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 500,000 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:

In DS questions when it asks for .. IS some data followed by ? given 2 statement below.

A definite yes or definite no is sufficient for the answer. or when we should use this method ???

In a Yes/No Data Sufficiency questions, statement(s) is sufficient if the answer is “always yes” or “always no” while a statement(s) is insufficient if the answer is "sometimes yes" and "sometimes no".

When a DS question asks about the value of some variable, then the statement(s) is sufficient ONLY if you can get the single numerical value of this variable.

2. Is the measure of one of the interior angles of quadrilateral ABCD equal to 60? (1) Two of the interior angles of ABCD are right angles. (2) The degree measure of angle ABC is twice the degree measure of angle BCD.

Please share your way of thinking, not only post the answers.

OA and explanations to follow.

Option 1 we know 2 angles are 90 degree each but we have no info about the other 2 angles which can be (60,120), (80,100) etc. hence insuff Option 2 says angle ABC = 2 * angle BCD again insuff as we can have an angle equal to 60 degree or none of the angles as 60degree

Taking 1 and 2 we know 2 angles are 90degree each and angle ABC is twice angle BCD lets assume angle ABC = 90 degree then angle BCD = 45. we know the 3rd angle is 90 degree so 4th angle becomes 135 (sum of angles of a quadrilateral = 360degrees) so answer will NO ( none of angles are 60degree) now lets consider that the 1st unknown angle is twice the 2nd unknown angle (other 2 are 90 degree each) we can get value for smaller angle as 60degree hence insuff

will go with E

Asterix from eeek?? to Eureka!!! to eeek???

One angle is x other is 2x, x+2x =180(360-180), this gives x = 60 degree....B is correct...
_________________

Like my post Send me a Kudos It is a Good manner. My Debrief: http://gmatclub.com/forum/how-to-score-750-and-750-i-moved-from-710-to-189016.html

2. Is the measure of one of the interior angles of quadrilateral ABCD equal to 60? (1) Two of the interior angles of ABCD are right angles. (2) The degree measure of angle ABC is twice the degree measure of angle BCD.

Please share your way of thinking, not only post the answers.

OA and explanations to follow.

Option 1 we know 2 angles are 90 degree each but we have no info about the other 2 angles which can be (60,120), (80,100) etc. hence insuff Option 2 says angle ABC = 2 * angle BCD again insuff as we can have an angle equal to 60 degree or none of the angles as 60degree

Taking 1 and 2 we know 2 angles are 90degree each and angle ABC is twice angle BCD lets assume angle ABC = 90 degree then angle BCD = 45. we know the 3rd angle is 90 degree so 4th angle becomes 135 (sum of angles of a quadrilateral = 360degrees) so answer will NO ( none of angles are 60degree) now lets consider that the 1st unknown angle is twice the 2nd unknown angle (other 2 are 90 degree each) we can get value for smaller angle as 60degree hence insuff

will go with E

Asterix from eeek?? to Eureka!!! to eeek???

One angle is x other is 2x, x+2x =180(360-180), this gives x = 60 degree....B is correct...

Bunuel, thanks for the wonderful set, i had a question

. Is x^4 + y^4 > z^4 ? (1) x^2 + y^2 > z^2 (2) x+y > z

x^4 + y^4 = ((x^2) + (y^2))^2 - 2*x^2*y^2

So basically we want to find out if ((x^2) + (y^2))^2 - 2*x^2*y^2 > z^4

OR if

((x^2) + (y^2))^2 > z^4 + 2*x^2*y^2

Now by (1), we know that x^2 + y^2 > z^2

Or

((x^2) + (y^2))^2 > z^4 ( squaring both sides, both sides are absolutely positive)...... (a)

We also know that 2*x^2*y^2 will be less than ((x^2) + (y^2))^2......(b)

Combining (a) and (b), can we not conclude that ((x^2) + (y^2))^2 > z^4 + 2*x^2*y^2 and hence making (1) sufficient

Is \(x^4+y^4>z^4\)?

The best way to deal with this problem is plugging numbers. Remember 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.

(1) \(x^2+y^2>z^2\) It's clear that we get YES answer very easily with big x and y (say 10 and 10), and small z (say 0).

For NO answer let's try numbers from Pythagorean triples: \(x^2=3\), \(y^2=4\) and \(z^2=5\) (\(x^2+y^2=7>5=z^2\)) --> \(x^4+y^4=9+16=25=z^4\), so we have answer NO (\(x^4+y^4\) is NOT more than \(z^4\), it's equal to it).

Not sufficient.

(2) \(x+y>z\). This one is even easier: again we can get YES answer with big x and y, and small z.

As for NO try to make z some big enough negative number: so if \(x=y=1\) and \(z=-5\), then \(x^4+y^4=1+1=2<25=z^4\).

Not sufficient.

(1)+(2) As we concluded YES answer is easily achievable. For NO try the case of \(x^2=3\), \(y^2=4\) and \(z^2=5\) again: \(x+y=\sqrt{3}+\sqrt{4}>\sqrt{5}\) (\(\sqrt{3}+2\) is more than 3 and \(\sqrt{5}\) is less than 3), so statement (2) is satisfied, we know that statement (1) is also satisfied (\(x^2+y^2=7>5=z^2\)) and \(x^4+y^4=9+16=25=z^4\). Not sufficient.

Ok Bunuel, i shall stick to the plugging number approach, please can you confirm if both terms in an equality are positive, we can square both sides or even raise to the power of 3, without changing the inequality sign?

Ok Bunuel, i shall stick to the plugging number approach, please can you confirm if both terms in an equality are positive, we can square both sides or even raise to the power of 3, without changing the inequality sign?

1. We can raise both parts of an inequality to an even power if we know that both parts of an inequality are non-negative (the same for taking an even root of both sides of an inequality)

2. We can always raise both parts of an inequality to an odd power (the same for taking an odd root of both sides of an inequality).

1. An integer greater than 1 that is not prime is called composite. If the two-digit integer n is greater than 20, is n composite? (1) The tens digit of n is a factor of the units digit of n. (2) The tens digit of n is 2.

Please share your way of thinking, not only post the answers.

OA and explanations to follow.

given n >20. let N be of the form 10x + y where x is ten's digit and y is units digit

1. x is a factor of y. Lets assume x = 2,3 or 4 then y can have values 2,4,6,8,9 . All possible combinations will give a composite number . hence suff 2. we do not have any info about units digit so insuff hence A

Answer is C. X can also be 1. 1 is factor of all the numbers.

gmatclubot

Re: Collection of 8 DS questions
[#permalink]
22 Jun 2014, 14:56

It’s quickly approaching two years since I last wrote anything on this blog. A lot has happened since then. When I last posted, I had just gotten back from...

Since my last post, I’ve got the interview decisions for the other two business schools I applied to: Denied by Wharton and Invited to Interview with Stanford. It all...