Book Online Demo

Check Your IQ

Try Test

Courses

Offline Centres

There is 60% increase in an amount in 6 years at simple interest. What will be the compound interest of Rs.12,000 after 3 years at the same rate?

see full answer

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

Rs. 2,160

Rs. 3,120

Rs. 3,972

Rs. 6,240

answer is C.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

Concept: Using formula for simple, final amount and compound interest which are,

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

C . I = A - P

and

A {= P (1 + \frac{R}{100})}^{N}

respectively.
where,

P = Initial amount

A = Final amount

R = Rate

T = Time

N = No . of years

Given is that:
There is a 60% increase in the amount in 6 years.
Therefore, Simple interest (S.I) will be 60% of the Initial amount P.

60 % of P

S . I = \frac{60 P}{100}

Now, the formula for S.I is:

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

where,

T = 6 years

Putting the values in, we get:

S . I = \frac{P \times R \times 6}{100}

S . I = \frac{6 PR}{100}

Now, the two values of S.I obtained will eventually be equal. Equating the two:

\frac{60 P}{100} = \frac{6 PR}{100}

∴ R = 10

Now, we have to find the compound interest after 3 years at the same rate.
Putting all the known values in the formula for of final amount A, we get:

A {= P (1 + \frac{R}{100})}^{N}

{= 12, 000 (1 + \frac{10}{100})}^{3}

{= 12, 000 \times (\frac{110}{100})}^{3}

{= 12, 000 \times (\frac{11}{10})}^{3}

= 12, 000 \times \frac{1331}{1000}

= 12 \times 1331

= 12 \times 1331

A = Rs . 15, 972

Therefore, C.I will be:

C . I = A - P

C . I = Rs . (15, 972 - 12, 000)

C . I = Rs . 3, 972

Hence, the correct option is 3.

Watch 3-min video & get full concept clarity