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A milk vendor has 2 cans of milk. The first contains 25% water and the
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05 Feb 2019, 07:52
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68% (02:35) correct 32% (02:50) wrong based on 184 sessions
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A milk vendor has 2 cans of milk. The first contains 25% water and the rest milk. The second contains 50% water. How much milk should he mix from each of the containers so as to get 12 litres of milk such that the ratio of water to milk is 3 : 5? A. 4 litres, 8 litres B. 6 litres, 6 litres C. 5 litres, 7 litres D. 7 litres, 5 litres E 8 litres, 7litres
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Re: A milk vendor has 2 cans of milk. The first contains 25% water and the
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05 Feb 2019, 08:15
To answer this question, let's match up the percentages of water in the two milk mixtures with the ratio that we want to achieve. The ratio that we want to achieve is 3:5, water to milk. So, we want 3/8 water and 5/8 milk. To answer the question, let's focus on the water, since, if 3/8 of the mixture is water, the other 5/8 will have be milk. The mixes that we have are 25% water and 50% water. Let's express them in 8ths as well. The first mix is 2/8 water. The second mix is 4/8 water. We want 3/8 water, and since 3/8 is halfway between 2/8 and 4/8, this question is turning out to be pretty easy. By using allegation, we can determine that we have to use equal quantities of the two mixtures. The correct answer is (B).
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Re: A milk vendor has 2 cans of milk. The first contains 25% water and the
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05 Feb 2019, 08:34
thx for the solution, so useful



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Re: A milk vendor has 2 cans of milk. The first contains 25% water and the
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05 Feb 2019, 21:41
dharam44 wrote: A milk vendor has 2 cans of milk. The first contains 25% water and the rest milk. The second contains 50% water. How much milk should he mix from each of the containers so as to get 12 litres of milk such that the ratio of water to milk is 3 : 5?
A. 4 litres, 8 litres B. 6 litres, 6 litres C. 5 litres, 7 litres D. 7 litres, 5 litres E 8 litres, 7litres Another quick way to solve could be 3:5 ratio of 12l water:milk ==>4.5l water and 7.5l milk Now see option Ait has 1l+2l=3l water this is not the solution Option B1.5l+3l water=4.5l water this is the answer (you can also check other options to confirm that they don't give the same answer,but they won't since the question has a unique solution)



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Re: A milk vendor has 2 cans of milk. The first contains 25% water and the
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19 Mar 2019, 12:47
MartyTargetTestPrep wrote: To answer this question, let's match up the percentages of water in the two milk mixtures with the ratio that we want to achieve.
The ratio that we want to achieve is 3:5, water to milk.
So, we want 3/8 water and 5/8 milk.
To answer the question, let's focus on the water, since, if 3/8 of the mixture is water, the other 5/8 will have be milk.
The mixes that we have are 25% water and 50% water. Let's express them in 8ths as well.
The first mix is 2/8 water. The second mix is 4/8 water.
We want 3/8 water, and since 3/8 is halfway between 2/8 and 4/8, this question is turning out to be pretty easy.
By using allegation, we can determine that we have to use equal quantities of the two mixtures.
The correct answer is (B). Hi MartyTargetTestPrep, Although the answer is B , are we getting 6 and 6 because I am getting a different equal proportion of the two mixtures . Can you please confirm if the two milk cans are indeed mixed in 6:6 proportion. Appreciate your help, Thank you.
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A milk vendor has 2 cans of milk. The first contains 25% water and the
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Updated on: 27 Jun 2019, 21:46
dharam44 wrote: A milk vendor has 2 cans of milk. The first contains 25% water and the rest milk. The second contains 50% water. How much milk should he mix from each of the containers so as to get 12 litres of milk such that the ratio of water to milk is 3 : 5?
A. 4 litres, 8 litres B. 6 litres, 6 litres C. 5 litres, 7 litres D. 7 litres, 5 litres E 8 litres, 7litres let x=total liters in first can .75x+.5(12x)=5/8*12→ x=6 126=6 liters in second can 6 liters, 6 liters B
Originally posted by gracie on 19 Mar 2019, 14:18.
Last edited by gracie on 27 Jun 2019, 21:46, edited 1 time in total.



