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

Question

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

(A) $750
(B) $1000
(C) $1500
(D) $2000
(E) $4000

broall · Answer

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

ashudhall · Answer

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

sonikavadhera · Answer

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

Leo8 · Answer

FullSizeRender (5).jpg +D

DavidTutorexamPAL · Answer

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.

mbafinhopeful · Answer

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.

mbafinhopeful · 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.

Kinshook · Answer

Given: 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.

Asked: What is the total sum that he had borrowed?

Let the total sum that Andrew had borrowed be 2x.

Interest on x @ 4% for y months = .04xy/12 = $1100 - x
Interest on x @5% for y-6 moths = .05x(y-6)/12 = $1100 - x
y/y-6 = 5/4
y = 30 months
.04x *30/12 = $1100 - x
.1x + x = $1100
x = $1000
The total sum that he had borrowed = 2x = $2000

IMO D

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

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