Question

The average of the first second and third of four numbers is 66 more than the average of the second third and fourth of these numbers. What is the fourth number if the average of first and fourth number is 129?​

Answer 1

Answer:

The fourth number is -20.

Step-by-step explanation:

Let the four numbers be a, b, c and d respectively.

The average of a, b and c is 66 more than average of b, c and d.

That is,

$\frac{a+b+c}{3}=\frac{b+c+d}{3} +66$

Taking LCM,

$\frac{a+b+c}{3} =\frac{198}{3}+ \frac{b+c+d}{3}$

$a+b+c = 198 + b+c+d$

$a=198+d$

$a-d =198$

Let the above equation be equation 1.

It is also given that the average of a and d is 129.

That is,

$\frac{a+d}{2} =129$

$a+d=158$

Now, let this equation be equation 2.

We have

$a+d=158$

$a-d =198$

Subtracting equation 1 from equation 2, we get:

$(a-a)+[d-(-d)]= -40$

$0+2d=-40$

$2d=-40\\d=\frac{-40}{2} \\d=-20$

Therefore the required number, that is, the fourth number is -20.