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A dishonest milk man tops up his bucket which is only 4/5 th full of m [#permalink]
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13 Jun 2016, 11:30
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A dishonest milk man tops up his bucket which is only 4/5 th full of milk with water. He again removes 5 liters of this mixture from the bucket and adds an equal quantity of water. If milk is now 60% of the mixture, what is the capacity of the bucket ? A. 15 lit B. 20 lit C. 22.5 lit D. 25 lit E. 30 lit Source : IIM Cat _________________________________ Consider Kudos if helpful
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A dishonest milk man tops up his bucket which is only 4/5 th full of m [#permalink]
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13 Jun 2016, 12:21
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msk0657 wrote: A dishonest milk man tops up his bucket which is only 4/5 th full of milk with water. He again removes 5 liters of this mixture from the bucket and adds an equal quantity of water. If milk is now 60% of the mixture, what is the capacity of the bucket ?
A. 15 lit B. 20 lit C. 22.5 lit D. 25 lit E. 30 lit
Source : IIM Cat _________________________________ Consider Kudos if helpful Solution: Let x be the capacity of the bucket.then, initial mixture has 80% milk. then after 5 lt replaced with water>\((80/100)*(x5)=(60/100)*x\) x=20 lt. Ans: B.
Last edited by AdlaT on 15 Jun 2016, 13:25, edited 1 time in total.



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Re: A dishonest milk man tops up his bucket which is only 4/5 th full of m [#permalink]
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14 Jun 2016, 05:57
I would have plugged in the answer choices !! for instance , if the capacity of the bucket = 20 litres milk = 4/5=16 litres and water = 4 litres he now removes 5 litres of this mixture , which is (4/5)*5 = 4 litres of milk milk remaining = 164 = 12 litres = 60% of the solution .



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Re: A dishonest milk man tops up his bucket which is only 4/5 th full of m [#permalink]
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14 Jun 2016, 06:23
msk0657 wrote: A dishonest milk man tops up his bucket which is only 4/5 th full of milk with water. He again removes 5 liters of this mixture from the bucket and adds an equal quantity of water. If milk is now 60% of the mixture, what is the capacity of the bucket ?
A. 15 lit B. 20 lit C. 22.5 lit D. 25 lit E. 30 lit
Source : IIM Cat _________________________________ Consider Kudos if helpful Hi, another way to look at the question.... when we are removing 5 litres of mix, it will contain\(5*\frac{4}{5} = 4\)litre of milk..... Now if capacity is x, \(\frac{4x}{5}  4 = \frac{60x}{100} = \frac{3x}{5}.........\frac{4x}{5}  \frac{3x}{5}= 4...............x = 4*5 = 20\) B
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Re: A dishonest milk man tops up his bucket which is only 4/5 th full of m [#permalink]
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21 Jun 2016, 21:34
AdlaT wrote: msk0657 wrote: A dishonest milk man tops up his bucket which is only 4/5 th full of milk with water. He again removes 5 liters of this mixture from the bucket and adds an equal quantity of water. If milk is now 60% of the mixture, what is the capacity of the bucket ?
A. 15 lit B. 20 lit C. 22.5 lit D. 25 lit E. 30 lit
Source : IIM Cat _________________________________ Consider Kudos if helpful Solution: Let x be the capacity of the bucket.then, initial mixture has 80% milk. then after 5 lt replaced with water>\((80/100)*(x5)=(60/100)*x\) x=20 lt. Ans: B. Hello  great quick solution, do you mind explaining how you got the (x5) portion of the equation. I'm having difficultly following the logic given the net amount of liquid is 0 (5 out, 5 in) Thanks!



