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

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

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02 Jul 2017, 01:26
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Difficulty:

55% (hard)

Question Stats:

63% (02:33) correct 38% (02:26) wrong based on 64 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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02 Jul 2017, 06:29
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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: $$1,000+1,000=2,000$$

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

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02 Jul 2017, 06:42
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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
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Re: Andrew borrows equal sums of money under simple interest at 5% and 4%  [#permalink]

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03 Jul 2017, 10:18
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
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Re: Andrew borrows equal sums of money under simple interest at 5% and 4%  [#permalink]

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03 Jul 2017, 11:02
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+D
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Re: Andrew borrows equal sums of money under simple interest at 5% and 4%  [#permalink]

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25 Dec 2018, 18:13
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Re: Andrew borrows equal sums of money under simple interest at 5% and 4% &nbs [#permalink] 25 Dec 2018, 18:13
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