GMAT Changed on April 16th - Read about the latest changes here

It is currently 26 Apr 2018, 16:09

Close

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

David used part of $100,000 to purchase a house. Of the remaining port

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

2 KUDOS received
Current Student
avatar
B
Status: Persevere
Joined: 09 Jan 2016
Posts: 122
Location: Hong Kong
GMAT 1: 750 Q50 V41
GPA: 3.52
Reviews Badge
David used part of $100,000 to purchase a house. Of the remaining port [#permalink]

Show Tags

New post 27 May 2016, 03:09
2
This post received
KUDOS
18
This post was
BOOKMARKED
00:00
A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

75% (02:01) correct 25% (02:41) wrong based on 415 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
[Reveal] Spoiler: OA
3 KUDOS received
Senior Manager
Senior Manager
User avatar
Joined: 18 Jan 2010
Posts: 256
Re: David used part of $100,000 to purchase a house. Of the remaining port [#permalink]

Show Tags

New post 27 May 2016, 03:31
3
This post received
KUDOS
4
This post was
BOOKMARKED
Let the Price of house was x.

Then

\(\frac{1}{3}\)(10000-x) * \(\frac{4}{100}\) --> 1st investment

\(\frac{2}{3}\)(10000-x) * \(\frac{6}{100}\) --> 2nd investment


\(\frac{4}{300}\)(10000-x) *(1+3) = 320


(10000-x) = 20 * 300 = 6000

x = 94000

Option B is correct
3 KUDOS received
Intern
Intern
avatar
B
Joined: 05 May 2016
Posts: 26
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

New post 14 Jun 2016, 11:57
3
This post received
KUDOS
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 :wink:

Happy solving!
Intern
Intern
avatar
B
Joined: 26 Sep 2017
Posts: 14
Re: David used part of $100,000 to purchase a house. Of the remaining port [#permalink]

Show Tags

New post 24 Nov 2017, 03:13
adiagr wrote:
Let the Price of house was x.

Then

\(\frac{1}{3}\)(10000-x) * \(\frac{4}{100}\) --> 1st investment

\(\frac{2}{3}\)(10000-x) * \(\frac{6}{100}\) --> 2nd investment


\(\frac{4}{300}\)(10000-x) *(1+3) = 320


(10000-x) = 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
Intern
User avatar
B
Joined: 16 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

New post Updated on: 25 Nov 2017, 23:29
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, 04:17.
Last edited by Vorovski on 25 Nov 2017, 23:29, edited 1 time in total.
1 KUDOS received
SC Moderator
avatar
D
Joined: 22 May 2016
Posts: 1574
Premium Member CAT Tests
David used part of $100,000 to purchase a house. Of the remaining port [#permalink]

Show Tags

New post 25 Nov 2017, 13:46
1
This post received
KUDOS
1
This post was
BOOKMARKED
lor12345 wrote:
adiagr wrote:
Let the Price of house was x.

Then

\(\frac{1}{3}\)(10000-x) * \(\frac{4}{100}\) --> 1st investment

\(\frac{2}{3}\)(10000-x) * \(\frac{6}{100}\) --> 2nd investment


\(\frac{4}{300}\)(10000-x) *(1+3) = 320


(10000-x) = 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.
_________________

In the depths of winter, I finally learned
that within me there lay an invincible summer.

-- Albert Camus, "Return to Tipasa"

Intern
Intern
avatar
B
Joined: 26 Sep 2017
Posts: 14
Re: David used part of $100,000 to purchase a house. Of the remaining port [#permalink]

Show Tags

New post 26 Nov 2017, 04:19
genxer123 wrote:
lor12345 wrote:
adiagr wrote:
Let the Price of house was x.

