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26 Oct 2020, 10:37
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
15%
(low)
Question Stats:
93% (01:58) correct
7%
(01:51)
wrong
based on 44
sessions
History
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Not Attempted Yet
Hector invested a total of $6,000. Part of the money was invested in a money market account that paid 9% simple annual interest, and the remainder of the money was invested in a fund that paid 7% simple annual interest. If the interest earned at the end of the first year from these investments was $490, how much did Hector invest at 9%?
A. $2,000
B. $2,500
C. $3,000
D. $3,500
E. $4,000
A. $2,000
B. $2,500
C. $3,000
D. $3,500
E. $4,000
26 Oct 2020, 10:43
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IMO D
Let's assume x and y as the part investment amounts.
Given : x + y = 6000 ........(1)
and 9/100 x + 7/100 y = 490 ....(2)
Solving eq 2, we get 9x + 7 y = 49000 and we need to find the value for x
Multiplying Eq 1 with 7 we get 7x+7y = 42000
Now (1) - (2) >>> 2x = 7000 and X = 3500
Let's assume x and y as the part investment amounts.
Given : x + y = 6000 ........(1)
and 9/100 x + 7/100 y = 490 ....(2)
Solving eq 2, we get 9x + 7 y = 49000 and we need to find the value for x
Multiplying Eq 1 with 7 we get 7x+7y = 42000
Now (1) - (2) >>> 2x = 7000 and X = 3500
26 Oct 2020, 11:00
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490 interest upon 6000 means 49/60 % interest
Using weighted average for the interests
7---- 49/60 ---- 9
5/60 -----------7/60
So the ratio of amounts invested with 7% and 9% interests is 5:7
So 12x =6000
7x=3500
So 3500 was invested at 9% interest rate , IMO
Posted from my mobile device
Using weighted average for the interests
7---- 49/60 ---- 9
5/60 -----------7/60
So the ratio of amounts invested with 7% and 9% interests is 5:7
So 12x =6000
7x=3500
So 3500 was invested at 9% interest rate , IMO
Posted from my mobile device
