${\displaystyle 6000[{\left(\frac{21}{20}\right)}^2-1]}$

= ${\displaystyle 6000[\frac{441}{400}-1]}$

= ${\displaystyle 6000\times \frac{41}{400}}$

= Rs. 615

Question 6

Simple interest on a sum of money for 2 years at 4% is Rs. 450. Find compound interest of the same sum and at the same rate for 2 years.

Sol:

SI = Rs. 450
R = 4%
T = 2 years
P = ?

P = ${\displaystyle \frac{\text{SI}\times 100}{\text{R}\times \text{T}}=\frac{450\times 100}{4\times 2}=Rs.5625}$

A = ${\displaystyle 5625{[1+\frac{4}{100}]}^2}$

= ${\displaystyle 5625{\left(\frac{26}{25}\right)}^2}$

= ${\displaystyle \frac{3802500}{625}}$

= Rs. 6084

CI = A - P = 6084 - 5625 = Rs. 459

Question 7

Simple interest on a certain sum of money for 4 years at 4% per annum exceeds the compound interest on the same sum for 3 years at 5 per cent per annum by Rs. 228. Find the sum.

Sol:

Let principal (P), R = 4%, T = 4 years

SI = ${\displaystyle \frac{\text{P}\times 4\times 4}{100}=\frac{\text{4P}}{25}}$

CI = ${\displaystyle \text{P}[1+\frac{5}{100}]-\text{P}}$

= ${\displaystyle \text{P}|{\left(\frac{21}{20}\right)}^3-1|}$

= ${\displaystyle \text{P}[\frac{9261}{8000}-1]}$

= ${\displaystyle \frac{1261}{8000}\text{P}}$

Given: SI = CI = Rs. 228

${\displaystyle \frac{\text{4P}}{25}-\frac{1261}{8000}\text{P}}$

${\displaystyle \frac{4\times 320\text{P}-1261\text{P}}{8000}=228}$

${\displaystyle 19\text{P}=228\times 8000}$

${\displaystyle \text{P}=\frac{228\times 8000}{19}}$ 

P = Rs. 96000

Thus, Principal = Rs. 96000 

Question 8

Compound interest on a certain sum of money at 5% per annum for two years is Rs. 246. Calculate simple interest on the same sum for 3 years at 6% per annum.

Sol:

CI = Rs. 246, R = 5%, T = 2 years

CI = A - P

${\displaystyle 246=\text{P}{[1+\frac{5}{100}]}^2-\text{P}}$

${\displaystyle 246=\text{P}|{\left(\frac{21}{20}\right)}^2-1|}$

${\displaystyle 246=\text{P}\left[\frac{61}{400}\right]}$

${\displaystyle \text{P}=\frac{246\times 400}{41}}$

= Rs. 2400

Now, P = Rs. 2400, R = 6%, T = 3 years

SI = ${\displaystyle \frac{2400\times 6\times 3}{100}}$

= Rs. 432.

Question 9

A certain sum of money amounts to Rs. 23,400 in 3 years at 10% per annum simple interest. Find the amount of the same sum in 2 years and at 10% p.a. compound interest.

Sol:

Let the sum (principle) = X
Given Amount = 23400, R = 10% and T = 3 years

⇒ Interest = ${\displaystyle \frac{\text{X}\times 10\times 3}{100}=\frac{3X}{10}}$

Amount = Principle + Interest

23400 = X + ${\displaystyle \frac{\text{3X}}{10}}$

X = 18000
Principle = 18000
Now,
Principle = 18000, r = 10% and n = 2 years

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

${\displaystyle \text{A}=18000{[1+\frac{10}{100}]}^2}$

${\displaystyle \text{A}=18000{\left[\frac{11}{10}\right]}^2}$

${\displaystyle \text{A}=18000\left[\frac{121}{100}\right]}$

A = 21780 

The amount of the same sum in 2 years and at 10% p.a. compound interest is 21780.

Question 10

Mohit borrowed a certain sum at 5% per annum compound interest and cleared this loan by paying Rs. 12,600 at the end of the first year and Rs. 17,640 at the end of the second year. Find the sum borrowed.

Sol:

For the payment of Rs. 12,600 at the end of first year :
A = Rs. 12,600; n = 1 year and r = 5%

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

⇒ 12,600 = ${\displaystyle \text{P}{(1+\frac{5}{100})}^1}$

⇒ 12,600 = ${\displaystyle \text{P}\left(\frac{21}{20}\right)}$

⇒ P = ${\displaystyle \frac{20}{21}\times 12,600=Rs.12,000}$

For the payment of Rs. 17,640 at the end of second year :
A = Rs. 17,640, n = 2 years and r = 5%

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

⇒ 17,640 = ${\displaystyle \text{P}{(1+\frac{5}{100})}^2}$

⇒ 17,640 = ${\displaystyle \text{P}{\left(\frac{21}{20}\right)}^2}$

⇒ P = ${\displaystyle \frac{20}{21}\times \frac{20}{21}\times 17,640=Rs.16,000}$

∴ Sum borrowed = Rs. (12,000 + 16,000 ) = Rs. 28,000.