The Compound Interest on an investment in 6 years is $400

Question

The Compound Interest on an investment in 6 years is $400 and Compound Interest on the same investment in 12 years is $ 960. Find the initial investment
A) 1250
B) 1200
C) 1000
D) 1500
E) 1100

chetan2u · Accepted Answer

Another way..

What is compound interest  ---  P(1+\frac{r}{100})^t-P=I

The Compound Interest on an investment in 6 years is $400 ----- --
  P(1+\frac{r}{100})^6-P=400........ ..
  P(1+\frac{r}{100})^6=P+400 ......
 (1+\frac{r}{100})^6=\frac{P+400}{P} ...(i)

Similarly, Compound Interest on the same investment in 12 years is $ 960 --------
  P(1+\frac{r}{100})^{12}-P=400........ ..
  P((1+\frac{r}{100})^6)^2=P+960 ...Substitute (i) in it.
 P(\frac{400+P}{P})^2=960+P.........(400+P)^2=P(960+P)...P=1000

C

nick1816 · Answer

We're getting 960-800=$160 interest on 400$ in 6 years

560=400(1+r/100)^6

(1+r/100)^6 =  \frac{560}{400}= 1.4

Let initial investment= P

P+400= P(1+r/100)^6

P+400= 1.4P

0.4P=400
P=1000

rajatchopra1994 · Answer

We're getting 960-800 =$160 interest on 400$ in 6 years

$560=400(1+r/100)^6$

$(1+r/100)^6$= $\frac{560}{400}= 1.4$

Let initial investment= P

$P+400= P(1+r/100)^6$

P+400= 1.4P

0.4P=400
P=1000
------------------------------------
Dear nick1816 ,
If Possible Pls explain properly. Many doubt creating mistake/wording are there in your Explanation.

960-400 =560 $
In 6 Years, Amount of 400$ as interest increased by 560$
n=6, A= 560, P= 400
A=P*(1+r/100)^n
560=400(1+r/100)^6
(1+r/100)^6=1.4.

As We Know for 1st 6years, A=P+C.I = 400+P
A=P*(1+r/100)^n
400+P = P*(1.4)
400+P=1.4P
P=400/0.4=1000

IMO-C

Please Give Kudos, If you find my explanation Good Enough

nick1816 · Answer

rajatchopra1994  You missed the logic i guess.

On P investment, we 're getting 400 bucks interest in 6 years, so we'll get the same interest in the next 6 years. Total is 400+400=800 bucks. 
Whatever extra we're getting is the interest on the interest of first 6 years.

rajatchopra1994 · Answer

Dear  nick1816  ,

I already understood the explanation. 
For others, it will be confusing to understand. That's why I have mentioned it. 
Hope you will get my Point.

Regards,
Rajat Chopra

nick1816 · Answer

rajatchopra1994  bro if someone will have a doubt, he/she can ask or can refer your solution lol. These kinda things happen because of time constraint; you can't write everything brother. I don't understand why you're even bothering about if you understood the solution. Anyways, i'm outta this thread.

rajatchopra1994 · Answer

nick1816 ,

I thought we explain every question properly for the other members of GMAT Club Community. 
That is the only reason for posting explained answers. I think i misunderstood the purpose of explanation.
Same here bro, outta from this thread. Regards!!

cabpri28 · Answer

chetan2u , Can you please expand (400+p)^2 = P(960+p), for some reason my solution isn't the same as yours.

Thanks.

chetan2u · Answer

(400+p)^2=p(960+p).....160000+800p+p^2=960p+p^2
160000=960p-800p=160p.....p=1000

bebs · Answer

This took me a little over 3 mins, but here is my approach:

Let P be the original amount and r = rate
There 400 + P = P (1+r)^{6}  ==> P( (1+r)^{6}  - 1) = 400 ==> Let's call this eq 1
Similarly, P( (1+r)^{12}  - 1) = 960 ==> Let's call this eq 2
Let x =  (1+r)^{6}  ==> eq 1: P(x - 1) = 400 and eq 2: P( x^{2}  - 1) = 960
Divide eq 1 by eq 2 (to cancel out P and x - 1) ==>  \frac{1}{x+1}  =  \frac{5}{12}  ==> x =  \frac{7}{5} 
Therefore, using eq 1, P =  \frac{400}{x - 1}  ==> P = 400/ \frac{2}{5}  ==> P = $1,000 (the answer is   C  )

GMATGuruNY · Answer

A $100 investment earns   20%   interest compounded every 10 years. 
In this case:
Interest earned after the FIRST 10-year interval = 20% of 100 = 20
Resulting amount after the first 10-year interval = 100+20 = 120
Interest earned in the SECOND 10-year interval = 20% of 120 = 24
Notice that the interest earned in the second 10-year interval ($24) is   20%   greater than the interest earned in the first 10-year interval ($20).
The underlined percentages in blue are the SAME.
Implication:
If the interest increases by x% from one interval to the next, then x% is the rate applied to the initial investment.

For the sake of clarity, I've revised the wording of the problem:

Interest earned in the second 6-year interval = (total earned after 12 years) - (interest earned in the first 6 years) = 960 - 400 = 560

The interest earned in the second 6-year interval ($560) is 40% greater than the interest earned in the first 6-year interval ($400).
Since the interest increases by 40% from one interval to the next, 40% is the rate applied to the initial investment.
Thus, the $400 earned in the first 6-year interval must be equal to 40% of the initial investment:
 400 = (0.4)x 
 x = \frac{400}{0.4} = \frac{4000}{4} = 1000

C

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The Compound Interest on an investment in 6 years is $400

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The Compound Interest on an investment in 6 years is $400

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