Question

Asif took a loan of Rs 80,000 from a bank. If the rate of interest is 10% per annum, find 1 the difference in amounts he would be paying after 1 and a half years if the interest is

(1) compounded annually

. (ii) compounded half yearly.​

Answer 1

Answer:

(i) = Rs.88000+Rs.4400 = 92400Rs.

(ii) = Rs.92610

Thus, the difference between the two amounts = Rs.92610−Rs.92400 =Rs.210

Step-by-step explanation:

Answer 2

Given :-

• Principal = ₹ 80,000 ( P )

• Rate = 10% per annum ( R )

$• \: Time \: = 1 \frac{1}{2} \: years \: ( \: T \: )$

To find :-

★ The difference in amounts he would be paying after 1 and a half years if the interest is

(1) compounded annually

(2) compounded half yearly.

Solution :-

Here we have ,

P = 80,000

R = 10%

$T \: = 1 \frac{1}{2}$

★ As we know ,

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

where ,

A = Compound interest

P = principal

R = rate

n = number of years

so , putting the values

$A = 80000 {(1 + \frac{10}{100} ) }^{1}$

$A = 80000 {(1 + \frac{1}{10} )}^{1}$

$A = 80000 \: {( \frac{10 + 1}{10} )}^{1} \: \: \: \:by \: LCM$

$A = 80000( \frac{11}{10} )$

$A = 8000 \times 11$

$A = 88000$

This compound interest was for 1 yrs but we have to take for 1 and a half years

So

Remaining half yrs

For remaining half yrs we will take simple interest

$SI = \frac{P× R× T}{100}$

where

SI = Simple interest

P = principal

R = rate of interest

T = Time

Note :- This time principal will be 88000

so , putting the values

$S.I = \frac{88000 \times 10 \times 1}{100\times 2 }$

$S.I = 880 \times 5 \: \: \: \: \: By \: cancellation$

$S.I = 4400$

Now total amount =

88000 + 4400

92400

So amount after one and a half years if the interest is compounded annually is ₹ 92400

Now,

Compounded half - yearly

P = 80000

$R = 10\% = \frac{10}{2} = 5$

$T = 1\frac{1}{2} = \frac{3}{2} = \frac{3}{2} \times 2 = 3$

Using

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

$A = 80000 {(1 + \frac{5}{100} )}^{3}$

$A = 80000 {(1 + \frac{1}{20} )}^{3}$

$A = 80000 {( \frac{20 + 1}{20} })^{3} \: \: \: \: by \: lcm$

$A = 80000 {( \frac{21}{20} })^{3}$

$A = 80000 \times \frac{21}{20} \times \frac{21}{20} \times \frac{21}{20}$

$A = 80 \times \frac{21}{2} \times \frac{21}{2} \times \frac{21}{2}$

$A = \frac{40 \times 21 \times 21 \times 21}{2 \times 2}$

$A = \frac{20 \times 21 \times 21 \times 21}{2}$

$A = 10 \times 21 \times 21 \times 21$

$A = 92610$

Now,

Difference

92610 - 92400

210 Ans

Formula used :-

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

• $SI = \frac{P× R× T}{100}$