Is x an integer ?

[Deepthi Subbu]

If m and n are both two digit numbers ,and m-n = 11x , is x an integer ?

a.The tens digit and units digit of m are same.
b.m+n is a multiple of 11.

OA to follow

[n@resh]

If m and n are both two digit numbers ,and m-n = 11x , is x an integer ?

a.The tens digit and units digit of m are same.
b.m+n is a multiple of 11.

OA to follow

from statement 1: number can be: 11,22,33,44...99 but you can't have control over n!
it can be 10 or 11..etc which lead to different values of x not restricted to integers...hence it's not sufficient.

from statement 2: As m and n values aren't fixed. eg: 10 + 12 ..then m-n = -2 that means x = -2/11 is not integer! but if you consider m+n as 11 + 22 ..then m-n = -11 => x ( = -1 )is an integer!
Clearly,not sufficient.

combine both, m must be multiple of 11, then from 2nd statement m + n is multiple of 11, leads to 'n' which must be multiple of 11.

Hence m - n will be always multiple of 11..hence x is integer!

Ans:C!

[shrivast]

IMO : C

[saketk]

If m and n are both two digit numbers ,and m-n = 11x , is x an integer ?

a.The tens digit and units digit of m are same.
b.m+n is a multiple of 11.

OA to follow

rephrase the question -- is (m-n) divisible by 11

statement 1: tens and units digit of m are same.

i.e. m can be 11, 22, 33 etc

not sufficient because we dont know anything about n yet.

Statement 2: m+n is a multiple of 11. since m and n are both 2 digits number

we can say m+n is 22,33,44,55,66 etc.

but this statement alone is also NS.

combine the 2 statements: -

we know that m is a multiple of 11.
and m+n is a multiple of 11.

this implies n is also a multiple of 11.

This also means that m-n wil also be a multiple of 11.

Answer C.

[navami]

IMO C