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Re: A milk vendor has 2 cans of milk. The first contains 25% water and the
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20 Mar 2019, 12:10
stne wrote: Hi MartyTargetTestPrep, Although the answer is B , are we getting 6 and 6 because I am getting a different equal proportion of the two mixtures . Can you please confirm if the two milk cans are indeed mixed in 6:6 proportion. Appreciate your help, Thank you. Yes. Since the total resulting quantity is to be 12 liters, the equal parts have to be 6 liters and 6 liters.
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Re: A milk vendor has 2 cans of milk. The first contains 25% water and the
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21 Mar 2019, 07:13
MartyTargetTestPrep wrote: To answer this question, let's match up the percentages of water in the two milk mixtures with the ratio that we want to achieve.
The ratio that we want to achieve is 3:5, water to milk.
So, we want 3/8 water and 5/8 milk.
To answer the question, let's focus on the water, since, if 3/8 of the mixture is water, the other 5/8 will have be milk.
The mixes that we have are 25% water and 50% water. Let's express them in 8ths as well.
The first mix is 2/8 water. The second mix is 4/8 water.
We want 3/8 water, and since 3/8 is halfway between 2/8 and 4/8, this question is turning out to be pretty easy.
By using allegation, we can determine that we have to use equal quantities of the two mixtures.
The correct answer is (B). This explanation is excellent  I started doing this problem with algebra but this method is much quicker. However, the way I read the question there are no correct answers; the question states that the target mixture has 12L of milk in it. Is the problem written incorrectly? The author must be intending to have a final solution of 12L total  otherwise it would be impossible to obtain a solution with 12L of milk using only 12L from each container...



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Re: A milk vendor has 2 cans of milk. The first contains 25% water and the
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22 May 2019, 08:09
5e3 wrote: MartyTargetTestPrep wrote: To answer this question, let's match up the percentages of water in the two milk mixtures with the ratio that we want to achieve.
The ratio that we want to achieve is 3:5, water to milk.
So, we want 3/8 water and 5/8 milk.
To answer the question, let's focus on the water, since, if 3/8 of the mixture is water, the other 5/8 will have be milk.
The mixes that we have are 25% water and 50% water. Let's express them in 8ths as well.
The first mix is 2/8 water. The second mix is 4/8 water.
We want 3/8 water, and since 3/8 is halfway between 2/8 and 4/8, this question is turning out to be pretty easy.
By using allegation, we can determine that we have to use equal quantities of the two mixtures.
The correct answer is (B). This explanation is excellent  I started doing this problem with algebra but this method is much quicker. However, the way I read the question there are no correct answers; the question states that the target mixture has 12L of milk in it. Is the problem written incorrectly? The author must be intending to have a final solution of 12L total  otherwise it would be impossible to obtain a solution with 12L of milk using only 12L from each container... Hi 5e3, You are Correct, it's 12L of milk and not 12L of total solution. From the question all we can tell is that equal proportions of both the cans have been used.
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Re: A milk vendor has 2 cans of milk. The first contains 25% water and the
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27 Jun 2019, 17:14
Is it possible to answer it with weighted averages?



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Re: A milk vendor has 2 cans of milk. The first contains 25% water and the
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27 Jun 2019, 17:30
dharam44 wrote: A milk vendor has 2 cans of milk. The first contains 25% water and the rest milk. The second contains 50% water. How much milk should he mix from each of the containers so as to get 12 litres of milk such that the ratio of water to milk is 3 : 5?
A. 4 litres, 8 litres B. 6 litres, 6 litres C. 5 litres, 7 litres D. 7 litres, 5 litres E 8 litres, 7litres Let us take x and 12x lt. of milk, so (5/8) = {.75x + .5(12x)}/12 x=6



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Re: A milk vendor has 2 cans of milk. The first contains 25% water and the
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28 Jun 2019, 07:49
12lite is the quantity of milk not water and milk mixture



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Re: A milk vendor has 2 cans of milk. The first contains 25% water and the
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28 Jun 2019, 09:42
Want to confirm if my solution (below) is right :
[ 75/100 x (milk in 1st can) + 50/100 x (milk in 2nd can) ] / [ x (total milk) + y (total water) ] = 12 / 19.2
after solving we get 2.4 x = 2.4 y > x = y
Hence, 6 = 6. Option B



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Re: A milk vendor has 2 cans of milk. The first contains 25% water and the
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26 Jul 2019, 01:20
allegation style: ((1/2)(3/8))/((3/8)(1/4))= 1/1 –> ratio 1:1
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Re: A milk vendor has 2 cans of milk. The first contains 25% water and the
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02 Aug 2019, 23:43
Given, 12 litres final solution in the ratio of water to milk is 3:5. So amount of water in 12L solution=4.5 L which is equal to 37.5% now using the weighted average formula:
2537.550
so, both are equally distant which is 1:1 ratio. Therefore, answer is 6 litres, 6 litres (B)



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Re: A milk vendor has 2 cans of milk. The first contains 25% water and the
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02 Aug 2019, 23:50
sameeruce08 wrote: 12lite is the quantity of milk not water and milk mixture Given that , final quantity of milk is 12 litres such that the ratio of water to milk is 3 :5> we can take the final mixture as 12 litres.




Re: A milk vendor has 2 cans of milk. The first contains 25% water and the
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