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# Collection of 12 DS questions

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

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20 Oct 2009, 23:47
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scoregmat wrote:
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 ?

highlighted part is incorrect. However the answer is correct
the equation in 1 is x^2 - y^2 = (x+5)(y-5) which is equal to
x^2 - y^2 = xy -5x +5y -25 ----eqn1
for x^2 = xy value of x and y needs to be same.
On taking values as +/-5 for both x and y we can show the statement to be insuff
 e-GMAT Discount Codes Magoosh Discount Codes EMPOWERgmat Discount Codes
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Re: NEW SET of good DS(4) [#permalink]

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21 Oct 2009, 15:53
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I'll try the ones that haven't been attempted -

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.
: E
if x = 29, it satisfies both (1) and (2) and the remainder when divided by 8 is 5.
if x = 65, it also satisfies both (1) and (2) but the remainder when divided by 8 is 1.
So both are Insuff.

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.
: A
(A) implies that both m and p are even since they have 2 as a common factor. If thats the case, and given that m is not a factor of p means that the remainder will never be 0 or 1 so r > 1. So (A) is suff

(B) says that LCM(m,p) = 30. If m=3 and p=10, r =1; but if m = 6, p = 10, r = 4. So (B) is Insuff

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.
: A - Need know at what intervals the population doubles.
A says that in 2 hours the population quadrupled so every hour it doubles. Also, say population was x when it got divided 2 hours ago, then at 2 hours population is 4x so increase is 3x = 3750. Solve for x and can find out what the population will be 4 hours from now. Hence Suff.

B only says that the population reached 40000 cells with one hour remaining and says nothing about the rate. Depending on the rate, population can multiply to different levels. Hence not suff.
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Re: NEW SET of good DS(4) [#permalink]

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23 Oct 2009, 05:12
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Thanks for excellent questions Bunuel. It's truly appreciated. Since I really need practice on DS I tried to do them all. Unquestionably, I'm going to be wrong on some of them - so I look forward to seeing your answers.

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.

