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Pls help me to solve this problem.(2)

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Pls help me to solve this problem.(2) [#permalink] New post 31 Jul 2011, 02:32
00:00
A
B
C
D
E

Difficulty:

  55% (hard)

Question Stats:

57% (03:47) correct 43% (03:05) wrong based on 7 sessions
The cost C of manufacturing a certain product can be estimated by the formular C=0.03rs(t^2), where r and s are the amounts, in pounds, oof the two major ingredients and t is the production time in hours. If r is increased by 50%, s is increased by 20%, and t is decreased by 30%, by approximately what percent will the estimated cost manufacturing the product change?

a) 40% increase
b) 12% increase
c) 4% incease
d) 12% decease
e) 24% decrease

pls explain me, thank you. I need it.
[Reveal] Spoiler: OA
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Re: Pls help me to solve this problem.(2) [#permalink] New post 31 Jul 2011, 04:30
Let r=s=t=100

C = 0.03*r*s(t^2)
=0.03*100*100*(100^2)
=3,000,000

r1 = 50% more than r = 150
s1 = 20% more than s =120
t1 = 30% less than t = 70

C1= 0.03 * 150 *120 * 70 * 70
=0.03 * 1800 * 4900
= 54*4900
=264600

Decrease = C-C1 = 3,000,000-264600=35400
%Decrease = 35400/300000 * 100 = 354/30 = 11.8 ~ 12% Decrease

OA D
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Re: Pls help me to solve this problem.(2) [#permalink] New post 31 Jul 2011, 05:15
tracyyahoo wrote:
The cost C of manufacturing a certain product can be estimated by the formular C=0.03rs(t^2), where r and s are the amounts, in pounds, oof the two major ingredients and t is the production time in hours. If r is increased by 50%, s is increased by 20%, and t is decreased by 30%, by approximately what percent will the estimated cost manufacturing the product change?

a) 40% increase
b) 12% increase
c) 4% incease
d) 12% decease
e) 24% decrease

pls explain me, thank you. I need it.


From the formula we deduce that C directly proportional to r, s and t^2
hence an increase or decrease in these values will directly change the value of C by that factor

There new r => 1.5r
new s => 1.2s
new t^2 = (0.7t)^2 = .49t^2
let the constant 0.03 be termed as A

Putting these new values: C1 = A*1.5r*1.2s*.49t^2
C1 = 0.882A*r*s*t^2 = 0.882(C)

There the net value decrease by 12% approx

Hence D
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Re: Pls help me to solve this problem.(2) [#permalink] New post 31 Jul 2011, 12:25
Hi

How did you guys determine if it was an increase or decrease?

is it because the value of c > c1?
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Re: Pls help me to solve this problem.(2) [#permalink] New post 31 Jul 2011, 14:19
Sudhanshuacharya wrote:
tracyyahoo wrote:
The cost C of manufacturing a certain product can be estimated by the formular C=0.03rs(t^2), where r and s are the amounts, in pounds, oof the two major ingredients and t is the production time in hours. If r is increased by 50%, s is increased by 20%, and t is decreased by 30%, by approximately what percent will the estimated cost manufacturing the product change?

a) 40% increase
b) 12% increase
c) 4% incease
d) 12% decease
e) 24% decrease

pls explain me, thank you. I need it.


From the formula we deduce that C directly proportional to r, s and t^2
hence an increase or decrease in these values will directly change the value of C by that factor

There new r => 1.5r
new s => 1.2s
new t^2 = (0.7t)^2 = .49t^2
let the constant 0.03 be termed as A

Putting these new values: C1 = A*1.5r*1.2s*.49t^2
C1 = 0.882A*r*s*t^2 = 0.882(C)

There the net value decrease by 12% approx

Hence D

I am using the same approach as above:

50% increase = 1.5r
20% incease = 1.2s
30% decrease = .7r

So new value = (1.5r)(1.2s)(.7r)^2=(15/10*12/10*49/100)(.03rst^2)
Old value = (.03rst^2)

Now I am approx. 49 to 50 for faster calculation -
New value = (15/10*12/10*50/100) = 180/100*1/2=(90/100)(.03rst^2)

New value is 90% of the old value
So 10% less ....Let me see any value near 10% decreases
B is the closest...So I go with B
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Re: Pls help me to solve this problem.(2) [#permalink] New post 31 Jul 2011, 14:21
arjun007 wrote:
Hi

How did you guys determine if it was an increase or decrease?

is it because the value of c > c1?


Hope it is clear to you how it is decreasing ... Taking a larger value like - 50 even decreased,so obviously a smaller value - 49 will always decrease
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Re: Pls help me to solve this problem.(2) [#permalink] New post 31 Jul 2011, 14:24
The reason for approximating is sometimes we make silly mistakes in calculation.
So the best way is to pick smart numbers when needed.........Again it is up to you to decide the approach that works nest for you.....
Some people can calculate this in head in (5-10) secs....then do not approximate...This is for the other folks who spend 20-30 secs on multiplication(calculation)
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Re: Pls help me to solve this problem.(2) [#permalink] New post 31 Jul 2011, 17:22
Thank you so much you guys, Why I haven't thought about using Plug-in strategy.

This question is very simple after I saw your explanation. Thank you again for the effot.


jamifahad wrote:
Let r=s=t=100

C = 0.03*r*s(t^2)
=0.03*100*100*(100^2)
=3,000,000

r1 = 50% more than r = 150
s1 = 20% more than s =120
t1 = 30% less than t = 70

C1= 0.03 * 150 *120 * 70 * 70
=0.03 * 1800 * 4900
= 54*4900
=264600

Decrease = C-C1 = 3,000,000-264600=35400
%Decrease = 35400/300000 * 100 = 354/30 = 11.8 ~ 12% Decrease

OA D
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Re: Pls help me to solve this problem.(2) [#permalink] New post 31 Jul 2011, 18:14
tracyyahoo - Give kudos to jamifahad if you liked his approach...

That is one way of saying thanks for the person spending time on your questions.....
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Re: Pls help me to solve this problem.(2) [#permalink] New post 31 Jul 2011, 23:44
Thank you guys.. :-D
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Re: Pls help me to solve this problem.(2) [#permalink] New post 01 Aug 2011, 17:04
C = 0.03 r s t^2
New C = C1 = 0.03(150/100)r(120/100)s((70/100)^2)t^2

=(882/1000)[0.03rst^2]
as C>C1 => its a decrease

% decrease = ((original - new) /original)*100

= (118/1000) *100
~= 12%

Answer is D.
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Re: Pls help me to solve this problem.(2) [#permalink] New post 12 Aug 2011, 11:30
I got it right, but a took a long time for calculation.
Re: Pls help me to solve this problem.(2)   [#permalink] 12 Aug 2011, 11:30
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