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:

Re: NEW SET of good DS(4) [#permalink]
18 Oct 2009, 14:03

6. How many perfect squares are less than the integer d? (1) 23 < d < 33 (2) 27 < d < 37

1) For this condition d = 25 is perfect square of 5. Can we say there are 9, 16... only 2 perfect squares are < d ...? so Suff.. 2) For this condition d = 36 is perfect square of 6. Can we say there are 9, 16 and 25.. only 3 perfect squares are < d ...? Suff..

D.

Is this explanation correct ?

9. Is x^2 equal to xy? (1) x^2 - y^2 = (x+5)(y-5) (2) x=y

Re: NEW SET of good DS(4) [#permalink]
18 Oct 2009, 14:15

1

This post was BOOKMARKED

scoregmat wrote:

6. How many perfect squares are less than the integer d? (1) 23 < d < 33 (2) 27 < d < 37

1) For this condition d = 25 is perfect square of 5. Can we say there are 9, 16... only 2 perfect squares are < d ...? so Suff.. 2) For this condition d = 36 is perfect square of 6. Can we say there are 9, 16 and 25.. only 3 perfect squares are < d ...? Suff..

D.

Is this explanation correct ?

I may be wrong but this is what I could make out from the question stem option1 - for this we have 23 < d < 33 where d can be 24 upto 32 so we can have 1,4,9,16 and 25 depending on value of d. If d = 24 then perfect squares should be 1,4,9,16. If d = 26 then we can have 1,4,9,16 and 25. Hence insuff option 2 -for this we have 27 < d < 37 where d can be 28 upto 36. So perfect squares less than d will be 1,4,9,16 and 25. even if d =36 we need to find squares less than 36 so it will have only 5 perfect squares. hence suff

will go with B (I may be totally wrong with my understanding)

Re: NEW SET of good DS(4) [#permalink]
18 Oct 2009, 14:21

Thanks for explanation.. However for condition 2) isn't same logic applies...If d = 35 or less then perfect squares should be 1,4,9,16, 25 and If d = 36 then we can have 1,4,9,16,25 and 36. In this scenario even this is insuff ?

Re: NEW SET of good DS(4) [#permalink]
18 Oct 2009, 14:28

scoregmat wrote:

Thanks for explanation.. However for condition 2) isn't same logic applies...If d = 35 or less then perfect squares should be 1,4,9,16, 25 and If d = 36 then we can have 1,4,9,16,25 and 36. In this scenario even this is insuff ?

E?

question says perfect squares less than so even if d=36 we cannot count 36.

Re: NEW SET of good DS(4) [#permalink]
18 Oct 2009, 19:31

asterixmatrix wrote:

GMAT TIGER wrote:

Bunuel wrote:

1. When the positive integer x is divided by 4, is the remainder equal to 3?

(1) When x/3 is divided by 2, the remainder is 1. (2) x is divisible by 5.

(1) (x/3) = 2k+1 where k is an integer x = 6k + 3 x is an odd integer; so x/4 may or may not have 3 reminder. NSF..

(2) x = 5m where m, a positive integer, could be 1, 2, 3, 4, 5, 6, 7 and so on.. If m = 1 or 2, 4, 5, reminder is not 0. If m = 3, reminder is 3 or 7. NSF..

From 1 and 2: 5m = 6k + 3 If m = 3, k = 2 and x = 15. Then reminder is 3. If m = 9, k = 7 and x = 45. Then reminder is 3. If m = 15, k = 12 and x = 75. Then reminder is 3.

Re: NEW SET of good DS(4) [#permalink]
19 Oct 2009, 06:51

Bunuel wrote:

New questions:

1. When the positive integer x is divided by 4, is the remainder equal to 3? (1) When x/3 is divided by 2, the remainder is 1. (2) x is divisible by 5.

statement 1: ========== x must be divisible by 3 but not by 2.So x must be odd multiple of 3 like 9,15,21,27..etc. Nt suff

statement 2: ========== x is perfectly divisible by 5. x can be 5,10,15 etc.When we take 5 the rem is 1 and for 10 we get 2 and for 15 we get 3.Nt suff

Re: NEW SET of good DS(4) [#permalink]
19 Oct 2009, 07:33

asterixmatrix wrote:

scoregmat wrote:

6. How many perfect squares are less than the integer d? (1) 23 < d < 33 (2) 27 < d < 37

1) For this condition d = 25 is perfect square of 5. Can we say there are 9, 16... only 2 perfect squares are < d ...? so Suff.. 2) For this condition d = 36 is perfect square of 6. Can we say there are 9, 16 and 25.. only 3 perfect squares are < d ...? Suff..

