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 3 of 10.
Harshit said:
4 years ago
How to solve (1 + 10/100) ^3?
Please explain it.
Please explain it.
(2)
Ganesh said:
7 years ago
@All.
1st find rate R of 6 yr Assume 100rs, for 6 yr.
R=(100 * s.i/p * n) = 10%.
After that find C.I using ratio proposition of 3 yr is 3:3:1 is fixed and 2yr is (2:1).
Then 1st yr c.i. is 3* 10% of 12000,
2nd yr C.I is 3* 10% of 1200,
3rd yr C.I is 1* 10% of 120 add all the yrs.
3600 + 360 + 12 = 3972 is the answer.
1st find rate R of 6 yr Assume 100rs, for 6 yr.
R=(100 * s.i/p * n) = 10%.
After that find C.I using ratio proposition of 3 yr is 3:3:1 is fixed and 2yr is (2:1).
Then 1st yr c.i. is 3* 10% of 12000,
2nd yr C.I is 3* 10% of 1200,
3rd yr C.I is 1* 10% of 120 add all the yrs.
3600 + 360 + 12 = 3972 is the answer.
(1)
Mohammad Gazali said:
8 years ago
Please tell me how R=100*60/100*6?
(1)
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)
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)
Rahul krishna said:
6 years ago
Let p=100 and si=60(as amount increase to 60%).
Si=(p*r*t)/100.
Find r=?.
Then, with the help of r find ci.
Si=(p*r*t)/100.
Find r=?.
Then, with the help of r find ci.
Deepthi said:
1 decade ago
How we get S.I=Rs.60 while P=Rs.100 ?
Vicky said:
1 decade ago
Formula for C.I,
Amount = p(1+R/100)^N, Where amount = Principal + C.I.
Principal + C.I = p(1+R/100)^N.
C.I = p(1+R/100)^N - Principal, ' p ' is a common factor.
C.I = p [(1+R/100)^N-1].
Amount = p(1+R/100)^N, Where amount = Principal + C.I.
Principal + C.I = p(1+R/100)^N.
C.I = p(1+R/100)^N - Principal, ' p ' is a common factor.
C.I = p [(1+R/100)^N-1].
Ramya c said:
5 years ago
Anyone can explain me, why we are calculating for 3 years?
Ronald said:
5 years ago
12000 in 10% is 1200 then,
12000+1200=13200 is 1st year.
And 13200 in 10% so 13200+3200= 16400.
Same next year we get 16400+6400=18040.
Then, 18040-12000=6040.
Why not like this? Anyone explain to me.
12000+1200=13200 is 1st year.
And 13200 in 10% so 13200+3200= 16400.
Same next year we get 16400+6400=18040.
Then, 18040-12000=6040.
Why not like this? Anyone explain to me.
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