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Re: NEW SET of good DS(4) [#permalink]
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20 Feb 2010, 10:14
honeyrai wrote: Bunuel wrote:
9. Is x^2 equal to xy? (1) x^2  y^2 = (x+5)(y5) (2) x=y
Answer: B.
Hi Bunuel I have a problem with this question. x^2=xy > either x=0 or x=yNo doubt that second option is sufficient because it clearly states that x=y. The problem is with the first option. Now x^2y^2= (x+5)(y5) > (x+y)(xy)=(x+5)(y5) Case I: x+y = x+5 > y=5, now xy=y5 put, y=5>x=5. Therefore, (x,y)=(5,5) Case II: xy=x+5, > y=5, now x+y=y5 >x=5. Therefore, (x,y)=(5,5) In both the cases, x=y which is nothing but same as the second statement. Hence, answer should be D. The part highlighted is not to be considered TRUE...... You need to prove that! .. Hence don't use that to solve the question!
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Re: NEW SET of good DS(4) [#permalink]
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20 Feb 2010, 11:17
honeyrai wrote: Bunuel wrote:
9. Is x^2 equal to xy? (1) x^2  y^2 = (x+5)(y5) (2) x=y
Answer: B.
Hi Bunuel I have a problem with this question. x^2=xy > either x=0 or x=y No doubt that second option is sufficient because it clearly states that x=y. The problem is with the first option. Now x^2y^2= (x+5)(y5) > (x+y)(xy)=(x+5)(y5) Case I: x+y = x+5 > y=5, now xy=y5 put, y=5>x=5. Therefore, (x,y)=(5,5) Case II: xy=x+5, > y=5, now x+y=y5 >x=5. Therefore, (x,y)=(5,5) In both the cases, x=y which is nothing but same as the second statement. Hence, answer should be D. Question: is \(x^2=xy\)? > \(x(xy)=0\) > Equation holds true if \(x=0\) or/and \(x=y\). So, basically question asks: Is \(x=0\) or/and \(x=y\) true? Obviously statement (2) is sufficient, as it gives directly that \(x=y\). (1) \(x^2y^2 = (x + 5)(y  5)\) > \((x+y)(xy)=(x + 5)(y5)\) If \(y=5\) > \((x+5)(x5)=(x+5)(55)\) > \((x+5)(x5)=0\) > Either \(x=5=y\) and in this case answer to the question is YES OR \(x=5\), hence \(x\) is not equal to \(y\) (nor to zero) and in this case answer to the question is NO. So two different answers. Not sufficient. Answer: B.
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Re: NEW SET of good DS(4) [#permalink]
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20 Oct 2010, 21:43
Thanks for the explanation above. 9 did confuse me as well Great sets!!
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Re: NEW SET of good DS(4) [#permalink]
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29 Oct 2010, 04:18
whichscore wrote: 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) 6. How many perfect squares are less than the integer d? (1) 23 < d < 33 (2) 27 < d < 37 First of all: a perfect square is a number which is the square of an integer. So, perfect squares are: 0=0^2, 1=1^2, 4=2^2, 9=3^2, ... (1) 23 < d < 33 > if \(d>25\) (for example 26, 27, ...) then there will be 6 perfect square less then d: 0, 1, 4, 9, 16, and 25 BUT if \(d\leq25\) (for example 25 or 24) then there will be only 5 perfect square less then d: 0, 1, 4, 9, and 16. Not sufficient. (2) 27 < d < 37 > no matter what the value of d is there will be 6 perfect square less then d: 0, 1, 4, 9, 16, and 25 (even for max and min values of d). Sufficient. Answer: B.
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Re: NEW SET of good DS(4) [#permalink]
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22 Feb 2011, 06:34
This one can only be E, as Pablo and Al could have driven for 50 hours at a very low speed or for 5 hours at a very high speed.. This would change the relations. asterixmatrix wrote: 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 S5 will be the speed with which Al drove we get S*T + (T+1)* (S5) =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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08 Aug 2011, 03:09
Quote: 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.
(1) m and p must both be even since p is not a multiple of m, remainder can't be 0. Can it be 1. For the remainder to 1, p will have to be odd, which is not possible. So, the remainder will always be more than 1.
Sufficient.
(2) Possible values for m and p are 2,15 > Remainder: 1 Ans: No 3,10 > Remainder: 1 Ans: No 5,6 > Remainder: 1 Ans: No
It can't be 1,30 because 30 is a multiple of 1. Sufficient.
Ans: "D" (2) Possible values for m and p are 2,15 > Remainder: 1 Ans: No 3,10 > Remainder: 1 Ans: No 5,6 > Remainder: 1 Ans: No could be also use the combination: 6,10 > remainder = 4 {thus, making option B insufficient}
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Re: NEW SET of good DS(4) [#permalink]
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02 Oct 2011, 05:08
asterixmatrix wrote: 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 I am not sure A is the answer. Because m= 4 and p = 8 then the remainder r will be 0.... IMO it should be C. See my logic below.... S1: GCF is 2 means at least 2 is present in m and p Not suff for above mentioned reason. S2. LCM is 30, means 2x3X5 are the common factors between these m and p As you have said earlier not suff because m and p can be (5,6) (10,15) (3,10) and remainder is not always > 1 Combining 2 statements LCM X GCF = m X P 30 X 2 = m X p so 60 can be ( 1 X 60, 2 X 30, 3 X 20, 4 X 15, 5 X 12, 6 X 10) given the condition 2<m<p remainder is always greater than 1.



