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PS: Commodity (m08q01)

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PS: Commodity (m08q01) [#permalink] New post 24 Nov 2008, 15:49
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The price of a certain commodity increased at a rate of X % per year between 2000 and 2004. If the price was M dollars in 2001 and N dollars in 2003, what was the price in 2002 in terms of M and N ?

(C) 2008 GMAT Club - m08#1

* \sqrt{MN}
* N\sqrt{\frac{N}{M}}
* N\sqrt{M}
* N\frac{M}{\sqrt{N}}
* NM^{\frac{3}{2}}

[Reveal] Spoiler: OA
A

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Re: PS: Commodity [#permalink] New post 24 Nov 2008, 18:34
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A.

let price in 2002 = a

then a = M + x/100M = M (1 + x/100) .... (I)

(note that none of the answer choice has x)

N = a + x/100 * a
N = a( 1 + x/100)

from I above 1+x/100 = a/M

N = a . a/M

or a^2 = MN

a = sqrt(MN)
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Re: PS: Commodity [#permalink] New post 24 Nov 2008, 20:46
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bigfernhead wrote:
The price of a certain commodity increased at a rate of X % per year between 2000 and 2004. If the price was M dollars in 2001 and N dollars in 2003, what was the price in 2002 in terms of M and N ?

a) sqrt (MN)

b) N * sqrt(N/M)

c) N * sqrt (M)

d) N * M/sqrt(N)

e) N * M^(3/2)


agree with A.
I would do this way:

price in 2001 = m
price in 2002 = k = m (1+x) ...........................1
price in 2003 = n = k (1+x)............................. 2

(1+x) = n/k .....................................................3

lets substitute the value of (1=x) = n/k in 1.

k = m (1+x)
k = m (n/k)
k^2 = mn
k = sqrt(mn)
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Re: PS: Commodity (m08q01) [#permalink] New post 10 Dec 2009, 18:07
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By picking numbers :

Since, question stem talks about relationship between year 2002, 2003 and 2004, let's assume that Price was 100 in 2002 and rate of increase X=10.

2001 : Ignore
2002 : 100 = M
2003 : 100 * 1.1 = 110
2004 : 110 * 1.1 = 121 = N

Now calculate for each option that qualifies for valid value of 2003.

A: sqrt (MN) = sqrt ( 121 * 100 ) = 11 * 10 = 110

This is correct. We don't need to check other options and stop here.
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Re: PS: Commodity (m08q01) [#permalink] New post 12 Dec 2009, 14:30
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The three prices are in a geometric progression, the answer is A,
\sqrt{MN} - the geometric mean of M and N.
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Re: PS: Commodity [#permalink] New post 25 Nov 2008, 06:19
Correct Answer is A).

Great explanations!
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Re: PS: Commodity (m08q01) [#permalink] New post 18 Dec 2009, 07:45
Good explanation. Picking numbers really helps to solve this problem faster. Thanks.

naheed wrote:
By picking numbers :

Since, question stem talks about relationship between year 2002, 2003 and 2004, let's assume that Price was 100 in 2002 and rate of increase X=10.

2001 : Ignore
2002 : 100 = M
2003 : 100 * 1.1 = 110
2004 : 110 * 1.1 = 121 = N

Now calculate for each option that qualifies for valid value of 2003.

A: sqrt (MN) = sqrt ( 121 * 100 ) = 11 * 10 = 110

This is correct. We don't need to check other options and stop here.
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Re: PS: Commodity (m08q01) [#permalink] New post 17 Dec 2010, 06:39
A.

