There are 2 circular cylinders X and Y, and both cylinders contain wat

total volume of water in both cylinders X and Y = 5pi*6 + 10pi*2 = 30pi + 20pi = 50pi

when the water drawn from cylinder X is poured into cylinder Y, the height of water becomes same in both X and Y, let the height be H

Therefore, we have
5pi*H + 10pi*H = 50pi (Since total volume will remain same)
15H = 50
H = 10/3

deepthit : hope this helps.

Hence option C is correct
Hit Kudos if you liked it

First let’s determine the volume of water in each of the two cylinders before water from one is poured into the other. Recall that the volume of a cylinder is V = Bh, in which B is the circular base area and h is the height of the cylinder. Thus we have:

Water in cylinder X: Volume = 5? x 6 = 30? in^3

Water in cylinder Y: Volume = 10? x 2 = 20? in^3

Now we can let w be the amount of water that should be poured from cylinder X to cylinder Y so that the water in both cylinders will be the same height. Notice that if V = Bh, then h = V/B.

Thus we have:

(30? - w)/5? = (20? + w)/10?

5?(20? + w) = 10?(30? - w)

100?^2 + 5w? = 300?^2 - 10w?

15w? = 200?^2

3w = 40?

w = 40?/3

Since w = 40?/3, the height of water in each cylinder is:

(30? - 40?/3)/5?

(90? - 40?)/15?

50?/15? = 10/3

Answer: C
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I did get the solution by brute force. It took me almost 3.5 minutes.
Is there any particular method to solve this.

I really like 0akshay0's method of doing this as it's fairly quick and mathematical. I did it slightly different to get a good quick estimate.

First look at the current volume:
X = 6*5*pi = 30pi
Y = 2*10*pi = 20pi

I then wondered what would happen if you move 10pi of volume from X to Y. The height on x moves down to 4 and the height on Y moves up to 3. Since they are not equal yet, you have to move some more water to where the height will be somewhere between 3 and 4. Since there is only one answer between 3 and 4, I didn't have to move into further calculations.

Just thought I'd share an alternative method. Took a little over a min with most of the time thinking how I wanted to approach it.

Solved it in 1.30 min.
Method : plugging in the ans
Sol :
Total water in both the cyl = 30pi + 20 pi
Total volume of water in both the cylinder should remain same.
Now plug in ans B . Height of both cylinders is 3
Total volume with new height =
15pi + 30pi
This is not equal to the initial volume.
Eliminate B
Try option C
Total vol = 50/3 pi + 100/3 pi = 50pi
Bingo. We got our ans
Mark it and move ahead to next question.

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The way I reasoned to the answer without using algebra

Since the Volume of a Cylinder = (Area of Circular Base) * (Height)

And

Cylinder X’s Circular Base = 5(pi), which is (1/2) Cylinder Y’s Circular Base of 10(pi)

This means, Y cylinder is 2 Times as Wide as X cylinder


This means that if both cylinders were completely empty:

To get IDENTICAL Water Heights for both cylinders,

for every 1 units of water we put in cylinder X ——->

we would have to put 2 units of water into cylinder Y(requires Twice as much water to get to the same water height as X because Y is Twice as wide)


Based on this proportional relationship, when transferring water from cylinder X to Cylinder Y———> every 2 units of water height we take from cylinder X will only fill up 1 unit of water height in cylinder Y


Now we need to determine at what height will both cylinders’ water heights be identical. We can use the above relationship to go through each answer choice, where each answer choice equals the Water Height of each cylinder, X and Y:

(A)2.5

This would lower cylinder X’s height 3.5 (from 6 to 2.5)

For each 1 unit of water taken from X’s height and put into Cylinder Y, we will only see a +(1/2) increase in T’s water height

This means 3.5 decrease in X’s Height———> put into Y will raise Y’s Water Height by only (3.5/2) = 1.75

Water Height of Y would = 2 + 1.75 = 3.75

Which is not equal to X’s 2.5 —— eliminate A

We can use the same logic for each answer

(B)3

X’s water height would drop 3 ———-> would result in (3/2) = 1.5 water height increase in Y

Y’s height = 2 + 1.5 = 3.5

Which is not equal to X’s height of 3

Eliminate B

Leave C for last

(D)4

We would be taking a water height of 2 out of X ———> added to Y, this would raise Y’s water height by +(2/2) = +1

Y’s height would = 3

X’s height would = 4 though

Eliminate D

(E)4.5
We would be removing 1.5 of water height from cylinder X ———> which would add (1.5 / 2) = .75 water height to Y

Y’s height would = 2.75

X’s height would = 4.5

Eliminate E

Answer has to be (C) 10/3

Just to be sure.

X’s water height now = 6 = 18/3

To lower it down to 10/3 ——-> take 8/3 water height out of cylinder X

and this would result in only (8/3)(1/2) = 8/6 water height added to Y or +(4/3)



Y’s water height would = (6/3) + (4/3) = 10/3

And

X’s water height would = 10/3

(C) both heights will be identical at 10/3

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