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PS: Commodity (m08q01) [#permalink]
24 Nov 2008, 15:49

2

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This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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 \(N\) ?

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

Re: PS: Commodity (m08q01) [#permalink]
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)

Re: PS: Commodity (m08q01) [#permalink]
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. _________________

KUDOS me if you feel my contribution has helped you.

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. _________________

Re: PS: Commodity (m08q01) [#permalink]
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)

gmatclubot

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