# The average of 5 numbers is 18.4. the average of first 3 numbers is 12 and the average of last three numbers is 30. find the third number.

2
by roshanben10

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by roshanben10

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average of 5 terms is 18.4

so

[x+y+z+a+b]/5=18.4

so

x+y+z+a+b=92................................1

average of first three terms is 12

[x+y+z]/3=12

so

x+y+z=36...................................2

average of last three terms is 30

so

[z+a+b]/3=30

so

z+a+b=90.....................................3

substituting the value of eq 2 in eq 1

a+b=92-36=56.........................4

substituting the value of a+b in eq 3

z=90-56=34

so the value of third term is 34

Let us consider those 5 numbers be a,b,c,d,e

The average of 5 numbers = 18.4

(a+b+c+d+e)/5 = 18.4

(a+b+c+d+e) = 18.4 * 5 = 92 --------------------> 1

The average of first three numbers is = 12

(a+b+c)/3 = 12

(a+b+c) = 12*3 = 36 ----------------------> 2

The average of last three numbers is = 30

(c+d+e)/3 = 30

(c+d+e) = 90 ------------------------------> 3

By adding 2 and 3 we get

a+b+c+c+d+e = 36+90 = 126

(a+b+c+d+e)+c = 126

but from it is clear (a+b+c+d+e) = 92

By substituting this in above we get

92+c = 126

c = 126 - 92 = 34