There are three vessels of equal capacity .Vessel A contains Milk and

Here is my approach, very similar to fluke's, just a different thought process of arriving at the same calculation.

The three vessels are of equal capacity (which we assume to be full since it is a PS question. I would prefer clearer direction though). The milk:water ratio is given for each as 4:3, 2:1 and 3:2. The reason we cannot just add the milk parts (4+2+3) and water parts (3+1+2) together is that they represent different fractions of the whole volume i.e. in the first vessel, milk is 4 parts out of a total of 7 parts while in the second vessel, milk is 2 parts out of a total of 3 parts etc. If we make the total number of parts in each ratio equal, then we can just add the 'milk parts' together and all the 'water parts' together.
Ratios:
4:3 (Total 7 parts)
2:1 (Total 3 parts)
3:2 (Total 5 parts)
How can we make the total number of parts equal in the 3 cases? By taking the LCM.
4*15 : 3*15 = 60:45 (Total parts 105)
2*35 : 1*35 = 70:35 (Total parts 105)
3*21 : 2*21 = 63:42 (Total parts 105)
When we mix the three solutions, the total number of parts of milk is 60+70+63 = 193
the total number of parts of water = 45+35+42 = 122
The required ratio = 193:122
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4:3=>4x+3x=7x
2:1=>2y+1y=3y
3:2=>3z+2z=5z

7x=3y=5z

Take LCM of 7,3,5=105

Or simply;
x=3*5=15
y=7*5=35
z=3*7=21

So,

Ratio of Milk:Water= (4x+2y+3z)/(3x+y+2z)=193/122

Ans: "D"

Perfect Fluke...Bang on!! This was the approach i was taking abt +1

Can someone tell what is wrong with my method, which is close to the OA:
Say A=B=C=100 ml
Therefore, Milk in A= 57 ml (57%)
B= 67 ml (66.66%)
C= 60 ml (60%)
Total Milk= 57+67+60= 184 ml out of 300 ml and the rest is Water = 116 ml
So, the ratio is 184:116

Nothing wrong. Just that this is an approximation.

To fluke and karishma,

why did you multiply 3*5 for x and so for x and z?? i just did not get it

thanks for your explanation

A--------B-----C
4X:3X---2Y:1Y---3Z:2Z


final mixture of A,B and C will have M and W in the ratio ((4/7)+(2/3)+(3/5)) : (3/7)+(1/3)+(2/3)

=(4*15+2*35+3*21)/(7*3*5) : (3*15+1*35+2*21)/(7*3*5)

= 193: 122

The basic idea here is to take the LCM.That's the reason 3*5 is multiplied for x and so on.Hope it's clear now.

I am guessing ones you get the LCM, you divide the LCM by the number of parts the ship carrys then multiply the resultant number to the single part.
For Ship 1,
4:3...7 parts
105/7=15
4*15=60
3*15=45
therefore 60:45

Thanks fluke and karishma

Vessel 1 -> 4:3 (Total 7 parts)
Vessel 2 -> 2:1 (Total 3 parts)
Vessel 3 -> 3:2 (Total 5 parts)

Right now, we can't just add the parts of milk together since the parts are not the same. 1 part of vessel 1 is smaller than 1 part of vessel 2 since the same volume in vessel 1 is divided into 7 parts but in only 3 parts in vessel 2. If we can make one part of each vessel equal, we can just add the parts.
What is the LCM of 7, 3 and 5? It will be 7*3*5 (=105) since all the numbers are prime numbers.

When we multiply the entire ratio by the same number, the ratio remains unchanged.

Vessel 1 -> We need a total of 105 parts instead of 7 so we multiply by 15. The ratio becomes 4*15 : 3*15 = 60:45
Vessel 2 -> We need a total of 105 parts instead of 3 so we multiply by 35. The ratio becomes 2*35 : 1*35 = 70:35
Vessel 3 -> We need a total of 105 parts instead of 5 so we multiply by 21. The ratio becomes 3*21 : 2*21 = 63:42

Now we can just add the parts.
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I solved this problem using smart numbers, it is not a lot faster but easier to imagine

m=portion of milk in solution
m=(4/7+2/3+3/5)/3➡193/315
315-193=122
122/315=portion of water in solution
ratio of milk to water=193/122

how can we write this?
"Or simply;
x=3*5=15
y=7*5=35
z=3*7=21"

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