Let a be the number of pencils costing 23 cents and b costing 21 cents

Stmt I

a +b = 6

insuff

Stmt II

23a+21b = 130

insuff

stmt I&II

solve for a and b

Suff

C

I get B.

23a+21b=130 where a & b must be integers >= 0 because they are pencils.

if a=o then 21b=130 so b does not = an integer
if a=1 then 21b=107 so b does not = an integer
if a=2 then 21b=84 so b=4
if a=3 then 21b=61 so b does not = an integer
if a=4 then 21b=38 so b does not = an integer
if a=5 then 21b=15 so b does not = an integer

if a=o then 21b=130 so b does not = an integer
if a=1 then 21b=107 so b does not = an integer
if a=2 then 21b=84 so b=4
if a=3 then 21b=61 so b does not = an integer
if a=4 then 21b=38 so b does not = an integer
if a=5 then 21b=15 so b does not = an integer

There is a flaw in your reasoning

As you said you are getting B as answer
This means (2) is suff for you

but as you have solved,you have used some value of a to get a value of b,am I right??

this means you need to have some value of a to solve this equation and you cannot solve solely on the equation given by statement (2)

Hope this helps

karan

Iam getting B as well , and for the same reasons as Ross showed in the previous posts.

gmat740 wrote:

if a=o then 21b=130 so b does not = an integer
if a=1 then 21b=107 so b does not = an integer
if a=2 then 21b=84 so b=4
if a=3 then 21b=61 so b does not = an integer
if a=4 then 21b=38 so b does not = an integer
if a=5 then 21b=15 so b does not = an integer

There is a flaw in your reasoning

As you said you are getting B as answer
This means (2) is suff for you

but as you have solved,you have used some value of a to get a value of b,am I right??

this means you need to have some value of a to solve this equation and you cannot solve solely on the equation given by statement (2)

Hope this helps

karan

This is not necessarily true Karan,

We know that "a" and "b" are VARAIBLES......except they are both INTEGERS as well (cannot have fractional pencils...a favorite GMAT trick on 600 + questions).

The only way to get 130 from a combination of 23 cents and 21 cents is to have 2 , 23 cent pencils and 4, 21 cent pencils.

Unique answer implies sufficiency.