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

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A man invested two equal sums of money in two banks at simple interest [#permalink]

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New post 02 Jul 2017, 02:34
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Question Stats:

62% (02:11) correct 38% (02:24) wrong based on 50 sessions

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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

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A man invested two equal sums of money in two banks at simple interest [#permalink]

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New post 02 Jul 2017, 07:11
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Bunuel 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


Assume that he invested \(X\) dollars in each bank.

For the 1st bank with the simple interest of 10% annually, after 2 years, he will earn the interests: \(2 \times X \times 10\% = 0.2X\)

For the 2nd bank with the simple interest of 20% annually, after 2 years, he will earn the interests: \(2 \times X \times 20\% = 0.4X\)

Hence, we have: \(120 < 0.4X-0.2X = 0.2X < 140 \implies 600 < X < 700\)

For the 1st bank with the compound interest of 10% annually, after 2 years, he will have total money: \(X \times 1.1^2 \)

For the 2nd bank with the compound interest of 20% annually, after 2 years, he will have total money: \(X \times 1.2^2\)

Hence, the difference between the amounts earned:

\(X \times 1.2^2 - X \times 1.1^2 = X \times (1.2^2 - 1.1^2) = X \times (1.2-1.1)(1.2+1.1) = X \times 0.1 \times 2.3 = 0.23X\)

Hence, we have \(600 < X < 700 \implies 138 < 0.23X < 161\)

We get the answer D as the correct answer
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Re: A man invested two equal sums of money in two banks at simple interest [#permalink]

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New post 02 Jul 2017, 07:32
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Difference of the amounts that earned simple interest= S.I earned for 2 yrs at 20% - S.I earned for 2 yrs at 10%
= Px2x20/100 - Px2x10/100 [ S.I = Pnr/100 ]
= 0.2P
Also given in the question,
the difference between the interests earned after two years is between $120 and $140, exclusive,

in other words.
120 < 0.2P < 140
or 600 < P < 700

Now,

Difference of the amounts that earned compound interest = P[1+ 20/100]^2- P [1+ 10/100]^2 [ Amount earned via C.I = P[1+ r/100]^n ]
= P[144-121/100]
= 0.23P

Now,
600 < P < 700
or 138 < 0.23P < 161

Only answer D matches our range
So Answer is D







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Re: A man invested two equal sums of money in two banks at simple interest [#permalink]

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New post 02 Jul 2017, 07:51
This is a long one:

It is given that 120 <= 2*20*P/100 - 2*10*P/100 < = 140

Solving: we get: 600 <= P <=700

Now, putting P as 600 and solving for compound interests:

10% : A = 726 after two years
20%: A = 864 after two years:
difference is 138.
This is the least value - hence option A, B, C

Repeating similar cases for P as 700 and solving for compound interests:

10%: A = 847 after two years
20%: A = 1008 after two years
difference is 161
That is the maximum value - hence option E is discarded

Answer is Option D.
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Re: A man invested two equal sums of money in two banks at simple interest [#permalink]

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New post 02 Jul 2017, 08:06
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)
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Re: A man invested two equal sums of money in two banks at simple interest [#permalink]

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New post 06 Aug 2017, 00:37
Sorry i have a question here. So when in the calculation of SI why did we not calculate the amount where as in the CI we did calculate the amount and then went forward . can someone explain
Re: A man invested two equal sums of money in two banks at simple interest   [#permalink] 06 Aug 2017, 00:37
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