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

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Joined: 12 Jul 2011
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A cylindrical tank of radius R and height H must be redesign [#permalink]  24 Oct 2011, 21:17
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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
[Reveal] Spoiler: OA
Veritas Prep GMAT Instructor
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Re: A cylindrical tank of radius R and height H must be redesign [#permalink]  18 Jul 2013, 22:24
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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.
_________________

Karishma
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Math Expert Joined: 02 Sep 2009 Posts: 30434 Followers: 5099 Kudos [?]: 57576 [2] , given: 8819 Re: MGMAT Geometry #13, Redesigned Tank [#permalink] 18 Apr 2012, 04:08 2 This post received KUDOS Expert's post 1 This post was BOOKMARKED 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 $$2\pi{r^2}h$$. _________________ Manager Joined: 16 Sep 2011 Posts: 182 Concentration: Strategy, Operations Schools: ISB '15 GMAT 1: 720 Q48 V40 GPA: 3.18 WE: Supply Chain Management (Manufacturing) Followers: 13 Kudos [?]: 48 [1] , given: 34 Re: MGMAT Geometry #13, Redesigned Tank [#permalink] 24 Oct 2011, 22:31 1 This post received KUDOS 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? Manager Joined: 17 Mar 2014 Posts: 92 Followers: 0 Kudos [?]: 33 [1] , given: 194 A cylindrical tank of radius R and height H must be redesign [#permalink] 03 Nov 2014, 10:20 1 This post received KUDOS I liked below solution. Just thought of posting here, it may help others. Credit goes to http://www.beatthegmat.com/new-design-r ... tml#231919 It is almost similar to Karishma and Bunuel's solution. The new design requirements need the final volume to be 2 π R^2 H from π R^2 H, or it needs to be doubled (2 times). QUICK CHECK is here A. 2^2 X 0.5 = 2, (2 - 2 = 0) B. 0.7^2 X 4 = 1.96, (2 - 1.96 = 0.04) C. 0.9^2 X 2.5 = 2.025, (2.025 - 2 = 0.025) D. 1.4^2 X 1 = 1.96, (2 - 1.96 = 0.04) E. 1.5^2 X 0.8 = 1.8, (2 - 1.8 = 0.2), this is the GREATEST DEVIATION Regards, Ammu Intern Joined: 07 Nov 2011 Posts: 31 Followers: 0 Kudos [?]: 1 [0], given: 9 Re: MGMAT Geometry #13, Redesigned Tank [#permalink] 04 Dec 2011, 08:04 easy but time consuming...ny 1 with any easy method?? Senior Manager Joined: 10 Nov 2010 Posts: 266 Location: India Concentration: Strategy, Operations GMAT 1: 520 Q42 V19 GMAT 2: 540 Q44 V21 WE: Information Technology (Computer Software) Followers: 5 Kudos [?]: 156 [0], given: 22 Re: MGMAT Geometry #13, Redesigned Tank [#permalink] 17 Apr 2012, 22:03 vaibhav123 wrote: easy but time consuming...ny 1 with any easy method?? Can any body explain it geometrically. Thanks _________________ The proof of understanding is the ability to explain it. Intern Joined: 09 Jun 2012 Posts: 31 Followers: 0 Kudos [?]: 11 [0], given: 13 Re: A cylindrical tank of radius R and height H must be redesign [#permalink] 18 Jul 2013, 03:52 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. Intern Joined: 09 Jun 2012 Posts: 31 Followers: 0 Kudos [?]: 11 [0], given: 13 Re: A cylindrical tank of radius R and height H must be redesign [#permalink] 18 Jul 2013, 21:09 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. Manager Joined: 07 May 2013 Posts: 109 Followers: 0 Kudos [?]: 14 [0], given: 1 Re: A cylindrical tank of radius R and height H must be redesign [#permalink] 01 Dec 2013, 04:35 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 . Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 6062 Location: Pune, India Followers: 1604 Kudos [?]: 8971 [0], given: 195 Re: A cylindrical tank of radius R and height H must be redesign [#permalink] 01 Dec 2013, 20:22 Expert's post madn800 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) _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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Re: A cylindrical tank of radius R and height H must be redesign   [#permalink] 01 Dec 2013, 20:22
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