Lesson: Mean, Median and Mode

Comment on Mean, Median and Mode

Hi, I just want to make sure about one information about the mode, I know that if all numbers in 1 set are different from each others there will be no mode, but you have said in the e.g at 2:20 mins all of the four numbesr we will consider them as mode so we have 4 modes.
greenlight-admin's picture

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


plz explain this sum
greenlight-admin's picture

You bet.
I've posted my solution here: http://gre.myprepclub.com/forum/course-of-an-experiment-95-measurements-...




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.

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.
greenlight-admin's picture

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 question: http://gre.myprepclub.com/forum/a-set-of-7-integers-has-a-range-of-2-an-average-of-3-and-a-mode-of-2675.html

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.
greenlight-admin's picture

You bet.
Here's my step-by-step solution: http://gre.myprepclub.com/forum/a-set-of-7-integers-has-a-range-of-2-an-...


2 3 3 3 3 3 4 is one possible set.

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.


Also for this question, http://gre.myprepclub.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 :)
greenlight-admin's picture

Hi Vinodhini,

Question link: http://gre.myprepclub.com/forum/x-y-3608.html
Yes, the weighted approach you describe is perfect. GREAT WORK!!

Question link: http://gre.myprepclub.com/forum/graph-above-shows-the-frequency-distribu...

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.


Hi Brent,
Thank you. Got my mistake in 2nd approach..

Hi Brent ! I wanted to ask if there is any way that i can find all the reinforcement problems in one place .Like there are very good problems that i have encountered but i tend to forget the tricks and complexity some problems involve as i have seen them only once .. but i wish to revise and go through all the problems once. Is there any such link available.
greenlight-admin's picture

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 help !

Thank you very much!


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

Question link: https://gre.myprepclub.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.


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?
greenlight-admin's picture

Question link: https://gre.myprepclub.com/forum/course-of-an-experiment-95-measurements...

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



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.
greenlight-admin's picture

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.


I don't understand this
greenlight-admin's picture

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

This question is not from the Greenlight bank of questions, it's one that I found online, but am still having difficulty answering it correctly. Can you help me?

"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?
greenlight-admin's picture

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://gre.myprepclub.com/forum/of-30-theater-tickets-sold-20-tickets-w...



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?
greenlight-admin's picture

Question link: https://gre.myprepclub.com/forum/qotd-5-the-sum-of-the-annual-salaries-o...

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?



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

greenlight-admin's picture


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!
greenlight-admin's picture

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.


please solve it brent
greenlight-admin's picture

Hello sir, in this question your answer confuses a and d at the very end.

greenlight-admin's picture

Good catch. Thanks!
I have edited my response accordingly.

Cheers, Brent


For this question why can't we say that a = a+11, since we want to minimize c wouldnt we want to maximize a, an according to the parameters a is less than or equal to b and b is less than or equal to c. so couldnt we say that a = a+11 to maximize the a value and then use the answer choices to solve for c?
greenlight-admin's picture

Question link: https://gmatclub.com/forum/three-positive-integers-a-b-and-c-are-such-th...

We can't say that a = a + 11, since such an equation is impossible.
It basically says that a number (a) is equal to that same number increased by 11.
Here's another analogy: Al's present age (a) is equal to his age 11 years from now (a + 11)

You can, however, say that b = a + 11, since it's clear that b = the median.


For this problem I found out the sum is 60 and set a +a+ 11 + c = 60 knowing that we want to minimize c we should maximize a and a+11. we can do this by setting a to a+11 given that, that's the median so it would become a+11 +a+11 +c = 60. Using the answer choices we can test numbers for c and see if it follows the rules that a<= b<=c and that they are all integers. I did it this way and landed on 24 as the answer choice. Unfortunately, that is wrong. Why?
greenlight-admin's picture

Question link: https://gmatclub.com/forum/three-positive-integers-a-b-and-c-are-such-th...

Your error occurs when you let a = a + 11
Since a ≤ b ≤ c, we know that b is the median.
Since a+11 is the median, we know that b = a + 11. In other words, b is 11 greater than a.
By letting a+11 be the a value, b is no longer 11 greater than a.


For this question I wanted to check if my approach is correct. I approached it by back solving and said the list in the following order x + x_2 + 55 + x_3 + 20+3x = 275. We then can find that the range is 20 +3x -x = 20 + 2x we want this to be the largest and so I set 20 + 2x = 78 solved for x and then I minimized x_2 and x_4 added the numbers and got the sum as 275. Would that be a good approach. What I cant wrap my head around in the algebraic way was how come we knew we wanted to minimize x?
greenlight-admin's picture

Question link: https://gmatclub.com/forum/set-r-contains-five-numbers-that-have-an-aver...
Your approach is great, Ravin.
This is a crazy hard question (harder than you will see on test day, in my opinion)

Ann $450,000
Bob $360,000
Cal $190,000
Dot $210,000
Ed $680,000

The table above shows the total sales recorded in July for the 5 salespeople at Acme Truck Sales. It was discovered that one of Cal's sales was incorrectly recorded as one of Ann's sales. After this error was corrected, Ann's total sales were still higher than Cal's total sales, and the median of the 5 sales totals was $330,000. What was the value of the incorrectly recorded sale?

Can you explain why Cal is 190 +x and Ann is 450 -x? From my understanding it comes from the sentence one of Cal's sales was incorrectly recorded as one of Ann's sales so that would mean that Ann's sales went down therefore 450-x and Cal got one of Ann's sales so he went up to 190 +x is that correct?
greenlight-admin's picture

Your analysis is correct. Consider the following example:

The ACTUAL sales were as follows:
Cal: 100K, 90K, 80K (total = 270K)
Ann: 370K (total = 370K)

However, the RECORDED sales were as follows:
Cal: 100K, 90K (total = 190K)
Ann: 370K, 80K (total = 450K)
Here, we can see that Cal's 80K sale was recorded as Ann's sale.

And this case, Cal's ACTUAL sales = 190K + 80K, and Ann's ACTUAL sales = 450K - 80K

Similarly, if we let x = the value of the Cal's sale that was incorrectly attributed to Ann, we get....
Cal's ACTUAL sales = 190K + x, and Ann's ACTUAL sales = 450K - x

Why aren't we considering negative integors? https://gre.myprepclub.com/forum/3-k-2-8-m-3-the-arithmetic-mean-of-the-list-of-numbers-19584.html
greenlight-admin's picture

Question link: https://gre.myprepclub.com/forum/3-k-2-8-m-3-the-arithmetic-mean-of-the-...

We definitely can consider negative integers. If we do, we'll still reach the same conclusion.
Since we already showed that k + m = 8, it could be the case that k = -4 and m = 12
In this case, the set becomes {3, -4, 2, 8, 12, 3}, which means the median is still 3.

In fact, if we take any distinct integer values of k and m such that k + m = 8, the median will always be 3.

Hey Brent can u please tell me differnce between these too in alegrbra and explain it a bit

The manufacturer sold twice as many units of Q as P.
There are twice as many employees in x as y
there are twice as many boys as girls

Like is there some hint when making alegbra eqautions....please and thnk u


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