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## Comment on

Mean, Median and Mode## I'm at cloud nine now! I'm a

## Thanks! Glad you like the

Thanks! Glad you like the videos,

## Hi, I just want to make sure

## In the case that all of the

In the case that all of the values are different, then we can say that there is a multi-way tie for the most common value.

So, for example, the set {2, 3, 6, 9} has four modes: 2, 3, 6 and 9

## http://greprepclub.com/forum

plz explain this sum

## You bet.

You bet.

I've posted my solution here: http://greprepclub.com/forum/course-of-an-experiment-95-measurements-wer...

Cheers,

Brent

## Thankyou.

## http://greprepclub.com/forum

Sometimes i get confused to put answers in 0.1, 0.01 similar like accuracies. Can you please help me in understanding those concepts well.

## We cover rounding here: https

We cover rounding here: https://www.greenlighttestprep.com/module/gre-arithmetic/video/1062

## Hi

Can you pl provide the soln to this - A set of 7 integers has a range of 2, an average of 3, and a mode of 3.

## Set A can be EITHER {2, 2, 3,

Set A can be EITHER {2, 2, 3, 3, 3, 4, 4} OR {2, 3, 3, 3, 3, 3, 4}

Both of these sets meet the given conditions.

Can you tell me how the question is worded?

## This is the link to the exact

I was able to come to the answer. But I wasn't sure till the end. Would be nice if you could help me think through.

## You bet.

You bet.

Here's my step-by-step solution: http://greprepclub.com/forum/a-set-of-7-integers-has-a-range-of-2-an-ave...

Cheers,

Brent

## 2 3 3 3 3 3 4 is one possible

Since they told mode 3 expect 1st and 7th, all member can be 3, we met mode restriction. Range has to be 2 so let 1st member be 2 and certainly 7th member will be 4.

## Hi Brent,

For the following question, can we say using the logic of weighted average that the qty B is greater ? Qty A has equal weight for x and y. So average has to be in the middle of x and y. However, Qty B has 2 weights of y and 1 weight of x. So average has to be closer to y than x. Since y > x, the average from qty B has to be greater.

http://greprepclub.com/forum/x-y-3608.html

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

Also for this question, http://greprepclub.com/forum/graph-above-shows-the-frequency-distribution-of-50-integer-1805.html, 23 values <5 (median), 16 values =5, 11 values >5. More weightage in < median. So average has to be < 5.

Thanks :)

## Hi Vinodhini,

Hi Vinodhini,

Question link: http://greprepclub.com/forum/x-y-3608.html

Yes, the weighted approach you describe is perfect. GREAT WORK!!

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

Question link: http://greprepclub.com/forum/graph-above-shows-the-frequency-distributio...

Your approach says that, since there are more values BELOW the median than there are values ABOVE the median, the average must be LESS THAN the median.

This approach can get you in trouble, because the mean (average) depends on the MAGNITUDE of the values above and below the median.

For example, if we replace one of the eleven 6's with 1,000,000 then the average of the 50 values will be greater than the median.

Cheers,

Brent

## Hi Brent,

Thank you. Got my mistake in 2nd approach..

## Hi Brent ! I wanted to ask if

## Sorry, but I don't have a

Sorry, but I don't have a list of all reinforcement problems in the course. Perhaps this is something I will add in the future.

## Ok np sir! Thanks for all the

## Thank you very much!

## https://greprepclub.com/forum

If x=0, and y= -2, then the A seems to be greater than the answer would be D, wouldn't it?

## Question link: https:/

Question link: https://greprepclub.com/forum/if-x-y-3281.html

If x = 0, and y = -2, then we get:

QUANTITY A: Average = [0 + 0 + 0 + (-2) + (-2)]/5 = -0.8

QUANTITY B: Average = [0 + 0 + (-2)]/3 = -0.66666...

So, -0.8 < -0.66666..., so Quantity B is greater.

## https://greprepclub.com/forum

I have a question here with the calculation of mean and determination of median. I calculated the mean by taking the mid-point of the bar chart i.e.: 15(3) + 35(8) + 15(13) + 13(18) + 10(23) + 5(28) + 3(33) = 1223 . To get mean 1223/ 95 = 12.87.

