In a series of six consecutive integers, the sum of the first three

Question

In a series of six consecutive integers, the sum of the first three integers is 93. What is the sum of the last three integers?

A 96
B 99
C 102
D 105
E 108

philipssonicare · Answer

six consecutive integers 
x+(x+1)+(x+2)+(x+3)+(x+4)+(x+5)

the sum of the first three integers is 93 
x+(x+1)+(x+2)=33
3x+3=93

x=30

What is the sum of the last three integers? 
(x+3)+(x+4)+(x+5)=?
(33)+(34)+(35)=102

C

Archit3110 · Answer

x x+1 x+2 x+3 x+4 x+5 
x+ x+1+x+2 = 93
x= 30

so 
3x+12 ; 90+12 ; 102 IMO C

ScottTargetTestPrep · Answer

Letting x = the first integer, we can create the equation:

x + (x + 1) + (x + 2) = 93

3x + 3 = 93

3x = 90

x = 30

The first integer is 30, and so the 4th integer is (30 + 3), the 5th integer is (30 + 4), and the 6th integer is (30 + 5). So the sum of the last 3 integers is:

(30 + 3) + (30 + 4) + (30 + 5) = 102

Alternate solution:

The 4th, 5th and 6th integers are 3 more than the 1st, 2nd and 3rd integers, respectively. Thus the sum of the last 3 integers is 3 x 3 = 9 more than the sum of the first 3 integers. Since the sum of the first 3 integers is 93, the sum of the last 3 integers is 93 + 9 = 102.

Answer: C

gracie · Answer

second integer is 93/3=31, mean of first three integers
fifth integer is 31+3=34, mean of last three integers
3*34=102, sum of last three integers
C

In a series of six consecutive integers, the sum of the first three

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In a series of six consecutive integers, the sum of the first three

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