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## Comment on

Mean, Median & Mode## Hi Brent,

One of my friends had his GMAT today and we were discussing some of the questions he remembered and Interestingly, I found one of the PS Question he told me about in GMATCLUB forum.

Is there a limit to the number of questions? or Is there any source we can refer to?

## That would be a problem if

That would be a problem if there were live GMAT questions floating around the GMAT Club forum. Several years ago, a test prep company called ScoreTop was selling live GMAT questions. They were promptly shut down AND their students had their test scores revoked.

More here: http://www.scoretop.com/

## omg!

That is so unfair then. The person who posted that question would have unknown it to be a GMAT question.I just checked, the posted date is of 2011.

## can you please explain the

What is the average of x, y, and z?

(1) 2x + y + 4z = 23

(2) 3x + 4y + z = 22

## Done! :-)

Done! :-)

see: https://gmatclub.com/forum/what-is-the-average-of-x-y-and-z-218923.html#...

## Sir, is there any easy way to

For the first 3 months of last year, the average daily rainfall in Lancaster was 6cm. For last year as a whole, the average daily rainfall was 12cm. What was the average daily rainfall for the last 9 months of last year?

## Here's my step-by-step

Here's my step-by-step solution: https://gmatclub.com/forum/for-the-first-3-months-of-last-year-the-avera...

Cheers,

Brent

## So many practice questions on

## Statistics questions

Statistics questions (especially questions about mean and median) are VERY COMMON on the GMAT.

Of course, you need not answer every question above, but be sure to understand the various types of questions the test-maker can create.

## I think this question has 2

If x is a positive, single-digit integer such that 4/3*x, 2x, x, and x + 2, and 3x – 2 form a non-ordered list of consecutive integers, which of the following could be the median of that list?

A. 3

B. 4

C. 5

D. 6

E. 8

## That's great that you

That's great that you recognized that x must be a multiple of 3. However, x = 6 does not yield five CONSECUTIVE integers.

If x = 6, then 4x/3 = 8, 2x = 12, x = 6, x + 2 = 8 and 3x - 2 = 14. So, the five numbers (in ascending order) are: 6, 8, 8, 12, 14 (since these aren't consecutive integers, x cannot equal 6.

Cheers,

Brent

## Guess I missed that small

## It happens to everyone :-)

It happens to everyone :-)

## Hi Brent,

The final answer of "A" in the answer box (to the problem below) does not match your simplified solution at the end of your problem solving.

The average (arithmetic mean) of y numbers is x. If z is added to the numbers, the new average (arithmetic mean) will be z-5. What is the value of z in terms of x and y?

## Question link: https:/

Question link: https://gmatclub.com/forum/the-average-arithmetic-mean-of-y-numbers-is-x...

Good catch - thanks!

I've edited my response accordingly.

Cheers,

Brent

## Hi Brent,

The problem below gives me some difficulty when dealing with how to solve/understand for the Absolute Value portion. In particular, I don't quite understand how Bunuel is manipulating the absolute values or what it is I'm not understanding.

Thanks.

A set S = {x, -8, -5, -4, 4, 6, 9, y} with elements arranged in increasing order. If the median and the mean of the set are the same, what is the value of |x|-|y|?

(A) -1

(B) 0

(C) 1

(D) 2

(E) Cannot be determined

## Question link: https:/

Question link: https://gmatclub.com/forum/a-set-s-x-8-5-4-4-6-9-y-with-elements-arrang-...

Nice question!

First off, the question tells us that the numbers are arranged in ascending order.

So, we know that x ≤ -8, and y ≥ 9

There are 8 elements in the set. So, the median = the average of the two middlemost values.

Here, the two middlemost values are -4 and 4

So, the median = (-4 + 4)/2 = 0/2 = 0

Since the median and the mean of the set are EQUAL, we know that the mean is also 0

That is, [x + (-8) + (-5) + (-4) + 4 + 6 + 9 + y]/8 = 0

Multiply both sides by 8 to get: x + (-8) + (-5) + (-4) + 4 + 6 + 9 + y = 0

Simplify: x + y + 2 = 0

This means x + y = -2

So, here's what we know:

x + y = -2

x ≤ -8

y ≥ 9

Let's find some values of x and y and see where this leads us....

x = -12 and y = 10

In this case, |x|-|y|= |-12|-|10| = 12 - 10 = 2

x = -13 and y = 11

In this case, |x|-|y|= |-13|-|11| = 13 - 11 = 2

x = -12.5 and y = 10.5

In this case, |x|-|y|= |-12.5|-|10.5| = 12.5 - 10.5 = 2

x = -100 and y = 98

In this case, |x|-|y|= |-100|-|98| = 100 - 98 = 2

As we can see, the answer will always be 2

Answer: D

Cheers,

Brent

## Thank you Brent! Great way to

Gmatify's description and method of dealing with Absolute value seems off to me, especially the bit about, "the modulus of a negative number opens with a negative sign." and the rest of the way he solves the equation seems "off!" What am I not seeing here??

