Invested interest

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Invested interest

by alex.gellatly » Mon May 21, 2012 7:03 pm
A total of $60,000 was invested for one year. Part of this amount earned simple annual interest at the rate of x percent per year, and the rest earned simple annual interest at the rate of y percent per year. If the total interest earned by the $60,000 for that year was $4,080, what is the value of x?

(1) x = 3y / 4
(2) The ratio of the amount that earned interest at the rate of x percent per year to the amount that earned interest at the rate of y percent per year was 3 to 2.

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by mathbyvemuri » Mon May 21, 2012 9:48 pm
Let us suppose the part of the amount that earned x% annual interest be P. This means, the remaining part that earned y% interest is 60000-P.
The total interest earned = 4080
=> Px/100 + (60000-P)y/100 = 4080
Px+(60000-P)y = 408000 ---(I)

stmt(1)=> gives relation b/w x and y,. Using this we can eliminate y and then by using equation(I), we land into an equation in two variables P and x. So, stmt(1) alone is not suff.

stmt(2)=> gives relation b/w P and (60000-P). This gives the value of P. Substituting the value of P in equation(I) leads us to land into an equation in two variables x and y. So, stmt(1) alone is not suff.

Using stmt(1) and (2) together: This approach gives us two equations and we can find the value of x.
Answer "C"[/spoiler]

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by GMATGuruNY » Tue May 22, 2012 2:57 am
alex.gellatly wrote:A total of $60,000 was invested for one year. Part of this amount earned simple annual interest at the rate of x percent per year, and the rest earned simple annual interest at the rate of y percent per year. If the total interest earned by the $60,000 for that year was $4,080, what is the value of x?

(1) x = 3y / 4
(2) The ratio of the amount that earned interest at the rate of x percent per year to the amount that earned interest at the rate of y percent per year was 3 to 2.
Since 4080 is between 5% and 10% of 60,000, it is likely that x and y are between 5 and 10.

Statement 1: x = (3/4)y
It's possible that x=6 and y=8 or that x=6.6 and that y=8.8.
To accommodate the desired combination of percentages, we could simply adjust the amount invested at each percentage so that the total amount of interest earned = 4080.
Since x can be different values, INSUFFICIENT.

Statement 2: The ratio of the amount that earned interest at the rate of x percent per year to the amount that earned interest at the rate of y percent per year was 3/2.

Sum of the parts of the ratio = 3+2 = 5.
Since the actual amount invested = 60,000, and 60,000/5 = 12000, the two parts of the ratio must be multiplied by 12,000:
12,000(3:2) = 36,000:24,000.
Thus, 36,000 earns x% interest and 24,000 earns y% interest.
It's possible that x=6 or that x=6.6.
To accommodate each value for x, we could simply adjust the value of y so that the total amount of interest earned = 4080.
Since x can be different values, INSUFFICIENT.

Statements 1 and 2 combined:
32,000 earns x% interest and 24,000 earns y% interest.
Since x=(3/4)y, for every 4% of interest earned by the $24,000, 3% interest is earned by the $36,000.
If x=3 and y=4:
(.03)(36,000) + (.04)(24,000) = 1080+960 = 2040.
Since 2040 is half the amount of interest needed, the percentages must be doubled to 6% and 8%.
Thus, x=6.
SUFFICIENT.

The correct answer is C.
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by heymayank08 » Tue May 22, 2012 6:20 pm
mitch
i got your method
but can you pls xplain with the actual algebraic procedure..coz when i tried i was unable to reach the answer

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by ronnie1985 » Wed May 23, 2012 7:08 am
(C)

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by GMATGuruNY » Wed May 23, 2012 7:15 am
heymayank08 wrote:mitch
i got your method
but can you pls xplain with the actual algebraic procedure..coz when i tried i was unable to reach the answer
Statement 1 indicates that y=4x/3.
Statement 2 indicates that 36,000 earns x% interest and 24,000 earns y% interest.
The total amount of interest earned = 4080.
Thus:
(x/100)(36,000) + (4x/3)/100 * (24,000) = 4080.
360x + 320x = 4080.
680x = 4080
x=6.
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