Mixture Question

[nadib002]

5 litres of water is added to a certain quantity of pure milk costing $3 per litre. If, by selling the mixture at the same price as before a profit of 20% is made, what is the amount of pure milk in the mixture?

My approach:

Quantity of milk = x lts

Cost of milk per lt = $3


Water added = 5 lt

Total quantity of mix = x+5 lts

Selling Price of milk = 3x


Mixture (x+5) lts is also sold at $3 per lt. Therefore SP of mixture = 3(x+5)

I am stuck after this. How do i set up the 20% profit?

Help needed.

Thank you

[smackmartine]

I think if you had posted the options, you would have gotten some reply. Anyways

Let the quantity of pure milk is x

so , total cost of x liter @ $3 per liter will be 3x
After adding 5 liters of water and selling @ same rate, New Total cost is 3(x+5)

As per question , profit is 20% => [3(x+5)-3x]/3x =1/5

15/3x = 1/5 => x = 25 . So, initial quantity of milk was 25 liters.

Now if you want to cross verify the answer :

Initial quantity of milk = 25 liters , Total cost = 3*25 = $ 75
After adding 5 liters of water and selling the solution @ same rate, the total cost will be 3*(25+5)= $90

Now, profit = [(90-75)/75 ]*100 = [15/75]*100 = [1/5]*100 = 20 (which is given)

5 litres of water is added to a certain quantity of pure milk costing $3 per litre. If, by selling the mixture at the same price as before a profit of 20% is made, what is the amount of pure milk in the mixture?

My approach:

Quantity of milk = x lts

Cost of milk per lt = $3


Water added = 5 lt

Total quantity of mix = x+5 lts

Selling Price of milk = 3x


Mixture (x+5) lts is also sold at $3 per lt. Therefore SP of mixture = 3(x+5)

I am stuck after this. How do i set up the 20% profit?

Help needed.

Thank you

[GMATGuruNY]

5 litres of water is added to a certain quantity of pure milk costing $3 per litre. If, by selling the mixture at the same price as before a profit of 20% is made, what is the amount of pure milk in the mixture?

My approach:

Quantity of milk = x lts

Cost of milk per lt = $3


Water added = 5 lt

Total quantity of mix = x+5 lts

Selling Price of milk = 3x


Mixture (x+5) lts is also sold at $3 per lt. Therefore SP of mixture = 3(x+5)

I am stuck after this. How do i set up the 20% profit?

Help needed.

Thank you

The cost of the milk is irrelevant.
Let m = volume of the pure milk.
To make a 20% profit, the added water must increase the volume of the pure milk by 20%.
Thus, the 5 liters of added water must be 20% of the volume of the milk:
5 = .2m
m = 5/.2 = 25.

For many test-takers, the easiest approach would be the plug in the answers, which would represent the amount of milk:

Answer choice: 25 liters of milk
Cost at $3 per liter = 3*25 = $75.
Volume of mixture = milk + water = 25+5 = 30 liters.
Selling price = 3*30 = $90.
Profit = 90-75 = 15, which is 20% of the cost of the milk.