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# If the price of a commodity is directly proportional to m^3

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Manager
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If the price of a commodity is directly proportional to m^3 [#permalink]  07 Apr 2013, 04:45
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Question Stats:

53% (02:03) correct 47% (00:50) wrong based on 144 sessions
If the price of a commodity is directly proportional to m^3 and inversely proportional to q^2, which of the following values of m and q will result in the highest price for the commodity?

A. m=3, q=2
B. m=12, q=12
C. m=20, q=20
D. m=30, q=36
E. m=36, q=72
[Reveal] Spoiler: OA

Last edited by Bunuel on 07 Apr 2013, 21:20, edited 1 time in total.
RENAMED THE TOPIC.
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Re: Price of a commodity [#permalink]  07 Apr 2013, 04:56
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If the price of a commodity is directly proportional to m^3 and inversely proportional to q^2, which of the following values of m and q will result in the highest price for the commodity?
A. m=3, q=2
B. m=12, q=12
C. m=20, q=20
D. m=30, q=36
E. m=36, q=72

Let the price is P & constant is K
So P = \frac{K m^3}{q^2}

Now Just use the values in the options and put in the before mentioned equation.
A) P = \frac{K m^3}{q^2} = \frac{K 3^3}{2^2} = \frac{K 27}{4} = 6.75 K
B) P = \frac{K m^3}{q^2} = \frac{K 12^3}{12^2} = 12 K
C) P = \frac{K m^3}{q^2} = \frac{K 20^3}{20^2} = 20 K
D) P = \frac{K m^3}{q^2} = \frac{K 30^3}{36^2} = 125/6 = >20K
E) P = \frac{K m^3}{q^2} = \frac{K 36^3}{72^2} = 9 K

So the answer is D. This is probably 600-700 level question.

Hope the explanation helps.
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Last edited by emmak on 07 Apr 2013, 05:00, edited 1 time in total.
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Re: Price of a commodity [#permalink]  07 Apr 2013, 04:57
2
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Dipankar6435 wrote:
If the price of a commodity is directly proportional to m^3 and inversely proportional to q^2, which of the following values of m and q will result in the highest price for the commodity?
A. m=3, q=2
B. m=12, q=12
C. m=20, q=20
D. m=30, q=36
E. m=36, q=72

Function = \frac{m^3}{q^2}

A=\frac{3*3*3}{2*2}=\frac{27}{4}=7 almost
B=\frac{12*12*12}{12*12}=12
C=\frac{20*20*20}{20*20}=20
D=\frac{30*30*30}{36*36}=125/6
E=\frac{36*36*36}{72*72}=9

It's down to C or D, because \frac{120}{6}=20, \frac{125}{6}>20
D
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If the price of a commodity is directly proportional to m^3 [#permalink]  07 Apr 2013, 21:22
Expert's post
Dipankar6435 wrote:
If the price of a commodity is directly proportional to m^3 and inversely proportional to q^2, which of the following values of m and q will result in the highest price for the commodity?

A. m=3, q=2
B. m=12, q=12
C. m=20, q=20
D. m=30, q=36
E. m=36, q=72

Similar questions to practice:

DS:
the-amount-of-coal-a-train-burns-each-mile-is-directly-93667.html

PS:
a-is-directly-proportional-to-b-when-a-8-b-88971.html
in-a-certain-formula-p-is-directly-proportional-to-s-and-80941.html
the-rate-of-a-chemical-reaction-is-directly-proportional-to-76921.html

Hope it helps.
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Re: If the price of a commodity is directly proportional to m^3 [#permalink]  09 Apr 2013, 03:56
Let price of the commodity be P = k *m^3 /Q^2

A. P = k * 27/4 = 6.75K
B. P = K * 12^3/12^2 = 12k
C. P = K * 20^3/20^2 = 20k
D. P = K * 30^3/36^2= k*5*5*30/6*6 = k * (125/6) = 21K (approx)
E. P = k * 36^3/72^2 = K*(36/2*2) = 9k

Ans : Option D.
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Re: If the price of a commodity is directly proportional to m^3 [#permalink]  10 Apr 2013, 17:09
Could you please explain how you get \frac{m^3}{q^2} or \frac{Km^3}{q^2}? Where am I wrong expressing this thing firstly as p=m^3k and p=\frac{k}{q^2}?
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Re: If the price of a commodity is directly proportional to m^3 [#permalink]  10 Apr 2013, 22:19
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obs23 wrote:
Could you please explain how you get \frac{m^3}{q^2} or \frac{Km^3}{q^2}? Where am I wrong expressing this thing firstly as p=m^3k and p=\frac{k}{q^2}?

