Math questions

Question

1)Jose invested 38,000 in two accounts one yielding 7 percent and another yielding 12 percent if she received a total of 3,560 in interest at the end of the year . How much did she invest in each account.

 

2) A man invested 10,000 in two business one business earned a profit of 5 percent while the other lost a profit of 7.5 percent. Find the amount invested if he earned a total of 25 dollars for a year.

Answer 1

Dolores,

 

(1) Let x be the amount invested at 7% and y be the amount invested at 12%.

 

First, we know that the sum of the two investments is $38,000, so:

 

x + y = $38000

 

Now, we also know that the sum of the interest gained on the two investments is $3560, so:

 

.07x + .12y = $3560

 

We now have two equations and two variables so we can use substitution to solve.

 

Rewrite x + y = $38000 as x = $38000 - y

 

Now, substitute this value of x into the interest equation:

 

.07(38000 - y) + .12y = 3560

 

Distribute:

 

2660 - .07y + .12y = 3560

 

2660 + .05y = 3560

 

Subtract 2660 from both sides:

 

.05y = 3560 - 2660

 

.05y = 900

 

Divide both sides by .05:

 

.05y/.05 = 900/.05

 

y = $18000 (the amount he invested at 12%)

 

To find x:

 

x + $18000 =$38000

 

Subtract $19800 from both sides:

 

x = $38000 - $18000

 

x = $20000 (the amount invested at 7%)

 

I'll let you try (2) ;-)

 

Hope this helps!