This is a geometric mean problem. if there is a sequence in which the next number is a multiple of the earlier number and so on(with the same multiplier of each next number) than these numbers are in Geometric Progression
2000 - A
2001 - (1 + x/100) * A = M
2002 - (1 + x/100)^2 * A = Y
2003 - (1+x/100) ^3 * A = N
Y is a geometric mean of M and N which is equal to SQRT (M *N)
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Re: PS: Commodity (m08q01) [#permalink] New post 28 Dec 2010, 02:20
GP(r = 1 + x/100):
T1: M.......take this as the 1st term (a)
T2: y (unknown)......2nd term
T3: N.....................3rd term
from Tn = ar^(n-1),
y = M(1 + x/100)
N = M(1 + x/100)^2..from here, we say (1 + x/100) = Sqrt(N/M)
==> y = sqrt(NM)

This approach could be clumsy if one is not conversant with GP series; In that case, plugin numbers
will help.
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Re: PS: Commodity (m08q01) [#permalink] New post 21 Dec 2011, 05:46
Maybe another way,

2000 - price is M
2002 - price is N
let's say 2001 - price is T

Knowing that X is the growth from one year to another

Then
+ T = M(1+X)
in that case (1+X)=T/M

also
+ N=T(1+X)
in that case N=T*T/M

then T^2 =NM

then T = NM^1/2

answer is A
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Re: PS: Commodity (m08q01) [#permalink] New post 22 Dec 2011, 03:16
Good question. I evaluated M and N, then started substituting in the answers.
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Re: PS: Commodity (m08q01) [#permalink] New post 27 Dec 2011, 23:10
let assume the price be Y for 2002.

so Y = M + X/100 M
Y = M(1+x/100).....................1
N= Y(1+x/100)
n = y*y/M ...... from 1
y^2 = MN
Y = Squrt MN

optionA
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Re: PS: Commodity (m08q01) [#permalink] New post 07 Jan 2012, 17:30
a.
picked numbers and plugged in to answers.
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Re: PS: Commodity [#permalink] New post 30 Sep 2012, 04:19
alpha_plus_gamma wrote:
A.

let price in 2002 = a

then a = M + x/100M = M (1 + x/100) .... (I)

(note that none of the answer choice has x)

N = a + x/100 * a
N = a( 1 + x/100)

from I above 1+x/100 = a/M

N = a . a/M

or a^2 = MN

a = sqrt(MN)


I would go with the answer A for the same reason as ALPHA-PLUS-GAMMA
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Re: PS: Commodity (m08q01) [#permalink] New post 30 Sep 2012, 04:22
Expert's post
The price of a certain commodity increased at a rate of X % per year between 2000 and 2004. If the price was M dollars in 2001 and N dollars in 2003, what was the price in 2002 in terms of M and N ?

A. \sqrt{MN}

B. N\sqrt{\frac{N}{M}}

C. N\sqrt{M}

D. N\frac{M}{\sqrt{N}}

E. NM^{\frac{3}{2}}

Use plug-in method for this problem.

Let the price in 2001 be 100 and the annual rate be 10%. Then:
2001 = 100 = M;
2002 = 110;
2003 = 121 = N;

Now, plug 100 and 121 in the answer choices to see which one gives 110:
A. \sqrt{MN}=\sqrt{100*121}=10*11=110, correct answer right away.

Answer: A.

P.S. For plug-in method it might happen that for some particular numbers more than one option may give "correct" answer. In this case just pick some other numbers and check again these "correct" options only.
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Re: PS: Commodity (m08q01) [#permalink] New post 21 Dec 2012, 06:56
A it is.

Plug-in is the fastest way to get to the answer.
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Re: PS: Commodity (m08q01) [#permalink] New post 16 Jan 2014, 10:41
The price of a certain commodity increased at a rate of X % per year between 2000 and 2004. If the price was M dollars in 2001 and N dollars in 2003, what was the price in 2002 in terms of M and N ?

Given:
Price in 2001 : M
Price in 2003 : N
Rate per year: X %

Price in 2002 = M ( 1 + X/100)........................................(1)
And, Price in 2003: N = M( 1 + X/100)^2

Or, (1 + x/100) =\sqrt{(N/M)}........(2)

Substituting the value of (2) in (1) above,
Price in 2002 = M \sqrt{(N/M)} = \sqrt{MN}

Answer: (A)
Re: PS: Commodity (m08q01)   [#permalink] 16 Jan 2014, 10:41
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