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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