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.
A man invested two equal sums of money in two banks at...
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Let the amount invested = $P sum in two bank accounts
Interest on the first investment at 10% rate; SI = PRT/100
$$=\frac{\left(P\cdot10\cdot2\right)}{100}=\ \frac{P}{5}$$
Interest on second investment at 20% rate ; $$=\frac{\left(P\cdot20\cdot2\right)}{100}=\ \frac{2P}{5}$$
$$difference\ of\ interest\ =\frac{2p}{5}-\frac{p}{5}=\frac{p}{5}$$
$$given\ that\ 120<\frac{p}{5}<140\ =\ 600<p<700$$
Amount received on first investment at compound interest = $$p\left(1+10\%\right)^2\ =\ p\left(1+\frac{10}{100}\right)^2$$
$$=\ p\left(1+0.1\right)^2\ =\ p\left(1.1\right)^2$$
= p(1.21) = 1.21p
Amount received on second installment at compound interest = p(1+20%)^2 = p (1 + 20/100)^2
$$=\ p\left(1+0.2\right)^2\ =\ p\left(1.2\right)^2$$
$$=\ p\ \left(\ 1.44\ \right)\ =\ 1.44p$$
$$Difference\ of\ amounts\ =\ 1.44p\ -\ 1.21p\ =\ 0.23p$$
= 0.23* 600 <Difference of amounts < 0.23 * 700
= 138 < 0.23p < 161
Only option D falls within the range .
Answer = Option D
Interest on the first investment at 10% rate; SI = PRT/100
$$=\frac{\left(P\cdot10\cdot2\right)}{100}=\ \frac{P}{5}$$
Interest on second investment at 20% rate ; $$=\frac{\left(P\cdot20\cdot2\right)}{100}=\ \frac{2P}{5}$$
$$difference\ of\ interest\ =\frac{2p}{5}-\frac{p}{5}=\frac{p}{5}$$
$$given\ that\ 120<\frac{p}{5}<140\ =\ 600<p<700$$
Amount received on first investment at compound interest = $$p\left(1+10\%\right)^2\ =\ p\left(1+\frac{10}{100}\right)^2$$
$$=\ p\left(1+0.1\right)^2\ =\ p\left(1.1\right)^2$$
= p(1.21) = 1.21p
Amount received on second installment at compound interest = p(1+20%)^2 = p (1 + 20/100)^2
$$=\ p\left(1+0.2\right)^2\ =\ p\left(1.2\right)^2$$
$$=\ p\ \left(\ 1.44\ \right)\ =\ 1.44p$$
$$Difference\ of\ amounts\ =\ 1.44p\ -\ 1.21p\ =\ 0.23p$$
= 0.23* 600 <Difference of amounts < 0.23 * 700
= 138 < 0.23p < 161
Only option D falls within the range .
Answer = Option D