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

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

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New post 27 May 2016, 03:09
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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
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Re: David used part of $100,000 to purchase a house. Of the remaining port  [#permalink]

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New post 27 May 2016, 03:31
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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
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Re: David used part of $100,000 to purchase a house. Of the remaining port  [#permalink]

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New post 14 Jun 2016, 11:57
3
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!
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Re: David used part of $100,000 to purchase a house. Of the remaining port  [#permalink]

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

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

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New post 25 Nov 2017, 13:46
1
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 (used the answer choices) in about 90 seconds. See how I used the answer choices, here.)

Hope that helps.
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Re: David used part of $100,000 to purchase a house. Of the remaining port  [#permalink]

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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!
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David used part of $100,000 to purchase a house. Of the remaining port  [#permalink]

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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
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Re: David used part of $100,000 to purchase a house. Of the remaining port  [#permalink]

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New post 30 Jan 2018, 13:01
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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
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David used part of $100,000 to purchase a house. Of the remaining port  [#permalink]

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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
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Re: David used part of $100,000 to purchase a house. Of the remaining port  [#permalink]

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New post 18 Mar 2018, 19:15
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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.
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David used part of $100,000 to purchase a house. Of the remaining port  [#permalink]

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New post 20 Mar 2018, 13:50
1
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 ?
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Re: David used part of $100,000 to purchase a house. Of the remaining port  [#permalink]

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

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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. :-)
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New post 20 Mar 2018, 19:05
1
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!


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

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New post 21 Mar 2018, 01:07
2
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 (the leftover money ) 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 :-)
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David used part of $100,000 to purchase a house. Of the remaining port &nbs [#permalink] 21 Mar 2018, 01:07
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