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Andrew borrows two equal sums of money under simple interest at 5% and
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23 Nov 2023, 02:46
23 Nov 2023, 02:46
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Andrew borrows two 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 date six months before the latter, he will have to pay the same amount of $1,100 in each case. What is the total sum that he had borrowed?
(A) $750
(B) $1,000
(C) $1,500
(D) $2,000
(E) $4,000
(A) $750
(B) $1,000
(C) $1,500
(D) $2,000
(E) $4,000
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Free Materials for the GRE General Exam - Where to get it!!
GRE Geometry Formulas
GRE - Math Book
Re: Andrew borrows two equal sums of money under simple interest at 5% and
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29 Nov 2023, 02:49
29 Nov 2023, 02:49
1
The answer is D.
Let each amount = x --> Total amount = 2x
Total repayment = 1100*2 = $2,200 ; months for the first amount = y
First amount:
1,100 = x + x*0.05*(y/12)
Second amount:
1,100 = x + x*0.04*(y+6)/12
Since both are equal:
x + (0.05xy/12) = x + 0.04x(y+6)/12
--> (0.05xy/12) = 0.04x(y+6)/12
--> 0.05xy = 0.04x(y + 6)
--> 0.05y = 0.04(y + 6)
--> 0.05y = 0.04y + 0.24
--> 0.01y = 0.24
--> y = 24 months
Again, first amount:
1,100 = x + x*0.05*(y/12)
--> 1,100 = x + x*0.05*(24/12)
--> 1,100 = x + 0.1x
--> 1,100 = 1.1x
--> x = $1,000
Hence, total amount borrowed = 2x = 2*1000 = $2,000
Let each amount = x --> Total amount = 2x
Total repayment = 1100*2 = $2,200 ; months for the first amount = y
First amount:
1,100 = x + x*0.05*(y/12)
Second amount:
1,100 = x + x*0.04*(y+6)/12
Since both are equal:
x + (0.05xy/12) = x + 0.04x(y+6)/12
--> (0.05xy/12) = 0.04x(y+6)/12
--> 0.05xy = 0.04x(y + 6)
--> 0.05y = 0.04(y + 6)
--> 0.05y = 0.04y + 0.24
--> 0.01y = 0.24
--> y = 24 months
Again, first amount:
1,100 = x + x*0.05*(y/12)
--> 1,100 = x + x*0.05*(24/12)
--> 1,100 = x + 0.1x
--> 1,100 = 1.1x
--> x = $1,000
Hence, total amount borrowed = 2x = 2*1000 = $2,000
gmatclubot
Re: Andrew borrows two equal sums of money under simple interest at 5% and [#permalink]
29 Nov 2023, 02:49
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