SOLUTION: Susan invested a total of $10000 in two accounts. One account pays 4 interest and the other account pays 6% Interest. The total retum from both accounts was $560.
Answer the fo
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Question 1183978: Susan invested a total of $10000 in two accounts. One account pays 4 interest and the other account pays 6% Interest. The total retum from both accounts was $560.
Answer the following questions.
How much was invested at 4%?
How much was invested at 6%?
Let x be the amount invested at 6%.
Then the amount invested at 4% is (10000-x) dollars.
The 6% investment generates 0.06x dollars as the annual interest.
The 4% investment generates 0.04*(10000-x) dollars as the annual interest.
The total annual interest is the sum 0.06 + 0.04*(10000-x) dollars.
It is equal to $560 (given).
So we have this equation for the total annual interest
0.06x + 0.04*(10000-x) = 560 dollars.
From the equation
x = = 8000.
ANSWER. $8000 invested at 6% and the rest, 10000-8000 = 2000 dollars invested at 4%.
CHECK. 0.06*8000 + 0.04*2000 = 560 dollars, total annual interest. ! Correct !
The other tutor showed a good formal algebraic solution using the standard method.
If a formal algebraic solution is not required, and if the speed of getting an answer is important (as in a timed math competition), then here is a very fast and easy way to solve this problem (and many similar problems) mentally.
(1) $10000 all invested at 4% would yield $400 interest; all at 6% would yield $600 interest.
(2) The actual interest $560 is 4/5 of the way from $400 to $600. (Picturing the three numbers 400, 560, and 600 on a number line might help: 400 to 600 is a difference of 200; 400 to 560 is a difference of 160; 160/200 = 4/5.)
(3) That means 4/5 of the $10000 was invested at the higher rate.
ANSWER: 4/5 of $10000, or $8000, was invested at 6%; the other $2000 at 4%.