chendawg 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.

Source: OG 12

Just wanted to see if anyone knew any other methods other than standard algebra!

Let the amount invested at x% per year = A, then amount invested at y% per year = 60,000 - x
Now, S.I = PRT/100, where P = amount invested, R = rate and T = time for which it is invested.
So, Ax/100 + (60,000 - x)y/100 = 4080, then we have to find the value of x.

(1) x = 3y/4 or y = 4x/3
Ax/100 + (60,000 - x)(4x/3)/100 = 4080, one equation, 2 variables. So, (1) is NOT SUFFICIENT.

(2) A: (60,000 - A) = 3: 2 implies 2A = 180000 - 3A or A = 36000
So, 36000x/100 + (60,000 - x)y/100 = 4080, again one equation, 2 variables. So, (2) is NOT SUFFICIENT.

Combining (1) and (2), 36000x/100 + (60,000 - x)(4x/300) = 4080, one equation, one variable, So SUFFICIENT.

The correct answer is C.