Then

\(\frac{1}{3}\)(10000-x) * \(\frac{4}{100}\) --> 1st investment

\(\frac{2}{3}\)(10000-x) * \(\frac{6}{100}\) --> 2nd investment


\(\frac{4}{300}\)(10000-x) *(1+3) = 320


(10000-x) = 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
Manager
avatar
B
Joined: 02 Jan 2017
Posts: 81
Location: Pakistan
Concentration: Finance, Technology
GMAT 1: 650 Q47 V34
GPA: 3.41
WE: Business Development (Accounting)
Reviews Badge
David used part of $100,000 to purchase a house. Of the remaining port [#permalink]

Show Tags

New post 16 Dec 2017, 21:53
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
1 KUDOS received
Intern
Intern
avatar
B
Joined: 17 Jun 2017
Posts: 12
CAT Tests
Re: David used part of $100,000 to purchase a house. Of the remaining port [#permalink]

Show Tags

New post 30 Jan 2018, 13:01
1
This post received
KUDOS
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
Manager
Manager
avatar
B
Joined: 11 Feb 2017
Posts: 204
David used part of $100,000 to purchase a house. Of the remaining port [#permalink]

Show Tags

New post 05 Mar 2018, 03:09
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

(100000-x)/3 * 4/100 + (100000-x)* 2/3 * 6/100 = 320?

INSTEAD OF

(100000-x)/3 * 104/100 + (100000-x)* 2/3 * 106/100 =320 ?





please help
1 KUDOS received
Director
Director
User avatar
G
Joined: 26 Oct 2016
Posts: 673
Location: United States
Concentration: Marketing, International Business
Schools: HBS '19
GMAT 1: 770 Q51 V44
GPA: 4
WE: Education (Education)
Re: David used part of $100,000 to purchase a house. Of the remaining port [#permalink]

Show Tags

New post 18 Mar 2018, 19:15
1
This post received
KUDOS
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

1 KUDOS received
Senior Manager
Senior Manager
User avatar
G
Joined: 09 Mar 2016
Posts: 450
David used part of $100,000 to purchase a house. Of the remaining port [#permalink]

Show Tags

New post 20 Mar 2018, 13:50
1
This post received
KUDOS
generis wrote:
lor12345 wrote:
adiagr wrote:
Let the Price of house was x.

Then

\(\frac{1}{3}\)(10000-x) * \(\frac{4}{100}\) --> 1st investment

\(\frac{2}{3}\)(10000-x) * \(\frac{6}{100}\) --> 2nd investment


\(\frac{4}{300}\)(10000-x) *(1+3) = 320


(10000-x) = 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 ?
Senior Manager
Senior Manager
User avatar
G
Joined: 09 Mar 2016
Posts: 450
Re: David used part of $100,000 to purchase a house. Of the remaining port [#permalink]

Show Tags

New post 20 Mar 2018, 13:55
adiagr wrote:
Let the Price of house was x.

Then

\(\frac{1}{3}\)(10000-x) * \(\frac{4}{100}\) --> 1st investment

\(\frac{2}{3}\)(10000-x) * \(\frac{6}{100}\) --> 2nd investment


\(\frac{4}{300}\)(10000-x) *(1+3) = 320


(10000-x) = 20 * 300 = 6000

x = 94000

Option B is correct




generis, how de got this :? \(\frac{4}{300}\)(10000-x) *(1+3) = 320[/m]
SC Moderator
avatar
D
Joined: 22 May 2016
Posts: 1574
Premium Member CAT Tests
David used part of $100,000 to purchase a house. Of the remaining port [#permalink]

Show Tags

New post 20 Mar 2018, 18:41
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

(100000-x)/3 * 4/100 + (100000-x)* 2/3 * 6/100 = 320?

INSTEAD OF

(100000-x)/3 * 104/100 + (100000-x)* 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 = $320

Interest 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. :-)
_________________

In the depths of winter, I finally learned
that within me there lay an invincible summer.

-- Albert Camus, "Return to Tipasa"

1 KUDOS received
SC Moderator
avatar
D
Joined: 22 May 2016
Posts: 1574
Premium Member CAT Tests
David used part of $100,000 to purchase a house. Of the remaining port [#permalink]

Show Tags

New post 20 Mar 2018, 19:05
1
This post received
KUDOS
1
This post was
BOOKMARKED
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!


