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A dishonest milk man tops up his bucket which is only 4/5 th full of m

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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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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


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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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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


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Solution:

Let x be the capacity of the bucket.then, initial mixture has 80% milk.
then after 5 lt replaced with water---->\((80/100)*(x-5)=(60/100)*x\)

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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New post 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 = 16-4 = 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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New post 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
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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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New post 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---->\((80/100)*(x-5)=(60/100)*x\)

x=20 lt.

Ans:- B.


Hello - great quick solution, do you mind explaining how you got the (x-5) 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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New post 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
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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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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: x-1

Now 5 L water is added. Bucket is now full.

Water volume: x-1+5

Milk : 4x-4
\(\frac{4x-4}{5x}\) = 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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New post 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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New post 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---->\((80/100)*(x-5)=(60/100)*x\)

x=20 lt.

Ans:- B.


Hello - great quick solution, do you mind explaining how you got the (x-5) 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 \(80/100*(x-5)\).

now the equation should look like this--->\(80/100*(x-5)+Percentage of the milk in water(which is 0)*5=60/100*(x-5+5)\).

\((80/100)*(x-5)+(0/100)*5=(60/100)*(x)---->(80/100)*(x-5)=(60/100)*x\)
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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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New post 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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New post 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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New post 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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New post 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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New post 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:(5-1), 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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Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html


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Re: A dishonest milk man tops up his bucket which is only 4/5 th full of m   [#permalink] 10 Feb 2018, 07:42
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