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16 Jan 2019, 06:08
x/3 *4% + 2/3x *6% = 320
solve for X ---> 6000

100,000 - 6000 --> Ans B
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Joined: 10 Apr 2018
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Location: United States (NC)

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22 Jun 2019, 05:13
Hello,

suppose the amount spent on the house is X$and the amount remaining is Y$

1/3 of Y were invested at 4% which gives (0.33*0.04)Y =0.0132Y (you can calculate this manually by multiplying 33 by 4 and adding the decimals later)
2/3 of Y were invested at 4% which gives (0.66*0.06)Y=0.0396Y

Given that: total investment amount=320$so 0.0132Y + 0.0396Y=320$

0.0528Y=320$y=6000$ approx. in this question you dont need the exact amount you need an approximate amount, so you can divide 300/0.05 which is 6000$now you deduct Y from the original amount which is 100000$

the answer is 94000$which is B _________________ HIT KUDOS IF YOU FOUND ME HELPFUL ! Intern Joined: 01 Jan 2019 Posts: 12 Re: David used part of$100,000 to purchase a house. Of the remaining port  [#permalink]

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23 Jul 2019, 21:44
Let the amount left after investing in the house be x.

(4/300)*x + (12/300)*x=320

x=6000

100000-6000= 94000

Ans: B
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Joined: 20 Apr 2019
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11 Sep 2019, 02:32
GMATINSECT wrote:
David used part of $100,000 to purchase a house. Of the remaining portion, he invested 1/3 of it at 4 percent simple annual interest and 2/3 of it at 6 percent simple annual interest. If after a year the income from the two investments totaled$320, what was the purchase price of the house?
Let X be the invested amount
Sol : 1x/3 * 0.04 + 2x/3 * 0.06 = 320 (total investment)
or 0.04x/3 + 0.12x/3 = 320
or 0.16x/3=320
or x = 320 * 3 * 100 /16 = 6000
David used part of $1,00,000(given) , hence = 100000-6000 =$ 94,000 QA

Thank you very much. I liked your simple approach more than any other. I made a slight change due to decimal aversion.

Let x be invested amount.
1x/3 * 4/100 + 2x/3 * 6/100 = $320 4x/300 + 12x/300 =$320
16x/300 = $320 16x =$320 * 300
16x = $96,000 x =$96,000/16
x = $6000$100,000 - $6,000 =$94,000