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. 1. A. (x/3)/2=k+1 <=> x=6k+3 <=> x= 9,15,21,... Plug in numbers: 9/4 gives remainder of 1, 15/4 gives remainder of 3 => Insufficient B. x/5 is an integer. Since x>0 that means that x=5k <=> x=5,10,15... Plug in numbers: 5/4 gives remainder of 1, 10/4 gives remainder of 2 => Insufficient C. Plug in numbers by finding mutual possible values for x, e.g.: 15, 45. 15/4 gives remainder of 3. 45/4 gives remainder of 1. => Insufficient So answer is E. 2. I have no idea what a gross revenue ratio is. Could someone tell me? 3. Lets formalize the information.The trucker paid a total of 155.76 including taxes of 10%, thus excluding taxes he paid 155.76/1.1 A. We are informed that the trucker paid$0.118 pr. gallon in taxes. Formalizing this
information we get that:
p*1.1 - p = $0.118 <=> p(1.1-1)=$0.118 <=> p= $0.118/0.1 = 1.18. To exclude taxes just divide by 1.18/1.1. Sufficient B. Formalize the information. We know the trucker paid a total of 155.76/1.1 excluding taxes. To get the price pr. gallon exluding taxes just divide by with 20. Sufficient Hence answer is D. 4. A. x/12 = k+5 <=> x=12k+60 <=> x=72, 84, 96,.... Plug in numbers: 72/8 gives a remainder of 0, 84/8 gives a remainder of 4. Insufficient B. x/18 = k+11 <=> x=18k+11 <=> x=29,47,65 Plug in numbers: 29/8 gives a remainder of 5, 47/8 gives a remainder of 1. Insufficient C. Since A will always be even while B will always be uneven and there are no mutual values. Insufficient Hence answer is E. 5.Remember the formula time*speed=distance <=> t*S=d A. We are only given relative information on two of the three drivers, hence Insufficient B. We are told that Marsha drives 9*50 = 450 < (1/3)*1500, so Marsha cannot have driven the longest distance. That leaves Al and Pablo and therefore this statement is Insufficient c. We know from (2) that Marsha cannot be the answer. The distance left is 1500-450 = 1050 miles. Moreover, we know from (1) that Al drove one hour longer than Pablo, 5 miles pr. hour slower. By formalizing information we have one equation in two unknowns which is Insufficient Answer is E. 6. A. Since there is a perfect square between 23 and 33 (25) the number of perfect squares cannot be determined. Insufficient B. Since there is a perfect square between 27 and 37 (26) the number of perfect squares cannot be determined. Insufficient C. Combining (1) and (2) we know that 27 < d < 33. There are no perfect squares in this interval and thus there are 5 perfect squares less than d (1, 4, 9, 16, 25). Sufficient So answer is C. 7. Formalize the information: 2 < m < p. p/m is not an integer. A. GCF of m and p is 2, implying they are even numbers. We know that m is not a factor of P. Plug in numbers: 6/4 gives remainder of 2, 6/10 gives remainder of 4. Insufficient B. LCM of m and p is 30. Break down 30 into its prime factors: 2,3,5. Since m < p and it is not a factor of p, it can be either 3, 5 implying that p can be 10 and 6. Plug in numbers: 10/3 gives a remainder of 1, 6/5 gives a remainder of 1. Thus the remainder r=1 and therefore r is not >1. Sufficient The answer is B. 8. We know that the population doubles at certain intervals and that it doubles (from an unknown quantity) at exactly 4 hours before being destroyed. A. We're told that the population has quadrupled in two hours. That means that the population has doubled 3 times in two hours and therefore it will double 6 times in 4 hours. Furthermore we are told that 4x - x = 3750, where x is population 2 hours ago <=> x=3750/3 = 1250. Thus the total population will be 1250*2^(3+6). Sufficient B. We cannot infer how frequent the population doubles using this information. Insufficient Answer is A. 9. Is x^2 equal to xy? (1) x^2 - y^2 = (x+5)(y-5) (2) x=y 9. Rewrite the expression X^2=xy <=> xx=xy <=> x=y, thus the question reduces to whether x=y. A. Rewrite (1): (x+y)(x-y) = (x+5)(y-5). This equation is solved for x=y=5, but not for x=y=0. Thus the information is insuficcient. B. Since the question reduced to whether x=y and we are told that x=y this is clearly Sufficient So answer is B. 10. Since there are 20 oranges to be distributed and all baskets contain one orange, there can be a maximum of 20 baskets. Furthermore since they are evenly distributed only these combinations (orange X baskets) remain: , 10*2, 5*4, 4*5, 2*10, and 1*20. A. This gives no additional information and is clearly Insufficient B. There can be either 20, 10, 5, or 4 baskets. Since there are 20 oranges there number of baskets can be doubled without problems for 10, 5 and 4 baskets. Thus there must be 20 baskets. Sufficient So answer is B. 11.Since P is a prime greater than 2 it must be odd. A. Since p+1 is even and >2 and thus is not a prime number, p must be the 99. (since we are excluding 2) prime number (have no idea what that is though). Sufficient B. Again this can be counted straightforward. Find out how many primes there are in the interval and subtract 1 (because of 2). Sufficient Answer is D. 12. The LCM of 6 and 9 is are both factors of 3, but e.g. 9 is not a factor of 2, so the LCM of 6 and 9 is 18. A. Since the LCM of x and 6 is 30, x must be a factor of 5. Thus the LCM is 18*5 = 90. Sufficient B. Since the LCM of x and 9 is 45, x must be a factor of 5. Thus the LCM is 18*5 = 90. Sufficient So answer is D. Math Expert Joined: 02 Sep 2009 Posts: 38848 Followers: 7721 Kudos [?]: 105952 [4] , given: 11602 Re: NEW SET of good DS(4) [#permalink] ### Show Tags 23 Oct 2009, 18:21 4 This post received KUDOS Expert's post 2 This post was BOOKMARKED ANSWERS (OAs): As most of the problems was solved correctly, I'm posting only OAs. Please let me know if anyone needs any clarification. 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. Answer: E. 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.

Answer: E

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.

Answer: D.

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.

Answer: E.

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

Answer: E.

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

Answer: B.

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.

Answer: A.

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.

Answer: C.

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

Answer: B.

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.

Answer: B.

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.

Answer: D.

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.

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

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02 Nov 2009, 02:02
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Hello.

I just wonder whether the question No.8 is A or C.

I got A. May anyone explain to me if C is the right answer?

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

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02 Nov 2009, 20:33
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Talinhuu wrote:
Hello.

I just wonder whether the question No.8 is A or C.

I got A. May anyone explain to me if C is the right answer?

thx

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.

This one is really tough.

Before considering the statements let's look at the stem:
A. Population doubles at constant intervals, but we don't know that intervals.
B. Experiment will end in 4 hours from now.
C. We don't know when bacteria divided last time, how many minutes ago.

