$60,000 was invested for a year. part of it earned simple annual interest at x percent per yr, and the rest earned simple annual int. at y percent per yr. if the total interest earned by the 60,000 for that yr was 4080, what is x?
1) x = 3y/4
2) the ratio of the amt that earned interest at x% to amt of interest earned at y% is 3:2
ratio prob
This topic has expert replies
To simplify the problem:
Let the amount invested at x% be A. So, 60,000-A is invested at y%
So, A*x%+(60,000-A)*y%=4080
We need to find x. The unknown variables are x,A and y
Stmt1
x=3y/4
This is insuff. No info about A
Stmt2
A/(60,000-A) = 3:2
This is insuff. No info about x and y
Combine stmt1 and stmt2
We will get the value of x. So the answer is C
Let the amount invested at x% be A. So, 60,000-A is invested at y%
So, A*x%+(60,000-A)*y%=4080
We need to find x. The unknown variables are x,A and y
Stmt1
x=3y/4
This is insuff. No info about A
Stmt2
A/(60,000-A) = 3:2
This is insuff. No info about x and y
Combine stmt1 and stmt2
We will get the value of x. So the answer is C