## David used part of $100,000 to purchase a house ##### This topic has expert replies Junior | Next Rank: 30 Posts Posts: 14 Joined: 24 Nov 2012 ### David used part of$100,000 to purchase a house

by ritumaheshwari02 » Thu Nov 29, 2012 8:40 am
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. $40,000 Newbie | Next Rank: 10 Posts Posts: 2 Joined: 29 Nov 2012 Location: Toronto, Canada by Avi_Gutman » Thu Nov 29, 2012 9:00 am When you invest different portions of your money and get different rates of return (in this case simple annual interest), your overall rate of return can be calculated as a weighted average, where the portion in each investment is the weight for that interest. In this case, the weighted average will be twice as close to 6% as it is to 4% (because 6% has double the weight). Take the distance between 4% and 6%, divide it into 3 equal parts (so each part is 2/3%) and pick 5 1/3 % (twice as close to 6% as it is to 4%). Rewrite 5 1/3 % as an improper fraction (you should always do this to make GMAT calculations easier) and you find the overall interest rate on the investments was 16/3 %. If David had$X left over after he purchased the house, then $16X/300 =$320 (I simply multiplied X by 16/3 % and translated % to 1/100).
Divide that equation by 16 and you'll get x/300=20
Multiply by 300 to get X=$6,000 So the house cost$100,000-$6,000=$94,000

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by GMATGuruNY » Thu Nov 29, 2012 12:42 pm
ritumaheshwari02 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 Of every 3 dollars invested,$1 earns 4% interest and $2 earns 6% interest. Average interest earned by every$3 = (1*4 + 2*6)/3 = 16/3% = 16/300.

From here we can plug in the answers, which represent the purchase price of the house.

Amount invested = 100,000 - 88,000 = 12000.
Interest earned = (16/300)(12000) = 640.
Since the interest earned is too great, the purchase price of the house must INCREASE, so that the amount invested DECREASES.

Amount invested = 100,000 - 94,000 = 6000.
Interest earned = (16/300)(6000) = 320.
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by kevincanspain » Thu Nov 29, 2012 2:47 pm
Suppose he invested x at 4%. Then he invested 2x at 6% and the price of the home was 100000 - 3x

320 = 0.04x + 0.06(2x) = 0.16x

x = 320/0.16 = 2000

The price of the home was 100000-3(2000) = 94000
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### Re: David used part of $100,000 to purchase a house by [email protected] » Fri Apr 08, 2022 6:30 am ritumaheshwari02 wrote: Thu Nov 29, 2012 8:40 am 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
In this case, we can easily test the answer choices.
Now let's give ourselves up to 20 seconds to identify a faster approach.
In this case, we can also solve the question algebraically. However, since I can already see we can immediately eliminate two answer choices, I know that I need only test ONE answer choice.
So I'm going to start testing answer choices

First recognize that the correct answer can't be A or D. Here's why:
If the correct answer were choice A ($96,000), then David invested the remaining$4000, which means he invested 1/3 of $4000 at an interest rate of 4% and 2/3 of$4000 at an interest rate of 6%.
The problem is that $4000 isn't divisible by 3. So, for example, 1/3 of$4000 = $1333.33333...., which means David invested$1333 and one-third of a penny at 4%.
Since we can't invest one-third of a penny, answer choice A is disqualified.
We can apply the same logic to eliminate answer choice D.

Of the remaining three answer choices (B, C and E), I'll test the middle answer...
(C) $88,000 If David spent$88,000 of his $100,000 on his house, then he invested the remaining$12,000
1/3 of $12,000 is$4000, which means David invested $4000 at a 4% simple interest rate, which means the interest after 1 year = 4% of$4000 = $160 2/3 of$12,000 is $8000, which means David invested$8000 at a 6% simple interest rate, which means the interest after 1 year = 6% of $8000 =$480
So, the TOTAL income from the two investments = $160 +$480 = $640 The question tells us that the total income from the two investments is$320, which means we can eliminate choice C.
Also, since we want the total income to be LESS THAN $640, we need David to invest less than$12,000.
In other words, we need the house price to be greater than \$88,000, which means we can also eliminate answer choice E.

By the process of elimination, the correct answer must be B.