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:
26 comments Page 1 of 3.
Harini said:
2 years ago
Here they said that the money they put out in compound interest is greater than double of the money.
Let the money they put out in CI at a 20% rate of interest be P.
P(1+(20/100)^n > 2P,
(1+(20/100))^n > P,
(1.2)^n > 2.
taking log on both sides;
log(1.2)^n >log 2.
n> log 2/log 1.2.
n>3.8.
n = 4 yrs.
Let the money they put out in CI at a 20% rate of interest be P.
P(1+(20/100)^n > 2P,
(1+(20/100))^n > P,
(1.2)^n > 2.
taking log on both sides;
log(1.2)^n >log 2.
n> log 2/log 1.2.
n>3.8.
n = 4 yrs.
(16)
Psit Kanpur said:
5 years ago
Can anyone explain about (6/5 * 6/5 * 6/5 * 6/5)>2?
(12)
Kapil dev said:
8 years ago
Let principal be P.
P(1+R/100)^T>2P
P(1+20/100)^T>2P
(1+20/100)^T>2
(120/100)^T>2
1.2^T>2.
Now let's find out the minimum value of T for which the above equation becomes true.
If T = 1, 1.2T = 1.21 = 1.2
If T = 2, 1.2T = 1.22 * 1.4
If T = 3, 1.2T = 1.23 * 1.7
If T = 4, 1.2T = 1.24 * 2.07 which is greater than 2
Hence T = 4
i.e., the required number of years = 4.
P(1+R/100)^T>2P
P(1+20/100)^T>2P
(1+20/100)^T>2
(120/100)^T>2
1.2^T>2.
Now let's find out the minimum value of T for which the above equation becomes true.
If T = 1, 1.2T = 1.21 = 1.2
If T = 2, 1.2T = 1.22 * 1.4
If T = 3, 1.2T = 1.23 * 1.7
If T = 4, 1.2T = 1.24 * 2.07 which is greater than 2
Hence T = 4
i.e., the required number of years = 4.
(8)
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)
Tanish said:
3 years ago
@All.
When you divide 72 by ROI you will get an approximate no of years in which the principal will get double.
When you divide 72 by ROI you will get an approximate no of years in which the principal will get double.
(6)
Sanjay Rajnedra Awate said:
7 years ago
For finding the time period in which a sum of money will double itself at R% rate of compound interest compounded annually, we generally use either of the following two formulas:
Time, T = 72 /R Years.
Time, T = 72 /R Years.
(4)
Rubi said:
7 years ago
Why we use amount formula instead of CISCO formula?
(4)
Muhammad Zunnoorain said:
9 years ago
We can also solve it as;
2P = P(1.2)^n.
2 = 1.2^n.
Applying lag on both side.
Log(2) = n*log(1.2).
n = log(2)/log(1.2).
n = 3.8 to nearest year 4.
2P = P(1.2)^n.
2 = 1.2^n.
Applying lag on both side.
Log(2) = n*log(1.2).
n = log(2)/log(1.2).
n = 3.8 to nearest year 4.
(2)
Kanishk said:
5 years ago
Instead of trial and error we can use logarithm.
(1)
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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