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# M08-01

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Math Expert
Joined: 02 Sep 2009
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16 Sep 2014, 00:36
1
16
00:00

Difficulty:

45% (medium)

Question Stats:

67% (02:39) correct 33% (02:22) wrong based on 103 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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04 Nov 2014, 13:27
16
5
Hi, Bunuel,
The official solution seems to be lengthy

Here is an Easy Way to solve the Problem......

Consider M as the price in 2001.

Then, M(1+x/100) will be the price in 2002 ----> We are asked to find this value

And, M(1+x/100)^2 will be the price in 2003, and this price is given as N

So,
M(1+x/100)^2 = N
(1+x/100) ^2 = N/M
(1+x/100) = sqrt. (N/M)
M(1+x/100) = sqrt. (N/M) x M
M(1+x/100) = sqrt. (NM) <------> This is the Answer... Thats it

PLease give kudos if you like the solution
##### General Discussion
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Joined: 02 Sep 2009
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16 Sep 2014, 00:36
2
1
Official Solution:

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.

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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08 Nov 2015, 21:37
1
Let 2002 = P.
So, M(1+X/100) = P = N/(1+X/100)
or, MN = P^2
or, P = sqrt(MN) -----> A.
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01 Dec 2015, 06:29
2
1
We have:
$$2002= M(1+x)$$
$$2003= M(1+x)^2 = N$$

Convert $$(1+x)^2$$ into M and N
$$(1+x)^2 = \frac{N}{M}$$

$$(1+x) = \sqrt{\frac{N}{M}}$$

So 2002 becomes:
$$2002= M*\sqrt{\frac{N}{M}}$$ which equals $$\sqrt{MN}$$
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03 Dec 2015, 18:06
1
Can be solved using simpler Geometric progression logic.
In 2000 : a
In 2001 : ax
In 2002 : ax^2
In 2003 : ax^3
In 2004 : ax^4

Then we can verify with the answer choices and find out that only (A) suffices.
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05 Jan 2016, 09:16
1
You could also do dimensionality check
consider dimensionality of N or M as 1

A. $$\sqrt{MN}$$ >> 1
B. N*$$\sqrt{NM}$$ >> 2
C. N*$$\sqrt{M}$$>>1.5
D. N*M/$$\sqrt{N}$$ >> 1.5
E. NM^ $$3/2$$>>1.5

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19 Feb 2016, 11:54
I thought that this was more of an understanding the question and staying organized type of problem.

2000 = P
2001 = M = Px
2002 = PxM
2003 = N = P(M^2)

So the question stem asks for 2002 (x*M) in terms of M and N

The square root of N = M The squareroot of M = X

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16 Jul 2016, 08:50
Since we know that the 2001 value time the constant rate should equal the 2003 value divided by the constant rate, we set the equation and solve for X
M*(X) = N / X

X^2 = N/M

X= sqrt(N/M)

Next we multiply the rate by the 2001 value to get the 2002 value, and rearrange to get it in terms similar to the answer choices

M* sqrt(N/M) = sqrt(NM^2/M) = sqrt(NM)
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05 Dec 2016, 11:06
1
since the resulting sequence will be a Geometric progression so the middle value will be the geometric mean of the prices of 2001 and 2003
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18 Dec 2016, 18:34
VinRag

Your method of solution has already been demonstrated by Bunuel on the similar question.
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08 Mar 2017, 18:54
Let I = the initial amount and u=(1+x/100).

Value in 2000: I
Value in 2001: I*u=M
Value in 2002: I*u^2
Value in 2003: I*u^3=N
Value in 2004: I*u^4

Plug the above into the answer choices. A gives sqrt(M*N) = sqrt(I^2 * u^4 ) = Iu^2

A
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25 May 2017, 08:40
vinraj wrote:
Hi, Bunuel,
The official solution seems to be lengthy

Here is an Easy Way to solve the Problem......

Consider M as the price in 2001.

Then, M(1+x/100) will be the price in 2002 ----> We are asked to find this value

And, M(1+x/100)^2 will be the price in 2003, and this price is given as N

So,
M(1+x/100)^2 = N
(1+x/100) ^2 = N/M
(1+x/100) = sqrt. (N/M)
M(1+x/100) = sqrt. (N/M) x M
M(1+x/100) = sqrt. (NM) <------> This is the Answer... Thats it

PLease give kudos if you like the solution

I´m a bit lost with roots properties, how sqrt. (N/M) x M = sqrt. (NM) ??
sorry for the basic question...
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25 May 2017, 19:54
1
Hi,
Because M = sqrt M * sqrt M
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04 Jun 2018, 01:31
Shouldn't we also consider 2000 and 2004 also? How do we know from the question that we don't have to consider it?
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04 Jun 2018, 06:32
Shouldn't we also consider 2000 and 2004 also? How do we know from the question that we don't have to consider it?

The question asks: 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? Why/how should we consider 2000 or 2004?
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07 Nov 2018, 09:47
M x X% = 2002 ---> 1
2002 x X% = N ---> 2

Now, we need 2002 in terms of X and M. So, eliminate the X%.

From equation 1,
X% = 2002/M

Substitute this into equation 2,
(2002^2)/M = N
2002^2= NM
2002 = sqrt(NM)

Posting my solution because for me I don't see how this question is difficult, or why Bunuel chose such a complicated method to solve. Cheers all.
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05 Apr 2019, 06:14
Bunuel wrote:
Official Solution:

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.

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.

Hi Bunuel

Can you please post the algebraic solution as well?

How should i translate the rate of increase as ? (1+x/100) or simply x%??

Thanks
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Joined: 02 Sep 2009
Posts: 59721

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05 Apr 2019, 06:20
JIAA wrote:
Bunuel wrote:
Official Solution:

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.

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.

Hi Bunuel

Can you please post the algebraic solution as well?

How should i translate the rate of increase as ? (1+x/100) or simply x%??

Thanks

You can find different solutions, including algebraic, here: https://gmatclub.com/forum/the-price-of ... 26464.html

Hope it helps.
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31 Oct 2019, 16:33
Somebody, please correct me if I'm wrong, but I believe there's an even quicker and dirtier way to solve this without doing any algebra. Every answer except A begins by multiplying N by some positive number. This will lead to a number larger than N. By simple logic and understanding the sequence, we can know that there is no way that the price in 2002 was MORE than the price in 2003, which all the answers have as N*pos numbers. The only answer that yields a chance of being more than M but less than N is A.
Re: M08-01   [#permalink] 31 Oct 2019, 16:33
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# M08-01

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