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Math Expert V
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Andrew borrows equal sums of money under simple interest at 5% and 4%  [#permalink]

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Question Stats: 61% (03:02) correct 39% (03:07) wrong based on 74 sessions

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Andrew borrows equal sums of money under simple interest at 5% and 4% rate of interest. He finds that if he repays the former sum on a certain date six months before the latter, he will have to pay the same amount of $1100 in each case. What is the total sum that he had borrowed? (A)$750
(B) $1000 (C)$1500
(D) $2000 (E)$4000

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Andrew borrows equal sums of money under simple interest at 5% and 4%  [#permalink]

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Bunuel wrote:
Andrew borrows equal sums of money under simple interest at 5% and 4% rate of interest. He finds that if he repays the former sum on a certain date six months before the latter, he will have to pay the same amount of $1100 in each case. What is the total sum that he had borrowed? (A)$750
(B) $1000 (C)$1500
(D) $2000 (E)$4000

Assume that Andrew borrows $$X$$ dollars with simple interest of 5% anually in $$n$$ months and $$X$$ dollars with simple interest of 4% annually in $$n+6$$ months.

Since Andreaw has to pay the same amount of $1100 in each case, we have: For the 1st loan, he has to pay total: $$X + X \times \frac{5\%}{12} \times n = X + \frac{0.05Xn}{12}$$ For the 2nd loan, he has to pay total: $$X + X \times \frac{4\%}{12} \times (n+6) = X + \frac{0.04X(n+6)}{12}$$ Hence $$X + \frac{0.05Xn}{12}=X + \frac{0.04X(n+6)}{12}=1,100$$ Since $$\frac{0.05Xn}{12}= \frac{0.04X(n+6)}{12} \implies 0.05n = 0.04(n+6) \implies n=24$$ Now $$X + \frac{0.05Xn}{12}=1,100 \implies X = 1,000$$ The total sum that he had borrowed is: $$1,000+1,000=2,000$$ The answer is D _________________ Manager  B Joined: 27 Mar 2016 Posts: 65 Re: Andrew borrows equal sums of money under simple interest at 5% and 4% [#permalink] Show Tags 2 let the amount borrowed in each case be x and time be t years In case of 5% interest rate time would be 6 months less: (x*5*(t-0.5))/100 +x =1100 In case of 4% interest : (x*4*t)/100 +x=1100 =>5xt+2.5x=4xt =>t=2.5 (4xt/100)+x=1100 substituting value of t 10x/100+x=1100 11x/10=1100 x=1000 total sum borrowed= x at 4% and x at 5% = 2x =2000 hence answer should be D Current Student S Joined: 06 Nov 2016 Posts: 103 Location: India GMAT 1: 710 Q50 V36 GPA: 2.8 Re: Andrew borrows equal sums of money under simple interest at 5% and 4% [#permalink] Show Tags 2 tricky one - slight! x is principal invested t is time at which 4% is invested hence t-1/2 is time at which 5% one is invested 1/2 as 6 months is 1/2 year Hence: 5x/100*(t-1/2) + x = 1100 = 4x/100*(t) + x Solving, xt will get cancelled and you will get: 11000 = 11x x is 1000 sum of both investments is 2x = 2000 which is Option D Manager  S Joined: 23 May 2017 Posts: 236 Concentration: Finance, Accounting WE: Programming (Energy and Utilities) Re: Andrew borrows equal sums of money under simple interest at 5% and 4% [#permalink] Show Tags Attachment: FullSizeRender (5).jpg [ 19.03 KiB | Viewed 1249 times ] +D _________________ If you like the post, please award me Kudos!! It motivates me examPAL Representative P Joined: 07 Dec 2017 Posts: 1072 Re: Andrew borrows equal sums of money under simple interest at 5% and 4% [#permalink] Show Tags Bunuel wrote: Andrew borrows equal sums of money under simple interest at 5% and 4% rate of interest. He finds that if he repays the former sum on a certain date six months before the latter, he will have to pay the same amount of$1100 in each case. What is the total sum that he had borrowed?

(A) $750 (B)$1000
(C) $1500 (D)$2000
(E) $4000 As the calculation we're asked to do is straightforward, we'll just do it. This is a Precise approach. Labeling x as the amount of money in each account and m as the number of months the first until the first sum was repaid, we have: x + (m/12)*(5/100) = x + (m/12 + 6/12)*(4/100). Canceling out x and multiplying by 12*100 gives 5m = 4m + 24 --> m = 24 So if after two years at 5% interest he paid$1100, then his original sum was $1000. Since he had two accounts with$1000, the total sum he borrowed was $2000. (D) is our answer. _________________ Intern  B Joined: 31 Dec 2017 Posts: 28 Concentration: Finance Re: Andrew borrows equal sums of money under simple interest at 5% and 4% [#permalink] Show Tags 1 Sum borrowed =$x.
Case I: SI-5%, Period = (n-6) months
Principal + Interest for (n-6) months = x + (5/100*(n-6)/12) = x + 5x-30/1200

Case II: SI-4%, Period = (n) months
Principal + Interest for (n) months = x + (4/100*(n)/12) = x + 4x/1200

x + 5n-30/1200 = x + 4n/1200
5n-4n = 30
n = 30

So, sum @ 5% was repaid in 24 months, and sum at 4% was repaid in 30 months.

If he repaid $1100 in 2 years at 5% SI, then original sum must be =$1100*100/110 = $1000; since he borrowed at 2 different rates, total sum =$2000.
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Re: Andrew borrows equal sums of money under simple interest at 5% and 4%  [#permalink]

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saurabh9gupta wrote:
Bunuel wrote:
Andrew borrows equal sums of money under simple interest at 5% and 4% rate of interest. He finds that if he repays the former sum on a certain date six months before the latter, he will have to pay the same amount of $1100 in each case. What is the total sum that he had borrowed? (A)$750
(B) $1000 (C)$1500
(D) $2000 (E)$4000

I got this question wrong but I am not sure what is wrong in my method

I assumed the amount to be P

( P * 5 * [t - .5])/ 100 = (P * 4 * t ) /100

t=2.5 yrs

But when I substitute in one of the equations. I get a wrong answer.

Hi, I think you are correct, I followed same method. No. of years is 2.5 @ 5% and 2 @ 4%... so 2($1100/1.1) =$2000.
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Re: Andrew borrows equal sums of money under simple interest at 5% and 4%  [#permalink]

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_________________ Re: Andrew borrows equal sums of money under simple interest at 5% and 4%   [#permalink] 25 Dec 2018, 19:13
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