When 55 is divided by 3x the remainder is 2x. What is the value of x?

I am not sure but will go with B..

x = 11 is the integer value.....

Any quick way... I just plugged in the values

ANSWER is B


FROM statement no 1 we can obtain 2 odd numbers that satisfy the condition . BOTH 5 an 11 can satisfy condition

if X=5 then 3X= 15 and 55/15 gives reminder 10 which is equal to twice the number (5)
also , if X=11 then 3X= 33 and 55/33 gives reminder 22 which is equal to twice the number (11)

so stat ONE is not sufficient ...


Statement 2 is Sufficient because ONLY number 11 is a TWO-DIGIT number which satisfies the condition as we know 55 is an ODD integer so 3X CAN NOT bigger than 55 because 55 CAN NOT EVENLY DIVIDE TO 2 and GET a FULL INTEGER (IF 3X is greater than 55 the reminder would be 55 ) so 3X MUST be less than 55 and 2 digit numbers which satisfy this condition would be

10, 11, 12, 13, 14, 15, 16, 17 , 18 and among these numbers ONLY 11 can satisfy the condition therefore X is 11 and the statement is sufficient...


SO answer is B.....

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...

How is the remainder when divided 55 by 30 15? It's 25.

A new approach
Dividend = Divisor x quotient + Remainder

3x (k) + 2x = 55
x (3k+2) = 55
if
x = 11 k = 1
x=5 k=3
st - 1 can't say
st - 2 true.
therfore only B is sufficient

Can x=-55? Then st. 2 is insuf. too.

x can't be -55

Please note:

Dividend = Divisor*Quotient + Remainder
e.g. when 17 is divided by 5(divisor) then the quotient is 3 and remainder is 2
i.e. 17 = 5*3 + 2

Here the statement 2 mentions that
" x is a 2-digit integer"
if you take x = -55 then it violates the condition "When 55 is divided by 3x the remainder is 2x"

I hope it helps!

NO, as per GMAT and OG, when you talk about divisbility and remainder, you are talking about positive divisors and not negative.

Also, if x = -55, then 3x = -165 and remainder will be 2*-55 = -110

Thus your divisibility euiqation will become: 3x * p + 2x = -165-110 = -275 and this is NOT equal to 55. Thus you can not use -55 or for that matter any negative value.

Bunuel, Souvik

Can you confirm the statement above?

Refer to the post by Bunuel: divisibility-multiples-factors-tips-and-hints-174998.html

Engr2012

Thanks for the reply, but I didn´t see your rule in the link - maybe you are referring to this one? https://gmatclub.com/forum/remainders-tips-and-hints-175000.html?fl=similar
If that´s the case, can I ALWAYS assume that when dealing with Remainder questions the quotient and remainer will never be negative?

Point 1 from the link that I posted mentions:

DIVISIBILITY1. Every GMAT divisibility question will tell you in advance that any unknowns represent positive integers (ALL GMAT divisibility questions are limited to positive integers only).

Your link as well talks about the same thing:

"DEFINITION
If x and y are positive integers, there exist unique integers q and r, called the quotient and remainder, respectively, such that y=divisor∗quotient+remainder=xq+r and 0≤r<x."

Yes, for GMAT quant, whenever you talk about divisbility and remainders, the divisors, quotient and remainders will be positive integers.

There is a small thing to remember if you encounter a question with a 'negative remainder'. The negative remainder can actually be modified to become a 'true remainder' as shown below;

If you say : You are given a number of the form: Q = 12P - 3 ----> Q = 12P + (-6+3) ----> Q = 12P -6 + 3 ----> Q = 6 (2P-1) + 3. Thus this shows that although you were given a 'negative' remainder, you were still able to convert the negative form to an actual positive remainder. Above, I have shown that the remainder will be 3 when Q is divided by 6 and not -3 when divided by 12.

\(3xq + 2x = 55\)

\(x = \frac{55}{(3q + 2)}\)

Statement (1) x is an odd integer

means that (3q + 2) is a factor of 55, two solutions, 11 and 5 (insufficient)

(2) x is a 2-digit integer: sufficient X equals 11

what if we divide 55 by 51 (3*17) and get remainder of 4 (2*2)??

The stem says: when 55 is divided by 3x the remainder is 2x. Your example does not satisfy this condition.

55=5*11
According to stem 55=3xq+2x. ==>55=x(3q+2)
So, x and 3q+2 will represent these two values i:e; 5 and 11. So, we have to find out which portion represent what.
statement 1 says x is odd. So, x can be 5 or 11 as both of these are odd.
Statement 2 says x is two digit number and hence it is 11. So it is sufficient.

Answer B