SOLUTION: Alvin has $760 less invested in 7% than he has at 5%. If his annual income from these two investments is $51.20, how much does he have invested at each rate?
Algebra ->
Customizable Word Problem Solvers
-> Finance
-> SOLUTION: Alvin has $760 less invested in 7% than he has at 5%. If his annual income from these two investments is $51.20, how much does he have invested at each rate?
Log On
Question 1114755: Alvin has $760 less invested in 7% than he has at 5%. If his annual income from these two investments is $51.20, how much does he have invested at each rate?
Let x be the amount invested at 5%, in dollars.
Then the amount invested at 7% is (x-760) dollars.
The annual income from the two amounts is
0.05*x + 0.07*(x-760) dollars
Which gives you an equation
0.05*x + 0.07*(x-760) = 51.20 dollars.
Simplify and solve for x
0.05x + 0.07x - 0.07*760 = 51.20,
0.12x = 51.20 + 0.07*760 = 104.40
x = = 870.
Answer. $870 was invested at 5%. 760 dollars less, or 870-760 = 110 dollars was invested at 7%.
Check. 0.05*870 + 0.07*110 = 51.20 dollars. ! Correct !
-----------------
It is a typical and standard problem on investment.
You will find there different approaches (using one equation or a system of two equations in two unknowns), as well as
different methods of solution to the equations (Substitution, Elimination).
