December 11, 2018 December 11, 2018 09:00 PM EST 10:00 PM EST Strategies and techniques for approaching featured GMAT topics. December 11 at 9 PM EST. December 13, 2018 December 13, 2018 08:00 AM PST 09:00 AM PST What people who reach the high 700's do differently? We're going to share insights, tips and strategies from data we collected on over 50,000 students who used examPAL.
Author 
Message 
TAGS:

Hide Tags

Current Student
Status: Persevere
Joined: 08 Jan 2016
Posts: 119
Location: Hong Kong
GPA: 3.52

David used part of $100,000 to purchase a house. Of the remaining port
[#permalink]
Show Tags
27 May 2016, 02:09
Question Stats:
75% (02:08) correct 25% (02:34) wrong based on 773 sessions
HideShow timer Statistics
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
Official Answer and Stats are available only to registered users. Register/ Login.




Senior Manager
Joined: 18 Jan 2010
Posts: 252

Re: David used part of $100,000 to purchase a house. Of the remaining port
[#permalink]
Show Tags
27 May 2016, 02:31
Let the Price of house was x.
Then
\(\frac{1}{3}\)(10000x) * \(\frac{4}{100}\) > 1st investment
\(\frac{2}{3}\)(10000x) * \(\frac{6}{100}\) > 2nd investment
\(\frac{4}{300}\)(10000x) *(1+3) = 320
(10000x) = 20 * 300 = 6000
x = 94000
Option B is correct




Intern
Joined: 05 May 2016
Posts: 20
Location: United States
WE: Web Development (Computer Software)

Re: David used part of $100,000 to purchase a house. Of the remaining port
[#permalink]
Show Tags
14 Jun 2016, 10:57
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 An alternative solution: "Of the remaining portion" is split two ways  1/3 & 2/3. The interest earned is an integer amount. So, we can pick some smart numbers for the remaining amount and calculate the interest rate. Use answer choices to pick a number that is divisible by 3. Of all the answer choices, (A) and (D) do not offer a number divisible by 3. (E) can be eliminated easily because when you invest 1/3 of $60,000 ($20,000) @ 4% interest rate, you earn $800 interest, which is definitely more than $320. But, if you calculate 2/3 of $60,000 ($40,000) @ 6% interest rate, you earn $2400 interest. Total interest when $60,000 is the remaining amount: $800 + $2400 = $3200. We need $320. It means the remaining amount should be $6000. Therefore, (B) is the answer. It may seem like a little bit lengthy approach, but with enough practice you can save yourself some trouble solving algebraic equations and save some time Happy solving!



Intern
Joined: 26 Sep 2017
Posts: 13

Re: David used part of $100,000 to purchase a house. Of the remaining port
[#permalink]
Show Tags
24 Nov 2017, 02:13
adiagr wrote: Let the Price of house was x.
Then
\(\frac{1}{3}\)(10000x) * \(\frac{4}{100}\) > 1st investment
\(\frac{2}{3}\)(10000x) * \(\frac{6}{100}\) > 2nd investment
\(\frac{4}{300}\)(10000x) *(1+3) = 320
(10000x) = 20 * 300 = 6000
x = 94000
Option B is correct Hi adiagr, Could you please clarify why there is no (1+%) in your base formula. As I remember simple interest formula should be interest*(1+years*%/100). What am I doing wrong? Thanks you in advance



Intern
Joined: 15 Oct 2017
Posts: 30
Location: Ireland
Concentration: Healthcare, Finance

David used part of $100,000 to purchase a house. Of the remaining port
[#permalink]
Show Tags
Updated on: 25 Nov 2017, 22:29
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 \(x\) be the amount of money left after purchase of the house. So \(\frac{1}{3} * x\) was invested under \(\frac{104}{100}\) percent and \(\frac{2}{3} * x\) was invested under \(\frac{106}{100}\). and \(x + 320\) we will get our investment back with the interests \(\frac{x}{3} * \frac{104}{100} + \frac{2}{3}x * \frac{106}{100}\) =\(x + 320\) \(\frac{104x + 2 * 106x}{300} = x + 320\) \(316x = 300x + 320 * 300\) \(16x = 32 * 10 * 300\) \(x = 6000\) To get the price of the house we should substract the money left after after purchasing from 100,000 : \(100000  6000 = 94000\) Hence, the answer is B
Originally posted by Vorovski on 24 Nov 2017, 03:17.
Last edited by Vorovski on 25 Nov 2017, 22:29, edited 1 time in total.



