A milk vendor sells milk at Cost Price but still gains 20%. Find the

Cost price=selling price =1
so there's a change in quantity which allows the gain of 20 percent
initial quantity 100 since same pricing water has been added becomes 120
20:100(w:M)1:5
hence option B

We can let the ratio of milk to water be 1 : n, and the cost price of 1 gallon of milk is $10. After mixing with water, we have 1 + n gallons of the milk mixture, and he sells every gallon of the mixture for $10.

In the 1 + n gallons of the milk mixture, 1/(1 + n) of it is milk. We can create the equation where x is the number of gallons of milk in 1 gallon of the milk mixture:

x/1 = 1/(1 + n)

x = 1/(1 + n)

So the selling price of milk by the vendor is 10/(1/(1 + n)) = 10(1 + n) per gallon (assuming water does not cost anything) and this is 120% of the cost price of $10 per gallon of milk. We can create the equation:

10(1 + n) = 1.2(10)

1 + n = 1.2

n = 0.2

So for every 1 gallon of milk, the vendor mixed it with 0.2 gallons of water. In other words, for every 5 gallons of milk, the vendor mixed it with 1 gallon of water.

Alternate Solution:

Suppose that the vendor bought 100 gallons of milk for c dollars. If the vendor made a profit of 20%, then it must be true that the vendor made a revenue of 1.2c dollars. Since the vendor sold the milk and water mixture at the same cost per gallon, it follows that the vendor sold 120 gallons of the mixture. Thus, the mixture contains 20 gallons of water for every 100 gallons of milk; which gives us a milk-to-water ratio of 5:1.

Answer: B

I solved this with picking numbers as well

Imagine:
Buys 100 liters of milk for $1 per liters, pays $100
He earns 20% gain while selling 1-liter concentrate for 1$, so his revenue is $120
We can deduce that he has sold 20 liters of water for $1 per liter

Milk=100 liters
Water=20 Liters

Ratio milk to water= 5/1

IMO
Ans: B

Somehow getting Ans B (5:1)...but not so sure !

Take Milk as 100 qty and let price be 1 rs per liter, and as profit is 20 more, water has to be 20

So, 100/20 = 5:1


Alternate method:

M = Milk; C = COst price

1.2MC = (M+W)C

1.2M = M +W;

0.2M=W can also be written as 2/10M = W

therefore m/w = 10/2

Therefore 5:1

To ensure that he makes 20% profit by selling same amount of milk at same selling price he would have to add 20% water to it.
Elaborating:
Say cost price of Milk is $ 100 per 100 ltr (CP= $1/ltr)
So he would have to sell Milk + Water (100+20) mixture at $120 per 120 ltrs for Selling price to be equal to Cost price of the milk. (SP = CP= $1/ ltr)

Thus Milk: Water = 100: 20 = 5:1

Answer : B

Note: The question must mention that water is for free. Apparently, at some place water costs more than milk

Use Cost Price per gallon of milk = CP
x is the amount of actual milk (in gallon) in the solution. So total cost price is CP*x
y is the amount of water (in gallon) mixed into the solution. Total volume of the solution is (x+y)

The seller sold the solution at cost price. So revenue is CP(x+y).
Profit is 20% of cost price. So CP*(x+y) = 1.2*CP*x
x+y = 1.2*x
y = 0.2*x so x/y = 1/0.2 = 5/1.

IMO B.

my crude thinking:

that he is making a profit of 20% means: SP = 1.2 x CP
therefore CP/SP = 1/1.2 = 10/12 = 5/6
this implies 5 parts is milk out of a total of 6 parts. and the ratios of the costs have to be the same as the ratio of milk to water since its a linear relationship.
so the ratio has to be 5:1

does this make sense?

As a matter of fact, I think this solution is missing a lot of details in the explanation. The question tells us that the cost price is the same as the sale price; so CP = SP. The vendor makes a profit because the milk and water mixture he is selling is greater than the quantity of milk he purchased. As far as I can tell, this solution would produce the same answer if the vendor didn't sell the mixture at the same cost price but instead he sold it at, say, 10% higher than the cost price.

