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 2 of 10.
Shalini said:
1 decade ago
I'm totally confused with these problems.
When compound interest is done by individual formulas the answer shows wrong.
When compound interest is done by individual formulas the answer shows wrong.
Poorvajain said:
1 decade ago
CI=AMT-PRINCIPLE
So in this question princple is 100 and u hve taken principle 1 why so is it wrong.
So in this question princple is 100 and u hve taken principle 1 why so is it wrong.
Yamini said:
1 decade ago
@Subash
There was no confusion in the given answer, here -1 is nothing but as you said Amount = {P x (1 + (R/100)^N)} - P so from this just take common as P then you get
Amount=[P x {(1 + (R/100)^N)-1}].
There was no confusion in the given answer, here -1 is nothing but as you said Amount = {P x (1 + (R/100)^N)} - P so from this just take common as P then you get
Amount=[P x {(1 + (R/100)^N)-1}].
Dibyajyoti hazarika said:
1 decade ago
I'm confused about 331 no. How it comes?
Neha said:
1 decade ago
@dibyajyoti Hazarika.
(1+ 10/100)^3 -1
After L.C.M. it will be
{(100+10)/100}^3 -1
(110/100)^3 -1
(11/10)^3 -1 //(11x11x11=1331)and (10x10x10=1000)
(1331/1000)-1
After L.C.M.
(1331-1000)/1000
331/1000
I think now you understand about 331
(1+ 10/100)^3 -1
After L.C.M. it will be
{(100+10)/100}^3 -1
(110/100)^3 -1
(11/10)^3 -1 //(11x11x11=1331)and (10x10x10=1000)
(1331/1000)-1
After L.C.M.
(1331-1000)/1000
331/1000
I think now you understand about 331
Sunita Sisodiya said:
1 decade ago
Can you help with shortcut method for the same.
Sandip said:
1 decade ago
C.I = Rs. (12000*(1+(10/100)^3))-12000 = 3972.
Deepthi said:
1 decade ago
How we get S.I=Rs.60 while P=Rs.100 ?
Vicky said:
1 decade ago
Formula for C.I,
Amount = p(1+R/100)^N, Where amount = Principal + C.I.
Principal + C.I = p(1+R/100)^N.
C.I = p(1+R/100)^N - Principal, ' p ' is a common factor.
C.I = p [(1+R/100)^N-1].
Amount = p(1+R/100)^N, Where amount = Principal + C.I.
Principal + C.I = p(1+R/100)^N.
C.I = p(1+R/100)^N - Principal, ' p ' is a common factor.
C.I = p [(1+R/100)^N-1].
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.
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