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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, 02:09
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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
[Reveal] Spoiler: OA

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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, 02:31
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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, 10:57
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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, 02:13
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 24 Nov 2017, 03:17
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

Last edited by Vorovski on 25 Nov 2017, 22:29, 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, 12:46
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.
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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, 03:19
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, 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

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David used part of $100,000 to purchase a house. Of the remaining port   [#permalink] 16 Dec 2017, 20:53
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