Senior SC Moderator
Joined: 22 May 2016
Posts: 2204

David used part of $100,000 to purchase a house. Of the remaining port
[#permalink]
Show Tags
25 Nov 2017, 12:46
lor12345 wrote: adiagr wrote: Let the Price of house was x.
Then
\(\frac{1}{3}\)(10000x) * \(\frac{4}{100}\) > 1st investment
\(\frac{2}{3}\)(10000x) * \(\frac{6}{100}\) > 2nd investment
\(\frac{4}{300}\)(10000x) *(1+3) = 320
(10000x) = 20 * 300 = 6000
x = 94000
Option B is correct Hi adiagr, Could you please clarify why there is no (1+%) in your base formula. As I remember simple interest formula should be interest*(1+years*%/100). What am I doing wrong? Thanks you in advance lor12345 , I think adiagr might not be around anymore. It looks as if you just got the formulas slightly mixed up. There is no (1 + %) in his formula because $320 is interest only. See his RHS: it's $320. That's interest only. (We weren't given $6,320, which in this case is principal + interest.) Simple interest with principal PLUS accrued interest ("A") formula is indeed A = P(1 + rt)But interest only for simple interest formula is I = (P)(r)(t)adiagr used this formula, and defined P as ($100,000  x)*(1/3) in one case, and ($100,000  x)*(2/3) in the other. True, both are multiplied by the respective interest rates, but without the 1 because his x is derived from subtraction from $100,000. If you use a multiplier with a 1, you include interest earned on $100,000. That is not how it worked. David earned interest on some small part of $100,000. If you want to include 1 in the multiplier, it seems easiest if you separate the principal (which earned interest) from the $100,000 (most of which did not). It's plenty hard either way, IMO. See Vorovski , above, who used a (1+%) multiplier by considering ONLY x as principal (and not $100,000  x, until the end). (Whew. I backsolved (used the answer choices) in about 90 seconds. See how I used the answer choices, here.) Hope that helps.



Intern
Joined: 26 Sep 2017
Posts: 13

Re: David used part of $100,000 to purchase a house. Of the remaining port
[#permalink]
Show Tags
26 Nov 2017, 03:19
genxer123 wrote: lor12345 wrote: adiagr wrote: Let the Price of house was x.
Then
\(\frac{1}{3}\)(10000x) * \(\frac{4}{100}\) > 1st investment
\(\frac{2}{3}\)(10000x) * \(\frac{6}{100}\) > 2nd investment
\(\frac{4}{300}\)(10000x) *(1+3) = 320
(10000x) = 20 * 300 = 6000
x = 94000
Option B is correct Hi adiagr, Could you please clarify why there is no (1+%) in your base formula. As I remember simple interest formula should be interest*(1+years*%/100). What am I doing wrong? Thanks you in advance lor12345 , I think adiagr might not be around anymore. It looks as if you just got the formulas slightly mixed up. There is no (1 + %) in his formula because $320 is interest only. See his RHS: it's $320. That's interest only. (We weren't given $6,320, which in this case is principal + interest.) Simple interest with principal PLUS accrued interest ("A") formula is indeed A = P(1 + rt)But interest only for simple interest formula is I = (P)(r)(t)adiagr used this formula, and defined P as ($100,000  x)*(1/3) in one case, and ($100,000  x)*(2/3) in the other. True, both are multiplied by the respective interest rates, but without the 1 because his x is derived from subtraction from $100,000. If you use a multiplier with a 1, you include interest earned on $100,000. That is not how it worked. David earned interest on some small part of $100,000. If you want to include 1 in the multiplier, it seems easiest if you separate the principal (which earned interest) from the $100,000 (most of which did not). It's plenty hard either way, IMO. See Vorovski , above, who used a (1+%) multiplier by considering ONLY x as principal (and not $100,000  x, until the end). (Whew. I backsolved in about 90 seconds. Much faster for me.) Hope that helps. Thank you for your help!



Manager
Joined: 02 Jan 2017
Posts: 73
Location: Pakistan
Concentration: Finance, Technology
GPA: 3.41
WE: Business Development (Accounting)

David used part of $100,000 to purchase a house. Of the remaining port
[#permalink]
Show Tags
16 Dec 2017, 20:53
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) $40,000 100,000 = 100 K h = house[100k  h]/3 * (4/100) + [100k  h]/3* (12/100) = 320 Taking [100k  h]/3 * (4/100) as a factor[100k  h]/3 * (4/100) *[ 1+3] = 320 [100k  h]/3 * (4/100)*4 = 320 [100k  h]*16/100 =320 *3 100k  h = (320*3*100)/16 h= 96000



