A milk vendor mixes water with milk and sells the mixture at the same

Should be D, but actually both statements give different answers for X. A gives 22.72 while B gives 25. This should not happen with an official question. Right? Bunuel

­Could you please show the method for both the values?
The value should always be same from both the staement.­ I believe it is true for this question too.

I was wrong. For statement 1, I was assuming that in the mixture, if water is halved, milk is increased accordingly. But of course, there is no basis for assuming this. Apologies!

­Why we haven't considered mark-up x% while doing calculations using statement 1? Because question does say "..The selling price per liter of the mixture is the vendor’s cost per liter of the milk plus a markup of x %..." shouldnt we consider this everytime? 

• Quantity of milk = m
• Quantity of water added = w
• Cost per liter of milk = c

Selling Price of the mixture = c*(1+\(\frac{x}{100}\))

• Revenue = Amount * Selling Price = (m+w)(c)(1+\(\frac{x}{100}\))
• Cost incurred = mc (because -> no cost for the water)
• Profit = (m+w)(c)(1+\(\frac{x}{100}\)) - mc

Profit % = 50%

So,

(m+w)(c)(1+\(\frac{x}{100}\)) - mc = \(\frac{1}{2}\) mc

=> (m+w)(c)(1+\(\frac{x}{100}\)) = \(\frac{3}{2}\) mc

Simplifying this, we get ->

\(\frac{3}{2}\) x (\(\frac{m}{m+w}\)) = \(\frac{(100 + x) }{ 100}\)

Insight: To find x, we need \(\frac{m}{m+w}\)

(1) If the vendor mixes half the intended quantity of water and sells every liter of the mixture at the cost price per liter of the undiluted milk, the vendor will get a 10% profit.

(m+\(\frac{w}{2}\))(c) - mc = \(\frac{1}{10}\) mc

Simplifying this, we get ->

m = 5w.

So,
\(\frac{m}{m+w}\) = 5/6

Clearly, this is sufficient.

\(\frac{(100 + x) }{ 100}\) = \(\frac{3}{2}\) x (\(\frac{5}{6}\)) = \(\frac{5}{4}\)

=> x = 25%

(2) The concentration of milk in the mixture after adding water is 5/6.­

\(\frac{m}{m+w}\) = 5/6

Clearly, this is also sufficient.

\(\frac{(100 + x) }{ 100}\) = \(\frac{3}{2}\) x (\(\frac{5}{6}\)) = \(\frac{5}{4}\)

=> x = 25%

So, choice D is the correct answer. Both statements are individually sufficient.

---
Harsha

Dumb question here. If the prompt says a "50% profit on the sale of the mixture" wouldn't it imply that P=1/2 Sale instead of P=1/2C?

KarishmaB Can you please explain this in a more simpler way?

­p: cost per liter of the milk
2w: amount of water to be mixed

\(p * \frac{p+2w}{p} * \frac{100 + x}{100} = \frac{3}{2} * p\)

=> \(\frac{p+2w}{p} * \frac{100 + x}{100} = \frac{3}{2}\)­ (i)

Question: x = ?


Statement 1: \(\frac{p+w}{p} * \frac{100 + x}{100} = \frac{11}{10}\)­ (ii)

(i) : (ii)

=> \(\frac{p+2w}{p} \div \frac{p+w}{p} = \frac{3}{2} \div \frac{11}{10}\)­

=> \(\frac{p + 2w}{p+w} = \frac{30}{22} = \frac{15}{11}\)

=>­ can find ratio of p to w

=> can find \(\frac{p+w}{p}\)

Combined with (ii)
=> can find x

=> Sufficient


Statement 2: \(\frac{p}{p + 2w} = \frac{5}{6}\)

Combined with (i)
=> can find x

=> Sufficient


Answer is D­

This question is a great example of why doing the math / solving is not always a great idea on hard Data Sufficiency questions.

Remember the definition of "sufficient": does this condition give me only one answer to the question? Many DS questions can be answered correctly--not by solving for the answer, but by understanding simply that you have enough information to get one answer only. In most cases, this means trying to simplify your math down to one remaining variable.

Thus, in order for a condition to be sufficient, it must give us enough information to make sure that the question being asked is the only variable left in an equation.

For example, if I only know that a+b=2, then I cannot solve for a or b. In order to solve, I must either be given the value of one of the variables--or another, different equation (system of equations) with the same variables, such as a-b=0. I could then add the two equations to eliminate b and solve for a.

