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

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There are 2 circular cylinders X and Y, and both cylinders contain wat  [#permalink]

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There are 2 circular cylinders X and Y, and both cylinders contain water inside. Cylinder X has 5π square inches as the base area and 6 inches as the height of the water inside, and cylinder Y has 10π square inches as the base area and 2 inches as the height of the water inside. If the height of the water becomes the same when the water drawn from cylinder X is poured into cylinder Y, what is the height of water in these cylinders, in inches?

A. 2.5
B. 3
C. \(\frac{10}{3}\)
D. 4
E. 4.5

Weekly Quant Quiz #3 Question No 1


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Originally posted by gmatbusters on 06 Oct 2018, 09:59.
Last edited by gmatbusters on 07 Oct 2018, 05:08, edited 3 times in total.
Renamed the topic, edited the question and added the OA.
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There are 2 circular cylinders X and Y, and both cylinders contain wat  [#permalink]

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New post 06 Oct 2018, 10:01

Official Explanation

Attachment:
1.jpg
1.jpg [ 16.33 KiB | Viewed 193 times ]


Volume of water = Area of Base * Height /level of water filled
As shown in the figure above, since the heights of both cylinders become the same, Initial total volume = final volume.
We get 6(5π) + 2(10π) = 5πk + 10πk or 50π = 15πk.
Therefore, k= \(\frac{10}{3}\).
The answer is C.

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Re: There are 2 circular cylinders X and Y, and both cylinders contain wat  [#permalink]

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New post 06 Oct 2018, 10:07
PFA solution in the image.

Answer should be C.
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Re: There are 2 circular cylinders X and Y, and both cylinders contain wat  [#permalink]

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New post 06 Oct 2018, 10:09
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
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Re: There are 2 circular cylinders X and Y, and both cylinders contain wat  [#permalink]

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New post 06 Oct 2018, 10:10
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
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Re: There are 2 circular cylinders X and Y, and both cylinders contain wat  [#permalink]

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New post 06 Oct 2018, 10:11
C, as it gives the same volume , as to when individual volumes are added with the respective heights

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Re: There are 2 circular cylinders X and Y, and both cylinders contain wat  [#permalink]

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New post 06 Oct 2018, 10:12
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
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Re: There are 2 circular cylinders X and Y, and both cylinders contain wat  [#permalink]

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New post 06 Oct 2018, 10:12
C, as it gives the same volume , as to when individual volumes are added with the respective heights

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Re: There are 2 circular cylinders X and Y, and both cylinders contain wat  [#permalink]

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New post 06 Oct 2018, 10:13
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.
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Re: There are 2 circular cylinders X and Y, and both cylinders contain wat  [#permalink]

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New post 06 Oct 2018, 10:14
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
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Re: There are 2 circular cylinders X and Y, and both cylinders contain wat  [#permalink]

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New post 06 Oct 2018, 10:17
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
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Re: There are 2 circular cylinders X and Y, and both cylinders contain wat  [#permalink]

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New post 06 Oct 2018, 10:19
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.
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Re: There are 2 circular cylinders X and Y, and both cylinders contain wat  [#permalink]

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New post 06 Oct 2018, 10:59
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.
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Re: There are 2 circular cylinders X and Y, and both cylinders contain wat  [#permalink]

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New post 06 Oct 2018, 11:06
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
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Re: There are 2 circular cylinders X and Y, and both cylinders contain wat &nbs [#permalink] 06 Oct 2018, 11:06
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