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Re: A dishonest milk man tops up his bucket which is only 4/5 th full of m [#permalink]
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21 Jun 2016, 22:23
msk0657 wrote: A dishonest milk man tops up his bucket which is only 4/5 th full of milk with water. He again removes 5 liters of this mixture from the bucket and adds an equal quantity of water. If milk is now 60% of the mixture, what is the capacity of the bucket ?
A. 15 lit B. 20 lit C. 22.5 lit D. 25 lit E. 30 lit
Source : IIM Cat _________________________________ Consider Kudos if helpful You can also think of it as a weighted average question. Initial solution has 4/5th milk and 1/5th water so 80% milk. We remove 5 litres of this and put 5 litres of water (0% milk) instead. Overall, we get 60% milk solution. Vol of 80% solution/ Vol of 0% milk solution = w1/w2 = (A2  Aavg)/(Aavg  A1) = (0  60)/(60  80) = 3/1 Since volume of 0% milk solution (pure water) added was 5 lt, the 80% solution was 15 lt (after removal of 5 lt). So original 80% solution was 15 + 5 = 20 lt Answer (B)
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A dishonest milk man tops up his bucket which is only 4/5 th full of m [#permalink]
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21 Jun 2016, 22:27
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msk0657 wrote: A dishonest milk man tops up his bucket which is only 4/5 th full of milk with water. He again removes 5 liters of this mixture from the bucket and adds an equal quantity of water. If milk is now 60% of the mixture, what is the capacity of the bucket ?
A. 15 lit B. 20 lit C. 22.5 lit D. 25 lit E. 30 lit
Milk : Water \(\frac{4}{5}\) :\(\frac{1}{5}\) 4: 1 say Milk is 4x, then water is x and total capacity of bucket is 5x. when the vendor removes 5 L of the mixture, then because Milk and water are in the ratio 4:1, he will actually remove 4L milk and 1 L water. So Now the bucket has: Milk: 4x  4 water: x1 Now 5 L water is added. Bucket is now full. Water volume: x1+5 Milk : 4x4 \(\frac{4x4}{5x}\) = 0.6 4x 4 = 3x x = 4. Capacity = 5x = 20 Litres. B is the answer



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Re: A dishonest milk man tops up his bucket which is only 4/5 th full of m [#permalink]
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22 Jun 2016, 04:20
This was a difficult one for me. I used answer choice 20 and it worked out ! May be I should learn the right way to solve it.



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A dishonest milk man tops up his bucket which is only 4/5 th full of m [#permalink]
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22 Jun 2016, 13:25
mdacosta wrote: AdlaT wrote: msk0657 wrote: A dishonest milk man tops up his bucket which is only 4/5 th full of milk with water. He again removes 5 liters of this mixture from the bucket and adds an equal quantity of water. If milk is now 60% of the mixture, what is the capacity of the bucket ?
A. 15 lit B. 20 lit C. 22.5 lit D. 25 lit E. 30 lit
Source : IIM Cat _________________________________ Consider Kudos if helpful Solution: Let x be the capacity of the bucket.then, initial mixture has 80% milk. then after 5 lt replaced with water>\((80/100)*(x5)=(60/100)*x\) x=20 lt. Ans: B. Hello  great quick solution, do you mind explaining how you got the (x5) portion of the equation. I'm having difficultly following the logic given the net amount of liquid is 0 (5 out, 5 in) Thanks! x is the capacity of the bucket. In first sentence we are given that milk is only 4/5 capacity of the bucket =\((80/100)*x\). In second sentence ,we are given that he replaced the 5 lt mixture with water, now milk percentage is \(80/100*(x5)\). now the equation should look like this>\(80/100*(x5)+Percentage of the milk in water(which is 0)*5=60/100*(x5+5)\). \((80/100)*(x5)+(0/100)*5=(60/100)*(x)>(80/100)*(x5)=(60/100)*x\)



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A dishonest milk man tops up his bucket which is only 4/5 th full of m [#permalink]
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22 Jun 2016, 19:25
let c=capacity look at water .2c.2*5+5=.4c c=20 liters