"directly proportional to m^3 and inversely proportional to q^2" is what the text says. So if m increases the price increases, if q increases the price decreases. The right formula here is \frac{m^3}{q^2} ( or with k, it doesnt change anything), if the numerator grows, the price does the same; and as in every fraction, if the denominator grows, the price drops.
Your formulas are right but the price depends on both m and q, you have to include them in one equation.

Let me know if it is clear
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Manager
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Re: If the price of a commodity is directly proportional to m^3 [#permalink]  11 Apr 2013, 02:51
Zarrolou wrote:
obs23 wrote:
Could you please explain how you get \frac{m^3}{q^2} or \frac{Km^3}{q^2}? Where am I wrong expressing this thing firstly as p=m^3k and p=\frac{k}{q^2}?

"directly proportional to m^3 and inversely proportional to q^2" is what the text says. So if m increases the price increases, if q increases the price decreases. The right formula here is \frac{m^3}{q^2} ( or with k, it doesnt change anything), if the numerator grows, the price does the same; and as in every fraction, if the denominator grows, the price drops.
Your formulas are right but the price depends on both m and q, you have to include them in one equation.

Let me know if it is clear

I see that makes sense. I guess I am making it more complicated than it really is. If you say that my formulas are correct, then \frac{m^3}{q^2} should somehow be extracted from p=m^3k and p=\frac{k}{q^2}, no? Technically speaking, I thought there was a way to combine them into \frac{m^3}{q^2}, just from pure algebraic manipulation (which I am very interested in here)... Or is it simply about common sense or the way you explained it?

P.S. Like your "kudos" tagline man!
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VP
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Kudos [?]: 1127 [1] , given: 219

Re: If the price of a commodity is directly proportional to m^3 [#permalink]  11 Apr 2013, 03:05
1
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obs23 wrote:
I see that makes sense. I guess I am making it more complicated than it really is. If you say that my formulas are correct, then \frac{m^3}{q^2} should somehow be extracted from p=m^3k and p=\frac{k}{q^2}, no? I thought there was a way to combine them into \frac{m^3}{q^2}, just from pure algebraic manipulation (which I am very interested in here)... Or is it simply about common sense or the way you explained it?

To create the formula we should refer to the text:
"If the price of a commodity is directly proportional to m^3 [and at this point we write down Price= m^3] and inversely proportional to q^2 [and at this point we complete the foumula adding the Denominator so Price=\frac{m^3}{q^2}], which of the following values of m and q will result in the highest price for the commodity?"
\frac{m^3}{q^2} is not extracted from p=m^3k and p=\frac{k}{q^2}, and you cannot obtain it from pure algebraic manipulation.
The idea behind your formulas is correct p=m^3 expresses the direct correlation between p and m; and also p=1/q^2 expresses the inverse correlation between p and q. But the text uses "and" so those ideas must be expressed in one formula, so to obtain this final formula you "complete" one with the other =>m^3/p^2
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Manager
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Re: If the price of a commodity is directly proportional to m^3 [#permalink]  11 Apr 2013, 03:30
Got it! Thanks much for help.
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Re: If the price of a commodity is directly proportional to m^3 [#permalink]  11 Apr 2013, 06:22
1
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Expert's post
Dipankar6435 wrote:
If the price of a commodity is directly proportional to m^3 and inversely proportional to q^2, which of the following values of m and q will result in the highest price for the commodity?

A. m=3, q=2
B. m=12, q=12
C. m=20, q=20
D. m=30, q=36
E. m=36, q=72

For more on direct, inverse and joint variation, check out these posts:

http://www.veritasprep.com/blog/2013/01 ... -directly/
http://www.veritasprep.com/blog/2013/02 ... inversely/
http://www.veritasprep.com/blog/2013/02 ... g-jointly/
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Re: If the price of a commodity is directly proportional to m^3   [#permalink] 11 Apr 2013, 06:22
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