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# A cylindrical tank of radius R and height H must be redesign

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Manager
Joined: 13 Jul 2011
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A cylindrical tank of radius R and height H must be redesign  [#permalink]

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24 Oct 2011, 22:17
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95% (hard)

Question Stats:

28% (02:51) correct 72% (02:42) wrong based on 625 sessions

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A cylindrical tank of radius R and height H must be redesigned to hold approximately twice as much liquid. Which of the following changes would be farthest from the new design requirements?

A. A 100% increase in R and a 50% decrease in H
B. A 30% decrease in R and a 300% increase in H
C. A 10% decrease in R and a 150% increase in H
D. A 40% increase in R and no change in H
E. A 50% increase in R and a 20% decrease in H
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Re: A cylindrical tank of radius R and height H must be redesign  [#permalink]

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18 Jul 2013, 23:24
8
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arjunbt wrote:
A cylindrical tank of radius R and height H must be redesigned to hold approximately twice as much liquid. Which of the following changes would be farthest from the new design requirements?

A. A 100% increase in R and a 50% decrease in H
B. A 30% decrease in R and a 300% increase in H
C. A 10% decrease in R and a 150% increase in H
D. A 40% increase in R and no change in H
E. A 50% increase in R and a 20% decrease in H

The solution I am going to give has already been provided but I am still writing it down to highlight the approximations you could use to save time.

Volume must be doubled.
We know volume of a cylinder $$= \pi * r^2 * h$$
To double the volume, either you double the height 'h' OR you multiply r by $$\sqrt{2}$$ i.e. by 1.41 i.e. increase radius by a little over 40%.
OR you use a combination.

A. A 100% increase in R and a 50% decrease in H
Volume $$= \pi * 4r^2 * \frac{h}{2} = 2* \pi * r^2 * h$$ (matches the design requirement)

B. A 30% decrease in R and a 300% increase in H
Volume $$= \pi * 0.49 r^2 * 4h = \pi * \frac{r^2}{2} * 4h = 2* \pi * r^2 * h$$ (matches the design requirement)
Note that .49 is approximately 0.5 i.e. 1/2

C. A 10% decrease in R and a 150% increase in H
Volume $$= \pi * .81 r^2 * \frac{5}{2}h = 2* \pi * \frac{4}{5}r^2 * \frac{5}{2} h = 2* \pi * r^2 * h$$ (matches the design requirement)
Notice that .81 is approximately .80 i.e. 80% i.e. 4/5

D. A 40% increase in R and no change in H
We discussed it above. A 40% increase in R matches our design requirement.

E. A 50% increase in R and a 20% decrease in H
Volume $$= \pi * 2.25 r^2 * 0.8 h$$
Compare with (C). 2.5*0.8 gives you 2 so 2.25*0.8 will give you less than 2. Hence doesn't match the requirement.
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Re: MGMAT Geometry #13, Redesigned Tank  [#permalink]

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24 Oct 2011, 23:31
2
1
ans is E
volume of cylinder=pi*r^2*H
=>reqd multiplication factor is 2
=>any change in R will cause volume to be multiplied by "square of (1+/- change in fractions)"
=>any change in H will cause vol to b multiplied by (1+/-change)
1. a 100% increase in R and a 50% decrease in H=>4*.5=2=>as required
2. a 30% decrease in R and a 300% increase in H=>.49*4=1.96
3. a 10% decrease in R and a 150% increase in H=>.81*2.5=2.025
4. a 40% increase in R and no change in H=>1.96
5. a 50% increase in R and a 20% decrease in H=>1.8=>max difference from required multiplication factor of 2

very easy qs...but my method took like 3 minutes....anyone with a faster solution?
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Re: MGMAT Geometry #13, Redesigned Tank  [#permalink]

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04 Dec 2011, 09:04
easy but time consuming...ny 1 with any easy method??
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Re: MGMAT Geometry #13, Redesigned Tank  [#permalink]

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17 Apr 2012, 23:03
vaibhav123 wrote:
easy but time consuming...ny 1 with any easy method??

Can any body explain it geometrically.
Thanks
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Posts: 60627
Re: MGMAT Geometry #13, Redesigned Tank  [#permalink]

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18 Apr 2012, 05:08
5
5
GMATD11 wrote:
vaibhav123 wrote:
easy but time consuming...ny 1 with any easy method??

Can any body explain it geometrically.
Thanks

What do you mean geometrically?