Intern
Joined: 17 Jun 2017
Posts: 33

Re: David used part of $100,000 to purchase a house. Of the remaining port
[#permalink]
Show Tags
30 Jan 2018, 12:01
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 = 1000006000 = $ 94,000 QA



Manager
Joined: 11 Feb 2017
Posts: 189

David used part of $100,000 to purchase a house. Of the remaining port
[#permalink]
Show Tags
05 Mar 2018, 02:09
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 Does the term income indicates that we need to use (100000x)/3 * 4/100 + (100000x)* 2/3 * 6/100 = 320? INSTEAD OF (100000x)/3 * 104/100 + (100000x)* 2/3 * 106/100 =320 ? please help



Director
Joined: 26 Oct 2016
Posts: 641
Location: United States
Concentration: Marketing, International Business
GPA: 4
WE: Education (Education)

Re: David used part of $100,000 to purchase a house. Of the remaining port
[#permalink]
Show Tags
18 Mar 2018, 18:15
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. Answer choice C: 88,000 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. Answer choice B: 94,000 Amount invested = 100,000  94,000 = 6000. Interest earned = (16/300)(6000) = 320. Success! The correct answer is B.
_________________
Thanks & Regards, Anaira Mitch



VP
Joined: 09 Mar 2016
Posts: 1213

David used part of $100,000 to purchase a house. Of the remaining port
[#permalink]
Show Tags
20 Mar 2018, 12:50
generis wrote: lor12345 wrote: adiagr wrote: Let the Price of house was x.
Then
\(\frac{1}{3}\)(10000x) * \(\frac{4}{100}\) > 1st investment
\(\frac{2}{3}\)(10000x) * \(\frac{6}{100}\) > 2nd investment
\(\frac{4}{300}\)(10000x) *(1+3) = 320
(10000x) = 20 * 300 = 6000
x = 94000
Option B is correct Hi adiagr, Could you please clarify why there is no (1+%) in your base formula. As I remember simple interest formula should be interest*(1+years*%/100). What am I doing wrong? Thanks you in advance lor12345 , I think adiagr might not be around anymore. It looks as if you just got the formulas slightly mixed up. There is no (1 + %) in his formula because $320 is interest only. See his RHS: it's $320. That's interest only. (We weren't given $6,320, which in this case is principal + interest.) Simple interest with principal PLUS accrued interest ("A") formula is indeed A = P(1 + rt)But interest only for simple interest formula is I = (P)(r)(t)adiagr used this formula, and defined P as ($100,000  x)*(1/3) in one case, and ($100,000  x)*(2/3) in the other. True, both are multiplied by the respective interest rates, but without the 1 because his x is derived from subtraction from $100,000. If you use a multiplier with a 1, you include interest earned on $100,000. That is not how it worked. David earned interest on some small part of $100,000. If you want to include 1 in the multiplier, it seems easiest if you separate the principal (which earned interest) from the $100,000 (most of which did not). It's plenty hard either way, IMO. See Vorovski , above, who used a (1+%) multiplier by considering ONLY x as principal (and not $100,000  x, until the end). (Whew. I backsolved in about 90 seconds. Much faster for me.) Hope that helps. Hi generis how is life ?:) well explained:) i actually had the same question but after reading your post got it, so in general 1+% means a principal account +acrrued percent right ? but we are interested only in finding accrued percent so that's why we use only 0.04 and 0.06 without "+1" , correct ?



VP
Joined: 09 Mar 2016
Posts: 1213

Re: David used part of $100,000 to purchase a house. Of the remaining port
[#permalink]
Show Tags
20 Mar 2018, 12:55
adiagr wrote: Let the Price of house was x.
Then
\(\frac{1}{3}\)(10000x) * \(\frac{4}{100}\) > 1st investment
\(\frac{2}{3}\)(10000x) * \(\frac{6}{100}\) > 2nd investment
\(\frac{4}{300}\)(10000x) *(1+3) = 320
(10000x) = 20 * 300 = 6000
x = 94000
Option B is correct generis, how de got this \(\frac{4}{300}\)(10000x) *(1+3) = 320[/m]



Senior SC Moderator
Joined: 22 May 2016
Posts: 2204

David used part of $100,000 to purchase a house. Of the remaining port
[#permalink]
Show Tags
20 Mar 2018, 17:41
rocko911 wrote: 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 Does the term income indicates that we need to use (100000x)/3 * 4/100 + (100000x)* 2/3 * 6/100 = 320? INSTEAD OF (100000x)/3 * 104/100 + (100000x)* 2/3 * 106/100 =320 ? please help rocko911 , yes, exactly. The word "income" tells us we are solving for interest only. Interest is defined in one way as the income earned on a principal amount of money. The original principal amount of money is NOT defined as "income." In addition, here, we are NOT told something such as: "after a year, the total amount of money from David's investment is $_____" So the word "income" lets us know that we need to use the "interest only" formula. The income earned from the money invested = interest = $320Interest only = (Principal Amt) * \(\frac{rate}{100}\) * time OR, Interest only = (Principal Amt * rate * time), if the interest rate is in decimal form Your first equation is correct. Hope that helps.