D.

Is this explanation correct ?

I may be wrong but this is what I could make out from the question stem option1 - for this we have 23 < d < 33 where d can be 24 upto 32 so we can have 1,4,9,16 and 25 depending on value of d. If d = 24 then perfect squares should be 1,4,9,16. If d = 26 then we can have 1,4,9,16 and 25. Hence insuff option 2 -for this we have 27 < d < 37 where d can be 28 upto 36. So perfect squares less than d will be 1,4,9,16 and 25. even if d =36 we need to find squares less than 36 so it will have only 5 perfect squares. hence suff

will go with B (I may be totally wrong with my understanding)

Re: NEW SET of good DS(4) [#permalink]
20 Oct 2009, 22:09

Bunuel wrote:

New questions:

5. Al, Pablo, and Marsha shared the driving on a 1500 mile trip, which of the three drove the greatest distance on the trip? (1) Al drove 1 hour longer than Pablo but at an average of 5 miles per hour slower than Pablo. (2) Marsha drove 9 hours and averaged 50 miles per hour

option 2 gives us that Marsha drives a total of 450 miles. No other info. hence insuff option 1 gives us details about Al and Pablo. No info about Marsha. hence insuff

together we get that Al and Pablo driive 1050 miles Let time be T for which Pablo drove then T+1 wil be the time that Al drove and S be the speed with which Pablo drove then S-5 will be the speed with which Al drove we get S*T + (T+1)* (S-5) =1050 this equation formed will give various answers will go with E

Re: NEW SET of good DS(4) [#permalink]
20 Oct 2009, 22:20

7. The integers m and p are such that 2 is less than m and m is less than p. Also, m is not a factor of p. If r is the remainder when p is divided by m, is r > 1. (1) The greatest common factor of m and p is 2. (2) The least common multiple of m and p is 30.

we have 2<m<p

1. we know m and p are factors of 2. So will be even. Lets take m = 8 and p=10 then remainder r>1 Let m = 6 and p = 12 then also remainder r>1. Hence suff

2. given least common multiple is 30 then we can have values for m and p as (5,6) (10,15) (3,10) for the above pairs r can be =1 or >1 hence insuff

Re: NEW SET of good DS(4) [#permalink]
20 Oct 2009, 22:39

Bunuel wrote:

New questions: 10. A number of oranges are to be distributed evenly among a number of baskets. Each basket will contain at least one orange. If there are 20 oranges to be distributed, what is the number of oranges per basket? (1) If the number of baskets were halved and all other conditions remained the same, there would be twice as many oranges in every remaining basket. (2) If the number of baskets were doubled, it would no longer be possible to place at least one orange in every basket.

we know each basket contains atleast 1 orange 1. we can have either 20 baskets having 1 orange each and when number of baskets are halved we will have 10 baskets with 2oranges each we can also have 10 baskets with 2 oranges each and number of baskets are halved we will have 5baskets with 4oranges each hence insuff

2. lets consider that we have 10 baskets having 2 oranges each. if the baskets are doubled then we will have 20 baksets each having 1 orange but option says that on doubling the criteria that each basket has one orange will not suffice. So number of baskets are more than 10. and 20 is the only number which fits the criteria that each baket has atleast one orange.We will have 20 baskets with 1 orange each Hence suff

Re: NEW SET of good DS(4) [#permalink]
07 Nov 2009, 21:02

less than the integer d made the trick..

I would also go with B.

asterixmatrix wrote:

scoregmat wrote:

6. How many perfect squares are less than the integer d? (1) 23 < d < 33 (2) 27 < d < 37

1) For this condition d = 25 is perfect square of 5. Can we say there are 9, 16... only 2 perfect squares are < d ...? so Suff.. 2) For this condition d = 36 is perfect square of 6. Can we say there are 9, 16 and 25.. only 3 perfect squares are < d ...? Suff..

D.

Is this explanation correct ?

