1.

(1) x=6(int)+2. the remainder for x/3 will also be 2.

ex. 14/6 = 2 with 2 remainder.
(2) x=15(int)+2. the remainder will also be 2 when x/3.

ex. 17/15 = 1 with 2 remainder.

[spoiler]hence (D).[/spoiler]



2. these are consecutive integers.

(a)

(n+(n+1)+(n+2)+......+(n+8) )/9 = 7

solve for n and n = 2. since these are consecutive integers, you can figure out the last 2 digits as well.

(b) do the same thing. ((n+11)+(n+10)......(n+3) )/9=9 and you get n=2.

[spoiler]hence (D)[/spoiler]

what is remainder when x is divided by 3?

A.when x is divide by 6 the remainder is 2

so possible x=6n+2 x=8,14,20,26 etc

so when these are divided by 3 remainder will be for x=8 r=2
x=14 r=2
x=20 r=2

so r is 2 suff

B. when x is divide by 15 r=2 so x = 17,32,47 etc
so when x is divided by 3 x=17 r=2
x=32 r=2 etc so suff

both are suff so answer is D


2.

what is the average of 11 consecutive integers

a. average of first nine integers is 7

suppose first interger is n then (n+n+1+n+2.......n+7++n+8.....n+10)/11 =?

given n+n+1....n+8/9=7 n+n+1+n+2......n+8 =63 9n+36=63 9n=27 n= 3 so suff

b. similarly we get the value of n and get the answer so suff

hence D

bkobilov wrote: 2. What is the average of eleven consecutive integers?

1. The average of the first nine integers is 7
2. The average of the last nine integers is 9

Memorize this rule:

average of consecutive integers = the median

Statement 1:
Average = 7 means median = 7.
So the first 9 integers are 3, 4, 5, 6, 7, 8, 9, 10, 11.
The 11 integers are 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13.
Median = 8, so average = 8.
Sufficient.

Statement 2:
Average = 9 means median = 9.
So the last 9 integers are 5, 6, 7, 8, 9, 10, 11, 12, 13.
The 11 integers are 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13.
Median = 8, so average = 8.
Sufficient.

The correct answer is D.