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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.

Answer

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

 

 

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

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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. 

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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

Answer: 55

 

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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”.

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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.

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