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A shopkeeper bought few oranges at 4 per dollar and an equal number of

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A shopkeeper bought few oranges at 4 per dollar and an equal number of  [#permalink]

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New post 06 May 2018, 03:34
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A Shopkeeper bought few oranges at 4 per dollar and an equal number of oranges, of a different quality, at 6 per dollar. If he sold the mix of the two qualities at a dozen per 3 dollars, what was his percentage profit on the investment?

A) 10%
B) 15%
C) 16.67%
D) 20%
E) 25%

Source : Experts Global

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Re: A shopkeeper bought few oranges at 4 per dollar and an equal number of  [#permalink]

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New post 06 May 2018, 04:48
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Though my answer was wrong in the first attempt but I found the right method after

For the first variety, price per orange is 1/4(4 per dollar)
For the second variety, price per orange is 1/6(6 per dollar)

so the combine value of the mix, i.e for 2 oranges( 1st orange first variety and 2nd other variety) will be 5/12
so for the dozen(12) it will be 6 x 5/12(multiply by 6 as price of 2 oranges is calculated above), price will be 5/2 i.e 2.5

Now a dozen of the mix is sold for 3 dollars, therefore profit = 3-2.5=0.5 and perc =20%
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A shopkeeper bought few oranges at 4 per dollar and an equal number of  [#permalink]

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New post 06 May 2018, 17:43
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doomedcat wrote:
A Shopkeeper bought few oranges at 4 per dollar and an equal number of oranges, of a different quality, at 6 per dollar. If he sold the mix of the two qualities at a dozen per 3 dollars, what was his percentage profit on the investment?

A) 10%
B) 15%
C) 16.67%
D) 20%
E) 25%

Source : Experts Global

Assign values. Use LCM of 3, 4 and 6. Cost and revenue will be in round numbers.

Let the # of oranges bought = 12 of each kind

Shopkeeper's cost

Cost of pricier oranges:
He bought 12 oranges at 4 oranges per $1
\(\frac{12}{4}=(3*$1)=$3\) spent

Cost of cheaper oranges:
12 oranges bought at 6 oranges per $1
\(\frac{12}{6}=(2*1$)=$2\) spent

Total cost: \($5\)
Total oranges bought: \((12+12)=24\)

Revenue
He sells 24 oranges at a dozen (12) per $3
\(\frac{24}{12}=(2*$3)=$6\) in revenue

Profit = (TR - TC) = ($6 - $5) = $1

What percent profit on investment?
\((\frac{1$}{5$}*100)=(0.20*100)=20\) percent

Answer D

*Example of full equation, F = fruit (oranges)
Revenue: \(\frac{24F}{(\frac{12F}{$3})}=(24F*\frac{$3}{12F})=(2*$3)=$6\)

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Re: A shopkeeper bought few oranges at 4 per dollar and an equal number of  [#permalink]

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New post 07 May 2018, 21:02
doomedcat wrote:
A Shopkeeper bought few oranges at 4 per dollar and an equal number of oranges, of a different quality, at 6 per dollar. If he sold the mix of the two qualities at a dozen per 3 dollars, what was his percentage profit on the investment?

A) 10%
B) 15%
C) 16.67%
D) 20%
E) 25%

Source : Experts Global


let x=number of oranges
total cost=25*x+16 2/3*x
total revenue=2x*25
50x/(41 2/3x)=1.2
20%
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Re: A shopkeeper bought few oranges at 4 per dollar and an equal number of  [#permalink]

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New post 08 May 2018, 11:25
doomedcat wrote:
A Shopkeeper bought few oranges at 4 per dollar and an equal number of oranges, of a different quality, at 6 per dollar. If he sold the mix of the two qualities at a dozen per 3 dollars, what was his percentage profit on the investment?

A) 10%
B) 15%
C) 16.67%
D) 20%
E) 25%


We can let the number of dozens of oranges bought at 4 per dollar = n = the number of dozens oranges bought at 6 per dollar. (In this case, the number of oranges bought at either quality is 12n.)

Thus the total number of dozens of oranges bought = n + n = 2n and the total cost = 12n/4 + 12n/6 = 3n + 2n = 5n.

Since he sold them at a dozen per 3 dollars, the total revenue = 2n x 3 = 6n.

Thus the profit = 6n - 5n = n and the percentage profit on the investment was:

n/(5n) = 1/5 = 20%

Answer: D
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A shopkeeper bought few oranges at 4 per dollar and an equal number of  [#permalink]

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New post 09 May 2018, 11:40
Total bought were 12, then 1/4x + 1/6x. Equal number of each type of orange, so 6 of each, then:
1/4*6 + 1/6*6 = 3/2 + 1 = 5/2= 2.5. The shopkeeper invested 2.5 dollar.
If the shopkeeper sold the dozen per 3 dollar so he had a profit of 0.5 dolar, then 0.5/2.5 equals 1/5 or 0.2 or 20%.
Option D.

P.s.: But I got wrong in my first attempt.
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Re: A shopkeeper bought few oranges at 4 per dollar and an equal number of  [#permalink]

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New post 25 May 2019, 09:27
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doomedcat wrote:
A Shopkeeper bought few oranges at 4 per dollar and an equal number of oranges, of a different quality, at 6 per dollar. If he sold the mix of the two qualities at a dozen per 3 dollars, what was his percentage profit on the investment?

A) 10%
B) 15%
C) 16.67%
D) 20%
E) 25%

Source : Experts Global

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Re: A shopkeeper bought few oranges at 4 per dollar and an equal number of   [#permalink] 25 May 2019, 09:27
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