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08 Mar 2021, 04:45
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
55% (03:01) correct
45%
(03:00)
wrong
based on 267
sessions
History
Date
Time
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Not Attempted Yet
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
30 May 2021, 02:45
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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!)
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!)
09 Mar 2021, 09:44
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Bunuel
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
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
