Determine the average of the first and last number in the row if...?

Question

Four numbers are displayed in a row. The average of the first two numbers is 8, the average of the middle two numbers is 11.8, and the average of the last two numbers is 7.7

Answer 1

I think there could be many answers for this.....here is one I have come up with....

Numbers-: A B C D

A = 4
B = 12
C = 11.6
D = 3.8

(A+B) / 2 = (12 + 4) / 2 = 8

(B + C) / 2 = (12 + 11.6) / 2 = 11.8

(C + D) / 2 = (11.6 + 3.8) / 2 = 7.7

So the average of (A+D) = (3.8 + 4) / 2 = 3.9

I'm pretty sure there are many more ways though... (although the average of the first and last will still be 3.9)

Answer 2

let w, x, y, z be the numbers in the row...

then,

(w + x) / 2 = 8 <<<<>>>> w + x = 16
(y + x) / 2 = 11.8 <<<<>>>> y + x = 23.6
(y + z) / 2 = 7.7 <<<<>>>> y + z = 15.4

so we have three equations to solve...

w + x = 16 (1)
x + y = 23.6 (2)
y + z = 15.4 (3)

subtract eq. 1 to eq. 2

w + x = 16
- ( x + y = 23.6)
-------------------------
w - y = -7.6 (4)

then add eq. 3

w - y = -7.6
+ y + z = 15.4
---------------------
w + z = 7.8

since your asked to find the average of the first and the last, just divide the whole thing by 2....

(w + z)/2 = 3.9

Answer 3

Call the four numbers w, x, y and z

(w+x)/2 = 8
(x+y)/2 = 11.8 -> x+y = 23.6
(y+z)/2 = 7.7 -> y+z = 15.4

The last two equations give x = z + 8.2

Subsititute into eq 1 to get

(w + z + 8.2)/2 = 8

(w+z)/2 = 8 - 8.2/2 = 3.9

Answer 4

Let the numbers be a,b,c,d respectively.
therefore,
(a+b)/2=8
(b+c)/2=11.8
(c+d)/2=7.7

Add the first and the third equation, and then subtract the second equation from it.

(a+b)/2+(c+d)/2-(b+c)/2=8+7.7-11.8
(a+b+c+d-b-c)/2=3.9
(a+d)/2=3.9

So the average of the first and the last numbers are 3.9

(a+d)/2=3.9

Answer 5

Let these numbers be, in order, a b c d. Then [a+b]/2 = 8, so a+b = 16. [b+c]/2 = 11.8, so b+c = 23.6. [c+d]/2 = 7.7, so c+d = 15.4 Now (a+b) - (b+c) + (c+d) = a+d = 7.8, so [a+d]/2 = 3.9.

Steve

Answer 6

abcd is the number

( a + b )/ 2 = 8
(b + c)/2 = 11.8
(c + d) / 2 = 7.7

find (a + d)/2

a+b = 16
b+c = 23.6
c+d = 15.4

a + (23.6 - c) = 16
a + 23.6 - (15.4 - d) = 16
a + 23.6 - 15.4 + d = 16

a + d = 7.8

(a+d)/2 = 3.9 <- the average of a and d

Answer 7

nubers are a,b,c and d
0.5a+0.5b=8
so a+b=16 ==>b=16-a
similarly
b+c=23.6 ==> 16-a+c=23.6 ==> c=a+7.6
c+d=15.4 ==> a+7.6+d=15.4 ==> a+d=7.8
average of the first and last number in the row =(a+d)/2=3.9

Answer 8

a,b,c,d
(a+b)/2 = 8 ------1
(b+c)/2 = 11.8 ----2
(c+d)/2 = 7.7-------3

(a-c)/2 = (1) - (2) = -3.8
(a+d)/2 = (a-c)/2 + (c+d)/2 = -3.8 + 7.7 = 3.9

Answer 9

1. a+b= 16
2. b+c= 23.6
3. c+d= 15.4

Subtract 1. from 2.
4. a - c = -7.6

Add 3. and 4.
5. a + d = 7.8

Average is 3.9

Answer 10

Let the no.s be a,b, c and d.

We have:-
a+b = 16 =>b = 16-a..(i)
b+c = 23.6 =>b = 23.6-c..(ii) =>c = 23.6-b..(iii)
c+d = 15.4 =>c= 15.4 -d..(iv)

From (i) and (ii),
a = c - 7.6

From (iii) and (iv)
d = b-8.2

:. a+d = b+c-15.8 = 23.6-15.8 = 7.8

Thus, the average is (a+d)/2 = 3.9