Last visit was: 19 Sep 2024, 03:25 |
It is currently 19 Sep 2024, 03:25 |

Customized

for You

Track

Your Progress

Practice

Pays

p is the remainder on dividing X by 2, while q is the remainder on div
[#permalink]
22 Aug 2024, 22:40

Expert Reply

1

Bookmarks

Question Stats:

p is the remainder on dividing X by 2, while q is the remainder on dividing Y by 2. To find the relation between p and q, whether p is equal to q or greater or less, which of the following options is alone sufficient?

A. X is an even integer.

B. Y is an odd integer.

C. Y is a non-negative number.

D. X + Y + XY is an odd integer and q is not 0

E. X + Y + XY is an even integer.

A. X is an even integer.

B. Y is an odd integer.

C. Y is a non-negative number.

D. X + Y + XY is an odd integer and q is not 0

E. X + Y + XY is an even integer.

Part of the project: GRE Quant & Verbal ADVANCED Daily Challenge 2024 Edition - Gain 20 Kudos & Get FREE Access to GRE Prep Club TESTS

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Free Materials for the GRE General Exam - Where to get it!!

Shorter GRE - Preparing for the Quantitative Reasoning Measure

Shorter GRE - Preparing for the Verbal Reasoning Measure

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Free Materials for the GRE General Exam - Where to get it!!

Shorter GRE - Preparing for the Quantitative Reasoning Measure

Shorter GRE - Preparing for the Verbal Reasoning Measure

Re: p is the remainder on dividing X by 2, while q is the remainder on div
[#permalink]
27 Aug 2024, 07:32

I think D is the correct answer and not E.

For option E, we can only say that either one of X or Y will be even and the other will be odd. So we have two cases, (p=0, q=1) and (p=1,q=0).

Option D is definitely correct because q=1 (given) and we derive p=0 (X is even and Y is odd).

For option E, we can only say that either one of X or Y will be even and the other will be odd. So we have two cases, (p=0, q=1) and (p=1,q=0).

Option D is definitely correct because q=1 (given) and we derive p=0 (X is even and Y is odd).

Re: p is the remainder on dividing X by 2, while q is the remainder on div
[#permalink]
29 Aug 2024, 09:52

i think both d & e

Re: p is the remainder on dividing X by 2, while q is the remainder on div
[#permalink]
29 Aug 2024, 13:03

Expert Reply

OE

We know that p is the remainder on dividing X by 2, while q is the remainder on dividing Y by 2; we need to check that which of the given statements is sufficient to find the relation between p and q, whether p is equal to q or greater or less.

As Dividend = Divisor x Quotient + Remainder , we get \(X = 2k_1 + p\) & \(Y = 2k_2 + q\) , where \(k_1\) & \(k_2\) are non negative integers i.e. X & Y are p & q greater, respectively, than the multiple of 2.

Note: - Since p & q are the remainders when X & Y are divided by 2, their values can be 0 or 1 only.

Now checking from the options we get

(A) X is an even integer — If X is an even integer the value of p must be 0 but as nothing is said about the value of Y we cannot say anything about the value of q, so the statement (A) is insufficient.

(B) Y is an odd integer - If Y is an odd integer the value of q must be an odd number i.e. 1 but as nothing is said about the value of X we cannot say anything about the value of p, so the statement (B) is insufficient.

(C) Y is a non-negative number — It gives nothing about the value of q or p, so the statement (C) is insufficient.

(D) X + Y + XY is an odd integer and q is not 0 — The expression X + Y + XY can be odd if exactly one of X and Y is even and the other one is odd or X & Y both are odd. If we consider exactly one of X & Y odd, as q is not 0, we must have X even & p = 0. So, the value of Y must be odd which gives q = 1. Thus, we get p < q. But if we take both X & Y odd, we must have p = q = 1. As different assumptions give different relations between p & q, statement (D) is insufficient.

(E) X + Y + XY is an even integer — which implies both X and Y are even, so the values of both p and q must be zero each. Hence, we get p = q = 0, so statement (E) is sufficient.

Hence only option (E) is correct.

We know that p is the remainder on dividing X by 2, while q is the remainder on dividing Y by 2; we need to check that which of the given statements is sufficient to find the relation between p and q, whether p is equal to q or greater or less.

As Dividend = Divisor x Quotient + Remainder , we get \(X = 2k_1 + p\) & \(Y = 2k_2 + q\) , where \(k_1\) & \(k_2\) are non negative integers i.e. X & Y are p & q greater, respectively, than the multiple of 2.

Note: - Since p & q are the remainders when X & Y are divided by 2, their values can be 0 or 1 only.

Now checking from the options we get

(A) X is an even integer — If X is an even integer the value of p must be 0 but as nothing is said about the value of Y we cannot say anything about the value of q, so the statement (A) is insufficient.

(B) Y is an odd integer - If Y is an odd integer the value of q must be an odd number i.e. 1 but as nothing is said about the value of X we cannot say anything about the value of p, so the statement (B) is insufficient.

(C) Y is a non-negative number — It gives nothing about the value of q or p, so the statement (C) is insufficient.

(D) X + Y + XY is an odd integer and q is not 0 — The expression X + Y + XY can be odd if exactly one of X and Y is even and the other one is odd or X & Y both are odd. If we consider exactly one of X & Y odd, as q is not 0, we must have X even & p = 0. So, the value of Y must be odd which gives q = 1. Thus, we get p < q. But if we take both X & Y odd, we must have p = q = 1. As different assumptions give different relations between p & q, statement (D) is insufficient.

(E) X + Y + XY is an even integer — which implies both X and Y are even, so the values of both p and q must be zero each. Hence, we get p = q = 0, so statement (E) is sufficient.

Hence only option (E) is correct.

gmatclubot

Re: p is the remainder on dividing X by 2, while q is the remainder on div [#permalink]

29 Aug 2024, 13:03
Moderators:

Multiple-choice Questions — Select One or More Answer Choices |
||

## Hi Generic [Bot],Here are updates for you: |