math problems?

Question

Bob invested $20,000, part at 14% and part at 13%. If the total interest at the end of the year is $2,720, how much did he invest at 14%?

A) $11,000 B) $12,000 C) $8,000 D) $13,000

Answer 1

Let x and y the amounts invested at 14% and 13% respectively.

14x + 13y = 2720,00

x + y = 20,000

14x + 14y = 280000

y = 280000 - 272000 = 8000 Subtracting the equation containing 13 y from that containing 14y)

y = 8000 and x = 12000

So, 12000 was invested at 14% and B is the answer

Answer 2

Let's set the investment at 14% as x, so the investment at 13% was 20000-x. The income was 2,720, so

0.14 x + 0.13 (20000 - x ) = 2720
0.14 x + 2600 - 0.13 x = 2720
subtracting 2600 from both sides of the equation,
0.14 x - 0.13 x = 120
0.01 x = 120
x = 12000

Hence, $12,000 was invested at 14% and $8,000 at 13%.

Answer 3

C)$8,000
Hope this Helps
Good Luck