A cylindrical tank of radius R and height H must be redesigned to hold approximately twice as much liquid. Which of the following changes would be farthest from the new design requirements?
A. A 100% increase in R and a 50% decrease in H
B. A 30% decrease in R and a 300% increase in H
C. A 10% decrease in R and a 150% increase in H
D. A 40% increase in R and no change in H
E. A 50% increase in R and a 20% decrease in H

The wording makes the question harder than it actually is.

The question basically asks: if we have a cylindrical tank of radius R and height H, then which of the following changes will give a cylindrical tank with the volume with greatest difference from the volume of $$2\pi{r^2}h$$ (twice the original).

Answer E ($$\pi{(1.5r)^2}*(0.8h)=1.8\pi{r^2}h$$) gives the greatest difference from $$2\pi{r^2}h$$.
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Re: A cylindrical tank of radius R and height H must be redesign  [#permalink]

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18 Jul 2013, 04:52
1
arjunbt wrote:
A cylindrical tank of radius R and height H must be redesigned to hold approximately twice as much liquid. Which of the following changes would be farthest from the new design requirements?
A. A 100% increase in R and a 50% decrease in H
B. A 30% decrease in R and a 300% increase in H
C. A 10% decrease in R and a 150% increase in H
D. A 40% increase in R and no change in H
E. A 50% increase in R and a 20% decrease in H

"Farthest from the new design requirement" denotes highest difference between "twice the volume" (- the actual requirement) and volume given by each of the answer choices (- the new design requirement). Actual required volume is 2*pi*r^2*h.
We need to calculate 2 metrics for each answer choice:
1) Volume of each of the new design requirements and
2) Difference from actual required volume (2*pi*r^2*h)
A)
Volume of new design = pi*(2r)^2 * (.50h) = 2*pi*r^2*h
Difference = 2*pi*r^2*h ~ 2*pi*r^2*h = 0
B)
Volume of new design = pi*(.7r)^2 * (4h) = 1.96 *pi*r^2*h
Difference = 1.96 *pi*r^2*h ~ 2*pi*r^2*h = 0.04 *pi*r^2*h
C)
Volume of new design = pi*(1.4r)^2 * (h) = 2.025*pi*r^2*h
Difference = 2.025*pi*r^2*h ~ 2*pi*r^2*h = 0.025 *pi*r^2*h
D)
Volume of new design = pi*(.9r)^2 * (2.5h) = 1.96*pi*r^2*h
Difference = 1.96 *pi*r^2*h ~ 2*pi*r^2*h = 0.04 *pi*r^2*h
E)
Volume of new design = pi*(1.5r)^2 * (.8h) = 1.8*pi*r^2*h
Difference = 1.8*pi*r^2*h ~ 2*pi*r^2*h = 0.2 *pi*r^2*h

Therefore, the largest difference (farthest design) is 0.2. so the correct choice is E.
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Re: A cylindrical tank of radius R and height H must be redesign  [#permalink]

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18 Jul 2013, 22:09
Also, we can follow the below technique for finding the percentage increase in volume/area.
Say if side of a square increased by 10%. The new area is (1.1*side)^2 = 1.21 (side)^2 = 1.21(Old Volume). Therefore volume increased by 21%.

Similarly, for this question, we can just multiply the numeric part before radius^2 and height after applying appropriate % increase/decrease.
The old volume is pi * r^2 *h. And the required volume is 2*pi * r^2 *h.
Working out option E -> If radius increased by 50% and height decreased by 20%, then new volume = (1.5)^2 * (.80) (Old Volume)=1.8(Old Volume). Whereas required volume is 2*(Old Volume). Hence new design is .2 farther from required design.
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Re: A cylindrical tank of radius R and height H must be redesign  [#permalink]

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01 Dec 2013, 05:35
Karishma, I took R=10 and H=10 and proceeded but I am unable to arrive at a constructive answer. Can you please explain how to proceed using the above values and using substitution method. Thanks in advance .
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Re: A cylindrical tank of radius R and height H must be redesign  [#permalink]

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01 Dec 2013, 21:22
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Karishma, I took R=10 and H=10 and proceeded but I am unable to arrive at a constructive answer. Can you please explain how to proceed using the above values and using substitution method. Thanks in advance .