Upon reading the stem, it is clear that there is more than one variable at play: 1) the x% markup of the diluted milk (the question being asked), and 2) the concentration of water being used to dilute the milk (w%), which we know leads to a 50% profit.

Let's imagine that the milk costs $1/liter, to make the math easy. If the vendor receives a 50% profit, then we can't still can't solve for the percent markup. We know that the vendor makes $1.50 for every $1 that they spend, but that is all: there are multiple markups (x%) and concentrations (w%) that would result in this profit figure, because there are still two variables remaining.

Condition #1 works because it gives us a second, different equation with the same two variables: in other words, a system of equations. It tells us that a mixture with half the intended quantity of water (.5w%) and the same markup (x%) will net a 10% profit ($1.10 per liter). Thus, we can combine the equations to eliminate variables and solve for each one individually. There is no need to do the (complicated) math that others have already demonstrated above, if you are confident that there is only one value of x that will work.

Condition #2 works for a more obvious reason: it provides one of the two missing variables directly, allowing us to solve for the other.­

Since statement 1 mentions that the mixture is sold at the "cost price", we can start from there. The vendor gets a 10% profit by selling the mixture at cost price, which basically means that he adds 10 liters of water per 100 liters of milk. Hence, the ratio of water : milk is 1 : 10, if the vendor mixes half the intended quantity of water. So, the intended ratio is 2 : 10 (or 1 : 5).

Let's assume that the cost of milk per liter is $2. Then, the cost of 5 liters of milk would be $10. A 50% profit would be $5. Therefore, revenue should be $15. This revenue is earned by selling 6 liters of the mixture (1 liter of water plus 5 liters of milk). Hence, the selling price per liter of mixture is $15/6 = $2.5. Therefore, the markup is 0.5/2 * 100 = 25%.

Now, let's consider statement 2. It indicates that the ratio of water to milk is 1 : 5, which is the same ratio inferred from statement 1. So, once again we can assume the cost of milk per liter and thereby calculate the markup as 25%.

Posted from my mobile device I found the math above to be extremely confusing for this question so I’ve decided to make a video to explain it. I hope I’ve done a good job, but I’m quite a novice and teaching so if anybody has any questions, please reach out to me as I think I’ve mastered this problem as well as other mixture problems like it

Bunuel can you explain? Didn't understand a single solution

Hehe, I fell for the same trap so had to take numbers to understand. Putting it for you the way I understood:

I will take 3 scenarios to explain some valuable points. Follow along:

Scenario 1: I take 1 litre of milk and add 500 ml additional water for free

Let Cost of 1 litre of milk is $100

Now READ VERY CAREFULLY - They have said selling price PER LITRE OF mixture NOT selling price PER mixture is equal to cost/litre of milk + mark up%. When you assumed profit = margin this is mistake you committed.

Ok so SP/Litre of my mixture = Cost/litre of milk + x%
I am taking x% = 30% here. Different x% will give different values
So
SP/Litre of my mixture= Cost/litre of milk + x%
= 100 (1+30/100)
=130

Ok great we have SP/litre of mixture

Now What is profit?
Total Revenue - Total Cost

COST OF MIXTURE:
Lets calculate cost of mixture = Cost of milk in mixture + Cost of water in mixture
= Cost of 1 litre milk + Cost of 500 ml water
= $100 + 0
Carefully observe cost is same everywhere in all the 3 scenarios I have taken because cost only depends on milk and my quantity of milk is same in every scenario. The water keeps changing but that doesn't affect my costs:)

REVENUE OF MIXTURE:
SP of mixture/litre * Total litres in mixture
130*1.5litres = $195
[Why 1.5 litres? 1 litre water + 500 ml water = 1.5L because he is selling water as if he was selling undiluted milk.]


Total profit = 195-100 = 95
Profit % = Total profit/ Total Costs = 95/100 = 95%




Scenario 2: I take 1 litre of milk and add 750 ml additional water for free
Everything upto revenue of mixture is same

REVENUE OF MIXTURE:
SP of mixture/litre * Total litres in mixture
130*1.75litres = $227.5
[Why 1.75 litres? 1 litre water + 750 ml water = 1.75L because he is selling water as if he was selling undiluted milk.]


Total profit = 227.5-100 = 127.5
Profit % = 127.5/100 = 127.5%

Scenario 3: I take 1 litre of milk and add 1 litre additional water for free
Everything upto revenue of mixture is same

REVENUE OF MIXTURE:
SP of mixture/litre * Total litres in mixture
130*2litres = $260
[Why 2 litres? 1 litre water + 1 litre water = 2L because he is selling water as if he was selling undiluted milk.]

