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 13 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.
Md.iftekhar said:
8 years ago
Hi, friends we can easily solve this problem with the help of this formula.
Formula :- Sum= Difference(100/r)^2.
Now put the value----- Sum= 1(100/4) ^2
= 1* 625/1
= 625 Ans.
Formula :- Sum= Difference(100/r)^2.
Now put the value----- Sum= 1(100/4) ^2
= 1* 625/1
= 625 Ans.
Baadshah said:
8 years ago
Shortcut method:
When the difference between SI and CI is x and rate of interest is R then,
x(100/R)^n where n is the number of years,
So 1(100/4)^2 = (25)^2 = 625.
When the difference between SI and CI is x and rate of interest is R then,
x(100/R)^n where n is the number of years,
So 1(100/4)^2 = (25)^2 = 625.
Ahana said:
8 years ago
Is there any formula for diff bw ci and si for 4 years ?
Vansh agre said:
8 years ago
Difference = P * R *R /100*100.
Shanti mishra said:
8 years ago
1/.16*100=625.
Arun said:
8 years ago
8.16-8=1;
0.16=1,
100=x,
x = 625.
0.16=1,
100=x,
x = 625.
Akshay said:
8 years ago
You can simple try this formula.
P=(100/R)^t *( C.I - S.I ).
P=(100/R)^t *( C.I - S.I ).
Jojo said:
8 years ago
For this use equation.
P=[(difference)*(100)^2]/(rate)^2.
P=[(difference)*(100)^2]/(rate)^2.
Puneet said:
7 years ago
Anyone can please tell why we have to subtract x from C.I?
and 51x/625-2x/25=1 so x=625 how?
and 51x/625-2x/25=1 so x=625 how?
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