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

Cylinder 1 total vol= 5 pie *6 = 30 pie
Cylinder 2 total vol = 10 pie*2= 20 pie

so total vol = 50 pie

Height becomes same in both so 5pie *h + 10 pie *h = 50 pie
15 pie *h = 50 pie
so h = 50/15 = 10/3

Option C fits

cylinder A contains 5Pi x 6 inches water, or 30Pi cubic inches water
cylinder B contains 10Pi x 2 inches water, or 20Pi cubic inches of water

total sum is 50Pi cubic inches.
in order to be the same the ratio of the volume between the cylinders needs to be the same after being divided by 5PI and 10Pi respectively

if we keep 15pi cubic inches in A, and 35 in B that results in 3 and 3,5
if we keep 20pi cubic inches in A, and 30 in B that results in 4 and 3
if we keep 25pi cubic inches in A, and 30 in B that results in 4 and 3

15-35 is the closest result we get with whole numbers
so the value has to be between 3 and 3,5

--> C

C, as it gives the same volume , as to when individual volumes are added with the respective heights

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Step 1:
Calculating the volume of water in each of the two cylinders before water from one is poured into the other.
Volume of cylinder = πr^2h
h is the height and r is the radius of the circular base.

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

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

Now we can let “A" 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.

So

(30π - A)/5π = (20π + A)/10π

15Aπ = 200π^2

A = 40 π/3

Since A = 40π/3, the height of water in each cylinder is:

(30 π - 40 π/3)/5π = 10/3

C, as it gives the same volume , as to when individual volumes are added with the respective heights

Posted from my mobile device

Total volume of water available is 50 π square inches.

With a height of 3, the total is less than 50 and with a height of 4, the total is more than 50. So, somewhere in between. The only option is C.

1 unit reduction in x = 0.5 unit increase in Y
2 unit decrease in X= 1 increase in y

6-2a = 2+ a

3a=4
a=4/3

resultant height = 2+4/3 =10/3...... ans C

Pls see my answer as below:

Vx = 5π x 6 = 30π
Vy = 10π x 2 = 20π

Let a is the volume of water drawn from X to Y

--> (30π - a)/5π = (20π + a)/10π --> a = 40π/3

--> New volume in Y = 40π/3 + 20π = 100π/3

--> New Height in cylinder Y = New volume Y/ Base area Y = (100π/3) /10π = 10/3 = Height in cylinder X

As the base of Cylinder X is = 5pie and Cylinder Y is = 10pie, assuming that if the Cylinder X would have base area of 10 pie than the water surface area of water inside it would become half thus after Cylinder X has base area of 10 pie the height would be 3. Once we balance the water inside both the cylinder the resultant height would be 2.5

Hope this make sense.
ANy comments much appreciated.

E:

its a ratio from 1:2 and so if you take away approximately greater than 1 from the first cylinder and put in a bit more than 2 into the second, they would even out to 4.5 each.

For first cylinder

V = 10pi*6

Second cylinder

V = 20pi*2

We need to find the transfer so that both have the same height

Thing to note is that V of second cylinder has twice as much surface are as first

Using smart numbers we see that we will need first cylinder to lose a height of 2 inches for every increase of 1 inch on second cylinder

Keeping that in mind, imagine this a problem of average weights

Y(2)----------(AV)--------------------X(6)

let height be t weight on Y side is 2 and on X is 6
Thus, (t-2)/(t-6) = 1/2
Solve, t = 10/3

gmatbusters
do you mean the total initial volume of the water in the cylinders is the same as the final volume of water?

As per the question, initially the level of water in the 2 cylinders are different ( 6 inches & 2 inches)
the liquid are transferred to make the liquid level same.

If it is still not clear, feel free to tag me again...

If you take two liters from X and put it into Y, is gonna be 4 vs 3... so you need to put more water so that 4 decreases and 3 increases... until they converge somewhere between 3 and 4. Where exactly? you don't care, because the only result included between 3 and 4 is 10/3

Thinking Conceptually about the Problem:

Cylinder X has a radius = sqrt(5)

Cylinder Y has a radius = sqrt(10)

This means cylinder Y is twice as wide/thick as Cylinder X.


If we start with Cylinder X with a height of 6 inches and we Doubled the width of the Cylinder’s Base, the Water in the Cylinder would now have TWICE as much room to “spread out” in the Cylinder and the Height would drop by 1/2 to 3 inches.

Thus, since Cylinder Y is TWICE as Wide as Cylinder X, in order for the water to be at the same Height in both Cylinders we would need to give Twice as much of the Total Volume of Water we have to Cylinder Y (whose base “spreads out” twice as much) than to Cylinder X.

Vol Water in X : Vol Water in Y : Total Amount of water in both cylinders


Need to be in the Ratio of:

1m : 2m : 3m

Therefore, 2/3rd of All the Water needs to go in Y and 1/3rd of All the Water needs to go in X. Only then will we have an equal Height of the Water in each cylinder.

Vol of water in X = 5(pi) * 6 = 30(pi)

Vol of water in Y = 10(pi) * 2 = 20(pi)

Total Amount of water = 50(pi)


(1/3) * 50(pi) = Volume of Water in X, which has a radius = sqrt(5)

(2/3) * 50(pi) = Volume of Water in Y, which has a radius = sqrt(10)


Setting each equation equal to the Volume Formula with the corresponding radius:

50(pi)/3 = (pi) * ( sqrt(5) )^2 * (Height of X)

100(pi)/3 = (pi) * ( sqrt(10) )^2 * (Height of Y)


Solving for the Height in each equation:

Height of X = Height of Y = 10/3

C

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