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The cost of a square slab is proportional to its thickness a [#permalink]
19 Jul 2008, 20:29

00:00

A

B

C

D

E

Difficulty:

35% (medium)

Question Stats:

57% (02:04) correct
42% (00:48) wrong based on 26 sessions

The cost of a square slab is proportional to its thickness and also proportional to the square of its length. What is the cost of a square slab that is 3 meters long and 0.1m thick.

(1) The cost of a square slab that is 2m long and 0.2 m thick is $160 more than the cost of a slab that is 2m long and 0.1 m thick

(2) The cost of a square slab that is 3 m long and 0.1 m thick is 200 more than the cost of a square slab that is 2m long and 0.1 m thick

The cost of a square slab is proportional to its thickness and also proportional to the square of its length. What is the cost of a square slab that is 3 meters long and 0.1m thick.

(1) The cost of a square slab that is 2m long and 0.2 m thick is $160 more than the cost of a slab that is 2m long and 0.1 m thick area1 = 2x2x0.2 = 0.4m^2; area2 = 2x2x0.1 = 0.2m^2 A1 - A2 = 0.2m^2 = $160; so you can calculate the area of 0.1m^2 and you know that the are of the salb in question is 3x3x.1 = 0.9m^2

(2) The cost of a square slab that is 3 m long and 0.1 m thick is 200 more than the cost of a square slab that is 2m long and 0.1 m thick Follow same logic as S1.

Re: Proportionality (Disagree with OA) [#permalink]
07 Sep 2009, 11:50

One clue for my view: the cost can be proportional to both thickness and length but with different proportionality constants. I mean, to me: Cost=a*thickness+b*length^2 not Cost=a*(thickness+length^2)

Re: Proportionality (Disagree with OA) [#permalink]
07 Sep 2009, 13:08

LenaA wrote:

I think the cost function is the following:

C=k\times t\times l^2 t-thickness l-length

then each statement alone is sufficient

stmt1 4\times 0.2\times k=4\times 0.1\times k +160 you can solve for k sufficient

stmt2 is basically similar to stmt 1... 9\times 0.1\times k=4\times 0.1\times k +200 you cansolve for k sufficient

In fact, that is the formula in order to be D the correct answer (as it is). But my point is, that in a very strict point of view, the proportionality constant (what you mean k), can be different for t and l, that is: Cost=k1*t+k2*l^2. So you need both statements to solve for k1 and k2, and correct answer is C.

To sum up, correct answer is C. OA is D.
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Re: Proportionality (Disagree with OA) [#permalink]
07 Sep 2009, 14:06

you are wrong. i would suggest to research about the jointly proportional functions. if z is proptional to x (when y is constant) and z is propotional to y (when x is constant), then z is propotional to the product xy and is of the form z=Kxy

Re: Proportionality (Disagree with OA) [#permalink]
07 Sep 2009, 14:20

LenaA wrote:

you are wrong. i would suggest to research about the jointly proportional functions. if z is proptional to x (when y is constant) and z is propotional to y (when x is constant), then z is propotional to the product xy and is of the form z=Kxy

Correct. Thank u. Dont know in what i was thinking about!

Re: Proportionality (Disagree with OA) [#permalink]
07 Sep 2009, 14:32

We have only one value missing (here it's K) so it's sure that each statement is sufficient.

noboru wrote:

LenaA wrote:

you are wrong. i would suggest to research about the jointly proportional functions. if z is proptional to x (when y is constant) and z is propotional to y (when x is constant), then z is propotional to the product xy and is of the form z=Kxy

Correct. Thank u. Dont know in what i was thinking about!

Re: The cost of a square slab is proportional to its thickness a [#permalink]
19 Jan 2013, 02:01

2

This post received KUDOS

kiyo0610 wrote:

The cost of a square slab is proportional to its thickness and also proportional to the square of its length. What is the cost of a square slab that is 3 meters long and 0.1 meter thick?

(1)The cost of a square slab that is 2 meters long and 0.2 meter thick is $160 more than the cost of a square slab that is 2 meters long and 0.1 meter thick. (2)The cost of a square slab that is 3 meters long and 0.1 meter thick is $200 more than the cost of a square slab that is 2 meters long and 0.1 meter thick.

Cost C 1) C proportional to Thickness t 2) C proportional to Length square l^2

C = K t l^2

We need to know constant K to find the answer.

Option 1: C1 and C2 difference is given for some thickness and length. We can find the constant Option 2: Same as option1
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Re: The cost of a square slab is proportional to its thickness [#permalink]
09 Sep 2013, 06:04

I formed equation for cost as :

C prop to l^2 C prop to t

C = kl^2 + rt l= for length t= for thickness. k and r constant of respective proportionality.

But in above mentioned solution it is taken as product.

I am not 100% satisfied with the derived proportionality as the product of length and thickness.

May be I am not able to identify the keyword in the question which governs product of two variables. Or lacking some basic concept, kindly help me to interpret the language of question into a equation. Please also share any theoretical stuff, which I should refer to understand concept of proportionality.

Thanks
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Piyush K ----------------------- Our greatest weakness lies in giving up. The most certain way to succeed is to try just one more time. ― Thomas A. Edison Don't forget to press--> Kudos

Re: The cost of a square slab is proportional to its thickness [#permalink]
11 Sep 2013, 04:12

When I first attempted to solve this problem I was a little thrown off by the question just saying proportional, and not directly proportional or indirectly proportional. I now realize that solving this problem is independent of the direct vs. indirect, you may get different values for the cost, but regardless you'll be able to get a value => sufficient.

My question is, can you assume that it's directly proportional from the question stem? Looking at a few of the answers above, it seems that some people have. If this was a P.S. problem instead of a D.S., the answer would depend on this assumption.

Re: Proportionality (Disagree with OA) [#permalink]
25 Nov 2013, 02:01

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