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Some amount of money was lent at 6% per annum simple interest. After

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rxs0005
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Some amount of money was lent at 6% per annum simple interest. After [#permalink] New post  01 Jan 2005, 16:19
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Some amount of money was lent at 6% per annum simple interest. After one year, 6800 is repaid and the rest of the amount is repaid at 5% per annum. If the second year's interest is 11/20 of the first year's interest,find what amount of money was lent out.

A. 12,000
B. 13,000
C. 15,000
D. 16,000
E. 17,000
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gayathri
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Re: Some amount of money was lent at 6% per annum simple interest. After [#permalink] New post  01 Jan 2005, 16:51
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Get 17,000.

Let money lent = p
Interest = npr (n=time, r= interest rate)
After one year new amount = p+npr-6800 = p+1*p*(6/100)-6800=1.06p - 6800

Given:
(1.06p-6800)(5/100) = (11/20)(6p/100)

=> 40p = 680,000; p = 17,000
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SaquibHGMATWhiz
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Joined: 23 May 2022
Posts: 474
Location: India
GMAT 1: 760 Q51 V40
Re: Some amount of money was lent at 6% per annum simple interest. After [#permalink] New post  15 Dec 2022, 02:01
Expert Reply
rxs0005 wrote:
Some amount of money was lent at 6% per annum simple interest. After one year, 6800 is repaid and the rest of the amount is repaid at 5% per annum. If the second year's interest is 11/20 of the first year's interest,find what amount of money was lent out.

A. 12,000
B. 13,000
C. 15,000
D. 16,000
E. 17,000


Solution:

  • First year:
    • Principal \(=x\)
    • rate \(=6\%\)
    • Interest \(=x\times \frac{6}{100}=\frac{6x}{100}\)
    • Amount \(=x+\frac{6x}{100}=\frac{106x}{100}\)

  • Second year:
    • Principal \(=\frac{106x}{100}-6800\)
    • rate \(=5\%\)
    • Interest \(=(\frac{106x}{100}-6800)\times \frac{5}{100}\)

  • Accordign to the question, \((\frac{106x}{100}-6800)\times \frac{5}{100}=\frac{11}{20}(\frac{6x}{100})\)
    \(⇒(\frac{106x}{100}-6800)\times 5=\frac{33x}{10}\)
    \(⇒\frac{106x-680000}{100}\times 5=\frac{33x}{10}\)
    \(⇒106x-680000=66x\)
    \(⇒x=17000\)

Hence the right answer is Option E
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adewale223
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Joined: 07 Oct 2022
Posts: 50
Location: Nigeria
Re: Some amount of money was lent at 6% per annum simple interest. After [#permalink] New post  15 Dec 2022, 11:14
First year

Let x be money borrowed.

I = prt/100

= 6x/100

0.06x


Second year

I = 1.06x-6800(5)/100

Second year interest

= 5.3x-34000/100

11/20(0.06x) = 5.3x-34000/100

55(0.06x) = 5.3x-34000

3.3x = 5.3x -34000

-2x = -34000

x = 17000

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Re: Some amount of money was lent at 6% per annum simple interest. After [#permalink]
15 Dec 2022, 11:14
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