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BTG Mean Problem

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BTG Mean Problem

by aleph777 » Tue Jan 25, 2011 12:22 pm
I started using the BTG practice program a couple weeks ago, and I'm getting quite a few supposedly low-level questions incorrect. In fact, I'm often not able to even figure out how to solve them. Has anyone else noticed they're generally more difficult than standard OG questions? Maybe it's the phrasing?

Anyhow, here's a problem I just saw.

If x is a prime number, and x-1 is the median of the set {x-1, 3x +3, 2x-4}, then what is the average (arithmetic mean) of the set?

1. 2
2. 5/3
3. 3
4. 10/3
5. 14/3

OA: D

Here's how I tried so solve, but it was wrong. Can someone explain why this approach is incorrect?

We know the set has 3 terms, and therefore the median is also the mean. So I set up mean equation: mean = sum of terms / number of terms:

x-1 = (x - 1 + 3x + 3 + 2x -4) / 3

3x-3 = 6x - 2
-1=3x
-1/3 = x

And therefore the mean, which is also the media x-1, must equal: -1/3 - 1 = -4/3

And that's not even an answer choice that you might typically see as a fake-out.

Why doesn't this approach work?
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by GMATGuruNY » Tue Jan 25, 2011 12:59 pm
aleph777 wrote:I started using the BTG practice program a couple weeks ago, and I'm getting quite a few supposedly low-level questions incorrect. In fact, I'm often not able to even figure out how to solve them. Has anyone else noticed they're generally more difficult than standard OG questions? Maybe it's the phrasing?

Anyhow, here's a problem I just saw.

If x is a prime number, and x-1 is the median of the set {x-1, 3x +3, 2x-4}, then what is the average (arithmetic mean) of the set?

1. 2
2. 5/3
3. 3
4. 10/3
5. 14/3

OA: D

Here's how I tried so solve, but it was wrong. Can someone explain why this approach is incorrect?

We know the set has 3 terms, and therefore the median is also the mean. So I set up mean equation: mean = sum of terms / number of terms:

x-1 = (x - 1 + 3x + 3 + 2x -4) / 3

3x-3 = 6x - 2
-1=3x
-1/3 = x

And therefore the mean, which is also the media x-1, must equal: -1/3 - 1 = -4/3

And that's not even an answer choice that you might typically see as a fake-out.

Why doesn't this approach work?
Mean = median only if the numbers are symmetrical about the median, as in the following sets:

{1,2,3}. Mean = (1+2+3)/3 = 2 = median.
{3,5,7}. Mean = (3+5+7)/3 = 5 = median.
{1,6,11}. Mean = (1+6+11)/3 = 6 = median.

For the problem above, I would plug in a value for x that satisfies the conditions in the problem.

Let x=2.
x-1 = 2-1 = 1.
3x+3 = 3*2 + 3 = 9.
2x-4 = 2*2 - 4 = 0.

This works because x-1 = 1 is the median of {0, 1, 9}.

Average = (1+9+0)/3 = 10/3.

The correct answer is D.
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by aleph777 » Tue Jan 25, 2011 1:11 pm
Oh, of course! Silly mistake. They weren't consecutive....
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by eyelikecheese » Tue Jan 25, 2011 3:34 pm
Hmm..How can the answer not be derived from the number 3

3-1=2

Therefore, 2,2,12. The average of this set is 16/3? or 5 1/3
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by GMATGuruNY » Tue Jan 25, 2011 3:59 pm
eyelikecheese wrote:Hmm..How can the answer not be derived from the number 3

3-1=2

Therefore, 2,2,12. The average of this set is 16/3? or 5 1/3
According to some, a set must contain distinct elements. The question above likely is relying on this definition; thus, {2,2,12} does not qualify as a set. Others disagree with this definition of a set. For this reason, if the question above were to appear on the GMAT, the wording would make clear that the numbers in the set were distinct.
Last edited by GMATGuruNY on Tue Jan 25, 2011 4:46 pm, edited 1 time in total.
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by eyelikecheese » Tue Jan 25, 2011 4:44 pm
What a horrible problem then
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by ontogenez » Mon Feb 14, 2011 5:33 am
there are two possible ordered sets

1. {3x+3 <= x-1 <= 2x-4} => x<=2 AND x>=3. No solution.
2. {2x-4 <= x-1 <= 3x+3} => x<=3 AND x>=-2. Only two primes - 2 and 3.

Average = (2x-4+x-1+3x+3)/3 = 2x - 2/3;

For x=2, average = 10/3
For x=3, average = 16/3

Choose D
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by thebigkats » Thu Feb 17, 2011 1:13 pm
HI:
We know that the set has {x - 1, 3x + 3, 2x - 4} and x is a prime no (means x = positive integer)

Mean of this set = (x-1+3x+3+2x-4)/3 = (6x-2)/3

M=(6x-2)/3 ==> x = (3M+2) / 6

Now we can easily plug in values to find which M results in x = integer value.

(A) 2
x = (3*2+2)/6 = 8/6 ------------- CAN'T BE

(B) 5/3
x = 7/6 --------------- CAN'T BE

(C) 3
x = 11/6 --------------- CAN"T BE


(D) 10/3
x = 2 -------------- POSSIBLE

(E) 14/3
x = 16/6 --------------- CAN"T BE

So we have D as answer.

