There are two identical containers A & B. The container A contains 1

Question

There are two identical containers A & B. The container A contains 1 Litre pure water and container B contains B 1 litre of pure milk. Now 5 cups of water container A taken out and is mixed well in container B. Then, 5 cups of this mixture is taken out and mixed in the container A. If M denotes the proportion of milk in the container A and N denote the proportion of water in container B, then:

A. M < N
B. M > N
C. M = N
D. 2M = N
E. Info insufficient, can’t be determined.

A. M < N
B. M > N
C. M = N
D. 2M = N
E. Info insufficient, can’t be determined.

pk123 · Answer

There are two identical containers A & B. The container A contains 1 Litre pure water and container B contains B 1 litre of pure milk. Now 5 cups of water container A taken out and is mixed well in container B. Then, 5 cups of this mixture is taken out and mixed in the container A. If M denotes the proportion of milk in the container A and N denote the proportion of water in container B, then:
Let a cup is equal to 100ml = 1/10 of the litre
hence A quantity after 5 cups of water is taken out = 1 - 5*100= 0.5 litre
B quantity after 5 cups of water is added = 1 + 0.5litre ( 1 litre milk & 0.5litre= water)
Now 5 cups of mixture is taken out from B, so B quantity becomes 1 litre but proportion of milk and water remains unchanged which is N= proportion of water = 0.5/(1.5)=1/3
A water quantity becomes = 0.5litre + 0.5/3 = 2/3 and milk quantity is 0.5*2/3 = 1/3
hence proportion of milk = (1/3)/1 = 1/3

hence M= if M denotes the proportion of milk in the container A = 1/3
hence M= N

lets say when 800ml = 0.8 litre is taken out so A quantity after 5 cups of water is taken out = 0.2
B quantity= 1.8 litre but proportion of water in mixture= 0.8/1.8= 4/9
N = proportion of water in container B= 4/9
A quantity after 0.8 litre is now taken from mixture B = 1 litre and proportion of milk = 0.8*5/9= 4/9
M= if M denotes the proportion of milk in the container A = 4/9

hence C is the answer

Raxit85 · Answer

Here the ratio of mixtures( i.e milk , water) doesnot matter. But the important point is that whether the total amount ( either pure or mixture ) being transferred is equal or not. Since the total amount ( i.e 5 cups) being transferred from each one to another , hence A =B.

So, C is the answer.

So, C is the answer.

shameekv1989 · Answer

There are two identical containers A & B. The container A contains 1 Litre pure water and container B contains B 1 litre of pure milk. Now 5 cups of water container A taken out and is mixed well in container B. Then, 5 cups of this mixture is taken out and mixed in the container A. If M denotes the proportion of milk in the container A and N denote the proportion of water in container B, then:
A. M<N
B. M>N
C. M=N
D. 2M= N
E. Info insufficient, can’t be determined.

Answer - C

We can have any proportion of 5 cups mixed in other container, the result will be container A having same % of milk as in container B having water

For eg. we take 500 ml (for 5 cups) of water mixed with milk -> Result will 1.5 ltrs having 0.5 ltr of water giving concentration of 0.5/1.5 => 1/3 => N = 33.33% of water

So 500 ml of this concentration will contain 2/3 of 500 ml of milk => 33.33% milk in 500 ml

When we add 500 ml of this concentration in 500 ml of water -> result will be 33.33% milk in 1 ltr of mixture => M

Therefore M = N

Similarly,

If we take 200 ml (for 5 cups) then 200 ml of water mixed in 1 ltr of milk -> 0.2 ltr water in 1.2 ltr of mixture -> 0.2/1.2 =1/6 concentration of water = N

When we mix 200 ml of this concentration => 5/6 of 200 ml added to 500 ml of water => 1000/6 in 1 ltr mixture => 1/6 of milk => M

Again M=N.

KhulanE · Answer

A 0% 1litre water
B 100% 1litre milk

if 5cups is equal to 0.9litre
proportion of milk in B is 100%*1+0%*0.9=x*1.9 x=52.6%, N=100%-52.6%=47.4%
proportion of milk in A is 0%*0.1+52.6%*0.9=M*1 M=47.4% N=M

if 5cups is equal to 0.1litre
proportion of milk in B is 100%*1+0%*0.1=x*1.1 x=90.9%, N=100%-90.9%=9.1%
proportion of milk in A is 0%*0.9+90.9%*0.1=M*1 M=9.1% N=M

in any measurement of 5cups M and N are equal so C is right

shridhar786 · Answer

container A (water) = 1000 ml
container B (milk) = 1000 ml

let the capacity of 1 cup = 100 ml

now 5 cups of water (that is 500 ml) is taken out from A and is mixed with B
the concentration of A and B after the process

container A = 500 ml
container B - milk = 1000 ml water = 500 ml---ratio of milk to water = 2:1

now 5 cups of mixture (that is 500 ml) is taken out and mixed with A 
milk and water taken out will be in ratio 2:1
let 2t of milk and t water is taken out from B

500 = 3t
t =  \frac{500}{3 } 
water taken out from B =  \frac{500}{3}  ml
milk taken out from B =  \frac{1000}{3}  ml

the concentration of A and B after the process

container A - milk =  \frac{1000}{3}  ml water =  \frac{2000}{3}  ml

ratio of milk : water = 1:2
proportion of milk in container A =  \frac{1}{3}  = M

container B - milk =  \frac{2000}{3}  ml water =  \frac{1000}{3}  ml

ratio of milk : water = 2:1
proportion of water in container B =  \frac{1}{3}  = N

from here we get M = N

C is the answer

aditibarjatya · Answer

Solution:

Question 6 :  There are two identical containers A & B. The container A contains 1 Litre pure water and container B contains B 1 litre of pure milk. Now 5 cups of water container A taken out and is mixed well in container B. Then, 5 cups of this mixture is taken out and mixed in the container A. If M denotes the proportion of milk in the container A and N denote the proportion of water in container B, then:

A. M<N 
 B. M>N 
 C. M=N 
 D. 2M= N 
 E. Info insufficient, can’t be determined.

In this question, we need to find out how the proportion of milk or N in A relates to the proportion of water in B after the transfer of mixture from B to A. .

