A man invests a certain amount of money at 2% interest and $800 more than that amount in another account at 4% interest. At the

Question

3 years ago

7

A man invests a certain amount of money at 2% interest and $800 more than that amount in another account at 4% interest. At the

end of one year he earned $112 in interest. How much money was invested in each account? Which of the following equations could be used to solve the problem? 0.02x + 0.04(800) = 112 0.02x + 0.04(x + 800) = 112 0.04x + 0.02(x + 800) = 112

Mathematics

2 answers:

Alex777 [14] · Answer 1 · 2021-11-30T03:16:43+03:00

Alex777 [14]3 years ago

6 0

112 dallors interest beca0.02x + 0.04(800) = 112 0.02x + 0.04(x + 800) = 112 0.04x + 0.02(x + 800) = 112 use

GaryK [48] · Answer 2 · 2021-11-30T04:56:04+03:00

Answer:

0.02x + 0.04(800) = 112

Step-by-step explanation:

Let the unknown amount be x

The amount is invested at 2 % interest so the intrest will be $ 0.02x

However, another amount, this time $ 800 is invested at 4 % interest. This gives:

0.04 (800)

The total amount at the end of the year will be = $ 112

Therefore, the sum of the two interests will be $\\$ 0.02x+ 0.04($800) = $ 112.00$