Algebra I -- Mixture Problem

Question

Mr. Graham invested some of his $18,000 in bonds that made a 5% profit, and the rest in bonds that made a 12% profit. If the profit of the 12% bonds was $885 more than the profit on the 5% bonds, how much did Mr. Graham invest in the 5% bonds?

Answer 1

Let's call the principals (pre-profit amounts) of the 5% profit bonds F and the 12% profit bonds T. The profit from the 5% bonds would thus be 0.05 * F, and the profit from the 12% bonds 0.12 * T

 

Then we have two equations, for two unknowns:

0.12 * T = 0.05 * F + 885  

T + F = 18000 (as the total amount invested into both those bonds is 18000) 

 

Then T = (0.05 * F + 885) / 0.12 = 5/12 * F + 7375 

Now, how do we introduce the information from the T+F equation into this problem? One strategy might be to subtract (T+F) from both sides, except on the other side (where the non-variable number is), we're going to subtract (T+F) by subtracting 18000. This will just give us a single equation in terms of one variable (F) 

 

T - (T+F) = (5/12)*F + 7375 - 18000

-F = (5/12)*F - 10625 

 

Let's solve for F by adding F to both sides of the last equation, getting us: 

-F + F = 0 = 17/12 * F - 10625 

 

This means 17/12 * F = 10625 

F = 12/17 * 10625 = 7500 

 

This F was the amount invested in the 5% bonds (before profit): 7500

 

Note that from here we can solve for the amount solved in the 12% bonds as well: T+F = 18000 so T = 18000-7500 = 10500. Note that 0.12 * 10500 - 0.05 * 7500 = 885 

 

 

Answer 2

CORRECTED SOLUTION

THANKS FELLOW TUTORS FOR CORRECTING ME 

 

 

 

Let x = the amount in 5% bonds

Let y = the amount in 12% bonds

 

Then y(0.12) = x(0.05) + 885

and we already know that x + y = 18000

 

By substitution:

 

(18000 - x)(0.12) = x(0.05) + 885

 

2160 -0.12x = 0.05x + 885

 

 1275 = 0.17x

 

x = 7500

 

Therefore y = 18000 - 7500 = 10500