Information given in the question: Water in A = Milk in B = 1 litre. 
 First transfer : First 5 cups of water from A to B makes remaining quantity in A = 1 litre - 5 cups of water
Let 1 cup measures x litres. Then, 5 cups = 5x litres. Thus, quantity in A =  1 - 5x  litres. 
Moreover, resulting mixture in B after the transfer = 1 litre milk + 5 cups of water (from A). Total quantity of mixture in B =  1 + 5x  litres.

Second transfer : 5 cups of mixture from B to A. Therefore, resulting quantity in B would be reduced to 1 litre again and quantity in A too would increase from  1 - 5x  to 1 litre. However, B is a mixture of water and milk before the second transfer. The mixture contains 1 litre milk and 5x litres water. Thus, if quantity of 5x is transferred from the mixture in B to A, then resulting ratios of milk to water can be calculated by:

Proportion of milk in B  =  \frac{1}{(1 + 5x)} .
 Proportion of water in B  =  \frac{5x}{(1 + 5x)} .
 Proportion of milk in B after transfer of 5x cups to A  =  1 - \frac{1*5x}{(1 + 5x)}  =  \frac{1}{(1 + 5x)} 
 Quantity of milk transferred  =  \frac{5x}{(1 + 5x)}

Proportion of water in B after transfer of 5x cups to A  =  N  =  5x - \frac{5x*5x}{(1 + 5x)}  = 5x/1 + 5x
 N  can be written as  \frac{5x}{1}  as the total quantity remaining in B after the transfer is 1 litres.

The proportion of milk in A after the transfer or  M  =  \frac{5x}{(1 - 5x + 5x)}  or  \frac{5x}{1 }  as the total quantity in A after the addition of 5x litres or 5 cups is  1 - 5x + 5x or 1  litres.

Thus,  M = N = \frac{5x}{1}  and  option C  is the right answer.

ajaymahadev · Answer

I had to think about this for some time.

HarshavardhanR · Answer

https://gmatclub.com/forum/download/file.php?id=89707 --- Harsha GMAT-Club-Forum-5zcvai9v.png

Gmatiseasyforall · Answer

Thanks all for answering this question! Is there any analytical or logical way to solve this question without using algebra i.e using the structure and logic of the question like symmetry and conservation of volumes in both directions? Request the top math expert Bunnuel to provide some insights in this regard! Thanks in advance!

Bunuel · Answer

You don’t need to use any equations for this one.

Assuming 5 cups means some amount less than 1 liter, the answer is C.

Initially, there is 1 liter of water in A and 1 liter of milk in B. After transferring the same amount from A to B and then back from B to A, each container again contains 1 liter.

Whatever amount of milk ends up in A must equal the amount of water that ends up in B. Since the containers also have equal final volumes, their proportions are equal.

Therefore, M = N.

Answer: C.

To elaborate:

The cleanest way to see it is to focus only on container A.

A starts with exactly 1 liter of water and no milk.

After everything is done, A again contains exactly 1 liter total, because the same amount was taken out and then put back in.

Now suppose A ends up containing 0.2 liters of milk.

Since A still contains exactly 1 liter total, it can contain only 0.8 liters of water.

So A has lost 0.2 liters of its original water.

Where can that missing 0.2 liters of water be? It must be in B.

Therefore:

milk that ends up in A = water that ends up in B.

The key is not really the mixing. It is that A has the same final volume as it had initially. Every bit of milk that replaces water in A corresponds to exactly the same amount of water that must have left A and remained in B.

Gmatiseasyforall

Gmatiseasyforall · Answer

Wow! That was awesome! Couldn't have thought about this problems on those lines myself. As usual you always remain the top math expert in this forum!
 One final question though(hope i am not bothering you with too many questions)! Would the answer remained the same i.e (c) M=N, if the volume of the two containers were not same but different? And could the problem still be solved using an analytical or logical approach, without using algebra in that case?

Bunuel · Answer

No. If the containers had different volumes, the answer would not necessarily be C.

The amount of milk that ends up in A would still equal the amount of water that ends up in B, because both containers return to their original volumes and whatever amount of water A loses is replaced by the same amount of milk.

But M and N are proportions, not amounts. The same amount makes up a larger proportion of the smaller container and a smaller proportion of the larger container.

So if A is smaller, M > N. If B is smaller, M < N. Only when the containers have equal volumes do we get M = N.

So yes, the unequal-volume case can still be handled logically without algebra.

Gmatiseasyforall · Answer

Bunuel-  YOU ROCK !!

There are two identical containers A & B. The container A contains 1

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There are two identical containers A & B. The container A contains 1

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