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# The price of a certain commodity increased at a rate of

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The price of a certain commodity increased at a rate of  [#permalink]

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Updated on: 24 Jan 2012, 15:09
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Difficulty:

55% (hard)

Question Stats:

66% (02:26) correct 34% (02:58) wrong based on 199 sessions

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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 $$N$$ ?

A. $$\sqrt{MN}$$

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

C. $$N\sqrt{M}$$

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

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

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Originally posted by manalq8 on 23 Jan 2012, 14:25.
Last edited by Bunuel on 24 Jan 2012, 15:09, edited 2 times in total.
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Re: The price of a certain commodity  [#permalink]

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23 Jan 2012, 16:23
5
1
Hi there! I'm happy to help with this!

Let's say, the price in 2000 is A. That the original amount. Each year it increases X%. To represent a percent increase (a) write the percent as a fraction/decimal, here X/100; (b) add one ---> 1 + X/100; (c) that's the multiplier -- multiplying a number by that multiplier results in a X% increase.

price in 2000 = A
price in 2001 = A*(1 + X/100)
price in 2002 = A*(1 + X/100)^2
price in 2003 = A*(1 + X/100)^3
price in 2004 = A*(1 + X/100)^4

For simplicity, I am going to define r = (1 + X/100). Then these equations become:

price in 2000 = A
price in 2001 = A*r
price in 2002 = A*r^2
price in 2003 = A*r^3
price in 2004 = A*r^4

Now, suppose we have M = 2001 price = A*r and N = 2003 price = A*r^3. How do we represent the 2002 prince (A*r^2) in terms of M and N?

There are two methods.

Method One: express r in terms of M and N

This is more a crank-it-out algebraic solution approach. We notice that N/M = (A*r^3)/(A*r) = r^2, so r = sqrt(N/M). Well,

2002 price = (2001 price)*(r) = M * sqrt(N/M) = [sqrt(M)*sqrt(M)]*[sqrt(N)/sqrt(M)] = sqrt(M)*sqrt(N) = sqrt(NM).

Through some fast-and-loose manipulation of the laws of squareroots, we arrive at answer .

Method Two: a more elegant solution for a more civilized age . . .

When you have an arithmetic sequence --- that is, adding the same number to get new terms (e.g. 8, 11, 14, 17, 20, 23, . . . ), when you take any three numbers in a row, the middle number is the mean, the arithmetic average, of the outer two. For example ---11, 14, 17 --- (11 + 17)/2 = 14. The arithmetic average is the ordinary average --- add the two numbers, and divide by two.

When you have a geometric sequence -- that is, multiplying the same ratio to get new terms (e.g. 2, 6, 18, 54, 162, . . . ), when you take any three numbers in a row, the middle number is the geometric mean of the outer two. The geometric mean of two numbers means multiply the two numbers and take the squareroot. For example --- 6, 18, 54 --- 6*54 = 324, and sqrt(324) = 18.

When you apply a fixed percentage increase from one term to the next, as we have in this problem, that's a geometric sequence. Thus, to find the 2002 price, all you have to do is take the geometric mean of the 2001 price and the 2003 price. 2002 price = sqrt(MN). Bam. Done. Again, answer = .

The ideas about arithmetic & geometric sequences, and the associated means, are good tricks to have up your sleeve for the more challenging GMAT math problems.

Does all this make sense? Please let me know if you have any questions.

Mike
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Re: The price of a certain commodity increased at a rate of  [#permalink]

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23 Jan 2012, 22:01
4
2
The price in 2001 is M
The price in 2002 is M(1+$$\frac{X}{100}$$) ----------------------(1)

The price in 2003 is M(1+$$\frac{X}{100}$$)(1+$$\frac{X}{100}$$) = N ---------(2)

Solving equation (2) we get
(1+$$\frac{X}{100}$$) = $$\sqrt{\frac{N}{M}}$$

Putting this value in equation (1) to get the desired answer
The price in 2002 is M$$\sqrt{\frac{N}{M}}$$ = $$\sqrt{MN}$$

Hence the option A
##### General Discussion
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Re: The price of a certain commodity  [#permalink]

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23 Jan 2012, 19:00
manalq8 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. $$N\sqrt{\frac{N}{M}}$$

C. $$N\sqrt{M}$$

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

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

I think that plug-in method is easiest 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.

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.

Hope it helps.
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Re: The price of a certain commodity increased at a rate of  [#permalink]

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08 Jul 2016, 21:44
manalq8 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. $$N\sqrt{\frac{N}{M}}$$

C. $$N\sqrt{M}$$

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

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

When no absolute values(or variables pertaining to them) are given, number plugging is the way to go..

Let the Value initially be = 1
and let X% = 100%(in other words..value doubles every year)

values are..
M = 2
N = 8

and we're looking for the value in 2002..which is equal to 4.

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The price of a certain commodity increased at a rate of  [#permalink]

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30 Mar 2018, 10:36
2001 - M

2002 - ? (Let it be 'x')

2003 - N

Now according to ques, since the rate of increase is equal both year and the rates should be calculated on the previous year's price,

$$\frac{x-M}{M} * 100= \frac{N-x}{x}*100$$

$$x^2 - Mx = MN - Mx$$

$$x = \sqrt{MN}$$

The price of a certain commodity increased at a rate of &nbs [#permalink] 30 Mar 2018, 10:36
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