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M08-01

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M08-01  [#permalink]

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New post 16 Sep 2014, 00:36
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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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Re M08-01  [#permalink]

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New post 16 Sep 2014, 00:36
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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.


Answer: A
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M08-01  [#permalink]

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New post 04 Nov 2014, 13:27
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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


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Re: M08-01  [#permalink]

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New post 08 Nov 2015, 21:37
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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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Re: M08-01  [#permalink]

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New post 01 Dec 2015, 06:29
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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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M08-01  [#permalink]

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New post 03 Dec 2015, 18:06
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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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Re: M08-01  [#permalink]

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New post 05 Jan 2016, 09:16
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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

correct answer A
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M08-01  [#permalink]

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New post 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

Answer A
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Re: M08-01  [#permalink]

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New post 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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Re: M08-01  [#permalink]

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New post 05 Dec 2016, 11:06
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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Re: M08-01  [#permalink]

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New post 18 Dec 2016, 18:34
VinRag

Your method of solution has already been demonstrated by Bunuel on the similar question.
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Re: M08-01  [#permalink]

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New post 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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Re: M08-01  [#permalink]

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New post 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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Re: M08-01  [#permalink]

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New post 25 May 2017, 19:54
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Hi,
Because M = sqrt M * sqrt M
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Re: M08-01  [#permalink]

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New post 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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Re: M08-01  [#permalink]

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New post 04 Jun 2018, 06:32
dishadesai wrote:
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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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: M08-01 &nbs [#permalink] 04 Jun 2018, 06:32
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