RATIO SOLUTION:

20 % of profit means 1/5 of the total.
It means that for 5 parts of milk there is 1 part of water.
Milk / Water = 5:1


Bunuel

Milk is M, water is W and P is price of milk without water per litre

(M+W)P=1.2M*P
0.2M=W
M/W=5:1
B:)

There are two ways to solve it, i.e., either by assuming convenient values or using variables to solve an equation.

Important Tip - When the question doesn't demand actual values but just a ratio or a percentage, it's usually safe to assume convenient values over solving equations. You'll arrive at the answer quicker.

Assume the milk and water to cost Rs. 1 per each unit. We know that the ratio of profit can also be written as the ratio of SP/CP, since CP * Multiplying Factor of Profit = SP.

We know that the seller is selling only 80% milk but gaining a profit for 100% of the milk. So, we take the ratio of SP/CP, which is:

[SP (Milk) + SP (Water) / CP (Milk)] = 6/5, and cross multiplying and solving the equation, we arrive at the ratio of milk to water as 5:1.

A milk vendor sells milk at Cost Price but still gains 20%. Find the ratio of milk and water in every gallon that he sells.

A. 4:1
B. 5:1
C. 5:2
D. 6:1
E. None of these

In the wise words of Ron Burgundy... "Milk was a bad choice"

This vendor is employing a strategy that most sketchy clubs today employ (ie. diluting liquor with water) The only way for the vendor to sell milk at the same cost price AND make a profit is if he cunningly diluting his milk with water.

We know he is making 20% profits or (6/5) more than his purchase price. If he is buying 1 part (5/5) milk, he is therefore adding (1/5) parts water to equal 6/5 more sellable product.

Therefore we know the ratio between milk and water is 5:1

Hope this helps!

I'll discuss 2 methods:

1.
CP 100 litres = 100
SP = 120 for 120 litres
120 litres = Milk(100L) + Water(20L)
Ratio: 5:1 (B)

1. Options
(a) M:W = 4:1
CP (for milk) = 4
SP (M+W) = 5
%P = 1/4 = 25%

(b) M:W = 5:1
CP for milk) = 5
SP (M+W) = 6
%P = 1/5 = 20% (OA)

Hi,
Basically, Vendor is selling water at milk price.

Because profit % is 20, we could easily guess the denominator(cost price).

Looking at answer choices, B & C has 5 for the parts of milk.
try B first,
5 parts milk + 1 part water = 6 parts
let cost price be 1$ per litre
profit % = (selling price - cost price)/cost price * 100
= (6-5)/5 * 100 = 20%

B it is.

Let the cost price of milk = the selling price of the mixture = $1 per gallon.
An alternate approach is to PLUG IN THE ANSWERS, which represent the ratio of milk to water.
When the correct answer is plugged in, profit = 20% \(= \frac{1}{5}\)

D: 6 gallons of milk : 1 gallon water, for a total of 7 gallons of mixture
Cost of 6 gallons of milk \(= 6*1 = 6\)
Selling price of 7 gallons of mixture \(= 7*1 = 7\)
\(\frac{profit}{cost} =\frac{7-6}{6} = \frac{1}{6}\)
The profit is too small.
Eliminate D.

B: 5 gallons of milk : 1 gallon water, for a total of 6 gallons of mixture
Cost of 5 gallons of milk \(= 5*1 = 5\)
Selling price of 6 gallons of mixture \(= 6*1 = 6\)
\(\frac{profit}{cost} = \frac{6-5}{5} = \frac{1}{5}\)
Success!

Let the cost price of milk = the selling price of the mixture = $1 per gallon.
We can PLUG IN THE ANSWERS, which represent the ratio of milk to water.
When the correct answer is plugged in, profit = 20% \(= \frac{1}{5}\)

D: 6 gallons of milk : 1 gallon water, for a total of 7 gallons of mixture
Cost of 6 gallons of milk \(= 6*1 = 6\)
Selling price of 7 gallons of mixture \(= 7*1 = 7\)
\(\frac{profit}{cost} =\frac{7-6}{6} = \frac{1}{6}\)
The profit is too small.
Eliminate D.

B: 5 gallons of milk : 1 gallon water, for a total of 6 gallons of mixture
Cost of 5 gallons of milk \(= 5*1 = 5\)
Selling price of 6 gallons of mixture \(= 6*1 = 6\)
\(\frac{profit}{cost} = \frac{6-5}{5} = \frac{1}{5}\)
Success!