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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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New post 02 Jul 2017, 02:26
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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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New post 02 Jul 2017, 07:29
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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: \(1,000+1,000=2,000\)

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

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New post 02 Jul 2017, 07:42
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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
hence answer should be D
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Re: Andrew borrows equal sums of money under simple interest at 5% and 4%  [#permalink]

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New post 03 Jul 2017, 11:18
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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New post 03 Jul 2017, 12:02
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Re: Andrew borrows equal sums of money under simple interest at 5% and 4%  [#permalink]

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New post 27 Sep 2018, 12:23
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.
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Re: Andrew borrows equal sums of money under simple interest at 5% and 4%  [#permalink]

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New post 30 Sep 2018, 21:46
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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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New post 30 Sep 2018, 22:16
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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New post 25 Dec 2018, 19:13
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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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