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Collection of 12 DS questions

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Math Expert
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Collection of 12 DS questions [#permalink]

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17 Oct 2009, 18:45
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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.

2. In 2003 Acme Computer priced its computers five times higher than its printers. What is the ratio of its gross revenue for computers and printers respectively in the year 2003?
(1) In the first half of 2003 it sold computers and printers in the ratio of 3:2, respectively, and in the second half in the ratio of 2:1.
(2) It sold each computer for \$1000.

3. Last Tuesday a trucker paid \$155.76, including 10 percent state and federal taxes, for diesel fuel. What was the price per gallon for the fuel if the taxes are excluded?
(1) The trucker paid \$0.118 per gallon in state and federal taxes on the fuel last Tuesday.
(2) The trucker purchased 120 gallons of the fuel last Tuesday.

4. What is the remainder when the positive integer x is divided by 8?
(1) When x is divided by 12, the remainder is 5.
(2) When x is divided by 18, the remainder is 11.

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

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

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.

8. A scientist is studying bacteria whose cell population doubles at constant intervals, at which times each cell in the population divides simultaneously. Four hours from now, immediately after the population doubles, the scientist will destroy the entire sample. How many cells will the population contain when the bacteria is destroyed?
(1) Since the population divided two hours ago, the population has quadrupled, increasing by 3,750 cells.
(2) The population will double to 40,000 cells with one hour remaining until the scientist destroys the sample.

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

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.

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.

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.

As always please share your way of thinking.

OA's (answers) are given in this post: collection-of-12-ds-questions-85441-20.html#p642315

Also you can check new set of PS problems: new-set-of-good-ps-85440.html
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Re: NEW SET of good DS(4) [#permalink]

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17 Oct 2009, 19:57
2
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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 1.
If m = 15, k = 12 and x = 75. Then reminder is 3.

E..
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17 Oct 2009, 20:18
Bunuel wrote:
4. What is the remainder when the positive integer x is divided by 8?

(1) When x is divided by 12, the remainder is 5.
(2) When x is divided by 18, the remainder is 11.

(1) x = 12k+5 ........ k is an integer that could be 1 or 2 or so on........NSF
(2) x = 18m+11 ........ m is an integer that could be 1 or 2 or so on......NSF

1 and 2:
12k + 5 = 18m + 11...............where k>m.
12k = 18m + 6
2k = 3m + 1

If k = 2, m = 1. x = 29 and reminder is 5.
If k = 5, m = 3. x = 60 and r = 4
If k = 8, m = 5................. r = 0. NSF..

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

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17 Oct 2009, 20:35
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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)/2 = k+1 where k is an integer
x = 6k + 6
x is an even integer; so x/4 has never 3 reminder. Suff...

When x/3 is divided by 2, the remainder is 1, should be "translated" x/3=2k+1 --> x=6k+3 and not x=6k+6
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Re: NEW SET of good DS(4) [#permalink]

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18 Oct 2009, 01:10
Bunuel 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)/2 = k+1 where k is an integer
x = 6k + 6
x is an even integer; so x/4 has never 3 reminder. Suff...

When x/3 is divided by 2, the remainder is 1, should be "translated" x/3=2k+1 --> x=6k+3 and not x=6k+6

will go with E
1. we can take x = 63 then (x/3)/2 = 21/2 leaves remainder 1 and remainder from 63/4 is 3
if we take x=45 then (x/3)/2 = 15/2 leaves a remainder 1 and remainder from 45/4 is 1
hence insuff
2. if x is 15 then its divisible by 5 and remainder is 3 (15/4 = 3 is remainder) but if x = 30 then remainder is 2 (30/4 gives 2 as remainder and divisible by 5)

if we take both then (x/3)/2 should give remainder 1 and x is divisible by 5
let x = 15 then it satisfies both the conditions and x/4 will leave remainder 3
let x= 45 then again both conditions are satisfied but x/4 will leave remainder 1 . Hence insuff

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

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18 Oct 2009, 01:40
Bunuel wrote:
New questions:

2. In 2003 Acme Computer priced its computers five times higher than its printers. What is the ratio of its gross revenue for computers and printers respectively in the year 2003?
(1) In the first half of 2003 it sold computers and printers in the ratio of 3:2, respectively, and in the second half in the ratio of 2:1.
(2) It sold each computer for \$1000.

3. Last Tuesday a trucker paid \$155.76, including 10 percent state and federal taxes, for diesel fuel. What was the price per gallon for the fuel if the taxes are excluded?
(1) The trucker paid \$0.118 per gallon in state and federal taxes on the fuel last Tuesday.
(2) The trucker purchased 120 gallons of the fuel last Tuesday.

As always please share your way of thinking.

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

2. Let cost of printer be x then cost of computer is 5x
option 1 -
given Nc1/Np1 = 3/2 and Nc2/Np2 = 2/1 (Nc1,Nc2,Np1 and Np2 are number of computers and printers sold in first and second half of the year)
here we do not know the exact price nor the numbers in each of 6months.
the ratios can be any number
Hence insuff
option2- cost of computer is 1000 we can calculate price of printer to be 200 but no further info on the number of units sold. hence insuff

taking both we have cost of computer and printer and ratios as well . Let ratio for first half be 3x/2x and for second half be 2y/y. Taking this also we will not be able to determine the gross revenue.hence E

3.given with tax the trucker paid \$ 155.76 . The value of diesel without tax will come out to be \$ 141.6.
1. Let number of gallons purchased is N. Then we have
(10/100)*141.6 = 0.118 *N we get N =120.
from this we can calculate price per gallon without tax. hence suff
2. number of gallons is given as 120 and we can calculate price per gallon without tax. hence suff
Hence D

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

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18 Oct 2009, 08:34
Bunuel 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)/2 = k+1 where k is an integer
x = 6k + 6
x is an even integer; so x/4 has never 3 reminder. Suff...

When x/3 is divided by 2, the remainder is 1, should be "translated" x/3=2k+1 --> x=6k+3 and not x=6k+6

Updated................
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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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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, 07:51
1
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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

combining both we x = 15,45,75....Still nt suff

So I will go with option E

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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.

As always please share your way of thinking.

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

Answer is D... I think

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

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19 Oct 2009, 10:38
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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.

As always please share your way of thinking.

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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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] 20 Oct 2009, 23:39

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