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18 Jun 2019, 11:18
The other method would be using allegation

Let the remaining amount be x then ,

$$\frac{1}{3}$$ x 4 % and $$\frac{2}{3}$$x 6 %

So average is$$\frac{16}{3}$$

so $$\frac{16}{3}$$ % of x =320

x =6000

So Amount invested in house is 94,000
_________________
Probus

~You Just Can't beat the person who never gives up~ Babe Ruth
Manager
Joined: 21 Jun 2019
Posts: 99
Concentration: Finance, Accounting
GMAT 1: 670 Q48 V34
GPA: 3.78
Re: David used part of $100,000 to purchase a house. Of the remaining port [#permalink] ### Show Tags 22 Jun 2019, 04:13 Hello, suppose the amount spent on the house is X$ 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
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Joined: 01 Jan 2019
Posts: 11

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02 Aug 2019, 09:25
1
nalinnair 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?

(A) $96,000 (B)$94,000
(C) $88,000 (D)$75,000
(E) $40,000 Let h be the cost of the house, remaining money for investment --> (100,000-h) (100,000-h)(1/3 * 4/100) + (100,000-h)(2/3 * 6/100) = 320 (100,000-h)(4/300 + 12/300) = 320 (100,000-h) = 320 * 300/16 100,000-h = 6,000 100,000 - 6,000 = h h = 94,000 Manager Joined: 20 Apr 2019 Posts: 112 Re: David used part of$100,000 to purchase a house. Of the remaining port  [#permalink]

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04 Aug 2019, 04:42
Probus wrote:
The other method would be using allegation

Let the remaining amount be x then ,

$$\frac{1}{3}$$ x 4 % and $$\frac{2}{3}$$x 6 %

So average is$$\frac{16}{3}$$

so $$\frac{16}{3}$$ % of x =320

x =6000

So Amount invested in house is 94,000

Why do we only calculate *0.04 and not *1.04?
Because usually we calculate principle*(1+%rate)^1 correct?
Intern
Joined: 28 Feb 2019
Posts: 3
David used part of $100,000 to purchase a house. Of the remaining port [#permalink] ### Show Tags 11 Sep 2019, 01:32 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 David used part of$100,000 to purchase a house. Of the remaining port   [#permalink] 11 Sep 2019, 01:32

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