A milkman cheats his customers by adding water to the milk he sells.

Question

A milkman cheats his customers by adding water to the milk he sells. He starts the day with 1000 litres of milk. However, after selling every 500 litres of milk, he adds and equal quantity of water to what he already has. The milkman can get away by cheating as long as the milk to water ratio does not fall below 1:7. He sells off all the milk if any addition of water will make the ratio fall below 1:7. Each customer buys exactly2 litres of milk a day from him. What is the total number of customers the milkman has cheated if he ends upselling all the milk ?

(A) 7550
(B) 7500
(C) 5500
(D) 1500
(E) 1000

(A) 7550 
(B) 7500 
(C) 5500 
(D) 1500 
(E) 1000

jjchhabra · Accepted Answer

Took a while, but here's my attempt!

1) Starts with 1000L of milk. 
Sells 500L to  250 customers , as each buy 2L.

2) Milkman has 500L of pure milk remaining, which he tops up with 500L of water. 
The ratio of the Milk:Water mixture is now 1:1 (within the realm of what he can sell) and the total volume is back up to 1000L. 
He sells 500L to  250 customers .

3) Milkman has 500L total of diluted milk remaining, consisting of 250L of milk and 250L of water. 
He tops up with 500L of water to a total of 1000L, so there's 250L of milk and 750L of water, for a ratio of 1:3 of Milk:Water (still within the realm of what he can sell). 
He sells 500L to  250 customers.

4) Milkman has 500L of diluted milk remaining, consisting of 125L of milk and 375L of water. 
He tops up with 500L of water to a total of 1000L, so there's 125L of milk and 875L of water, for a ratio of 1:7 of Milk:Water (just within the realm of what he can sell, the 1:7 limit). 
Since he won't be able to dilute it any further, he sells all remaining 1000L of diluted milk, to  500 customers.

Now comes the trap to avoid - add all the customers in bold above and you get 1250... which is not one of the answers.

The question is asking  how many customers he cheated , not how many he sold to. He didn't actually cheat the first 250 as they bought pure milk, so the correct answer IMO is E, 1000 cheated customers.

Let me know if you can see any shortcuts, as this took me way too long to figure out (and yes, I fell into the trap so didn't even get it right the first time!)

TestPrepUnlimited · Answer

Because we refill 500 L water (or after taking away 500 L of the liquid), the milk to total volume ratio is decreased by  \frac{500}{1000} = \frac{1}{2}  every time we refill (hence multiply by 50% every time).

After 1st refill, we have Milk:Total = 1:2, and after the 2nd refill we have Milk:Total = 1:4 etc.

We do not want to fall below Milk:Total of = 1:8. After the 3rd refill we hit 1:8 exactly. Therefore we can refill 3 times, which means we can use up 500*4 = 2000 L of milk. Then we can get away with cheating 2000/2 = 1000 customers.

Ans: E

Yogananda · Answer

TestPrepUnlimited 
I have a question.
Since, we are refilling 3 times, the target customers should be 750 because for first(500 litres) 250 customers, they got real milk without mixture.

Please tell me, what am I missing here.
Thanks.

karnavora1 · Answer

The most important part of the question is to understand that we need to find out what is the number of people that the milkman has CHEATED, not the total number of people he sold to.

He first has 1000L of pure milk, of which he sells 500L to customers.
2L per customer means he sells to 250 customers. We don't need to count these customers since they were not cheated because they got pure milk.

He is now left with 500L and adds another 500L of water, making the ratio of Milk to water 1:1
He sells 2L per customer ie 250 customers that have been cheated.

He stops at 500L and adds another 500L of water. The earlier mixture had a ratio 1:1, hence the 500L has 250L of milk and 250L of water. Adding another 500L water makes it 250L of milk and 750L of water which is a ratio 1:3.
He again sells to 2L per customer and cheats 250 customers.

He is left with 500L of adulterated milk and adds another 500L of water to it. The earlier ratio was 1:3, hence now we have 125L of milk and 875L of water. This is exactly 1:7 and hence he needs to sell of all the milk because anymore adulteration will breach the limit of the ratio. 2L of milk per customer for 1000L will yield 500 customers.

Total number of customers cheated

250+250+500=1000

rohanc8790 · Answer

No that will not be the case. In this approach, we are not considering the first 500L

When water is mixed for 3 times, then 1500 L of water is added. On top of that, there is 500 L left which will also be sold since the solution cant be diluted any further. Hence it will a total of 2000 L of diluted milk which when divided by 2 gives us the correct answer.

This approach is brilliant but tough to perceive.

TestPrepUnlimited · Answer

Great question  Yogananda  , I overlooked that part and the amount of milk sold in total would be 500 + 500 from the first 2 refills, then on the 3rd refill we still have 1000 L of milk to sell resulting in 2000 L of diluted milk sold in total. So I just happened to get the right answer but the logic and the method still applies!

adewale223 · Answer

He started with 1000L

First time
500 removed did not cheat in this case

Remaining mixture

M W

500 500

Second time

500 cheated, ratio of milk to water 1:1

Remaining mixture

M W

250 750

Third time

500 cheated ratio 1;3

Remaining mixture

M W

125 875

Fourth time

500 cheated, ratio 1;7

Remaining mixture

M W

62.5 437.5

Since he cant sell if ratio falls below 1;7 remaining volume 62.5 + 437.5 = 500.

Cheated volume 500 x 3 = 1500

Total = 1500 + 500 = 2000

Therefore number of people cheated = 2000/2 = 1000

Answer E.

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A milkman cheats his customers by adding water to the milk he sells.

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A milkman cheats his customers by adding water to the milk he sells.

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