Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 350,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

stmt1: if both x and y are negative such that the product of their absolute values is > 6 then xy>6. eg. x=-8, y=-3

stmt2: x is +ve, -8<y<8. Let us take max value of x =2/3 and max value of y=8, the product is <6. Now, let us take max value of x=2/3 and a negative value of y=-8, the product is -ve and hence < 6. One more try, take x=2/3 and y=5, the product is <6.

stmt1: if both x and y are negative such that the product of their absolute values is > 6 then xy>6. eg. x=-8, y=-3

stmt2: x is +ve, -8<y<8. Let us take max value of x =2/3 and max value of y=8, the product is <6. Now, let us take max value of x=2/3 and a negative value of y=-8, the product is -ve and hence < 6. One more try, take x=2/3 and y=5, the product is <6.

no sure about -8<y<8 y^2<64 either becomes y<+8 or y<-8 OR mod y < 64, meaning -64<y<64

stmt1: if both x and y are negative such that the product of their absolute values is > 6 then xy>6. eg. x=-8, y=-3

stmt2: x is +ve, -8<y<8. Let us take max value of x =2/3 and max value of y=8, the product is <6. Now, let us take max value of x=2/3 and a negative value of y=-8, the product is -ve and hence < 6. One more try, take x=2/3 and y=5, the product is <6.

no sure about -8<y<8 y^2<64 either becomes y<+8 or y<-8 OR mod y < 64, meaning -64<y<64

what am i missing here?

y^2 < 64 mean that -8 < y < 8 because min y^2 is 0 and max is close to 64. In that case, y<8 or y>-8. If y<-8, y could be -10, resulting y^2 = 100, which is >64 and violets the given info in the question. Whenever you have y^2< 64, y is either +ve or 0, or -ve.

i. y^2 < 64 doesnot mean that -64 < y^2 < 64. ii. "Mod y < 64, meaning -64<y<64" is not correct. _________________

Re: Problem : Is xy <6 (1) x < 3 and y < 2 (2) 1/2 < [#permalink]
09 Apr 2012, 11:31

3

This post received KUDOS

Expert's post

1

This post was BOOKMARKED

mymbadreamz wrote:

I didn't understand this one. could someone please explain why C is not the answer? thanks.

Is \(xy<6\)?

(1) \(x<3\) and \(y<2\) --> now, if both \(x\) and \(y\) are equal to zero then \(xy=0<6\) and the answer will be YES but if both \(x\) and \(y\) are small enough negative numbers, for example -10 and -10 then \(xy=100>6\) and the answer will be NO. Not sufficient.

(2) \(\frac{1}{2}<x<\frac{2}{3}\) and \(y^2<64\), which is equivalent to \(-8<y<8\) --> even if we take the boundary values of \(x\) and \(y\) to maixmize their product we'll get: \(xy=\frac{2}{3}*8\approx{5.3}<6\), so the answer to the question "is \(xy<6\)?" will always be YES. Sufficient.

Re: Problem : Is xy <6 (1) x < 3 and y < 2 (2) 1/2 < [#permalink]
28 May 2013, 14:14

Bunuel wrote:

mymbadreamz wrote:

I didn't understand this one. could someone please explain why C is not the answer? thanks.

Is \(xy<6\)?

(1) \(x<3\) and \(y<2\) --> now, if both \(x\) and \(y\) are equal to zero then \(xy=0<6\) and the answer will be YES but if both \(x\) and \(y\) are small enough negative numbers, for example -10 and -10 then \(xy=100>6\) and the answer will be NO. Not sufficient.

(2) \(\frac{1}{2}<x<\frac{2}{3}\) and \(y^2<64\), which is equivalent to \(-8<y<8\) --> even if we take the boundary values of \(x\) and \(y\) to maixmize their product we'll get: \(xy=\frac{2}{3}*8\approx{5.3}<6\), so the answer to the question "is \(xy<6\)?" will always be YES. Sufficient.

Answer: B.

i dont understand i put A....i could answer the question with that information. N why are we letting x times y = to 0 why cant we let it be equal to 1 or 2?

if a number times a number is less than 6......cant we just say use 1?

Re: Problem : Is xy <6 (1) x < 3 and y < 2 (2) 1/2 < [#permalink]
28 May 2013, 14:23

1

This post received KUDOS

Expert's post

madzstar wrote:

Bunuel wrote:

mymbadreamz wrote:

I didn't understand this one. could someone please explain why C is not the answer? thanks.

Is \(xy<6\)?

(1) \(x<3\) and \(y<2\) --> now, if both \(x\) and \(y\) are equal to zero then \(xy=0<6\) and the answer will be YES but if both \(x\) and \(y\) are small enough negative numbers, for example -10 and -10 then \(xy=100>6\) and the answer will be NO. Not sufficient.

(2) \(\frac{1}{2}<x<\frac{2}{3}\) and \(y^2<64\), which is equivalent to \(-8<y<8\) --> even if we take the boundary values of \(x\) and \(y\) to maixmize their product we'll get: \(xy=\frac{2}{3}*8\approx{5.3}<6\), so the answer to the question "is \(xy<6\)?" will always be YES. Sufficient.

Answer: B.

i dont understand i put A....i could answer the question with that information. N why are we letting x times y = to 0 why cant we let it be equal to 1 or 2?

if a number times a number is less than 6......cant we just say use 1?

On DS questions when plugging numbers, goal is to prove that the statement is not sufficient. So we should try to get a YES answer with one chosen number(s) and a NO with another.

Now, for x=y=0 we got an YES answer and for x=y=-10 we got a NO answer, thus the statement is NOT sufficient.

Of course we could use some other numbers to get an YES and a NO answers to prove that the statement is not sufficient: x=y=0 and x=y=-10 are just examples of many possible sets.

Re: Is xy < 6 (1) x < 3 and y < 2 (2) 1/2 < x < 2/3 and y^2 < 64 [#permalink]
24 Jul 2014, 22:34

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

Re: Is xy < 6 (1) x < 3 and y < 2 (2) 1/2 < x < 2/3 and y^2 < 64 [#permalink]
15 Aug 2015, 02:45

Bunuel wrote:

i dont understand i put A....i could answer the question with that information. N why are we letting x times y = to 0 why cant we let it be equal to 1 or 2?

if a number times a number is less than 6......cant we just say use 1?

On DS questions when plugging numbers, goal is to prove that the statement is not sufficient. So we should try to get a YES answer with one chosen number(s) and a NO with another.

Now, for x=y=0 we got an YES answer and for x=y=-10 we got a NO answer, thus the statement is NOT sufficient.

Of course we could use some other numbers to get an YES and a NO answers to prove that the statement is not sufficient: x=y=0 and x=y=-10 are just examples of many possible sets.

On September 6, 2015, I started my MBA journey at London Business School. I took some pictures on my way from the airport to school, and uploaded them on...

When I was growing up, I read a story about a piccolo player. A master orchestra conductor came to town and he decided to practice with the largest orchestra...

I’ll start off with a quote from another blog post I’ve written : “not all great communicators are great leaders, but all great leaders are great communicators.” Being...