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. |
|
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|
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| = 3972. |
Discussion:
94 comments Page 10 of 10.
Anurag said:
1 decade ago
Thanks a lot. Now it became very easy.
Priya said:
2 decades ago
Thanks Jeeva. Now I also clear abut this problem.
Javeed said:
2 decades ago
Dude here principle is Rs. 12000, so amount = P(1+r/100)^t
C.I = amount - principle i.e., p((1+r/100)^t )-p
Removing 'p' common factor we get:
p((1+(R/100)^T)-1)
HERE 1200 IS REMOVED AS COMMON FACTOR SINCE IT IS THE VALUE OF P.
C.I = amount - principle i.e., p((1+r/100)^t )-p
Removing 'p' common factor we get:
p((1+(R/100)^T)-1)
HERE 1200 IS REMOVED AS COMMON FACTOR SINCE IT IS THE VALUE OF P.
Subash said:
2 decades ago
Please explain the use of -1 in C.I calculation. I think here we have to subtract the principle at the back. But u use the value '1' here, I can't understand Wat u r did.
But i got the same answer by using the formula.
C.I = (Amount - Principle)
Where
Amount = {P x (1 + (R/100)^N)} - P
P = 12000.
Therefore, C.I = Rs. (12000*(1+(10/100)^3))-12000 = 3972
But i got the same answer by using the formula.
C.I = (Amount - Principle)
Where
Amount = {P x (1 + (R/100)^N)} - P
P = 12000.
Therefore, C.I = Rs. (12000*(1+(10/100)^3))-12000 = 3972
(1)
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