Compound Interest Questions and Answer Quiz 8 for IBPS PO, SBI Clerk Mains, IBPS RRB PO, IBPS RRB Clerk, CET and RBI Assistant Mains Exams with Detailed Solution

Important for :
1
₹ 6100 was partly invested in Scheme A at 10% pa compound interest (compounded annually) for 2 years and partly in Scheme B at 10% pa simple interest for 4 years. Both the schemes earn equal interests. How much was invested in Scheme A ?
» Explain it
C
Let the amount invested in Scheme A is ₹ x.

Then, the amount invested in Scheme B be ₹ (6100 – x)

Now, according to the question,

 x ( 1 + 10 ) 2 – x  = (6100 – x) × 10 × 4 100 100

 ⇒  x ( 121 – 1 ) = (6100 – x) × 40 100 100

 ⇒ 21x = (6100 – x) × 40 100 100

⇒  21x = 6100 × 40 – 40x

⇒  61x = 6100 × 40

 ⇒  x = 6100 × 40 =  ₹ 4000 61

∴  The amount invested in Scheme A is ₹ 4000.

Hence, option C is correct.

2
A sum of ₹ 198 deposited at CI doubles itself after 4 years. After 20 years it will become
» Explain it
A
 P ( 1 + R ) 4 = 2P 100

 ⇒ ( 1 + R ) 4 = 2     ...(i) 100

After 20 years,

 ( 1 + R ) 20 = [ ( 1 + R ) 4 ] 5 = 25 = 32 100 100

Thus, the amount becomes 32 times.

So, amount = 198 × 32  =  ₹ 6336

Hence, option A is correct.

3
Lata had ₹ 40000. She invested some amount in scheme A at CI at 15% and the remaining amount in scheme B at SI at 10%. If she got the same interest from both the investments at the end of one year. How much in ₹ did she invest in scheme B?
» Explain it
B
Let the amount invested in Scheme A is ₹ x.

Then, the amount invested in Scheme B therefore will be ₹ (40000 – x)

We know that for the 1st year both Simple Interest and Compound Interest on a sum remains the same. Now, according to the question,

⇒ 15% of x = 10% of (40000 – x)

⇒ 15x = 400000 – 10x

⇒ 25x = 400000

⇒ x = 16000

Amount invested in scheme B = 40000 – 16000 = 24000.

Hence, option B is correct.

4
An amount of ₹ 110000 is invested at compound interest payable annually. If the rate of interest is 11% pa, what will be the total interest after two years ?
» Explain it
B
Method I :

P = ₹ 110000; R = 11%; n = 2 years

 CI = P ( 1 + R ) n – P 100

 CI = 110000 ( 1 + 11 ) 2 – 110000 100

 = 110000 ( 111 ) 2 – 110000 100

= 135531 – 110000  =  ₹ 25531

_________________________________________

Method II :

To solve this question, we can apply the net% effect formula

 Net% effect = ( x + y + xy ) % 100

Here, x = y = 11%         (because rate of interest is same for both the years)

By the net% effect, we get effective rate of interest

 = ( 11 + 11 + 11 × 11 ) % = 23.21% 100

Therefore,  23.21% of 110000  =  ₹ 25531

Hence, option B is correct.

5
Find the compound interest on ₹ 12000 for 2 years, the rate of interest being 3% per annum.
» Explain it
D
Method I :

P = ₹ 12000; R = 3%; n = 2 years

 CI = P ( 1 + R ) n – P 100

 CI = 12000 ( 1 + 3 ) 2 – 12000 100

 = 12000 × 103 × 103 – 12000 100 100

= 12730.8 – 12000  =  ₹ 730.8

_____________________________________________

Method II :

To solve this question, we can apply the net% effect formula

 Net% effect = ( x + y + xy ) % 100

Here, x = y = 3%            (because rate of interest is same for both the years)

By the net% effect, we get effective rate of interest

 = ( 3 + 3 + 3 × 3 ) % = 6.09% 100

Therefore, 6.09% of 110000  =  ₹ 730.8

Hence, option D is correct.