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# Collection of 12 DS questions

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Manager
Joined: 12 Oct 2009
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Re: NEW SET of good DS(4) [#permalink]

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18 Oct 2009, 09:03
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.

C.

45/4 will give remainder 1

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Re: NEW SET of good DS(4) [#permalink]

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18 Oct 2009, 15:03
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

1) x^2 - y^2 = x^2 - 25
=> - y^2 = -25...=> y^2 = 25....y = 5.....Insuff...

2) x = y....Suff..

B...Is there some trap ?

Last edited by scoregmat on 18 Oct 2009, 15:15, edited 1 time in total.

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Manager
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Re: NEW SET of good DS(4) [#permalink]

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18 Oct 2009, 15:15
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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)

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Re: NEW SET of good DS(4) [#permalink]

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18 Oct 2009, 15:21
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 ?

E?

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Re: NEW SET of good DS(4) [#permalink]

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18 Oct 2009, 15:28
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.

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Re: NEW SET of good DS(4) [#permalink]

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18 Oct 2009, 20:31
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.

C.

45/4 will give remainder 1

Thanks. Thats a good catch..
_________________

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Re: NEW SET of good DS(4) [#permalink]

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19 Oct 2009, 08:33
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)

This is how I came up with B also

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Re: NEW SET of good DS(4) [#permalink]

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19 Oct 2009, 10:34
Bunuel wrote:
New questions:

12. If x is a positive integer, what is the least common multiple of x, 6, and 9?
(1) The LCM of x and 6 is 30.
(2) The LCM of x and 9 is 45.

Also you can check new set of PS problems: new-set-of-good-ps-85440.html

Prime factors of 6 are 2 & 3; and prime factors of 9 are 3 & 3

1) 30 has prime factors of 2,5, & 3
6 has prime factors of 2 & 3

Therefore X must have one 5, one or zero 2s, and one or zero 3s.

Possibilities of X are 30 (2 x 5 x 3); 15 (5 x 3), & 10 (2 x 5)

We know LCM has to have one 5, two 3s and one 2. So LCM is 5 x 3 x 3 x 2 = 90

2) 45 has prime factors of 5, 3, 3
9 has prime factors of 3 & 3

X must have one 5 and could have two 3s, one 3, or zero 3s

Possiblities are 45 (5 x 3 x 3); 15 (5 x 3); and 5

We know LCM has to have two 3s, one 2, and one 5.

So LCM is 3 x 3 x 2 x 5 = 90

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Manager
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Re: NEW SET of good DS(4) [#permalink]

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20 Oct 2009, 23:09
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

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Re: NEW SET of good DS(4) [#permalink]

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20 Oct 2009, 23:20
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

will go with A

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Re: NEW SET of good DS(4) [#permalink]

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20 Oct 2009, 23:39
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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

will go with B

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Re: NEW SET of good DS(4) [#permalink]

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02 Nov 2009, 23:54
Bunuel wrote:

Thank you buddy.

Thanks a lot!

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Re: NEW SET of good DS(4) [#permalink]

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07 Nov 2009, 22:02
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)

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Re: NEW SET of good DS(4) [#permalink]

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07 Nov 2009, 22:32
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.

Also you can check new set of PS problems: new-set-of-good-ps-85440.html

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

D

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Re: NEW SET of good DS(4) [#permalink]

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08 Nov 2009, 02:14
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).
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Re: NEW SET of good DS(4) [#permalink]

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08 Nov 2009, 04:25
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).

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Re: NEW SET of good DS(4) [#permalink]

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09 Nov 2009, 04:09
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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:

10 / 2
2 / 10

INSUFFICIENT

Statement 2

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

INSUFFICIENT

Taken together, they are still insufficient... so

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Re: NEW SET of good DS(4) [#permalink]

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19 Dec 2009, 10:55
first the ans ..(1)E(2)E(3)D(4)E(5)E(6)B(7)A(8)A(9)B(10)B(11)D(12)D...
almost all have been tried, ill try the ones not yet tried..
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Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

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Re: NEW SET of good DS(4) [#permalink]

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19 Dec 2009, 11:06
i think gone wrong on 8.... it shud be C
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Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

Last edited by chetan2u on 19 Dec 2009, 11:43, edited 1 time in total.

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Re: NEW SET of good DS(4) [#permalink]

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19 Dec 2009, 11:28
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.
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Re: NEW SET of good DS(4)   [#permalink] 19 Dec 2009, 11:28

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