Aptitude - Compound Interest - Discussion
Discussion Forum : Compound Interest - General Questions (Q.No. 8)
8.
The least number of complete years in which a sum of money put out at 20% compound interest will be more than doubled is:
Answer: Option
Explanation:
| P |  |
1 + | 20 |  |
n | > 2P |  |
|
6 | |
n | > 2. |
| 100 | 5 |
| Now, | |
6 | x | 6 | x | 6 | x | 6 | |
> 2. |
| 5 | 5 | 5 | 5 |
So, n = 4 years.
Discussion:
27 comments Page 2 of 3.
Shubham said:
9 years ago
What is the least number of complete years in which sum will become more than double itself at 12% per annum?
Given me the answer with clear explanation.
Given me the answer with clear explanation.
(1)
Dilip said:
9 years ago
@Kalyani
Suppose double principle then 2p.
Triple principle then 3p.
I hope you can understand.
Suppose double principle then 2p.
Triple principle then 3p.
I hope you can understand.
Dhairya adhikari said:
9 years ago
Easy method:
n = 72/r,
n = 72/20 = 3.6 years,
n = 4 years approximately.
n = 72/r,
n = 72/20 = 3.6 years,
n = 4 years approximately.
Ashok said:
10 years ago
@Kalyani.
In question, they asked how much of year will give the principal amount doubled so the only 2p is used.
In question, they asked how much of year will give the principal amount doubled so the only 2p is used.
Kalyani said:
10 years ago
Why are we using 2p?
Pankaj said:
10 years ago
Very simple.
Suppose amount is Rs. 100.
Rate of interest is 12% annual.
Interest after 1st year-100*1.20 = 120.
Interest after 2nd year-120*1.20 = 144.
Interest after 3rd year-144*1.20 = 172.8.
Interest after 4th year-172.8*1.20 = 207.36.
So means to say it will take 4 years to double money.
Suppose amount is Rs. 100.
Rate of interest is 12% annual.
Interest after 1st year-100*1.20 = 120.
Interest after 2nd year-120*1.20 = 144.
Interest after 3rd year-144*1.20 = 172.8.
Interest after 4th year-172.8*1.20 = 207.36.
So means to say it will take 4 years to double money.
(7)
Ritesh said:
1 decade ago
Hello @Anjali,
First take Amount = Rs. 5000 and Rate = 2%, Years = 1.
Then calculate SI, SI = PNR/100.
So, SI = 5000*1*2/100 = RS. 100.
[Here you are getting only interest].
Now calculate CI, CI = P(1+r)^n.
So, CI = 5000*(1+2/100)^1 = RS. 5100.
[Here we are getting Interest+Principal amount].
First take Amount = Rs. 5000 and Rate = 2%, Years = 1.
Then calculate SI, SI = PNR/100.
So, SI = 5000*1*2/100 = RS. 100.
[Here you are getting only interest].
Now calculate CI, CI = P(1+r)^n.
So, CI = 5000*(1+2/100)^1 = RS. 5100.
[Here we are getting Interest+Principal amount].
Anjali said:
1 decade ago
I am unable to understand why are taking amount formula in place of CI. Please explain it.
Swathi said:
1 decade ago
Hi,
We are using > 2. Can you explain it?
We are using > 2. Can you explain it?
Narayana Rao Gartam said:
1 decade ago
Dear Jahnavi,
Amount = P(1+R/100)^n.
Sum of money is more than the double of Amount.
So, 2P < P(1+20/100)^n , Since R= 20%
P and P cancelled each other. So,
2< (1+20/100)^n.
2< (120/100)^n.
2<(6/5)^n.
Try n=1, 2, 3, 4, 5.
What least number can satisfy the 2<(6/5)^n.
Lets n=1 ( 2<1.2 so, wrong).
Lets n=2 ( 2<1.44 so, wrong).
Lets n=3 ( 2<1.728 so, wrong).
Lets n=4 ( 2<2.1036 so, right).
Answer is 4 years.
Amount = P(1+R/100)^n.
Sum of money is more than the double of Amount.
So, 2P < P(1+20/100)^n , Since R= 20%
P and P cancelled each other. So,
2< (1+20/100)^n.
2< (120/100)^n.
2<(6/5)^n.
Try n=1, 2, 3, 4, 5.
What least number can satisfy the 2<(6/5)^n.
Lets n=1 ( 2<1.2 so, wrong).
Lets n=2 ( 2<1.44 so, wrong).
Lets n=3 ( 2<1.728 so, wrong).
Lets n=4 ( 2<2.1036 so, right).
Answer is 4 years.
(1)
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