Installment problem

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Installment problem

by hemant_rajput » Sat Jan 26, 2013 2:12 am
Q25 . Some amount was lent at 6% per annual SI. After one year, rs 6800 us repaid and the rest of the amount is repaid at 5% per annual. If the second year's interest is 11/20 if the first year's interest, find what amount of money was lent out.

a. 17000
b. 16800
c. 16500
d. 17500
e. None of these
[spoiler]OA:C[/spoiler]
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by GMATGuruNY » Sun Jan 27, 2013 5:05 am
hemant_rajput wrote:Q25 . Some amount was lent at 6% per annual SI. After one year, rs 6800 us repaid and the rest of the amount is repaid at 5% per annual. If the second year's interest is 11/20 if the first year's interest, find what amount of money was lent out.

a. 17000
b. 16800
c. 16500
d. 17500
e. None of these
We can plug in the answers, which represent the amount invested.
GMAT problems about percentages tend to involve VERY ROUND NUMBERS.
Thus, we should start with A:

A: 17,000
First-year interest = .06(17000) = 1020.
After 1020 of interest is added and 6800 is repaid, remaining amount = 17,000 + 1020 - 6800 = 11220.
Second-year interest = .05(11220) = 561.
If A is correct, the second-year interest (561) will be equal to 11/20 of the first-year interest (1020).
(11/20)1020 = 561.
Success!

The correct answer is A.
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by mparakala » Sun Jan 27, 2013 6:10 am
let the principal be "P"

1st year's interest: I = PTR/100 = 6P/100
1st year's amount = P+I = P + 6P/100 = 106P/100 = 1.06P

now, 6800 is paid! => 1.06P-6800 is remaining

2nd year's interest = PTR/100 = (1.06P-6800) * 1 * 5 / 100
But, given that this interest is equal to 11/20 of 1st year's INTEREST

=> (1.06P-6800) * 5 /100 = 11/20 * (6P/100)
SOLVE FOR P

ANSWER: P = 17000 OPTION A