If y is the average (arithmetic mean) of 15 consecutive positive

Question

If y is the average (arithmetic mean) of 15 consecutive positive integers, which of the following must be true?

I. y is an integer.
II. y > 7
III. y < 100

A. I only
B. II only
C. III only
D. I and II
E. II and III

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Bunuel · Accepted Answer

Firstly, the average of any evenly spaced set is equal to its median, and the median of an odd number of integers is the middle number when arranged in order.

Therefore, the average/median will always be an integer, making I always true.

The lowest set of 15 consecutive positive integers is {1, 2, 3, 4, 5, 6, 7,  8 , 9, 10, 11, 12, 13, 14, 15}. The average/median in this case is 8, so it is greater than 7. Hence, all other sets will also have averages greater than 7, making II also true.

As for III, it's not always true. We can make the set as large as we want, so the upper limit of the average is not restricted at all.

Answer: D.

Kinshook · Answer

guddo

Asked: If y is the average (arithmetic mean) of 15 consecutive positive integers, which of the following must be true?

I. y is an integer. For odd number of integers (15), the average is middle value (8th integer) MUST BE TRUE
II. y > 7: If starting integer = 1(lowest); y=8>7. For all other starting integers > 1; y > 7; MUST BE TRUE
III. y < 100; If starting integer =1 ; No; If starting integer = 100; Yes. COULD BE TRUE

A. I only
B. II only
C. III only
D. I and II
E. II and III

IMO D

DanTheGMATMan · Answer

The median equals the mean in an evenly spaced set, so the mean will just be the middle number of the 15 integers (importantly there are an ODD number of integers in the set, so the median is just the middle one)

At the very least, the median is 8

No reason it needs to be less than 100

At the very least, the median is 8

No reason it needs to be less than 100

https://www.youtube.com/watch?v=3bZVUHyz7iM

aditipr · Answer

hey isn't 0 an integer? the lowest list could also start from 0 right? then x, being the first term, could also be 0. and x+7=y so y could be 7

Bunuel · Answer

﻿If y is the average (arithmetic mean) of 15 consecutive  positive  integers, which of the following must be true?

0 IS an integer but it's not positive.

aditipr · Answer

oh my bad. glad to have corrected this now. thank you!

lucky21 · Answer

Since the integers must be positive, the smallest the first term can be is 1.

Let the smallest set: {1,2,3,4,...,15)

Method1:
1. Average of Evenly Spaced Sets formula: (First Term + Last Term)/2= (1+15)/2=8
2. Y=8

Method 2:
1. The sum of 15 consecutive positive integer = Y*15
2. The sum of First n consecutive positive integer = n(n+1)/2
3. 15*Y= (15*16)/2 (n=15 as 15 terms)
4. 15*Y = 15*8
5. Y=8

I. Y is an integer = true
II. Y>7 = true
III. Y<100 = true/false (as start and end range of set is no defined)

MartyMurray · Answer

﻿If y is the average (arithmetic mean) of 15 consecutive positive integers, which of the following must be true?

I. y is an integer.

Consecutive integers are evenly spaced.

The average of the integers in a set of evenly spaced integers is the average of the lowest and highest integers in the set.

In this case, we can call the lowest integer  x  and the highest  x + 14 .

So,  y = \frac{x + x + 14}{2} .

We can see that, since x is an integer,  \frac{2x + 14}{2}  must be an integer.

Must be true.

II. y > 7

Using  y = \frac{2x + 14}{2}  again, we can find the lowest possible value of  y  by using the lowest possible  x , which is  1  since the integers in the set are positive.

So, the lowest possible  y = \frac{(2 × 1) + 14}{2} = 8  .

Must be true.

III. y < 100

Since  x  can be any positive integer, there is no upper limit to the value of  y = \frac{2x + 14}{2} .

Thus,  y  could be greater than  100 .

Does not have to be true.

A. I only
B. II only
C. III only
D. I and II
E. II and III

Correct answer:   D

If y is the average (arithmetic mean) of 15 consecutive positive

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If y is the average (arithmetic mean) of 15 consecutive positive

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

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