_________________

In the depths of winter, I finally learned
that within me there lay an invincible summer.

-- Albert Camus, "Return to Tipasa"

1 KUDOS received
SC Moderator
avatar
D
Joined: 22 May 2016
Posts: 1574
Premium Member CAT Tests
Re: David used part of $100,000 to purchase a house. Of the remaining port [#permalink]

Show Tags

New post 21 Mar 2018, 01:07
1
This post received
KUDOS
dave13 wrote:
adiagr wrote:
Let the Price of house was x.
Then
\(\frac{1}{3}\)(10000-x) * \(\frac{4}{100}\) --> 1st investment

\(\frac{2}{3}\)(10000-x) * \(\frac{6}{100}\) --> 2nd investment

\(\frac{4}{300}\)(10000-x) *(1+3) = 320

(10000-x) = 20 * 300 = 6000

x = 94000

Option B is correct

generis, how de got this :? \(\frac{4}{300}\)(100,000-x) * (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 factor

Original equation (not in solution above)
(\(\frac{1}{3}*($100,000-x)*\frac{4}{100}) + (\frac{2}{3}*($100,000-x)*\frac{6}{100})= $320\)

\((\frac{$100,000-x}{3} * \frac{4}{100} * 1) + ((\frac{1}{3}*\frac{2}{1})*($100,000-x)*\frac{6}{100})= $320\)

\((\frac{$100,000-x}{3} * \frac{4}{100} * 1) = (\frac{1}{3}*($100,000-x)*\frac{2}{1}*\frac{6}{100})
= $320\)


\((\frac{$100,000-x}{3} * \frac{4}{100} * 1) + (\frac{$100,000-x}{3}*\frac{12}{100}) = $320\)

\((\frac{$100,000-x}{3} * \frac{4}{100} * 1) + (\frac{$100,000-x}{3}*\frac{4*3}{100}) = $320\)

\((\frac{$100,000-x}{3} * \frac{4}{100} * 1) + (\frac{$100,000-x}{3}*\frac{4}{100} * 3) = $320\)

COMMON FACTOR in both terms: \(\frac{$100,000-x}{3}*\frac{4}{100}\)

COMMON FACTOR, factored OUT of blue equation above
\((\frac{$100,000-x}{3} * \frac{4}{100} * 1) + (\frac{$100,000-x}{3}*\frac{4}{100} * 3) = $320\)
\((\frac{$100,000-x}{3} *\frac{4}{100})\) * \((1 + 3) = $320\)

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:
\((\frac{$100,000-x}{3} *\frac{4}{100})* (1 + 3) = $320\)

\((\frac{$100,000-x}{3} *\frac{4}{100})* (4) = $320\)

\(\frac{$100,000-x}{3} * (\frac{4}{100} * 4) = $320\)

\((\frac{$100,000-x}{3} * \frac{16}{100})= $320\)

Clear the 3 in the denominator. Multiply both sides by 3

\($100,000-x * \frac{16}{100}= $960\)

\($100,000 - x = $960 * \frac{100}{16}\)

\($100,000 - x = \frac{$960}{16}*100\)

\($100,000 - x = $60 * 100\)

\($100,000 - x = $6,000\)

\($100,000 - $6,000 = x\)

\($94,000 = x\)
:dazed

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) $88,000

If 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 is too great.
The house needs to be more expensive.

Try B) $94,000

If 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 :-)
_________________

In the depths of winter, I finally learned
that within me there lay an invincible summer.

-- Albert Camus, "Return to Tipasa"

Re: David used part of $100,000 to purchase a house. Of the remaining port   [#permalink] 21 Mar 2018, 01:07
Display posts from previous: Sort by

David used part of $100,000 to purchase a house. Of the remaining port

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  


GMAT Club MBA Forum Home| About| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.