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

Question

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

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

Source : Experts Global

generis · Accepted Answer

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

varundixitmro2512 · Answer

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%

gracie · Answer

let x=number of oranges
total cost=25*x+16 2/3*x
total revenue=2x*25
50x/(41 2/3x)=1.2
20%
D

ScottTargetTestPrep · Answer

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

devlin · Answer

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.

Maxximus · Answer

Happy to see this discussion.

Here is the official video explanation by us-

https://www.youtube.com/watch?v=i4DGNs7qpJk

Hope this helps.

All the best!
Experts' Global

Kinshook · Answer

Given:  A Shopkeeper bought few oranges at 4 per dollar and an equal number of oranges, of a different quality, at 6 per dollar.

Asked: If he sold the mix of the two qualities at a dozen per 3 dollars, what was his percentage profit on the investment?

Let us assume that oranges bought be 2x in quantities.

Cost price = $x/4 + $x/6 = $10x/24 = $5x/12
Sales price = $3/12 *2x = $ x/2

Profit % = (x/2 - 5x/12)*100/(5x/12) = (x/12)*100/(5x/12) = 20%

IMO D

Goldenfuture · Answer

price of per orange is 1/4 and for others is 1/6
so 1/4 + 1/6 = 5/12 
Thus this is price of 2 oranges and price of 1 orange is 5/24
Now we know the cost of 12 oranges is 3
=> 4 oranges is 1 and thus 1 orange is 1/4 or 6/24
so the profit is 1/5 = 20%

I feel the wording of the question is confusing

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

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