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

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