Math Is Fun Forum

tonyz1949

Member

Registered: 2007-08-04

Posts: 20

Averages

The average (arithmetic mean ) of four different positive integers is 9. If the first of these integers is 3 times the second integer and the second integer is 2 less than the third integer, what is the least possible value of the fourth integer?

        (A)  5
        (B)  4
        (C)  3
        (D)  2
        (E)  1

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: Averages

If the arithmetic mean of four numbers is 9, then their sum is 4*9 = 36.

The first three numbers are all related to each other, and so can all be defined by one variable.
That is, if the 2nd is called x, then the first is 3x and the third is x+2.

The fourth number is independent from the other three, so it gets a letter to itself, y.

They all add to give 36, so 3x+x+x+2+y = 5x+2+y = 36.

5x+y = 34.

x has to be an integer, so the highest it can go is 6. That means the lowest possible y is 4.

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Why did the vector cross the road?
It wanted to be normal.

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Averages

Suppose the second and fourth integers are x and y respectively. Then

Hence y ≥ 4.

EDIT: Grrr, beaten to it by Mathsy.

Last edited by JaneFairfax (2007-08-14 06:11:46)

tonyz

Guest

Re: Averages

Thanks mathsyperson and Jane Fairfax.