Aptitude - Compound Interest - Discussion
Discussion Forum : Compound Interest - General Questions (Q.No. 3)
3.
There is 60% increase in an amount in 6 years at simple interest. What will be the compound interest of Rs. 12,000 after 3 years at the same rate?
Answer: Option
Explanation:
Let P = Rs. 100. Then, S.I. Rs. 60 and T = 6 years.
 R = |
 |
100 x 60 |  |
= 10% p.a. |
| 100 x 6 |
Now, P = Rs. 12000. T = 3 years and R = 10% p.a.
| C.I. |
|
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|
||||||||||||||
| = 3972. |
Discussion:
94 comments Page 5 of 10.
Sanjay said:
8 years ago
60% is equal to 60/100 i.e., S.I means adding the same rate of interest every year, So here we knew that 60% is increased in 6 years, i.e., Every year it is increased by 10%.
Then, caculate net % for 3yrs.
i.e.
Calculte 1st (10%)and 2nd yr(10%).then result with 3rd yr(10%).
i.e 10+10+(10 * 10)/100 = 21.
Next 21+10+(21 * 10)/100 = 33.1%.
So now just calculate 33.1%12000.
i.e 3972.
Then, caculate net % for 3yrs.
i.e.
Calculte 1st (10%)and 2nd yr(10%).then result with 3rd yr(10%).
i.e 10+10+(10 * 10)/100 = 21.
Next 21+10+(21 * 10)/100 = 33.1%.
So now just calculate 33.1%12000.
i.e 3972.
Sammy said:
9 years ago
Thank you so much for explaining the solution.
Victor said:
9 years ago
Please help me to solve this question.
Formula:
Fp=Pp(1+r) raise to power n.
if Pp=100, r=3%,n = 10.
Formula:
Fp=Pp(1+r) raise to power n.
if Pp=100, r=3%,n = 10.
Malinimadhavan said:
9 years ago
@Angad.
Formula to find the rate of interest from ci is r= n[(A/P)^1/nt -1].
Now take p=12000; amount (A) = C.I + PRINCIPLE (3972+12000) = 15972; time(t) = 3years n=1.
By substituting this values in the above equation.
r = 1[(15972/12000)^1/3 -1],
= [(1.331)^1/3 -1],
r = [1.1-1],
r = 0.1.
To calculate rate of interest in percentage we use R = r*100.
Therefore R = 0.1*100 = 10%.
Thus rate of interest is 10% p.a.
I hope it will help you.
Formula to find the rate of interest from ci is r= n[(A/P)^1/nt -1].
Now take p=12000; amount (A) = C.I + PRINCIPLE (3972+12000) = 15972; time(t) = 3years n=1.
By substituting this values in the above equation.
r = 1[(15972/12000)^1/3 -1],
= [(1.331)^1/3 -1],
r = [1.1-1],
r = 0.1.
To calculate rate of interest in percentage we use R = r*100.
Therefore R = 0.1*100 = 10%.
Thus rate of interest is 10% p.a.
I hope it will help you.
Angad said:
9 years ago
How to find the rate of interest if compound interest and principle and time is already given but the rate is missing? I need the solution.
Renu said:
9 years ago
10% successive of 3 years is 33.1%
33.1% of 12000.
= 3972.
33.1% of 12000.
= 3972.
BLESSING said:
9 years ago
The formula for compound interest:.
A = P (1 + r) ^n.
Now, P = 12000.
A = ?
N = 3.
From the first statement, there was 60% increase in an amount for 6 years simple interest.
Let the amount (P) be 100 then SI=60, SI = PIN.
60 = 100 * I * 6,
60 = 600 I,
I = 0.1.
Since we have 0.1 x 100 = 10%.
Now A = 12000 (1 + 1/10)^3
= 12000 (11/10)^3,
= 12000 (11/10 x 11/10 x 11/10),
= 12000 (1331/1000) = 15,972.
So, A = 15972, P = 12000 then,
CI = 15972 - 12000 = 3972.
A = P (1 + r) ^n.
Now, P = 12000.
A = ?
N = 3.
From the first statement, there was 60% increase in an amount for 6 years simple interest.
Let the amount (P) be 100 then SI=60, SI = PIN.
60 = 100 * I * 6,
60 = 600 I,
I = 0.1.
Since we have 0.1 x 100 = 10%.
Now A = 12000 (1 + 1/10)^3
= 12000 (11/10)^3,
= 12000 (11/10 x 11/10 x 11/10),
= 12000 (1331/1000) = 15,972.
So, A = 15972, P = 12000 then,
CI = 15972 - 12000 = 3972.
(1)
Arvindsamy said:
9 years ago
Guys your discussions and solutions are very useful. Thank you all guys.
Malleswar said:
9 years ago
Beautifully explained. Thanks to all.
Prabhakaran said:
9 years ago
Hi, can anyone explain to me why CI = Amount - P if its P we can substitute the actual principal.
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