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09 Dec 2019, 10:06
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Difficulty:

55% (hard)

Question Stats:

50% (01:53) correct 50% (02:59) wrong based on 20 sessions

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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
VP
Joined: 19 Oct 2018
Posts: 1293
Location: India

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09 Dec 2019, 22:52
[quote="nick1816"]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 VP Joined: 19 Oct 2018 Posts: 1293 Location: India The Compound Interest on an investment in 6 years is$400  [#permalink]

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09 Dec 2019, 23:07
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 wrote:
nick1816 wrote:
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 Senior Manager Joined: 16 Feb 2015 Posts: 258 Location: United States Concentration: Finance, Operations Schools: INSEAD, ISB Re: The Compound Interest on an investment in 6 years is$400  [#permalink]

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09 Dec 2019, 23:39
Dear nick1816 ,

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

Regards,
Rajat Chopra
VP
Joined: 19 Oct 2018
Posts: 1293
Location: India

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09 Dec 2019, 23:58
nick1816 wrote:
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 wrote:
Dear nick1816 ,

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,

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!!
Math Expert
Joined: 02 Aug 2009
Posts: 8341
The Compound Interest on an investment in 6 years is $400 [#permalink] ### Show Tags 10 Dec 2019, 00:08 1 1 Ansh777 wrote: 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 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 _________________ Intern Joined: 12 Sep 2019 Posts: 5 Re: The Compound Interest on an investment in 6 years is$400  [#permalink]

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13 Dec 2019, 10:21
chetan2u, Can you please expand (400+p)^2 = P(960+p), for some reason my solution isn't the same as yours.

Thanks.
Math Expert
Joined: 02 Aug 2009
Posts: 8341

### Show Tags

13 Dec 2019, 20:41
1
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) The Compound Interest on an investment in 6 years is$400   [#permalink] 13 Dec 2019, 20:41
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