R = 10
H = 10

We know volume of a cylinder$$= \pi * r^2 * h = 1000\pi$$

A. A 100% increase in R and a 50% decrease in H (R = 20, H = 5)
Volume =$$\pi * 400 * 5 = 2000\pi$$(matches the design requirement)

B. A 30% decrease in R and a 300% increase in H (R = 7, H = 40)
Volume =$$\pi * 49 * 40 = 1960\pi$$ (approximately matches the design requirement)
Note that 49 is approximately 50

C. A 10% decrease in R and a 150% increase in H (R = 9, H = 25)
Volume$$= \pi * 81 * 25 = 2025\pi$$ (approximately matches the design requirement)
Notice that 81 is approximately 80

D. A 40% increase in R and no change in H (R = 14)
Volume$$= \pi * 196 * 10 = 1960 \pi$$ (approximately matches the design requirement)

E. A 50% increase in R and a 20% decrease in H (R = 15, H = 8)
Volume $$= \pi * 225 * 8 = 1800\pi$$ (Does not match)
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Re: A cylindrical tank of radius R and height H must be redesign  [#permalink]

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04 Dec 2015, 10:00
Can't you just use the formula, and multiply it pretty quickly? Took me less than 2 minutes if you just use the percentage increases.

A) 100% increase to R = 2R and 50% decrease to H = .5 H so formula is pi*r^2*h
so 2^2*.5=2

Repeat with each solution and you get E at 1.8.

This worked for me, but is this correct?
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Re: A cylindrical tank of radius R and height H must be redesign  [#permalink]

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02 Jun 2017, 17:55
Bunuel wrote:
GMATD11 wrote:
vaibhav123 wrote:
easy but time consuming...ny 1 with any easy method??

Can any body explain it geometrically.
Thanks

What do you mean geometrically?

A cylindrical tank of radius R and height H must be redesigned to hold approximately twice as much liquid. Which of the following changes would be farthest from the new design requirements?
A. A 100% increase in R and a 50% decrease in H
B. A 30% decrease in R and a 300% increase in H
C. A 10% decrease in R and a 150% increase in H
D. A 40% increase in R and no change in H
E. A 50% increase in R and a 20% decrease in H

The wording makes the question harder than it actually is.

The question basically asks: if we have a cylindrical tank of radius R and height H, then which of the following changes will give a cylindrical tank with the volume with greatest difference from the volume of $$2\pi{r^2}h$$ (twice the original).

Answer E ($$\pi{(1.5r)^2}*(0.8h)=1.8\pi{r^2}h$$) gives the greatest difference from $$2\pi{r^2}h$$.

Let the radius be 1 and the height be 2- now, the algebraic translation of twice the volume would be

2( (pi)(r)^2(h))
2( (pi)(1)^2(2))

We could simply construct an equation for each of the answer choices- for example

pi * 2(1)^2 (h-1/2h)

Eventually we would be able to arrive at the answer. Bunuel is there a quicker method- perhaps I am not reading into some of the other answers deeply enough- but it seems too time-consuming and inefficient to construct an equation for each answer choice and calculate which has an answer choice has a volume furthest away from the volume of twice the original cylinder's volume- is there a more efficient way of solving this problem?
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A cylindrical tank of radius R and height H must be redesign  [#permalink]

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21 Jul 2018, 10:07
KarishmaB wrote:
Karishma, I took R=10 and H=10 and proceeded but I am unable to arrive at a constructive answer. Can you please explain how to proceed using the above values and using substitution method. Thanks in advance .

R = 10
H = 10

We know volume of a cylinder$$= \pi * r^2 * h = 1000\pi$$

A. A 100% increase in R and a 50% decrease in H (R = 20, H = 5)
Volume =$$\pi * 400 * 5 = 2000\pi$$(matches the design requirement)

B. A 30% decrease in R and a 300% increase in H (R = 7, H = 40)
Volume =$$\pi * 49 * 40 = 1960\pi$$ (approximately matches the design requirement)
Note that 49 is approximately 50

C. A 10% decrease in R and a 150% increase in H (R = 9, H = 25)
Volume$$= \pi * 81 * 25 = 2025\pi$$ (approximately matches the design requirement)
Notice that 81 is approximately 80

D. A 40% increase in R and no change in H (R = 14)
Volume$$= \pi * 196 * 10 = 1960 \pi$$ (approximately matches the design requirement)

E. A 50% increase in R and a 20% decrease in H (R = 15, H = 8)
Volume $$= \pi * 225 * 8 = 1800\pi$$ (Does not match)

Life saver, right here! I almost postponed this one for later. Thanks a lot!
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Re: A cylindrical tank of radius R and height H must be redesign  [#permalink]

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30 Jul 2019, 05:37
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Re: A cylindrical tank of radius R and height H must be redesign   [#permalink] 30 Jul 2019, 05:37
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