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Re: A dishonest milk man tops up his bucket which is only 4/5 th full of m [#permalink]
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09 Feb 2018, 20:17
msk0657 wrote: A dishonest milk man tops up his bucket which is only 4/5 th full of milk with water. He again removes 5 liters of this mixture from the bucket and adds an equal quantity of water. If milk is now 60% of the mixture, what is the capacity of the bucket ?
A. 15 lit B. 20 lit C. 22.5 lit D. 25 lit E. 30 lit
Source : IIM Cat _________________________________ Consider Kudos if helpful After filling the bucket with water: Milk  4x/5, Water  1x/5. After removing 5 litres of mixture, quantity removed: Milk  4, Water  1 Resultant mixture = Milk: 4x/5  4, Water: x/5  1 After adding 5 litres of water, resultant mixture = Milk: 4x/5  4, Water: x/5  1 + 5 4x/5  4 = 0.6x 4x  20/5 = 0.6x 4x  20 = 3x x = 20. (B).



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Re: A dishonest milk man tops up his bucket which is only 4/5 th full of m [#permalink]
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09 Feb 2018, 23:03
I solved by plugging in values. Initial composition is 80/20. I chose the option B to verify. 20 liters = 16 liters of milk and 4 liters of water. Now remove 5 liters of the mixture, which by 80/20 mix ratio gives 4 lit milk/1 lit water. Now we are left with 12 liters of milk and 8 liters of water. The milk is now 60% of mixture which matches the question stem. Ans: B
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A dishonest milk man tops up his bucket which is only 4/5 th full of m [#permalink]
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10 Feb 2018, 07:22
chetan2u wrote: msk0657 wrote: A dishonest milk man tops up his bucket which is only 4/5 th full of milk with water. He again removes 5 liters of this mixture from the bucket and adds an equal quantity of water. If milk is now 60% of the mixture, what is the capacity of the bucket ?
A. 15 lit B. 20 lit C. 22.5 lit D. 25 lit E. 30 lit
Source : IIM Cat _________________________________ Consider Kudos if helpful Hi, another way to look at the question.... when we are removing 5 litres of mix, it will contain\(5*\frac{4}{5} = 4\)litre of milk..... Now if capacity is x, \(\frac{4x}{5}  4 = \frac{60x}{100} = \frac{3x}{5}.........\frac{4x}{5}  \frac{3x}{5}= 4...............x = 4*5 = 20\) B Hello chetan2u, why do you you assume when we subtract 5 litres of mixture from \(4/5\) we get 4 litres of milk we subtract the the mixture of both water and milk from 4/5 why do you put x next to only 4 ? \(\frac{4x}{5}\) shouldnt it be \(\frac{4x}{5x}\) thank you



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Re: A dishonest milk man tops up his bucket which is only 4/5 th full of m [#permalink]
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10 Feb 2018, 07:42
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dave13 wrote: chetan2u wrote: msk0657 wrote: A dishonest milk man tops up his bucket which is only 4/5 th full of milk with water. He again removes 5 liters of this mixture from the bucket and adds an equal quantity of water. If milk is now 60% of the mixture, what is the capacity of the bucket ?
A. 15 lit B. 20 lit C. 22.5 lit D. 25 lit E. 30 lit
Source : IIM Cat _________________________________ Consider Kudos if helpful Hi, another way to look at the question.... when we are removing 5 litres of mix, it will contain\(5*\frac{4}{5} = 4\)litre of milk..... Now if capacity is x, \(\frac{4x}{5}  4 = \frac{60x}{100} = \frac{3x}{5}.........\frac{4x}{5}  \frac{3x}{5}= 4...............x = 4*5 = 20\) B Hello chetan2u, Quote: why do you you assume when we subtract 5 litres of mixture from \(4/5\) we get 4 litres of milk we subtract the the mixture of both water and milk from 4/5 milk is 4/5 of total which means in 5x litres, 4x are milk and 1x is water so when we take out 5 litres and when both milk and water are mixed in that ratio 4:(51), 4 litres taken out is milk and 1 is water Quote: why do you put x next to only 4 ? \(\frac{4x}{5}\) shouldnt it be \(\frac{4x}{5x}\) thank you 4x/5x will be used when I say there are 5x litres and 4x is milk out of it.. But here I am taking x as total MIX and 4/5 of total is milk, therefore \(\frac{4}{5} * x = \frac{4x}{5}\)
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