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26 Nov 2018, 23:20
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
79% (02:32) correct
21%
(02:11)
wrong
based on 201
sessions
History
Date
Time
Result
Not Attempted Yet
An investor placed a total of $6,400 in two accounts for one year. One of the accounts earned simple annual interest at a rate of 5 percent, and the other earned simple annual interest at a rate of 3 percent. The investor made no deposits or withdrawals from the accounts. If each account earned the same amount of interest after one year, what was the total amount of interest earned from both accounts?
A. $128
B. $144
C. $240
D. $256
E. $512
A. $128
B. $144
C. $240
D. $256
E. $512
27 Nov 2018, 00:04
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Accounts=X and Y
equals interest; (3/100)X=(5/100)Y
X:Y=5:3
X=(5/8)*6400=4000
Y=(3/8)*6400=2400
Interest=4000*(3/100)+2400*(5/100)=$240
C.
equals interest; (3/100)X=(5/100)Y
X:Y=5:3
X=(5/8)*6400=4000
Y=(3/8)*6400=2400
Interest=4000*(3/100)+2400*(5/100)=$240
C.
Archit3110
27 Nov 2018, 07:29
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Bunuel
SI= P*R*T-(1)
P1+P2=$6400; P1=6400-P2
since both interests are equal so
P1*.03*1=P2*.05*1
(6400-P2)*.03=P2*.05
Solve we get P2= 2400 and P1= 4000
Find interest using formula in (1)
4000*.03*1= 120 and 2400*.05*1 = 120
total interest = $ 240 option C
