A man invested two equal sums of money in two banks at simple interest

If x is the principal, Clearly,
SI in 20% - 2x/5
SI in 10% - x/5
Difference - x/5 Also 120 < x/5 < 140
600 < x < 700

Now CI = P(1+r/n)^nt - P
So difference is 0.23 x

if, 600 < x < 700
138 < 0.23x < 161

Given data : The range of simple interests is between 120$ and 140$(with 10% and 20% interest)

Minimum range(Simple interest difference : 120$)
At 10% interest, Amount is 600$ to achieve 120$ over a period of 2 years(60$ * 2)
At 20% interest, For amount = 600$, we will have 240$ interest for 2 years(120$ * 2)
Given the difference in the simple interests will be 120$.

When charging 10% compound interest
For this amount the compound interest will be 60$(for 1st year) and 66$(for 2nd year)
Total Compound interest is 126$
When charging 20% compound interest
For this amount the compound interest will be 120$(for 1st year) and 144$(for 2nd year)
Total Compound interest is 264$

Difference(Compound interest) is 264$ - 126$ = 138$

Maximum range(Simple interest difference : 140$)
At 10% interest, Amount : 700$ to achieve 140$ over a period of 2 years(70$ * 2)
At 20% interest, For amount = 700$, we will have 280$ interest for 2 years(140$ * 2)
As given the difference in the simple interests will be 140$.

When charging 10% compound interest
For this amount the compound interest will be 70$(for 1st year) and 77$(for 2nd year)
Total Compound interest is 147$
When charging 20% compound interest
For this amount the compound interest will be 140$(for 1st year) and 168$(for 2nd year)
Total Compound interest is 308$

Difference(Compound interest) is 308$ - 147$ = 161$

Hence, compound interest must be in the range $138 and $161(Option D)

Difference of interest after 1 year = 20%-10% = 10% of principal
Difference of interest after 2 years = 40%-20% = 20% of principal

i.e. 20% of half of the sum is between 120 and 140
i.e. Sum is between 1200 and 1400

For 1200 distributed in two accounts with sum 600 each
Interest @ 10% after two years \(= (1.1)^2*600-600 = 126\)
Interest @ 20% after two years \(= (1.2)^2*600-600 = 264\)
Difference = 138

For 1400 distributed in two accounts with sum 700 each
Interest @ 10% after two years \(= (1.1)^2*700-700 = 147\)
Interest @ 20% after two years \(= (1.2)^2*700-700 = 161\)

Only value available in options that falls between 147 and 161 is 154 hence

Answer: Option D

Responding to a pm:
The difference between the interests earned in two years is in the range $120 - $140. So the answer will be obtained in a range too. We need to find the value that falls in this range.
Let's take the middle value and find the value closest to it.

Take the difference in SI of two years to be say $130. So each year extra SI will be $65 (because rate in second case is 20%, 10% more than the rate of 10%).
$65 is 10% of the principal so principal would be $650.

Diff in interest earned in CI on $650 at 10% interest vs 20% interest:
On $650, CI earned at 10% in 2 years = 65 + 65 + 6.5
On $650, CI earned at 20% in 2 years = 130 + 130 + 26
The difference is about $150.
$150 is going to lie in the middle of our acceptable range so $154 should certainly lie in this range.

Answer (D)

METHOD 1
given: invested same amount @ 10% and @ 20% for 2 years and the difference between interest earned was $120 to $140;
difference simple interest: 0.2x(2)-0.1x(2)=2x(0.1)=0.2x… range: 120<0.2x<140… 600<x<700
difference compound interest: [0.2x+(x+0.2x)0.2]-[0.1x+(x+0.1x)0.1],…[0.2(2.2x)]-[0.1(2.1x)],…
=2.2x/5-2.1x/10=4.4x-2.1x/10=2.3x/10… range: 2.3(600)/10 to 2.3(700)/10… $138 to $161

Answer (D).

METHOD 2
difference simple interest 2 years: x(2)(0.2)-x(2)(0.1)=x2(0.1)=0.2x; range: 120<0.2x<140, 600<x<700;
difference compound interest 2 years: x(1.2)(1.2)-x(1.1)(1.1)… x(1.2^2-1.1^2)… x(1.2+1.1)(1.2-1.1)… x(2.3)(0.1)… x(0.23);
range: (600)(0.23) to (700)(0.23) = $138 to $161

Answer (D).

simple interest earned: principal•years•rate_p.a.%
compound interest earned: principal(1+rate.a.%)^years-principal
compound interest earned 2 years:
[interest earned 1 year] + [interest earned 2 year]
[p(r)]+[p(r)+p(r)(r)] = [pr]+[pr(1+r)] = pr(2+r)

find principal range: 600<p<700
120<p2(20%)-p2(10%)<140… 120<p2(20%-10%)<140… 600<p<700

find low compound range: 264-126=138
[600*.2] + [600*.2+120*.2] = 120+120+24 = 264
[600*.1] + [600*.1+60*.1] = 60+60+6 = 126

find high compound range: 308-147=161
[700*.2] + [700*.2+140*.2] = 140+140+28 = 308
[700*.1] + [700*.1+70*.1] = 70+70+7 = 147

difference must be between: 138 to 161

Answer (D)

\(120<\frac{(A*20*2)}{100}-\frac{(A*10*2)}{100}<140\)

\(600<A<700\)

\(600(1.44-1.21)<X<700(1.44-1.21)\)

\(6*23<X<7*23\)

Another way:

\(600<A<700\)

\(600*0.2*2+600*0.2*0.2-600*0.1*2+600*0.1*0.1=23*6=138\)

\(700*0.2*2+700*0.2*0.2-700*0.1*2+700*0.1*0.1=23*7=161\)

\(138<X<161\)

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