Bunuel wrote:
When 55 is divided by 3x the remainder is 2x. What is the value of x?
(1) x is an odd integer
(2) x is a 2-digit integer
Kudos for a correct solution.
VERITAS PREP OFFICIAL SOLUTION:This problem is a good candidate for the "Pick Numbers / Play Devil's Advocate" strategy, as if you can come up with two different values of x, you can prove that a statement is not sufficient.
For statement 1, one combination is that 55/33 is 1 remainder 22, so x could be 11 and satisfy this statement. Another possibility is x = 5, as that would be 55/15 which would be 3 remainder 10. Since you've now found two possible values for x (11 and 5), you can prove that statement 1 is not sufficient.
Moving on from statement 1 to statement 2, you may also want to look at what you've learned from working with statement 1. x = 11 is a valid solution given statement 2 (x is 2 digits), so you have one solution in hand. And you should also recognize that x cannot be even, as 55 is odd and you couldn't have an even quotient plus an even remainder equal 55.
What you've also learned from trying x = 11 above is that that makes 3x + 2x = 55, meaning that x cannot be any bigger than 11. Accordingly, x = 11 is the only two-digit number that will work with statement 1, and the answer is B.
Hi i think am wrong ...plz justify......Why every one tried the 2nd statment with 11......they have stated its a 2 digit integer ...then if we try with 10,...55/3(10) ---remainder is 15 and its not equal to 2(x) rite....then the answer must be "E".......am confused...