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PS: Commodity (m08q01) [#permalink]
 24 Nov 2008, 16:49
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}}Source: GMAT Club Tests - hardest GMAT questions
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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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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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Correct Answer is A).
Great explanations!
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Re: PS: Commodity (m08q01) [#permalink]
10 Dec 2009, 19:07
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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]
12 Dec 2009, 15:30
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 (m08q01) [#permalink]
18 Dec 2009, 08:45
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]
17 Dec 2010, 07:39
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]
28 Dec 2010, 03:20
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]
21 Dec 2011, 06:46
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]
22 Dec 2011, 04:16
Good question. I evaluated M and N, then started substituting in the answers.
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Re: PS: Commodity (m08q01) [#permalink]
28 Dec 2011, 00:10
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]
07 Jan 2012, 18:30
a.
picked numbers and plugged in to answers.
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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]
30 Sep 2012, 05:22
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]
21 Dec 2012, 07:56
A it is.
Plug-in is the fastest way to get to the answer.
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Re: PS: Commodity (m08q01)
[#permalink]
21 Dec 2012, 07:56
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