Senior SC Moderator
Joined: 22 May 2016
Posts: 2204

David used part of $100,000 to purchase a house. Of the remaining port
[#permalink]
Show Tags
20 Mar 2018, 18:05
dave13 wrote: generis wrote: There is no (1 + %) in his formula because $320 is interest only. See his RHS: it's $320. That's interest only. (We weren't given $6,320, which in this case is principal + interest.)
Simple interest with principal PLUS accrued interest ("A") formula is indeed
A = P(1 + rt)
But interest only for simple interest formula is
I = (P)(r)(t)
(Whew. I backsolved in about 90 seconds. Much faster for me.)
Hope that helps.
Hi generis how is life ?:) well explained:) i actually had the same question but after reading your post got it, so in general 1+% means a principal account +acrrued percent right ? but we are interested only in finding accrued percent so that's why we use only 0.04 and 0.06 without "+1" , correct ?Hi dave13  life is good. You? Yes, you are 100% correct.If you see a 1.xx for an interest rate, it almost always means the principal has been included...If you see a 0.xx for an interest rate, it means the principal has not been included.Nice work. Below is an explanation for WHY the "1" usually means that we're talking about principal, too. The 1 means 100% of Principal= \(\frac{100}{100} P= 1P\) For SIMPLE INTEREST (not compound)
• interest only formula is \(i = P * r * t\) \(i\) = interest \(P\) = principal \(r\) = rate in decimal form \(t\) = time
• Total Amount(principal plus interest) has a different formula. But it borrows from the interest only formula.
The "1" stands for 100% of original Principal, or \(\frac{100}{100}P\) or \(1P\)
Total amount (principal plus interest) earned formula is: \(A = P(1 + rt)\) Breaking it down, Total Amount, A A = Principal + Interest A = Principal + (P*r*t) (from above for interest only) A = P + (P*r*t) Now factor out P A = P (1 + rt) If this makes no sense to you, ignore it. You have the basic concept!



