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.
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)
VAMSI said:
5 months ago
@All.
There is a direct formula to solve the difference between CI and SI is;
Difference = Principal * Rate^2 /100^2,
1 = P*16/10000,
10000/16 = P,
P = 625.
There is a direct formula to solve the difference between CI and SI is;
Difference = Principal * Rate^2 /100^2,
1 = P*16/10000,
10000/16 = P,
P = 625.
(9)
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)
Kurdush said:
4 years ago
Shortcut to solve :
(100*100*Diff)/Rate * Rate.
here diff =1.
rate=4.
Substitute in above formula u get Rs.625.
(100*100*Diff)/Rate * Rate.
here diff =1.
rate=4.
Substitute in above formula u get Rs.625.
(5)
Joa said:
2 years ago
Thanks everyone for explaining the answer.
(5)
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)
Anondo said:
9 months ago
Profit of two is the same for C and S but it is a different profit of profit last year.
So, let first-year profit x then
4%of x = 1,
x= 25.
If.
Principle y.
Then 4%of y = 25,
y= 625.
So, let first-year profit x then
4%of x = 1,
x= 25.
If.
Principle y.
Then 4%of y = 25,
y= 625.
(4)
Sumit said:
5 years ago
Diff = p(R/100) ^n.
By using this formula you can directly find the amount i.e. p.
1=p(4/100) ^2.
=> p=625.
By using this formula you can directly find the amount i.e. p.
1=p(4/100) ^2.
=> p=625.
(3)
Mohit said:
5 years ago
How 51/625x came? Please explain.
(3)
Hema said:
2 years ago
How does 51P come in CI? Please explain to me.
(3)
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