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.
Anonymous said:
1 decade ago
In that why -15000 is done in compound interest. As per formula its not there na?
Priyanka P. said:
1 decade ago
@Anonymous
for C.I. formula is=>
p(1-r/100)^n-p
without subtracting, which amount we get is
"Principle amount+Interest"
so to get C.I. we have to subtract it..
for C.I. formula is=>
p(1-r/100)^n-p
without subtracting, which amount we get is
"Principle amount+Interest"
so to get C.I. we have to subtract it..
Swap said:
1 decade ago
Please tell something more about compound interest.
Ravin said:
1 decade ago
Kindly explain in 3rd step howcum -1 would be showing...is this problem solve throuGh the LCM?
NARENDRA said:
1 decade ago
SIMPLE FORMULA FOR TWO YEARS
C.I.-S.I.=P(R/100)*(R/100)
96=15000(R/100)(R/100)
R*R=960/15=64
R=8
C.I.-S.I.=P(R/100)*(R/100)
96=15000(R/100)(R/100)
R*R=960/15=64
R=8
PRAWIN said:
1 decade ago
shortcut:
p = d (100/r)^2
15000 = 96*100*100/r^2
r^2 = 96*100*100/15000
r = 8.
p = d (100/r)^2
15000 = 96*100*100/r^2
r^2 = 96*100*100/15000
r = 8.
Vimal said:
1 decade ago
Thanks prawin.
Mudasir said:
1 decade ago
It's simply p(r/100)^2
p=1500,r=x,t=2yrs
Thus
15000(x/100)^2 it finally comes out like
x^2=64
x^2=8^2
x=8
Therefore rate=8%
p=1500,r=x,t=2yrs
Thus
15000(x/100)^2 it finally comes out like
x^2=64
x^2=8^2
x=8
Therefore rate=8%
Lucky said:
1 decade ago
In 4th step how they done R^2=64 ?
If we solve it there should be R^2-200R=64 and this is a quadratic equation. Can anyone explain me how to solve this equation ?
If we solve it there should be R^2-200R=64 and this is a quadratic equation. Can anyone explain me how to solve this equation ?
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)
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