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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|
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| = 3972. |
Discussion:
94 comments Page 1 of 10.
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)
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.
Gaichalung kamei said:
2 years ago
160% increase..
160/100 = 8/5.
Let's assume 5 is the amount.
60% increase of 5 gives 8. There is a difference of 3 between 5 and 8. The given period is 6 yrs.
So, 6/3 = 2.
Now we can calculate the rate.
P × R × T/100 = 5 × R × 2/100 which gives 10%.
Now compound interest for 3 yrs with the same rate.
(11/10)³= 1331/1000.
The given principle is 12000. We get 12 as a multiplying factor.
now, CI = 331×12 = 3972.
160/100 = 8/5.
Let's assume 5 is the amount.
60% increase of 5 gives 8. There is a difference of 3 between 5 and 8. The given period is 6 yrs.
So, 6/3 = 2.
Now we can calculate the rate.
P × R × T/100 = 5 × R × 2/100 which gives 10%.
Now compound interest for 3 yrs with the same rate.
(11/10)³= 1331/1000.
The given principle is 12000. We get 12 as a multiplying factor.
now, CI = 331×12 = 3972.
(6)
Dearlie Beloved said:
10 years ago
'I' means increase, since I=F-P. Taking 60 as SI while letting P=100 in i=I/Pn assumes that the future amount will be 60 greater than 100.
We can let P as any number, say P=210, but we need to change I=150 to signal 60 in increase of future amount. We only put P=100 to give justice to the percent sign after 60.
And compound interest is what was asked and not CI rate? Hope this helped.
We can let P as any number, say P=210, but we need to change I=150 to signal 60 in increase of future amount. We only put P=100 to give justice to the percent sign after 60.
And compound interest is what was asked and not CI rate? Hope this helped.
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.
Vattikuti shivaram said:
2 years ago
We can solve in a simple way also if we have 12000 as the principal amount then we can.
Divide with a 10% rate
By dividing 12000. With we get 1200 in three times because of three years of interest as SI;
And now doing 1200 we get 3 times 120 for three years CI
We know that ci is recursive with year one 120 divided by 1/10
Then we get 12;
Now 3*1200+3*120+12 = 3972.
Divide with a 10% rate
By dividing 12000. With we get 1200 in three times because of three years of interest as SI;
And now doing 1200 we get 3 times 120 for three years CI
We know that ci is recursive with year one 120 divided by 1/10
Then we get 12;
Now 3*1200+3*120+12 = 3972.
(14)
Anmol said:
1 decade ago
@Karthi & Vasanth: you can take any value of P instead of 100.
100 is taken just for simplicity of calculation.
Now, as there is 60% rise in principal in 6 years
therefore, Net amount after 6 years, A= P+ 60% of P =1.6P
therefore interest,I= 1.6P-P= .6P
using SI formula, I= PTR/100
so, 0.6P= P*6*R/100
-> 60=6R
-> R=10%
100 is taken just for simplicity of calculation.
Now, as there is 60% rise in principal in 6 years
therefore, Net amount after 6 years, A= P+ 60% of P =1.6P
therefore interest,I= 1.6P-P= .6P
using SI formula, I= PTR/100
so, 0.6P= P*6*R/100
-> 60=6R
-> R=10%
Bhushan said:
7 years ago
SI for 6 yrs increase by 60%.
Means;
For 1st yr 10%.
1st yr , 10% of 12000 = 1200 - - - >(1)
Add 1200 ... ie. = 12000+1200= 13200.
For 2nd yr, 10% of 13200 = 1320 - - - > (2)
So, 13200+1320= 14520.
For 3rd yr, 10% of 14520 = 1452. - - - > (3)
So add the interest amount of all 3 equations ie....
1200 + 1320 + 1452 = 3972 is the right answer.
Means;
For 1st yr 10%.
1st yr , 10% of 12000 = 1200 - - - >(1)
Add 1200 ... ie. = 12000+1200= 13200.
For 2nd yr, 10% of 13200 = 1320 - - - > (2)
So, 13200+1320= 14520.
For 3rd yr, 10% of 14520 = 1452. - - - > (3)
So add the interest amount of all 3 equations ie....
1200 + 1320 + 1452 = 3972 is the right answer.
(2)
Subash said:
2 decades ago
Please explain the use of -1 in C.I calculation. I think here we have to subtract the principle at the back. But u use the value '1' here, I can't understand Wat u r did.
But i got the same answer by using the formula.
C.I = (Amount - Principle)
Where
Amount = {P x (1 + (R/100)^N)} - P
P = 12000.
Therefore, C.I = Rs. (12000*(1+(10/100)^3))-12000 = 3972
But i got the same answer by using the formula.
C.I = (Amount - Principle)
Where
Amount = {P x (1 + (R/100)^N)} - P
P = 12000.
Therefore, C.I = Rs. (12000*(1+(10/100)^3))-12000 = 3972
(1)
AKASH KADAM said:
2 weeks ago
For simple interest, 60% for the 6 years.
So, for one year it is - 10 % p.a.
Now the principle amount is- 12000.
We have to calculate the compound interest for 3 years.
On the 10% per annum rate.
So for the first year it's - 1200.
For the second year it's - 1320.
For the third year, it's - 1452.
Now add all of them - 1200 + 1320 + 1452.
= 3972.
So, for one year it is - 10 % p.a.
Now the principle amount is- 12000.
We have to calculate the compound interest for 3 years.
On the 10% per annum rate.
So for the first year it's - 1200.
For the second year it's - 1320.
For the third year, it's - 1452.
Now add all of them - 1200 + 1320 + 1452.
= 3972.
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