Note
(A) If we had found more than one values satisfying the equation then we could have gone to the next step and see if x was a prime

(B) If more than one values were prime then we would have to check if {x-1, 3x +3, 2x-4} has median of (x-1)

Rgds,
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by sanju09 » Fri Feb 18, 2011 3:07 am
aleph777 wrote:I started using the BTG practice program a couple weeks ago, and I'm getting quite a few supposedly low-level questions incorrect. In fact, I'm often not able to even figure out how to solve them. Has anyone else noticed they're generally more difficult than standard OG questions? Maybe it's the phrasing?

Anyhow, here's a problem I just saw.

If x is a prime number, and x-1 is the median of the set {x-1, 3x +3, 2x-4}, then what is the average (arithmetic mean) of the set?

1. 2
2. 5/3
3. 3
4. 10/3
5. 14/3

OA: D

Here's how I tried so solve, but it was wrong. Can someone explain why this approach is incorrect?

We know the set has 3 terms, and therefore the median is also the mean. So I set up mean equation: mean = sum of terms / number of terms:

x-1 = (x - 1 + 3x + 3 + 2x -4) / 3

3x-3 = 6x - 2
-1=3x
-1/3 = x

And therefore the mean, which is also the media x-1, must equal: -1/3 - 1 = -4/3

And that's not even an answer choice that you might typically see as a fake-out.

Why doesn't this approach work?
Choices are not in right order, not an ideal look to be a GMAT question, but it's still a good question



Two methods...

Method 1

Here we don't need the info that x - 1 is the median of the set {x - 1, 3 x + 3, 2 x - 4}, because average of the set {x - 1, 3 x + 3, 2 x - 4} is (6 x - 2)/3 or 3 times an answer choice is 6 x - 2, where x is a prime.

1. 6 x - 2 = 6 doesn't prove x a prime.
2. 6 x - 2 = 5 doesn't prove x a prime.
3. 6 x - 2 = 9 doesn't prove x a prime.
4. 6 x - 2 = 10 [spoiler]proves x = 2, a prime. SUCCESS
5. Needless, 6 x - 2 = 14 doesn't prove x a prime.

Choice 4[/spoiler]

Method 2

If x is prime then x - 1 is positive and 3 x + 3 > 2 x - 4, and if x - 1 is the median of the set {x - 1, 3 x + 3, 2 x - 4}, then 2 x - 4 < x - 1 < 3 x + 3, with this 2 x - 4 < x - 1 gives x < 3 and x - 1 < 3 x + 3 gives x > -2, overall x is a prime such that -2 < x < 3. Nothing EXCEPT x = 2 fits here. Once x = 2, the set {x - 1, 3 x + 3, 2 x - 4} becomes {1, 9, 0} whose average is otherwise [spoiler]10/3

Choice 4[/spoiler]
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by thebigkats » Fri Feb 18, 2011 9:26 am
Hi Aleph777:

you mentioned: "We know the set has 3 terms, and therefore the median is also the mean. So I set up mean equation: mean = sum of terms / number of terms". This is NOT a true assumption. Consider {2, 3, 5} and {2, 3, 11}. Both these sets have 3 elements, median =3 and yet different mean values.

Probably this was the reason for you to get to wrong conclusion.

rgds
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by Night reader » Fri Feb 18, 2011 3:13 pm
what a quest :) x-1+3x+3+2x-4=3*answer --> x=(3*answer+2)/6. Instantly note, answer>3 for x=2 (the least prime number). Eliminate A and B. Try x=2, --> answer=10/3 and choice 4(D) is correct.

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aleph777 wrote:I started using the BTG practice program a couple weeks ago, and I'm getting quite a few supposedly low-level questions incorrect. In fact, I'm often not able to even figure out how to solve them. Has anyone else noticed they're generally more difficult than standard OG questions? Maybe it's the phrasing?

Anyhow, here's a problem I just saw.

If x is a prime number, and x-1 is the median of the set {x-1, 3x +3, 2x-4}, then what is the average (arithmetic mean) of the set?

1. 2
2. 5/3
3. 3
4. 10/3
5. 14/3

OA: D

Here's how I tried so solve, but it was wrong. Can someone explain why this approach is incorrect?

We know the set has 3 terms, and therefore the median is also the mean. So I set up mean equation: mean = sum of terms / number of terms:

x-1 = (x - 1 + 3x + 3 + 2x -4) / 3

3x-3 = 6x - 2
-1=3x
-1/3 = x

And therefore the mean, which is also the media x-1, must equal: -1/3 - 1 = -4/3

And that's not even an answer choice that you might typically see as a fake-out.

Why doesn't this approach work?
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by LalaB » Mon Oct 24, 2011 10:28 pm
well, I seem to use the stupidest and easiest method hehe

I just dully applied the fist prime number, e.g. 2 and got the answer 10/3
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by rohit_gmat » Tue Oct 25, 2011 1:07 am
jus plug in...

i tried with 7 first and saw that x-1 doesnt become the median, so i went smaller (5 also doesnt)
and got 3.. but the avg wasn't in the list (also the Q doesnt mention "distinct") so i went lower to
2 and that worked (10/3)
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by ronnie1985 » Sat Apr 07, 2012 8:38 am
This is really a stupid question and the test taker is left with no choice except for trial and error method.
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by keepintouch_7 » Sat Jun 30, 2012 10:24 pm
Mean = Median only when the elements in a set are evenly spaced i.e the difference between two consecutive elements is constant.
Here is how I did the problem.
Since X is a prime number so x is positive i.e x>= 2 ( 2 is smallest prime)
Arrange the elements in increasing order i.e 2x-4 ,x-1, 3x+3
now 2x-4<x-1=>x<3 ; prime number less than 3 is 2 so x=2
now we can find the mean as 2x-4+x-1+3x+3/3 = 6x-2/3 = 10/3 (putting x=2)
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