The mean of twenty-five consecutive positive integers. . .

Question

The mean of twenty-five consecutive positive integers numbers is what percent of the total?

(A) 4%

(B) 5%

(C) 20%

(D) 25%

(E) Cannot be determined by the information provided.

The OA is A.

I can't understand why the correct answer is A. It shouldn't be D? Experts, can you clarify this for me?

EconomistGMATTutor · Answer

Hello M7MBA.

The mean of 25 consecutive positive integers is: $$M=\frac{a_1+a_2+.\ .\ .\ +\ a_{25}}{25}.$$

The total is: $$T=a_1+a_2+.\ .\ .\ +\ a_{25}.$$

So, the mean can be write as follows:

$$M=\frac{1}{25}\cdot T=\frac{4}{100}\cdot T=4\%\cdot T.$$

So, the mean is the 4% of the total.

So, the correct answer is A.

I hope this can help you.

Feel free to ask me again if you have any doubt.

Regards.

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Brent@GMATPrepNow · Answer

M7MBA wrote:The mean of twenty-five consecutive positive integers numbers is what percent of the total?

(A) 4%

(B) 5%

(C) 20%

(D) 25%

(E) Cannot be determined by the information provided.

The question stem asks for the following value:
(mean)/(sum) * 100.

For any evenly spaced set:
Sum = (count)(mean).

Here, there are 25 consecutive integers, implying a count of 25.

Case 1: mean = 20
In this case:
sum = (count)(mean) = (25)(20) = 500.
(mean/sum) * 100 = (20/500) * 100 = 4%.

Case 2: mean = 40
In this case:
sum = (count)(mean) = (25)(40) = 1000.
(mean/sum) * 100 = (40/1000) * 100 = 4%.

In each case, the mean is 4% of the sum.

The correct answer is A.

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GMATGuruNY · Answer

M7MBA wrote:The mean of twenty-five consecutive positive integers numbers is what percent of the total?

(A) 4%

(B) 5%

(C) 20%

(D) 25%

(E) Cannot be determined by the information provided.

----------ASIDE----------------------
There's a nice rule that says, "In a set where the numbers are equally spaced, the mean will equal the median."
For example, in each of the following sets, the mean and median are equal:
{7, 9, 11, 13, 15}
{-1, 4, 9, 14}
{3, 4, 5, 6}
------------------
For this question, let x = first value
So, x+1 = second value
x+2 = third value
.
.
.
x+24 = last value

This means x+12 is the median AND the mode

Since x+12 = the mean of the 25 numbers, we can conclude that (25)(x+12 ) = the SUM of the 25 numbers

The mean of twenty-five consecutive positive integers numbers is what percent of the total?
We must convert (x+12)/(25)(x+12) to a PERCENT
(x+12)/(25)(x+12) = 1/25
= 4/100
= 4%
Answer: A

Brent Hanneson - Creator of GMATPrepNow.com

Scott@TargetTestPrep · Answer

M7MBA wrote:The mean of twenty-five consecutive positive integers numbers is what percent of the total?

(A) 4%

(B) 5%

(C) 20%

(D) 25%

(E) Cannot be determined by the information provided.

----------ASIDE----------------------
There's a nice rule that says, "In a set where the numbers are equally spaced, the mean will equal the median."
For example, in each of the following sets, the mean and median are equal:
{7, 9, 11, 13, 15}
{-1, 4, 9, 14}
{3, 4, 5, 6}
------------------
For this question, let x = first value
So, x+1 = second value
x+2 = third value
.
.
.
x+24 = last value

This means x+12 is the median AND the mode

Since x+12 = the mean of the 25 numbers, we can conclude that (25)(x+12 ) = the SUM of the 25 numbers

The mean of twenty-five consecutive positive integers numbers is what percent of the total?
We must convert (x+12)/(25)(x+12) to a PERCENT
(x+12)/(25)(x+12) = 1/25
= 4/100
= 4%
Answer: A

Brent Hanneson - Creator of GMATPrepNow.com

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