To calculate the median of 95. I divided it by 2 and got 47.5 and then went for cumulative frequencies to determine the range 6-10. As it was lower than the mean I went for A.

Is my calculation correct or I got lucky?

## Question link: https:/

Question link: https://greprepclub.com/forum/course-of-an-experiment-95-measurements-we...

Unfortunately, that strategy won't always work, since we need to consider possible ranges of each value (median and mean)

Cheers,

Brent

## https://www.youtube.com/watch

When calculating the answer I got c+d = 100. Choice A states the greatest possible NUMBER for d. I did not mention the word integer. On the gre when the question mention number should I not think of it as integer?

I assumed the value for c = 1 when got me 99 for d, therefore C answer which is not correct.

## Question link: https://www

Question link: https://www.youtube.com/watch?v=AgL5IjNNrzY

It's a common mistake to assume that numerical values are integers, but we should never make this assumption unless the question specifically states that a number is an INTEGER.

Cheers,

Brent

## https://m.youtube.com/watch?v

I don't understand this

## Can you give me an idea of

Can you give me an idea of which part you don't understand?

## This question is not from the

"Of 30 theater tickets sold, 20 tickets were sold at prices between $10 and $30 each and 10 tickets were sold at prices between $40 and $60 each."

Quantity A: the average (arithmetic mean) of the prices of the 30 tickets

Quantity B: $50

The explanation says the answer is B, however wouldn't the answer be D due to the varying price ranges of the 30 tickets?

## Yes, the average price does

Yes, the average price does vary.

In fact, it varies from $20 to $40

However, since Quantity B = $50, we can see that Quantity B is greater in all possible cases.

Here's my full solution: https://greprepclub.com/forum/of-30-theater-tickets-sold-20-tickets-were...

Cheers,

Brent

## https://greprepclub.com/forum

I was the reading the explanation of this question answer and it stated that "when the 21 salaries are listed in increasing order, the first 12 salaries in the list are less than $37,000"

How can we assume that the salaries as listed in increasing order?

## Question link: https:/

Question link: https://greprepclub.com/forum/qotd-5-the-sum-of-the-annual-salaries-of-t...

In order to find the median salary (Quantity B), we need to arrange the 21 salaries in ASCENDING order.

Let's say the salaries, when listed in ASCENDING order, are:

a, b, c, d, e, f, g, h, i, j, K, l, m, n, o, p, q, r, s, t, u

We can see that the median salary = K

Since we're told that "Twelve of the 21 teachers have an annual salary that is less than $37,000.", we know that salaries a to l are each less than $37,000

This tells us that K is less than $37,000

So, Quantity B is less than $37,000

Meanwhile, Quantity A = $781,200/21 = $37,200

So, we have:

QUANTITY A: Some number less than $37,000

QUANTITY B: $37,200

Answer: B

Does that help?

Cheers,

Brent

## https://greprepclub.com/forum

I am unsure about how we should deal with the defective part? Could you please help me?

Thanks!

## Here's my full solution:

Here's my full solution: https://greprepclub.com/forum/50-boxes-each-containing-30-machine-parts-...

Cheers,

Brent

## https://www.youtube.com/watch

Hi Brent! In the above question, can't the value of C be 0? That means 'd' could be 100.

Could be please clarify this point.

Thanks for the help!

## Question link: https://www

Question link: https://www.youtube.com/watch?v=AgL5IjNNrzY

The question tells us that a, b, c, and d are positive, and c is not positive.

Cheers,

Brent

## https://greprepclub.com/forum

please solve it brent

## Here's my solution: https:/

Here's my solution: https://greprepclub.com/forum/graph-above-shows-the-frequency-distributi...

Cheers,

Brent

## Hello sir, in this question

https://greprepclub.com/forum/the-average-arithmetic-mean-of-a-b-c-and-d-is-12-the-a-19323.html

## Good catch. Thanks!

Good catch. Thanks!

I have edited my response accordingly.

Cheers, Brent

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