Aptitude - Compound Interest - Discussion
Discussion Forum : Compound Interest - General Questions (Q.No. 2)
2.
The difference between simple and compound interests compounded annually on a certain sum of money for 2 years at 4% per annum is Re. 1. The sum (in Rs.) is:
Answer: Option
Explanation:
Let the sum be Rs. x. Then,
| C.I. = |  |
x |  |
1 + | 4 |  |
2 | - x |  |
= | |
676 | x | - x | |
= | 51 | x. |
| 100 | 625 | 625 |
| S.I. = | |
x x 4 x 2 | |
= | 2x | . |
| 100 | 25 |
 |
51x | - | 2x | = 1 |
| 625 | 25 |
x = 625.
Discussion:
149 comments Page 9 of 15.
Bunny_xavi said:
8 years ago
(R/100)^2= D/P
D=difference bw si and ci.
100^2/4*4 = 1/P.
P= 10000/16 =>625.
D=difference bw si and ci.
100^2/4*4 = 1/P.
P= 10000/16 =>625.
Sagy said:
10 years ago
I did not understand this sum please help give me a simple way to solve this sum.
Eshwar said:
10 years ago
CI = A-P.
A = P(1+r/100)n.
CI = P(1+r/100)n-P.
Where in above problem P = x.
A = P(1+r/100)n.
CI = P(1+r/100)n-P.
Where in above problem P = x.
Balaji said:
2 years ago
CI - SI = P(R/100)^2,
1 = P(4/100)^2,
1 = P(1/25)^2,
1 = P /625.
625 = P.
1 = P(4/100)^2,
1 = P(1/25)^2,
1 = P /625.
625 = P.
(40)
Pratik patel said:
1 decade ago
But. It is for only two years. I think. Not for 4;5;6 years. Try it. @Chetan.
Santhosh P said:
10 years ago
Difference for 2 years = P*(R/100)^2.
=> P*(4/100)^2 = 1.
=> P = 625.
=> P*(4/100)^2 = 1.
=> P = 625.
Amit said:
9 years ago
Principal for two years = pr2/100*100.
Three years = pr2(300+r)/100*100*100.
Three years = pr2(300+r)/100*100*100.
Revant Borawat said:
9 years ago
Assume principal 100.
Then, CI=8.16 & SI=8.00,
1÷ 0.16 * 100 = 625.
Then, CI=8.16 & SI=8.00,
1÷ 0.16 * 100 = 625.
Hema said:
1 decade ago
Shortcut formula P = (100/R)^T*D, D = Difference here.
= (100/4)^2*1 = 625.
= (100/4)^2*1 = 625.
Manjunath Reddy said:
9 months ago
@All.
Principle = difference(100/r)^2.
=> p = 1(100/4)^2 = (25)^2 = 625.
Principle = difference(100/r)^2.
=> p = 1(100/4)^2 = (25)^2 = 625.
(11)
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