A retailer adds freely available water to an orange concentrate

Question

A retailer adds freely available water to an orange concentrate, costing $50 per liter, to prepare an orange drink. He sells the drink at $30 per liter and earns 50% profit on his investment. How many liters of water does he add to each liter of the orange concentrate?

A. 0.5
B. 0.60
C. 0.67
D. 1
E. 1.5

Source: Experts Global

A. 0.5
B. 0.60
C. 0.67
D. 1
E. 1.5

Source: Experts Global

gracie · Accepted Answer

sale price=1.5*$50=$75
$75/$30=2.5 liters sold
2.5 liters-1 liter oc=1.5 liters water added 
E

generis · Answer

Find cost from revenue, then volume from cost.

• Find cost from revenue
Suppose he sells 10 liters for $30 per liter.
(10 * $30) = $300 Total revenue

That revenue earns him 50% profit on his investment
Revenue = 150% Cost = 1.5(Cost)
1.5(Cost) = $300
Cost =  \frac{$300}{1.5}  =$200

• Liters of OJ used in mix sold?

Cost = $200 
Cost per liter of OJ = $50
Cost per liter of water = $0

\frac{TotalCost}{CostPerLiter} =   # of liters of OJ

For 10L of mix he used
  \frac{$200}{$50}=4   liters of OJ concentrate

• Liters of water used in mix?
(10L mix - 4L OJ) = 6L of water used

• Liters of water added to liter of OJ to make mix?
Volume ratio is

\frac{Water}{OJ}=\frac{6}{4}=\frac{3}{2}=\frac{1.5}{1}=1.5   L of water per L of OJ

Answer E

Alternatively,  ONE  liter sold.
Revenue = 1.5C
$30 = 1.5C

Cost:  \frac{$30}{1.5} = $20

Volume of $20 worth of OJ at   \frac{$50}{L}  
  \frac{$20}{$50}=\frac{2}{5}   L of OJ

1 L mix sold.   \frac{2}{5}   L is OJ
1L -   \frac{2}{5}   L =   \frac{3}{5}   L of water

\frac{Water}{OJ}=\frac{\frac{3}{5}}{\frac{2}{5}}=\frac{3}{2}=1.5   liters of water added per L of OJ

Answer E

sandeepkonda · Answer

First, we need to find the final concentration of orange concentrate in the drink. Which is 40% ( ($30-$c)/$c=0.5)-----> c= $20 ----> we can buy 0.4L of the concentrate with $20). The question now boils down to how much water is to be added to a 100% concentrate to make it a 40% concentrate. This is 1/(1+W) = 0.4 ----> w =1.5L.

Puja priya · Answer

we can use the allegation general formula here..
Quantity of cheaper/quantity of derar=(price of dearer-price mean price)/
(mean price-price of cheaper)
here mean price = 20(as sp is 30 and profit is 50%)
hence, Qty of water/qty of orange concentrate=(50-20)/(20-0)=30/20-3/2
now, since orang concentrate is 1l, qty of water = 3/2=1.5L.

Deyoz · Answer

we can use this as well:
Total Revenue = Total Cost + Profit 
30 (X) = 50+25 (x Being the quantity sold in Liter; 50 is the cost and 25 profit), we are given that cost per liter is 50 and profit 50 % of the cost.

Solving, gives X 2.5, in which 1 is Orange Drink and 1.5 is water.

Dereno · Answer

using alligations
Price of orange juice is $: 50/litre
Price of Water : $: 0
Both the prices are Cost price.
Given , selling price is $ 30. Hence CP* 3/2 = 30 , CP =20

50 0
 \ /
 20
 / \
20 30
Ratio of 2:3

1 litre is 1000ml , so 2*500 == 1000 ml
therefore 3*500 = 1500 ml = 1.5L

Kinshook · Answer

Given: A retailer adds freely available water to an orange concentrate, costing $50 per liter, to prepare an orange drink. He sells the drink at $30 per liter and earns 50% profit on his investment.

Asked: How many liters of water does he add to each liter of the orange concentrate?

Let the water add to each liter of hte organge concentrate be x litres.

Cost of the orange drink = $ 50/(x+1) per liter = $30 / 1.5 = $20
50 = 20(x+1) = 20x + 20
20x = 30
x = 1.5 liters

IMO E

AnishTrivedi99 · Answer

Water is free here, Orange conc costs $50/L.

He sells at 30/L, making a 50% profit, hence his cost is 20/L.

So by looking at it, he uses 0.4L of orange concentrate (0.4*50 =20)

0.6L water used.

Hence, for each 0.4 L conc, 0.6L water is used

Therefore, for 1L of concentrate, he uses 1.5L of water (by unitary method)

MS26 · Answer

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A retailer adds freely available water to an orange concentrate

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