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Aptitude - Compound Interest - Discussion

Discussion Forum : Compound Interest - General Questions (Q.No. 3)
Compound Interest
3.
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?
Rs. 2160
Rs. 3120
Rs. 3972
Rs. 6240
None of these
Answer: Option
Explanation:

Let P = Rs. 100. Then, S.I. Rs. 60 and T = 6 years.

R = 100 x 60 = 10% p.a.
100 x 6

Now, P = Rs. 12000. T = 3 years and R = 10% p.a.

C.I.
= Rs. 12000 x 1 + 10 3 - 1
100
= Rs. 12000 x 331
1000
= 3972.

Discussion:
94 comments Page 2 of 10.
Vishwanath said:   3 years ago
@Keerti.

12000 * 331/1000 = 12 * 331/1 = 3972.
(5)
Vishwanath said:   3 years ago
@Harshit.

First solve brackets, then multiply them into 3 times.

Then, You will get 1331/1000.
(4)
Harshit said:   4 years ago
How to solve (1 + 10/100) ^3?

Please explain it.
(2)
Keerthi said:   4 years ago
1200*{(1+10/100)^3 - 1}, and next step how will become?

Especially- (1200*331/1000).

Please explain to me.
(3)
Sriram Manikanth said:   5 years ago
60 is interest 100 is the principle.
SI = pnr/100.
Therefor 60 = 100 * 6*r/100,
r = 10%.
(3)
Banshika said:   5 years ago
If the amount is increased by 60%. It should be 160 right?

Then why they took 60 straight to find the answer?
Pratiksha said:   5 years ago
@Katherin,

Instead of -1 you can take principal amount( in this case -12000/-), you will get the same answer.
(8)
Katherin said:   5 years ago
I'm confused on using -1. Why it is used?

Please anyone explain.
(16)
Taraka said:   5 years ago
A = p+si
P(160/100) = p + (p*6*R).
R = 10%.

Next;
A = p+ci.
A = 12000 * (110/100) ^3
P = 12000.
Then substituting a and p then;
Ci = 3972.
(2)
Ramya c said:   5 years ago
Anyone can explain me, why we are calculating for 3 years?
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