(1) Population divided 2 hours ago and increased by 3750 cells. Note that this statement is talking that bacteria quadrupled during 2 hours before NOW. So, starting point 2 hours ago, end of experiment 4 hours from now. Total 6 hours.
This statement gives ONLY the following info:
A. population of bacteria TWO hours ago - 1250.
B. population of bacteria now - 5000.

But we still don't know the interval of division. It can be 45 min, meaning that bacteria divided second time 30 min ago OR it can be 1 hour, meaning that bacteria just divided. Not sufficient.

(2) An hour before the end of experiment bacteria will double 40.000. Clearly insufficient.

(1)+(2) We can conclude that in 5 hours (2 hours before now+3 hours from now) population of bacteria will increase from 1250 to 40.000, will divide 5 times, so interval is 1 hour. The population will contain 40.000*2=80.000 cells when the bacteria is destroyed. Sufficient.

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

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02 Nov 2009, 23:54
Bunuel wrote:

Answer: C.

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.

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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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:

# of oranges per basket / # of baskets
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

Answer: E?
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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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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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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]

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19 Dec 2009, 11:35
2)

Price of Computer = CP
Price of printer = PP
Number of computers sold = CN
Number of Printers sold = PN

We need CP*CN/PP*PN
Given CP=5PP
Now we need 5*CN/PN

Using Statement 1 CN = 3x+2y & PN=2x+y
We can't get the ratio from this.

Statement 2 is irrelevant as the price factor has been knocked off our equation, hence E.
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Re: NEW SET of good DS(4) [#permalink]

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19 Dec 2009, 11:43
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chetan2u wrote:
hi bunuel,
it seems the OA have already been given.. however i differ on one ans..i.e. 8..
i feel A is sufficient... SI tells us that the cells doubled two hour back and now are four times, ... it means that cells have doubled twice ..x then 2x... then 4x so they are doubling once every hr..
(although the SII doesnt have to relate to SI..... but just to check if it doubles once every hr..... change in the no 4x-x=3750 ..x=1250 so str presently is 5000... after 1 hr 10000..2 hr-20000...3hr-40000.... that is what SII tells) not a part of sol although..

I think this one was the toughest. I've already given the solution for this one previously, so here it is:

Before considering the statements let's look at the stem:
A. Population doubles at constant intervals, but we don't know that intervals.
B. Experiment will end in 4 hours from now.
C. We don't know when bacteria divided last time, how many minutes ago.

(1) Population divided 2 hours ago and increased by 3750 cells. Note that this statement is talking that bacteria quadrupled during 2 hours before NOW. So, starting point 2 hours ago, end of experiment 4 hours from now. Total 6 hours.
This statement gives ONLY the following info:
A. population of bacteria TWO hours ago - 1250.
B. population of bacteria now - 5000.

But we still don't know the interval of division. It can be 45 min, meaning that bacteria divided second time 30 min ago OR it can be 1 hour, meaning that bacteria just divided. Not sufficient.

(2) An hour before the end of experiment bacteria will double 40.000. Clearly insufficient.

(1)+(2) We can conclude that in 5 hours (2 hours before now+3 hours from now) population of bacteria will increase from 1250 to 40.000, will divide 5 times, so interval is 1 hour. The population will contain 40.000*2=80.000 cells when the bacteria is destroyed. Sufficient.

Answer: C.

The point is that from (1) we can not say what the interval of division is, hence it's not sufficient.
Please, tell me if you find this explanation not convincing and I'll try to answer your doubts.
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Re: NEW SET of good DS(4) [#permalink]

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19 Dec 2009, 11:45
thanx bunuel.... i missed the pt
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Re: NEW SET of good DS(4) [#permalink]

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19 Dec 2009, 11:46
3)
Let Total price of gas = P & N is number of gallons filled.
We need P/N

Given 1.1P=155.76
P can be calculated
What is N?

Statement 1)Given 0.1P/N=.118
P/N = 1.18 Sufficient (Note you don't even need to use the equation given in the question)

Statement 2 Gives N hence sufficient.

Answer is D.
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Re: NEW SET of good DS(4) [#permalink]

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19 Dec 2009, 11:53
4)

Statement 1)
Possible numbers 17,29,41,53,65
All give different remainders when divided by 8.

Statement 2)
Possible numbers 29,47,65
All give different remainders when divided by 8.

Together
29,65
All give different remainders when divided by 8.

Hence E.
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Re: NEW SET of good DS(4)   [#permalink] 19 Dec 2009, 11:53

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