Question

Out of six consecutive integers, if the sum of first three is 33, what is the sum of the remaining three numbers?
A) 48
B) 42
C) 54
D) 46

Answer 1

let the six consecutive integers be

a-3,a-2,a-1,a,a+1,a+2,a+3

given sum of first three numbers is 33

(a-3)+(a-2)+(a-1) =33

3a-6=33

3a=39

a=39/3=13

then the next remaining numbers be

a,a+1,a+2

13,14,15

sum of remaining numbers be 13+14+15=42