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 6 of 10.
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?
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.
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.
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].
Deepthi said:
1 decade ago
How we get S.I=Rs.60 while P=Rs.100 ?
Sandip said:
1 decade ago
C.I = Rs. (12000*(1+(10/100)^3))-12000 = 3972.
Sunita Sisodiya said:
1 decade ago
Can you help with shortcut method for the same.
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
Dibyajyoti hazarika said:
1 decade ago
I'm confused about 331 no. How it comes?
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}].
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