stmt 1 - the number ab is a multiple of 3...straightfwd..if a number has to be a multiple of 3 then the sum of digits has to be a multiple of 3...so a+b is a multiple of 3.......Suff

stmt 2 - a-2b is a multiple of 3. now if a+b is a multiple of 3 then either the sum or the difference or the product HAS to be a multiple of 3.You can verify with examples...here we'll take the difference of 2 numbers. Difference = (a-2b) - (a+b) = -3b....definitely a multiple of 3 and so a+b has to be a multiple of 3....Suff

IMO ans is D

Hi Nisha,
I dont know whether this is the best way, but I would solve it like this

(1) The two-digit number "ab" (where a is in the tens place and b is in the ones place) is a multiple of 3.

now there is a mathematical rule that "A nos is divisible by 3 if the sum of its digits is divisible by 3" (this is true always)

i.e a+b is divisible by 3

SUFF

(2) a – 2b is a multiple of 3.

when taking multiples & fators we only use natural nos i.e 1 & above

hence a-2b cannor be -ve

also a - 2b = a + b - 3b

i.e (a+b) - 3b

here 3b is divisible by 3 i.e it is a multiplr of 3

now since it is substacted from (a+b) hence a+b must be a multiple of 3

i.e 3* somenos - 3*somenos will be a multiple of 3

henece SUFF

D

Thanks ur explanations were really helpful jango and samir!

Can we do it this way:-

1) The two-digit number "ab" (where a is in the tens place and b is in the ones place) is a multiple of 3.

This is clear for explanations provided earlier.

(2) a – 2b is a multiple of 3.

(a-2b)/3 = k (k is any integer)
(a-2b+3b-3b)/3 = k( adding & subtractoing 3b)
( (a+b) - 3b ) /3 = k
(a+b) / 3 - b = k
(a+b) /3 = b + k (since b is given as single digit integer, so b + k will be another integer, say n)

(a+b)/3 = n, hence a+b is divisible by 3.