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01 Jan 2005, 16:19
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
48% (02:47) correct
52%
(02:50)
wrong
based on 106
sessions
History
Date
Time
Result
Not Attempted Yet
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
A. 12,000
B. 13,000
C. 15,000
D. 16,000
E. 17,000
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
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
15 Dec 2022, 02:01
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rxs0005
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
