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# A bank employee invested \$2500

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Intern
Joined: 10 Nov 2017
Posts: 12
A bank employee invested \$2500  [#permalink]

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Updated on: 18 Jul 2018, 11:01
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(N/A)

Question Stats:

50% (00:57) correct 50% (00:31) wrong based on 15 sessions

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A bank employee invested \$2500 in January 2000 at 5% rate of interest p.a. The bank revised the rate of interest to 6% for the whole year of 2001. If he invested the money in January 2000, which of the following options represents the amount of money he will get in January 2002.

A. 2500(1.05)(1.06)
B. 2500(1.113)
C. 2500(1.03)
D. 2500(1.5)(1.6)
E. 2500(1+11/100)^2

P.S - how to find out whether this sum is Simple interest or compound interest. In the question, they have not mentioned it explicitly.

Originally posted by Arun1994 on 18 Jul 2018, 08:15.
Last edited by Arun1994 on 18 Jul 2018, 11:01, edited 2 times in total.
Math Expert
Joined: 02 Aug 2009
Posts: 7036
Re: A bank employee invested \$2500  [#permalink]

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18 Jul 2018, 08:28
1
Arun1994 wrote:
A bank employee invested \$2500 in January 2000 at 5% rate of interest p.a. The bank revised the rate of interest to 6% for the whole year of 2011. If he invested the money in January 2000, which of the following options represents the amount of money he will get in January 2002.

A. 2500(1.05)(1.06)
B. 2500(1.113)
C. 2500(1.03)
D. 2500(1.5)(1.6)
E. 2500(1+11/100)^2

Ans is A and B. But how ?

2500 @5% = 2500(1+5/100)=2500(1+0.05)=2500*1.05
Now this amount is at 6% so 2500*1.05*(1+6/100)=2500*1.05*1.06...
So A

Now let's work on it further
2500(1.05)(1.06)=2500(1+.05)(1+0.06)=2500(1*1+1*0.05+1*0.06+0.05*0.06)=2500(1+0.05+0.06+0.003)=2500(1.113)..
Hence B
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Intern
Joined: 10 Nov 2017
Posts: 12
A bank employee invested \$2500  [#permalink]

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Updated on: 18 Jul 2018, 11:06
chetan2u wrote:
Arun1994 wrote:
A bank employee invested \$2500 in January 2000 at 5% rate of interest p.a. The bank revised the rate of interest to 6% for the whole year of 2011. If he invested the money in January 2000, which of the following options represents the amount of money he will get in January 2002.

A. 2500(1.05)(1.06)
B. 2500(1.113)
C. 2500(1.03)
D. 2500(1.5)(1.6)
E. 2500(1+11/100)^2

Ans is A and B. But how ?

2500 @5% = 2500(1+5/100)=2500(1+0.05)=2500*1.05
Now this amount is at 6% so 2500*1.05*(1+6/100)=2500*1.05*1.06...
So A

Now let's work on it further
2500(1.05)(1.06)=2500(1+.05)(1+0.06)=2500(1*1+1*0.05+1*0.06+0.05*0.06)=2500(1+0.05+0.06+0.003)=2500(1.113)..
Hence B

Hi chetan2u,

Thanks for answering. But how did you find that this problem is a compound interest problem?

Originally posted by Arun1994 on 18 Jul 2018, 10:37.
Last edited by Arun1994 on 18 Jul 2018, 11:06, edited 2 times in total.
Board of Directors
Status: QA & VA Forum Moderator
Joined: 11 Jun 2011
Posts: 4220
Location: India
GPA: 3.5
Re: A bank employee invested \$2500  [#permalink]

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18 Jul 2018, 10:49
Arun1994 wrote:
A bank employee invested \$2500 in January 2000 at 5% rate of interest p.a. The bank revised the rate of interest to 6% for the whole year of 2011. If he invested the money in January 2000, which of the following options represents the amount of money he will get in January 2002.

A. 2500(1.05)(1.06)
B. 2500(1.113)
C. 2500(1.03)
D. 2500(1.5)(1.6)
E. 2500(1+11/100)^2

P.S. I know the answers are A and B. But how to find out whether this sum is Simple interest or compound interest. In the question, they have not mentioned it explicitly.

Dear Arun1994

Please check the highlighted part IMHO it must be 2001 instead of 2011, further request you to kindly follow the rules for posting in the forum, rules in my signature.
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Thanks and Regards

Abhishek....

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Re: A bank employee invested \$2500 &nbs [#permalink] 18 Jul 2018, 10:49
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