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# The cost C of manufacturing a certain product can be

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Director
Joined: 01 May 2007
Posts: 793
The cost C of manufacturing a certain product can be [#permalink]

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13 Apr 2008, 16:01
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

The cost C of manufacturing a certain product can be estimated by the formula C = 0.03rs(t^2) where r and s are the amounts, in pounds of the two major ingredients and t is the production time, in hours. If r is increased by 50 percent, s in increased by 20 percent, and t is decreased by 30 percent, by approximately what percent will the estimated cost of manufacturing the product change?

40% increase
12% increase
4% increase
12% decrease
24% decrease
Senior Manager
Joined: 18 Feb 2008
Posts: 494
Location: Kolkata
Re: Looking for fast way to do this problem... [#permalink]

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13 Apr 2008, 20:23
took me about 2-3 min....but maybe some faster method is there...after all gmat is all abt timing!!!
C=0.03rst^2
C'=0.03(1.5r1.2s0.49t^2)=0.02646rst^2

C-C'/C=0.00354/0.03=11.8%decrease=12%decreaseHence d.
SVP
Joined: 04 May 2006
Posts: 1892
Schools: CBS, Kellogg
Re: Looking for fast way to do this problem... [#permalink]

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13 Apr 2008, 21:52
jimmyjamesdonkey wrote:
The cost C of manufacturing a certain product can be estimated by the formula C = 0.03rs(t^2) where r and s are the amounts, in pounds of the two major ingredients and t is the production time, in hours. If r is increased by 50 percent, s in increased by 20 percent, and t is decreased by 30 percent, by approximately what percent will the estimated cost of manufacturing the product change?

40% increase
12% increase
4% increase
12% decrease
24% decrease

D
C1 =0.03*1.5r*1.2s* (0.7t)^2 = 1.5*1.2*0.7^2 *C = 0.882*C
The estimated cost decreased =(1-0.882)*100% =12%
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Manager
Joined: 09 Apr 2008
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Concentration: Strategy, Operations
Schools: CBS '15 (A)
Re: Looking for fast way to do this problem... [#permalink]

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13 Apr 2008, 22:06
1
KUDOS
Keeping in mind we don't have calculators I would go this route...

C = 0.03 * r * s * t^2
C' = 0.03 * 1.5r * 1.2s * (0.7t)^2
= 0.03 * 1.5r * 1.2s * 0.49t^2
= 0.054rs * 0.49t^2 (1.5 times 0.03 is 0.045 then do 0.045*1.2 on paper)

NOW, since no calculator... estimate 0.49 to 0.5 to get 0.027 * r * s * t^2

(0.030 - 0.027) / (0.030) = 0.003 / 0.030 = 1/10 = 10%

10% ~ 12% from the choices [D] is the only viable answer.
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Re: Looking for fast way to do this problem... [#permalink]

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13 Apr 2008, 22:19
we can express the change as 1.5*1.2*0.7^2

1.5*1.2 = 1.5+1.5*0.2=1.8

0.7^2=0.49~0.5

1.8*0.5=1.8/2=0.9 --> decrease ~10% --> D
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SVP
Joined: 29 Aug 2007
Posts: 2473
Re: Looking for fast way to do this problem... [#permalink]

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13 Apr 2008, 23:09
walker wrote:
we can express the change as 1.5*1.2*0.7^2

1.5*1.2 = 1.5+1.5*0.2=1.8

0.7^2=0.49~0.5

1.8*0.5=1.8/2=0.9 --> decrease ~10% --> D

same process here.
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Senior Manager
Joined: 18 Feb 2008
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Location: Kolkata
Re: Looking for fast way to do this problem... [#permalink]

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14 Apr 2008, 06:03
thank you gmattiger and last but not the least...walker you always rock
Re: Looking for fast way to do this problem...   [#permalink] 14 Apr 2008, 06:03
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