Opening the modulus with appropriate signs:

Modulus of any number is the absolute value of the number, or simply the positive value

Always remember that the modulus of a negative number opens with a negative sign and of a positive number opens with a positive sign

We have ,

|x| - |y| = -x - y = -(x + y) = -(-2) = 2

Hence Option D

## Gmatify is trying to explain

Gmatify is trying to explain how/why he/she wrote: |x| - |y| = -x - y

Before stating a rule about absolute values, let's look at some examples of the absolute value of a NEGATIVE number.

|-3| = 3

|-8| = 8

|-11| = 11

|-1.2| = 1.2

In general, if x is NEGATIVE, then |x| = -x

For example, if x = -6, then |x| = |-6| = 6 = -(-6) = -x

-------------------------------------

Now, let's look at some examples of the absolute value of a POSITIVE number.

|3| = 3

|9| = 9

|5.2| = 5.2

|19| = 19

In general, if y is POSITIVE, then |y| = y

For example, if y = 7, then |y| = |7| = 7 = y

-------------------------------------

In the original question, we're told (indirectly) that x is NEGATIVE and y is POSITIVE

So, it must be the case that |x| = -x and |y| = y

So, |x| - |y| = -x - y = -(x + y)

Does that help?

Cheers,

Brent

## Hi Brent,

I need a little clarification on the challenge problem posted below.

Here is my question:

How do you figure that “3” is the median of set “A” when “J” is unknown? It just assumes J will somehow be negative, but what if it is positive?

Set A consists of integers -9, 8, 3, 10, and J; Set B consists of integers -2, 5, 0, 7, -6, and T. If R is the median of Set A and W is the mode of set B, and R^W is a factor of 34, what is the value of T if J is negative?

(A)-2

(B)0

(C)1

(D)2

(E) 5

## Question link: http://www

Question link: http://www.veritasprep.com/blog/2010/10/gmat-challenge-question-solve-this/

The answer to your question ("How do you figure that 3 is the median of set A when J is unknown?) lies at the very end of the question stem, where it says "......what is the value of T if J is NEGATIVE?"

So, even though we don't know the actual value of J, we do know that J is some NEGATIVE number.

To determine the median of set A, we must arrange the values in ASCENDING order.

We know that J is NEGATIVE, and we also have another negative value (-9) in set A.

This gives us two possible cases: J < -9 OR -9 < J

It turns out that both cases yield the SAME MEDIAN for set A.

Case i: J < -9

When we arrange set A in ASCENDING order, we get: {J, -9, 3, 8, 10}. The MEDIAN = 3

Case ii: -9 < J

When we arrange set A in ASCENDING order, we get: {-9, J, 3, 8, 10}. The MEDIAN = 3

In both cases, the median is 3.

In other words, R = 3

Does that help?

Cheers,

Brent

## Well, I guess that was right

" now?

'

## Set B:{-2, 5, 0, 7, -6, T}

Set B:{-2, 5, 0, 7, -6, T}

W is the mode of set B

The key word here is "THE."

This means there is ONLY ONE mode.

So, for example, if T = 7, then set B is {-2, 5, 0, 7, -6, 7}, in which case the mode is 7

Conversely, if T = 11, then set B is {-2, 5, 0, 7, -6, 11}, in which case the modes are -2, 5, 0, 7, -6 and 11

Since the key word "THE" tells us that there is ONLY ONE mode, we know that T is one of the 5 values: -2, 5, 0, 7 or -6

We're also told that R^W is a factor of 34

We know (from my post above) that R = 3

So how can 3^W be a factor of 34, when the factors of 34 are: 1, 2, 17 and 34?

We know that, if W is an integer, 3^W CANNOT equal 2, 17 or 34

HOWEVER, 3^W CAN equal 1, when W = 0

So, it MUST be the case that W = 0

If W = 0, then 0 is the mode of the set {-2, 5, 0, 7, -6, T}

If 0 is THE mode of the set {-2, 5, 0, 7, -6, T}, then it must be the case that T = 0

Answer: B

Cheers,

Brent

## CLICKED! Thank you so much

## Yes, it's definitely a tough

Yes, it's definitely a tough one!

## What if w=-2? 3^-2 is equal

(2*17)/(1/9)=34*9=306 which is an integer. So W could equal -2 also.

What is wrong with this?

## Hi Kaori,

Hi Kaori,

Which question are you referring to? Please provide a link.

Cheers,

Brent

## This the link. Question link:

I was talking about R^W

## Link: http://www.veritasprep

Link: http://www.veritasprep.com/blog/2010/10/gmat-challenge-question-solve-this/

In your suggested case (r = 3 and w = -2), r^w = 1/9

However, the question tells us that r^w is a factor of 34, and 1/9 is NOT a factor of 34.

Important: The factors (aka divisors) of a certain integer are always integers.

So, for example, the positive factors of 34 are: 1, 2, 17 and 34

Cheers,

Brent

## Great. I was confused as I

Factors can't be fractions and should be integers.

Things are clear now!

## Set Q consists of 6

A. 3

B. 2.5

C. 0.8

D. 0.3

E. 0.25

Hi Brent, I feel the question here is incomplete.

Set Q can have two possibilities: -4, -6, -8....... or -4, -2, 0......

Similarly Set P will have two possibilities: 1,3,5,7 or 1,-1,-3,-5

Am I understanding this correctly?

## Question link: https:/

Question link: https://gmatclub.com/forum/set-q-consists-of-6-consecutive-even-integers...

Perhaps the question could/should be worded differently to avoid ambiguity.

"beginning with" assumes we're listing the values in ascending order. So, we get: -4, -2, 0, 2, etc.

Perhaps the question should read as follows:

"Set Q consists of 6 consecutive even integers, WITH -4 AS THE SMALLEST (LEAST) INTEGER."

Cheers,

Brent

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