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A rectangle with its two sides as 'x' and 'y' units respectively is

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A rectangle with its two sides as 'x' and 'y' units respectively is  [#permalink]

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New post 21 Jul 2018, 21:02
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A rectangle with its two sides as 'x' and 'y' units respectively is rotated along its side 'x' to form a cylinder C1, such that the circumference of base of C1 is 'x' and the height of C1 is 'y'. Another rectangle with its two sides as 'p' and 'q' units respectively is rotated along its side 'p' to form another cylinder C2, which now has circumference of base as 'p' and height as 'q'. Is the volume of C1 lesser than that of C2?

(1) Ratio of x to p is 3:2

(2) Ratio of y to q is 2:3
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Re: A rectangle with its two sides as 'x' and 'y' units respectively is  [#permalink]

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New post 21 Jul 2018, 21:43
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Volume of C1, V1 = πr1^2h1
Volume of C2, V2 = πr2^2h2
We need both r1,r2 and h1,h2 to find out which volume is greater.

Statement 1:
Circumference of C1 = X, 2πr1=X
Circumference of C2 = P, 2πr2=P

Ratio of X:P = 3:2
Consider X = 3c and P = 2c, where C is constant

Circumference C1,
2πr1 = 3c --> r1 = 3c/2π
Circumference C2,
2πr2 = 2c --> r2 = 2c/2π --> C/π

Volume V1 = πr1^2h1, we still miss h1 here so insufficient.

Statement 2:
Ratio of y to q is 2:3
Height of C1, y = 2h
Height of C1, q= 3h, where h is constant

We don't have r1 and r2 here hence insufficient.

Statement 1&2:
We have both r1,r2 from (1) and h1,h2 from (2).

V1/V2 = π*(3c/2π)^2*2h/π*(c/π)^2*3h

Reducing the above equation we get

V1/V2 = 9/4

Hence V1 is not lesser than V2. Sufficient

Method 2:
We can solve this without using any calculations. We know that to find volume of a cylinder we require both height and radius. This tell us that each statement alone is not sufficient and we require both statement 1 and statement 2 to find the volume and determine which volume is greater. Hence C (But to be sure you can solve and find volumes)

Please give kudos if you like the explanation.

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A rectangle with its two sides as 'x' and 'y' units respectively is  [#permalink]

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New post 21 Jul 2018, 21:50
amanvermagmat wrote:
A rectangle with its two sides as 'x' and 'y' units respectively is rotated along its side 'x' to form a cylinder C1, such that the circumference of base of C1 is 'x' and the height of C1 is 'y'. Another rectangle with its two sides as 'p' and 'q' units respectively is rotated along its side 'p' to form another cylinder C2, which now has circumference of base as 'p' and height as 'q'. Is the volume of C1 lesser than that of C2?

(1) Ratio of x to p is 3:2

(2) Ratio of y to q is 2:3



Volume is dependent on both the variables for a cylinder so each statement just giving ratio of one side is insufficient
Combined both should be sufficient as the volume of both cylinders would be in terms of the product of these two variables X&y for one and p&q for other..
Therefore we will get the ratio of two volumes ...
Although we can answer C with out further work but let us see how algebraically we get the answer..

Volume of cylinder = \(πr^2h\)
For cylinder with base X....2πr=X.....r=X/2π and H =y
\(V_x=π*(\frac{X}{2π})^2*y\).....
Similarly for other one \(V_p=π*(\frac{p}{2π})^2*q\)..
\(\frac{V_x}{V_p}=\frac{x^2y}{p^2q}=(\frac{X}{p)})^2*\frac{y}{q}=(9/4)*(2/3)=3/2\)
So V_x is more
Sufficient

C
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Re: A rectangle with its two sides as 'x' and 'y' units respectively is  [#permalink]

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New post 21 Jul 2018, 21:56
A rectangle with its two sides as 'x' and 'y' units respectively is rotated along its side 'x' to form a cylinder C1, such that the circumference of base of C1 is 'x' and the height of C1 is 'y'.

The circumference of base of C1 is 'x'

height of C1 is 'y'

=> \(2\pi r_{1} = x\)

=> \(r_{1} = \frac{x}{2\pi}\)

volume of this cylinder C1 is \(\pi (r_{1})^2 y\) = \(\pi \frac{x^2}{4(\pi)^2}y\) = \(\frac{x^2 y}{4\pi}\)

Another rectangle with its two sides as 'p' and 'q' units respectively is rotated along its side 'p' to form another cylinder C2, which now has circumference of base as 'p' and height as 'q'.

The circumference of base of C2 is 'p'

height of C2 is 'q'

=> \(2\pi r_{2} = p\)

=> \(r_{2} = \frac{p}{2\pi}\)

volume of this cylinder C2 is \(\pi (r_{2})^2 q\) = \(\pi \frac{p^2}{4(\pi)^2}\) = \(\frac{p^2 q}{4\pi}\)

We need to compare the volume of cylinders C1 and C2

=> we need to compare \(\frac{x^2 y}{4\pi}\) and \(\frac{p^2 q}{4\pi}\)

=> We need to compare \(x^2 y\) and \(p^2 q\)

Statement 1

Ratio of x to p is 3:2 => \(\sqrt{x^2 /p^2} = \frac{9}{4}\)

This doesn't tell anything about y and q

Statement 1 is insufficient

Statement 2

Ratio of y to q is 2:3 => \(\sqrt{y/q} = \frac{2}{3}\)

This doesn't tell anything about x and p

Statement 2 is insufficient

Combining statements 1 and 2

=> \(\frac{C1 volume}{C2 volume}\) = \(\frac{x^2}{p^2} * \frac{t}{q}\)

= \(\frac{9}{4} * \frac{2}{3} = \frac{3}{2}\) > 1

Since the \(\frac{C1 volume}{C2 volume}\) > 1 => C1 volume is > C2 volume

Hence option C
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Re: A rectangle with its two sides as 'x' and 'y' units respectively is &nbs [#permalink] 21 Jul 2018, 21:56
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A rectangle with its two sides as 'x' and 'y' units respectively is

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