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 2 of 15.
Kumaresh Panjabrao Ghate said:
2 years ago
Suppose that we take the sum 100.
2 years simple interest of 100 is 108.
And 2 years compound interest of 100 is;
108.16.
The difference between two interests is 108.16-108 = 0.16;
Now we can say that;
0.16 difference than 100 sum.
1 difference then what will be the sum?
Hence
0.16 : 100 :: 1 : ?
100×1
--------
0.16
= 10000/16.
= 625.
2 years simple interest of 100 is 108.
And 2 years compound interest of 100 is;
108.16.
The difference between two interests is 108.16-108 = 0.16;
Now we can say that;
0.16 difference than 100 sum.
1 difference then what will be the sum?
Hence
0.16 : 100 :: 1 : ?
100×1
--------
0.16
= 10000/16.
= 625.
(22)
Hema said:
2 years ago
How does 51P come in CI? Please explain to me.
(3)
Nitin Tiwari said:
2 years ago
C.I=A-P ------> (1).
So the amount is, A = P(1+R/100)^n.
Put the value of A in equation (1).
Where n = year = 2, r = rate = 4%,
C.I = P(1+4/100)^2-P,
= P(1+1/25)^2-P,
= P(26/25)^2- P.
Taking 'P' common.
= P[(26/25)^2-1].
= P[((26*26)-(25*25))/(25*25)].
= P[(676-625)/625],
= P[51/625].
So the amount is, A = P(1+R/100)^n.
Put the value of A in equation (1).
Where n = year = 2, r = rate = 4%,
C.I = P(1+4/100)^2-P,
= P(1+1/25)^2-P,
= P(26/25)^2- P.
Taking 'P' common.
= P[(26/25)^2-1].
= P[((26*26)-(25*25))/(25*25)].
= P[(676-625)/625],
= P[51/625].
(5)
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)
SICI said:
2 years ago
In C. I = where -x is come from? Please explain me.
(24)
Joa said:
2 years ago
Thanks everyone for explaining the answer.
(5)
Saudagar Rokade said:
2 years ago
Diff = PR^²/10000.
1 = P×4^²/10000,
10000 = P×16,
P = 10000/16,
P = 625.
1 = P×4^²/10000,
10000 = P×16,
P = 10000/16,
P = 625.
(9)
Vanhel said:
2 years ago
SI = 8%.
CI = 8.16%.
CI - SI = 1.
8.16% - 8% = 1.
0.16% = 1.
Sum of the difference principal = 1/0.16 * 100 = 625.
CI = 8.16%.
CI - SI = 1.
8.16% - 8% = 1.
0.16% = 1.
Sum of the difference principal = 1/0.16 * 100 = 625.
(15)
P manoj said:
3 years ago
It is simply derived by using this formula => Diff =PR^2/100^2 -> It is only for 2 years difference.
Given -> Diff=1 , R =4, Tp find P?
By Formula ,
Diff = PR^2/100^2.
1 = P * 4 * 4/100 * 100.
1= P * 1/20 * 1/20.
Note : (Also for three years difference use this formula -> Diff = PR^2(300 + R)/100^3).
Given -> Diff=1 , R =4, Tp find P?
By Formula ,
Diff = PR^2/100^2.
1 = P * 4 * 4/100 * 100.
1= P * 1/20 * 1/20.
Note : (Also for three years difference use this formula -> Diff = PR^2(300 + R)/100^3).
(8)
Naveen said:
4 years ago
1=10000*x/4 * 4.
= 625.
= 625.
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