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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

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)

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 ?

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.

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.

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

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)

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

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

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

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)

(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

(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). _________________

(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).

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.

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, 11:28

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