Aptitude - Compound Interest - Discussion
Discussion Forum : Compound Interest - General Questions (Q.No. 3)
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?
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. |
|
|||||||||||||
|
||||||||||||||
| = 3972. |
Discussion:
94 comments Page 4 of 10.
Rajata said:
1 decade ago
Its totally wrong hear as compound interest 1200(1+r/1000)n.
As the given formula it should be 12000*1331/1000.
So the answer is..15972.
As the given formula it should be 12000*1331/1000.
So the answer is..15972.
Shri said:
6 years ago
Thanks for explaining @Anmol.
Hari said:
6 years ago
Why we are calculating for 3 years?
Rahul krishna said:
6 years ago
@Sangavi we use - p because.
CI = Amount - principal.
So, CI=p(1+r/100)^n-p.
CI = Amount - principal.
So, CI=p(1+r/100)^n-p.
Sandip said:
1 decade ago
C.I = Rs. (12000*(1+(10/100)^3))-12000 = 3972.
Sangavi said:
7 years ago
When will we use -P in the compound interest formula?
Steve said:
1 decade ago
You make your calculation too complicated for beginners, it would have been much more easier if you just convert it to decimal and ignore the fraction thing. Good job though.
Aayushi said:
1 decade ago
Can anyone help me that why do we have to subtract the P. And when to subtract and when not to?
Shub said:
7 years ago
Amount after 1st year =1200.
After 2nd year=1200 + 120.
After 3rd year= 1200 + 120+120+12.
Total = 3972.
After 2nd year=1200 + 120.
After 3rd year= 1200 + 120+120+12.
Total = 3972.
Rajeev rathor said:
7 years ago
Using pascal traingle.
= 3*10%of12000+3*10%of1200+1*10%of120,
= 3600+360+12,
= 3972.
= 3*10%of12000+3*10%of1200+1*10%of120,
= 3600+360+12,
= 3972.
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