# Question

The answer given was such that the avg for X, Y and Z is 43. But reading the qns , X , Y and Z seems to be avg 1. Someone please enlighten me.

eg. 1

a=1

b=85

a+b+x+y+z=43 x 5 = 215

a+b = 86 = 43 x 2 = 1+85

x+y+z=43 x 3 = 129

a+b+x+y+z = 129+86 = 215

eg. 2

a=1, b=2, c=126

a+b+c+x+y+z=43 x 6 = 258

a+b+c =43 x 3 = 129 = 1+2+126

a+b+c+x+y+z = 129+129 = 258

eg. 3 ……………………………………………………………………

0 Replies 0 Likes ✔Accepted Answer

x  + y + z = 43 x 3 =129

x  + y  = 74

z = 129 – 74 = 55

I have the answer, but I am seeking to understand why the group avg and x+y+z avg is the same?

The problem states, “The average of the remaining numbers remains as 43”. This condition is possible only if the average of XYZ, is same as the whole average.

0 Replies 0 Likes

Total value of X and Y = 37 x 2

= 74

Let number of numbers in group = n

Let value of Z = z

Value of numbers in group = 43n

43n ÷ n = 43

(43n – z – 74) ÷ (n – 3) = 43

43n – z – 74 = 43n – 129

43n – z = 43n – 55

-z = -55

z = 55

0 Replies 0 Likes

The average of the “three numbers that are removed from the group” would remain the same if “the average of the remaining numbers remains the same”.

0 Replies 0 Likes

Always replace with simple numbers to explain to child

1+2+3+4+5 = 15

15/5 = 3 (avg of group of 5 numbers is 3)

When 1 and 5 left , 2,3,4 remains.

1+5=6

6/2 = 3 (avg of 2 numbers is 3)

2+3+4=9

9/3 = 3 (avg remains the same as initial avg when avg of numbers taken away is the same )

If avg remains the same after a group of numbers has been removed, it means that the average of the group of numbers is always the same average.

0 Replies 0 Likes