A student asked me to elaborate on this question, so here goes....

Since the numbers in the set are INTEGERS, and since the range = 2, we can conclude that some integers in the set will be repeated. I say this because we can't have a range of 2 with 7 DIFFERENT integers.

Also, the fact that the mode is 3 means that 3 is repeated the MOST TIMES.

Since the MEAN = 3, then the repeated 3's must appear in the middle of the set.
To show why this is, let's look at a situation in which the repeated 3's do NOT appear in the middle of the set.
For example, let's say that the 3's are the SMALLEST values in the set.
This is no good, since some of the other numbers would have to be greater than 3.
So, we'd have a set that looks something like {3, 3, 3, 3+, 3+, 3+, 3+}, where 3+ represents a number greater than 3.
This set will have a mean that's greater than 3. I know this because the set {3, 3, 3, 3, 3, 3, 3} has a mean of 3. So, if we replace one or more 3 with 3+, the mean must be greater than 3.
Given this, the 3's cannot be the SMALLEST values.
Using similar logic, we can conclude that the 3's cannot be the GREATEST values.

So, the 3's must be the MIDDLEMOST values.
This means 2 must be the smallest value and 4 must be the greatest value.
So, we have something like {2, some 3's, 4}

So, one option might be {2, 3, 3, 3, 3, 3, 4}
This set meet all of the given conditions.

Of course, we could also have {2, 2, 3, 3, 3, 4, 4}

Are there any other options? No.
Keep in mind that the sum of the 7 integers must be 21 (since the mean is 3, the sum must be such that we get 3 when we divide the sum by 7)
In the last set I listed, I took two 3's (with a sum of 6) from the first set and replaced them with a 2 and a 4 (which also has a sum of 6).
Doing this means the mean remains 3.

From here, we MIGHT consider taking {2, 2, 3, 3, 3, 4, 4} and replacing two 3's with a 2 and a 4 to get: {2, 2, 2, 3, 4, 4, 4}.
HOWEVER, the mode of this new set is no longer 3 (the modeS are 2 and 4).
So, this new set does not meet the given conditions.

So, there are only 2 possible sets that meet the given conditions:
Set 1: {2, 3, 3, 3, 3, 3, 4}
Set 2: {2, 2, 3, 3, 3, 4, 4}

Now onto the question...
Quantity A: The third number in the set when the numbers are arranged in ascending order
Quantity B: The fifth number in the set when the numbers are arranged in ascending order

Let's consider each possible set separately...

Set 1: {2, 3, 3, 3, 3, 3, 4}
We get:
Quantity A: 3
Quantity B: 3
The two quantities are EQUAL

Set 2: {2, 2, 3, 3, 3, 4, 4}
We get:
Quantity A: 3
Quantity B: 3
The two quantities are EQUAL

Answer:

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the average = 3
the mode = 3
range = 2
the sum = 7*3 = 21
we will find the maximum and minimum that makes the range = 2
and with a multiple of 3 that makes the sum = 21
we know the average is 3 so we will have these choices for the min value = {-2,-1,1,2}
1 will work because it makes the maximum = 3
so the range will be 3-1 = 2 .
and the set will be
{1 ,2,3,3,3,3,3,3}
mode 3
average : 3
sum =21
range : min-max = 3-1 = 2
the ans : c .

Hello
I got another set but I don't know what is my mistake:
3, 3, 3, 3, 4, 4, 5

GIVEN: A set of 7 integers has a range of 2, an average of 3, and a mode of 3.
Your seven numbers don't have an average of 3.
I have a feeling you and may have confused average with median.

You can't select this data as the sum of the number is 25, and the average not equals to 3.

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