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A man invested two equal sums of money in two banks at...

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A man invested two equal sums of money in two banks at...

by swerve » Tue Mar 27, 2018 8:43 am
A man invested two equal sums of money in two banks at simple interest, one offering annual rate of interest of 10% and the other, at a rate of 20%. If the difference between the interests earned after two years is between $120 and $140, exclusive, which of the following could be the difference between the amounts earned for the same amounts of money, invested at the same rates of interest as above, but at compound interest?

A. $130
B. $135
C. $137
D. $154
E. $162

The OA is D.

Please, can any expert explain this PS question for me? I tried to solve it but I can't get the correct answer. I need your help. Thanks.
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by Jay@ManhattanReview » Thu Apr 05, 2018 1:01 am
swerve wrote:A man invested two equal sums of money in two banks at simple interest, one offering annual rate of interest of 10% and the other, at a rate of 20%. If the difference between the interests earned after two years is between $120 and $140, exclusive, which of the following could be the difference between the amounts earned for the same amounts of money, invested at the same rates of interest as above, but at compound interest?

A. $130
B. $135
C. $137
D. $154
E. $162

The OA is D.

Please, can any expert explain this PS question for me? I tried to solve it but I can't get the correct answer. I need your help. Thanks.
Say the man invested $P sums in two bank accts.

Interest on the first investment @ Simple interest = (P*10*2)/100 = P/5
Interest on the second investment @ Simple interest = (P*20*2)/100 = 2P/5

Difference of Interest = 2P/5 - P/5 = P/5

It is given that 120 < P/5 < 140

=> 600 < P < 700

Amount received on the first investment @ Compound interest = P(1 + 10%)^2 = P(1.1)^2 = 1.21P
Amount received on the second investment @ Compound interest = P(1 + 20%)^2 = P(1.2)^2 = 1.44P

Difference of Amounts = 1.44P - 1.21P = 0.23P

=> 0.23*600 < Difference of Amounts < 0.23*700

138 < Difference of Amounts < 161

Only Option D: $154 is within the range.

The correct answer: D

Hope this helps!

-Jay
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RE:

by deloitte247 » Sun Apr 15, 2018 1:24 pm
Let the equal sums of money be x
$$simple\ interest,\ I=\frac{PRT}{100}$$
$$For\ 10\%\ interest\ annually,$$
$$I=\frac{\left(x\cdot10\cdot2\right)}{100}=0.4x$$
$$the\ difference\ between\ interest\ earned=0.4x-0.2x=0.2x$$
Since the difference is between $120 and $140,
$$when\ 0.2x=120,\ we\ have\ x=\frac{120}{0.2}=\text{600}$$
$$when\ 0.2x=140,\ we\ have\ x=\frac{140}{0.2}=\text{700}$$
This means that the amount invested is between $600 and $700.
Now, say amount invested is $600,
The compound interest for two years is given as
$$first\ year,\ I=\frac{\left(600\cdot10\right)}{100}=60$$
$$Amount\ =\ P+I=600+60=660$$
$$\sec ond\ year,\ I=\frac{\left(660\cdot10\right)}{100}=66$$
$$Amount=P+I=660+66=726$$
$$At\ 20\%\ interest,$$
$$First\ year,\ I=\frac{\left(600\cdot20\right)}{100}=120$$
$$Amount=P+I=720$$
$$Second\ year,\ I=\frac{\left(720\cdot20\right)}{100}=144$$
$$Amount\ =P+I=864$$
Therefore, the difference between amount earned for the same amount of money (i.e $600) = $864 - $726 = $138
Now, say amount invested is $700,
The compound interest for two years is;
$$At\ 10\%\ interest;$$
$$First\ year,\ I=\frac{\left(700\cdot10\right)}{100}=70$$
$$Amount=P+I=770$$
$$Second\ year,\ I=\frac{\left(770\cdot10\right)}{100}=77$$
$$Amount=P+I=847$$
$$At\ 20\%\ interest;$$
$$First\ year,\ I=\frac{\left(700\cdot20\right)}{100}=140$$
$$Amount=P+I=840$$
$$Second\ year,\ I=\frac{\left(840\cdot20\right)}{100}=168$$
$$Amount=P+I=1008$$
Therefore, the difference between amount earned for the same amount of money (i.e $700) = $1008 - $847 = $161
Hence, the answer is any value between $138 and $161.
Answer is $154 which thus makes option D the correct option
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