Total profit = 260-100 = 160
Profit % = 160/100 = 160%

Now here is why I put in so much effort.
- The more the % of water the more my profit increases for the same qty of milk added at the same cost/litre of milk
- The more the margin% the more my profit increases for the same qty of milk added at the same cost/litre of milk


So what do we learn? We need these two variables to arrive adequately at the profit %. Since profit is already given we can sole 2 linear equations with two variables and we will be able to find the margin. This is what statement A does.

After not finding any satisfactory answers I figured it my own way
Let cost of milk/litre be c and
Let there be 1 litre of milk in mixture and y litres of water in mixture
{I have assumed 1 litre of milk to simplify the calculations because in statement 1 only water is changing milk is same so even if I would have taken a variable nothing would have been different - taking 1 helps us reduce 1 variable}

SP/litre of mixture is= Cost of milk/liter + x% margin
= c(1+x/100)...................... (1)

Now SP of mixture is = 1.5* Cost of mixture........ (2)
{Profit is given as 50% of total cost}

Using (1) and (2)
SP of mixture = SP/litre of mixture * Total litres in mixture
1.5*Cost of mixture = c(1+x/100) * (1 litre milk + y litres water)

Cost of mixture = Cost of milk = c
{Water is free of cost}

1.5*c = c(1+x/100)*(1+y)
C gets cancelled from both sides
1.5 = (1+x/100)*(1+y)............. Equation 1 from prompt
We have to find X. If we can find y we can plug it here and it is solved!

Statement 1 tells us
SP/litre of mixture = c
{Inference to be drawn here is margin % is 0}

SP of mixture is = 1.1* Cost of mixture
SP of mixture = SP/litre of mixture * Total litres in mixture
1.1*Cost of mixture = c(1+x/100) * (1 litre milk + y litres water)
Cost of mixture = Cost of milk = c
1.1*c = c* (1 litre milk + y/2 litres water)
1.1=1+y/2
y = 0.2
This tells us that ratio of water/mixture is 0.2/1.2 = 16.6% {If you had not assumed milk to be 1 litre in mixture you would have still arrived at it}

You don't need to solve further because you already got one variable you can plug here and you have the answer. But just in case this was a PS Q
1.5 = (1+x/100)*(1+0.2)
X/100 = 1.5/1.2 - 1
= 0.25
X = 25%

b) The concentration of milk in the mixture after adding water is 5/6.
Milk/Total mixture =
1/1+y = 5/6
6 = 5+5y
y = 0.2

I again got value of Y! Yay so D

This question confused the living daylights out of me. As I finally managed to figure it out in a way that makes sense in my brain, I will try to share my thought process in hopes it can be helpful, even if just to one person:

Initial Conditions

w = quantity of water
m = quantity of milk
M = price of milk

Mixture Selling Price = M (we are selling our diluted mixture at the standard price of milk)
Revenue = (m+w)*M
Cost = m*M

Thus, profit = (m+w)*M - m*M = w*M

(1) If the vendor mixes half the intended quantity of water and sells every liter of the mixture at the cost price per liter of the undiluted milk, the vendor will get a 10% profit.

We are given profit = 10% of costs, which translates to:

Profit = w*M --> (W/2)*M = 10%(m*M)
w*M = 20% (m*M)
w/m = 20% = 1/5

Therefore, our ratio of water to milk is 1:5. This means that for every 1 part water, we use 5 parts milk. Therefore, our dilution is 5/6 milk.

So, our cost is 5/6*M per liter, and we sell at a 50% profit margin, meaning we sell at 5/6*1.5 = 7.5/6 per liter. This means that our markup is 1.5/6 = 25%.

(2) The concentration of milk in the mixture after adding water is 5/6.

This is the same as statement (1) but skipped to the end! Again, this means we sell at 7.5/6 per liter, representing a 25% markup.

Therefore, both statements INDEPENDENTLY can provide us X. The answer is therefore (D).

Bunuel,

Since you're an incredible and competent teacher, can you please do your version of this one?

Hi Zach1188

This is not a dumb question. Please consider this into a practical scenario by putting yourself as a greedy businessman selling milk bottles

PROFIT is always calculated on the TOTAL COST price; TOTAL COST price here is QUANTITY*COST of producing undiluted milk
Thus Profit is 50/100(M*C) or MC/2

Hope this clears your confusion.