I may be wrong but this is what I could make out from the question stem option1 - for this we have 23 < d < 33 where d can be 24 upto 32 so we can have 1,4,9,16 and 25 depending on value of d. If d = 24 then perfect squares should be 1,4,9,16. If d = 26 then we can have 1,4,9,16 and 25. Hence insuff option 2 -for this we have 27 < d < 37 where d can be 28 upto 36. So perfect squares less than d will be 1,4,9,16 and 25. even if d =36 we need to find squares less than 36 so it will have only 5 perfect squares. hence suff

will go with B (I may be totally wrong with my understanding)

Re: NEW SET of good DS(4) [#permalink]
07 Nov 2009, 21:32

I think I differ in this..

To me the ans seems to B..

Please correct me if I am wrong...

(1) Says that there are 100 primes between 1 and p+1. This only says that P lies anywhere between the 100th prime and 101st (prime-1). P could be anywhere between that limit. So not sufficient.

(2)There are P primes between 1 and 3912. Clearly countable and leads to a distinct number. So sufficient.

Ans: B

cdowwe wrote:

Bunuel wrote:

New questions:

11. If p is a prime number greater than 2, what is the value of p? (1) There are a total of 100 prime numbers between 1 and p+1 (2) There are a total of p prime numbers between 1 and 3912.

1) Sufficient ... you can figure out the 100th prime # (from 1 to p+1) 2) Sufficient... you can count how many prime #s are between 1 & 3912 and that is P

Re: NEW SET of good DS(4) [#permalink]
08 Nov 2009, 01:14

Expert's post

mbaquestionmark wrote:

I think I differ in this..

To me the ans seems to B..

Please correct me if I am wrong...

(1) Says that there are 100 primes between 1 and p+1. This only says that P lies anywhere between the 100th prime and 101st (prime-1). P could be anywhere between that limit. So not sufficient.

(2)There are P primes between 1 and 3912. Clearly countable and leads to a distinct number. So sufficient.

Ans: B

Statement (1) is saying that there are 100 primes in the range between 1 and P+1 is basically telling us that P is the 100th prime . We can determine the single numerical value of P. So it's sufficient.

So D is the answer (as statement 2 is also sufficient). _________________

Re: NEW SET of good DS(4) [#permalink]
08 Nov 2009, 03:25

Ops... missed that P is a prime..

Thanks.

Bunuel wrote:

mbaquestionmark wrote:

I think I differ in this..

To me the ans seems to B..

Please correct me if I am wrong...

(1) Says that there are 100 primes between 1 and p+1. This only says that P lies anywhere between the 100th prime and 101st (prime-1). P could be anywhere between that limit. So not sufficient.

(2)There are P primes between 1 and 3912. Clearly countable and leads to a distinct number. So sufficient.

Ans: B

Statement (1) is saying that there are 100 primes in the range between 1 and P+1 is basically telling us that P is the 100th prime . We can determine the single numerical value of P. So it's sufficient.

So D is the answer (as statement 2 is also sufficient).

Re: NEW SET of good DS(4) [#permalink]
09 Nov 2009, 03:09

1

This post was BOOKMARKED

10. A number of oranges are to be distributed evenly among a number of baskets. Each basket will contain at least one orange. If there are 20 oranges to be distributed, what is the number of oranges per basket? (1) If the number of baskets were halved and all other conditions remained the same, there would be twice as many oranges in every remaining basket. (2) If the number of baskets were doubled, it would no longer be possible to place at least one orange in every basket.

I know it's dangerous to assume, but I interpreted the term "distributed evenly" as basically saying that each basket has to have the same number of oranges.

Statement 1

If so, then based on Statement 1 we can have the following:

# of oranges per basket / # of baskets 10 / 2 2 / 10

INSUFFICIENT

Statement 2

For Statement 2, the same rations would meet the requirement.

Re: NEW SET of good DS(4) [#permalink]
19 Dec 2009, 10:28

1) The possible numbers are 7,11,15,... Statement 1. Works for 15 but not 7,11 not sufficient. Statement 2. Works for 15 but not 7,11 Together they work for 15 but not 45. Hence E. _________________

Rock On

gmatclubot

Re: NEW SET of good DS(4)
[#permalink]
19 Dec 2009, 10:28

My last interview took place at the Johnson School of Management at Cornell University. Since it was my final interview, I had my answers to the general interview questions...