Senior SC Moderator
Joined: 22 May 2016
Posts: 2204

David used part of $100,000 to purchase a house. Of the remaining port
[#permalink]
Show Tags
21 Mar 2018, 00:07
dave13 wrote: adiagr wrote: Let the Price of house was x. Then \(\frac{1}{3}\)(10000x) * \(\frac{4}{100}\) > 1st investment \(\frac{2}{3}\)(10000x) * \(\frac{6}{100}\) > 2nd investment
\(\frac{4}{300}\)(10000x) *(1+3) = 320
(10000x) = 20 * 300 = 6000
x = 94000
Option B is correct generis, how de got this \(\frac{4}{300}\)(100,000x) * (1+3) = 320[/m] dave13  important steps were not spelled out. This one is a bear to explain. I tacked (*1) to the first expression in the equation because (*1) changes nothing but makes the factoring out easier to see. I have broken the steps down below. That said, I would not and did not use much algebra to solve. \(x\) = price of house \((100,000  x)\)= amount invested End game: Find a common factorOriginal equation (not in solution above) ( \(\frac{1}{3}*($100,000x)*\frac{4}{100}) + (\frac{2}{3}*($100,000x)*\frac{6}{100})= $320\)\((\frac{$100,000x}{3} * \frac{4}{100} * 1) + ((\frac{1}{3}*\frac{2}{1})*($100,000x)*\frac{6}{100})= $320\)\((\frac{$100,000x}{3} * \frac{4}{100} * 1) = (\frac{1}{3}*($100,000x)*\frac{2}{1}*\frac{6}{100}) = $320\)\((\frac{$100,000x}{3} * \frac{4}{100} * 1) + (\frac{$100,000x}{3}*\frac{12}{100}) = $320\)\((\frac{$100,000x}{3} * \frac{4}{100} * 1) + (\frac{$100,000x}{3}*\frac{4*3}{100}) = $320\)\((\frac{$100,000x}{3} * \frac{4}{100} * 1) + (\frac{$100,000x}{3}*\frac{4}{100} * 3) = $320\)COMMON FACTOR in both terms: \(\frac{$100,000x}{3}*\frac{4}{100}\)COMMON FACTOR, factored OUT of blue equation above\((\frac{$100,000x}{3} * \frac{4}{100} * 1) + (\frac{$100,000x}{3}*\frac{4}{100} * 3) = $320\)\((\frac{$100,000x}{3} *\frac{4}{100})\) * \((1 + 3) = $320\)When you factor it out of the first expression, you "leave" 1; out of the second expression, you "leave" 3. (Try multiplying it back the other way, e.g. A(b+c) = Ab + Ac) The rest: \((\frac{$100,000x}{3} *\frac{4}{100})* (1 + 3) = $320\)\((\frac{$100,000x}{3} *\frac{4}{100})* (4) = $320\)\(\frac{$100,000x}{3} * (\frac{4}{100} * 4) = $320\)\((\frac{$100,000x}{3} * \frac{16}{100})= $320\)Clear the 3 in the denominator. Multiply both sides by 3 \($100,000x * \frac{16}{100}= $960\)\($100,000  x = $960 * \frac{100}{16}\)\($100,000  x = \frac{$960}{16}*100\)\($100,000  x = $60 * 100\)\($100,000  x = $6,000\)\($100,000  $6,000 = x\)\($94,000 = x\) If you cannot follow that algebra, do not worry. It is not necessary to solve the problem with that particular version of algebra. You can use another kind of algebra. Or very little. I backsolved.Quote: 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 Start with C) $88,000If the house cost $88,000, then (100,000  88,000) = $12,000 is left over 1/3 of that $12,000 is invested at 4 percent $4,000 earns .04 2/3 of that $12,000 is invested at 6 percent $8,000 earns .06 Combined, do they earn $320 in a year? \(.04(4,000) + .06(8,000) =\) \(160 + 480 = $600\)NO. If the house costs $88,000, the leftover money is $12,000. It earns $600 in one year. Too much. The base principal (the leftover money ) is too great. The house needs to be more expensive. Try B) $94,000If the house cost $94,000, then (100,000  94,000) = $6,000 is left over 1/3 of that $6,000 is invested at 4 percent $2,000 earns .04 2/3 of that $6,000 is invested at 6 percent $4,000 earns .06 Combined, do they earn $320 in a year? \(.04(2,000) + .06(4,000) =\) \(80 + 240 = $320\)BINGO Answer B Hope that helps, dave13



Intern
Joined: 02 Mar 2017
Posts: 4
Location: United States
GPA: 3.65

Re: David used part of $100,000 to purchase a house. Of the remaining port
[#permalink]
Show Tags
30 Sep 2018, 08:16
I started with the question. x = price of the house. 100k = 100,000 I didn't calculate anything until the very end of the equation.
100k  x = the amount left after house purchase.
[(100k  x)/3]*0.04 + 2[(100k  x)/3]*0.06 = 320 (I litterally intepret everything from the question to create this equation).
=> 0.04*100k  0.04x + 0.12*100k  0.12x = 960 0.16(100kx) = 960 100k  x = 960/0.16 = 6000 => x = 100k  6,000 = 100,0006,000 = 94,000 => B



Director
Joined: 19 Oct 2013
Posts: 509
Location: Kuwait
GPA: 3.2
WE: Engineering (Real Estate)

Re: David used part of $100,000 to purchase a house. Of the remaining port
[#permalink]
Show Tags
04 Nov 2018, 10:34
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 I found it easier to start with backsolving. If the house costs 88,000 that means the remaining is 12,000 12,000 * 1/3 * 4/100 = 4,000 * 4/100 = 160. Without further calculation we can tell that the value from the 8,000 * 6/100 + 160 > 320 so we pick another choice. Pick B 94,000 means 6,000 is the remaining balance. 6,000 * 1/3 = 2,000 2,000 * 4/100 = 80 4,000 * 6/100 = 240 80 + 240 = 320



Intern
Joined: 02 Jun 2018
Posts: 3

Re: David used part of $100,000 to purchase a house. Of the remaining port
[#permalink]
Show Tags
11 Dec 2018, 10:35
Let y be the house and 6x the left over, then y+6x=100000
Given 2x * 4% + 4x* 6% =320 8x + 24x = 320*100 32x=320*100 Therefore x=1000 We need 6x= 6000 Therefore y=Price of house =94000




Re: David used part of $100,000 to purchase a house. Of the remaining port &nbs
[#permalink]
11 Dec 2018, 10:35






