Aptitude - Compound Interest - Discussion
Discussion Forum : Compound Interest - General Questions (Q.No. 14)
14.
The difference between compound interest and simple interest on an amount of Rs. 15,000 for 2 years is Rs. 96. What is the rate of interest per annum?
Answer: Option
Explanation:
![]() |
15000 x | ![]() |
1 + | R | ![]() |
2 | - 15000 | ![]() |
- | ![]() |
15000 x R x 2 | ![]() |
= 96 | |
100 | 100 |
![]() |
![]() |
![]() |
1 + | R | ![]() |
2 | - 1 - | 2R | ![]() |
= 96 |
100 | 100 |
![]() |
![]() |
(100 + R)2 - 10000 - (200 x R) | ![]() |
= 96 |
10000 |
![]() |
![]() |
96 x 2 | ![]() |
= 64 |
3 |
R = 8.
Rate = 8%.
Discussion:
32 comments Page 1 of 4.
Naveen said:
4 years ago
CI-SI = P(R/100)^2.
CI-SI = 96.
P = 15000.
96 = 15000(R/100)^2,
96/15000 = R^2/100^2,
0.0064 = R^2/10000,
64 = R^2,
R = 8.
CI-SI = 96.
P = 15000.
96 = 15000(R/100)^2,
96/15000 = R^2/100^2,
0.0064 = R^2/10000,
64 = R^2,
R = 8.
(7)
Ash said:
4 years ago
Hi guys,
I will tell you the simplest method for the difference between Si and Ci sums.
Use the formula:
Difference = P*(R^2)/100^2.
I will tell you the simplest method for the difference between Si and Ci sums.
Use the formula:
Difference = P*(R^2)/100^2.
(5)
Diwagar said:
2 years ago
Difference between S.I and C. I is;
D = P(R/100) ^2,
96 =15000(R/100) ^2,
(96 * 2)/3 =R^2,
64 = R^2,
Ans. R = 8.
D = P(R/100) ^2,
96 =15000(R/100) ^2,
(96 * 2)/3 =R^2,
64 = R^2,
Ans. R = 8.
(4)
Ratan said:
8 years ago
Hi All ,
There is a shortcut method for this -
C.I. - S.I. = P*R^2/(100)^2.
so if we will put the values here it will be like this;
96 = 15000*R*R/10000,
64 = R*R,
So R =8.
There is a shortcut method for this -
C.I. - S.I. = P*R^2/(100)^2.
so if we will put the values here it will be like this;
96 = 15000*R*R/10000,
64 = R*R,
So R =8.
(3)
Wrick said:
7 years ago
Simply, the solution is;
15000 * [{1+(r/100)}^2-1]-(15000 * 2 * r)/100 = 96,
=> 15000[{1+(r/100)}^2-1-(2 * r)/100]= 96,
=> 15000[1+(r^2)/10000-1] = 96 where [(a+b)^2-2ab = a^2 + b^2],
=> 15000[r^2/10000] = 96,
=> r^2 = 64,
=> r = 8.
15000 * [{1+(r/100)}^2-1]-(15000 * 2 * r)/100 = 96,
=> 15000[{1+(r/100)}^2-1-(2 * r)/100]= 96,
=> 15000[1+(r^2)/10000-1] = 96 where [(a+b)^2-2ab = a^2 + b^2],
=> 15000[r^2/10000] = 96,
=> r^2 = 64,
=> r = 8.
(2)
Tamalika Roy said:
5 years ago
@All.
The solution is;
Difference = sum(R/100)^n => formula.
Here;
96 = 15000(r/100)^2 = 8.
The solution is;
Difference = sum(R/100)^n => formula.
Here;
96 = 15000(r/100)^2 = 8.
(1)
Himani said:
6 years ago
Simply:
sum = difference * (100/R)^2 for 2 year.
sum = {difference * (100^3)}/{R^2*(300+R)} for 3 year.
sum = difference * (100/R)^2 for 2 year.
sum = {difference * (100^3)}/{R^2*(300+R)} for 3 year.
(1)
Muthu said:
1 decade ago
C.P - S.I
[p(1-r/100)^n-p ]-[(P*R*N)/100] = 96.
Where,
p=1500, n=2, R=?, C.P and S.I difference is 96.
[15000(1-r/100)^2-15000 ]-[(1500*R*2)/100] = 96 ...(1).
Now, 15000 common in above so take it out,
15000[1(1-r/100)^2 - 1 ]-[(1*R*2)/100] = 96 ...(2).
Take L.C.M. we get,
15000[(100 + R)^2 - 10000 - (200 x R)/10000] = 96 ...(3).
Where (100 + R)^2 is like (a + b)^2 = (a^2 + b^2 + 2*a*b).
(100^2 + R^2 + 2*100*R) = (10000 + R^2 + 200R).
Sub (10000 + R^2 + 200R) in eqn(3) we get,
15000[ (10000 + R^2 + 200R) - 10000 - (200 x R)/10000] = 96.
15000[ (10000 + R^2 + 200R - 10000 - 200R) /10000] = 96.
Now 10000 and 200R is cancelled, remaining.
15000[ R^2 /10000] = 96.
[15000R^2/10000] = 96.
[15R^2/10] = 96.
divide by 5 we get,
[3R^2/2] = 96.
Now keep the R^2 in left hand side and move 3/2 to right side
R^2 = 96(2/3) // 96/3 = 32.
R^2 = 32*2.
R^2 = 64 // square root of 64 is 8.
R = 8%.
[p(1-r/100)^n-p ]-[(P*R*N)/100] = 96.
Where,
p=1500, n=2, R=?, C.P and S.I difference is 96.
[15000(1-r/100)^2-15000 ]-[(1500*R*2)/100] = 96 ...(1).
Now, 15000 common in above so take it out,
15000[1(1-r/100)^2 - 1 ]-[(1*R*2)/100] = 96 ...(2).
Take L.C.M. we get,
15000[(100 + R)^2 - 10000 - (200 x R)/10000] = 96 ...(3).
Where (100 + R)^2 is like (a + b)^2 = (a^2 + b^2 + 2*a*b).
(100^2 + R^2 + 2*100*R) = (10000 + R^2 + 200R).
Sub (10000 + R^2 + 200R) in eqn(3) we get,
15000[ (10000 + R^2 + 200R) - 10000 - (200 x R)/10000] = 96.
15000[ (10000 + R^2 + 200R - 10000 - 200R) /10000] = 96.
Now 10000 and 200R is cancelled, remaining.
15000[ R^2 /10000] = 96.
[15000R^2/10000] = 96.
[15R^2/10] = 96.
divide by 5 we get,
[3R^2/2] = 96.
Now keep the R^2 in left hand side and move 3/2 to right side
R^2 = 96(2/3) // 96/3 = 32.
R^2 = 32*2.
R^2 = 64 // square root of 64 is 8.
R = 8%.
(1)
Khyati mehta said:
4 years ago
We can directly do this by using the formula.
Difference between SI and CI for 2 years = principle *(Rate of Interest)^2
= P(R%)^2.
So here the difference is 96, principle=15000.
96 = 15000*(R/100)^2.
r = 8%.
Difference between SI and CI for 2 years = principle *(Rate of Interest)^2
= P(R%)^2.
So here the difference is 96, principle=15000.
96 = 15000*(R/100)^2.
r = 8%.
(1)
Shine said:
8 years ago
C.I - S.I (FOR 2 YEARS)= P (R/100)^2.
HENCE,
96 = 15000(R/100)^2.
Solve for R.
HENCE,
96 = 15000(R/100)^2.
Solve for R.
(1)
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