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Question

stuck on these problems
averages (arithmetic mean)

What is the average of 9 numbers if the sum of any 4 of them is 20?
a) 20/ 9 b) 4 c) 5 d) 6 e) 36

If a = 2b and b = 3c and the average of a, b, and c is 40, what is the value of a?
a) 6 b) 12 c) 24 d) 36 e) 72

plz show work or explain

Answer 1

#1
Let's say that the numbers are a,b,c,d,e,f,g,h,and i.

We have a+b+c+d=a+b+c+i=20

So we can get d=i. The same logic can show that all the numbers are equal, so each is 5.

The average is (20+20+5)/9=5


#2

a=2b, b=3c
(a+b+c)/3=40

a+b+c=120

b=a/2, c=b/3=a/6

a+a/2+a/6=120

6a+3a+a=720

10a=720

a=72

Answer 2

1. Since any 4 have a sum of 20, the average of these numbers is 5. However, all such sums have the same avereage. So the average is 5.

2. a=2b b= 3c and a+b+c= 120
Then, eliminating a and c, 2b+b+b/3=120
10/3 *b = 120 and b= 36
From the other stipulations, a=72 and c=12

Answer 3

a) If the sum of ANY 4 of them equals 20, then ALL of the numbers must equal 5, therefore the average is 5.


b)
a+b+c= 3(40)= 120; get rid of a by substituting 2b for it-

2b + b + c = 120 ==

3b + c = 120; now get rid of b by substituting 3c for it-

3(3c) + c = 120 ==

10c = 120

c = 12, so then

b = 3c = 36, and

a = 2b = 72

Check for average of 3 -

a+b+c = 12 + 36 + 72 = 120, divided by 3 = 40

Izkewl?

Answer 4

Sum of any 4 of them is 20, means all are equal & each is 20/4=5.
Sum of 9 numbers = 5*9=45
Average of 9 numbers =45/9=5
Ans. (c)

a=2b, b=3c
b=a/2
c=b/3=a/6
Average =(a+b+c)/3
=(a+a/2+a/6)/3
=(6a+3a+a)/18
=10a/18=5a/9=40 (given)
5a=40*9
a=40*9/5
a=72
Ans.(e)