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Re: New questions: 1. When the positive integer x is divided by [#permalink]
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09 Mar 2012, 20:58
Quote: These may be the lousy way to solve the questions, but this is how I attempted to solve the questions.
The answers may be incorrect. I didn't match them with OA.
******************************************************************************************************** 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.
Soln:
Q: will "x/4" leave a remainder of 3?
(1) Rephrase: (x/3)/2= x/6 leaves remainder 1.
Values of x: 1,7,13,19 x%4 : 1,3 Remainder may be 3 or NOT 3. Not Sufficient. Hey Bunuel, Please let me know where I am going wrong. I also made the equation  (x/3)/2= x/6 leaves remainder 1 i.e x = 6a+ 6 However, as per your previous posts it should be 6a+3. What I am doing wrong. Please help Thanks
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Re: New questions: 1. When the positive integer x is divided by [#permalink]
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10 Mar 2012, 01:14
imhimanshu wrote: Quote: These may be the lousy way to solve the questions, but this is how I attempted to solve the questions.
The answers may be incorrect. I didn't match them with OA.
******************************************************************************************************** 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.
Soln:
Q: will "x/4" leave a remainder of 3?
(1) Rephrase: (x/3)/2= x/6 leaves remainder 1.
Values of x: 1,7,13,19 x%4 : 1,3 Remainder may be 3 or NOT 3. Not Sufficient. Hey Bunuel, Please let me know where I am going wrong. I also made the equation  (x/3)/2= x/6 leaves remainder 1 i.e x = 6a+ 6 However, as per your previous posts it should be 6a+3. What I am doing wrong. Please help Thanks Positive integer \(a\) divided by positive integer \(d\) yields a reminder of \(r\) can always be expressed as \(a=qd+r\), where \(q\) is called a quotient and \(r\) is called a remainder, note here that \(0\leq{r}<d\) (remainder is nonnegative integer and always less than divisor).Hence, when x/3 is divided by 2, the remainder is 1 should be expressed as x/3=2q+1 > x=6q+3. So, x can be 3, 9, 15, ... For all those values, x/3 (1, 3, 5, ...) divided by 2 yields the remainder of 1. As for your formula x can be 6, 12, 18, ... For those values, x/3 (2, 4, 6, ...) divided by 2 yields the remainder of 0 not 1. Hope it's clear.
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Re: New questions: 1. When the positive integer x is divided by [#permalink]
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10 Mar 2012, 05:02
Thanks Bunuel, I can see the mistake.
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Re: New questions: 1. When the positive integer x is divided by [#permalink]
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27 Mar 2012, 23:47
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.
its E
I. 7, 13, 19, 25...... (No./6 = remainder 1) II. 5,10,15,25....( No. divisible by 5)
I & II . 25, 55,85,.... (LCM of 5 & 6 =30)
hence 1 3 1 is pattern of remainders for I & II together......
Hence E



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Re: Collection of 12 DS questions [#permalink]
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28 Mar 2012, 02:31
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.
IMO A.
since m=2*x p=2* y and y should be greater than x hence p/m will always have remainder >1



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Re: Collection of 12 DS questions [#permalink]
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28 Mar 2012, 02:51
vdbhamare wrote: 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.
IMO A.
since m=2*x p=2* y and y should be greater than x hence p/m will always have remainder >1 OA's (answers) are given in this post: collectionof12dsquestions8544120.html#p642315Solution for #7: The integers m and p are such that 2<m<p, and m is not a factor of p. If r is the remainder when p is divided by m, is r>1?Given: \(2<m<p\) and \(\frac{p}{m}\neq{integer}\). \(p=xm+r\). q: \(r=?\) (1) The greatest common factor of m and p is 2 > \(p\) and \(m\) are even (as both have 2 as a factor) > even divided by even can give only even remainder (0, 2, 4, ...), since the remainder is not zero (as \(\frac{p}{m}\neq{integer}\)), then the remainder must be more than 1. Sufficient. (2) The least common multiple of m and p is 30 > if \(m=5\) and \(p=6\), then remainder=1 =1 and thus the answer to the question will be NO. BUT if \(m=10\) and \(p=15\), then remainder=5 >1 and thus the answer to the question will be YES. Two different answers. Not sufficient. Answer: A.
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Re: Collection of 12 DS questions [#permalink]
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08 Apr 2012, 12:21
Hey Bunuel,
I am a little confused by question 2.
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.
========================================================
CPrice of a computer P Price of a Printer xnumber of computers sold ynumber of printers sold
C=5p
were trying to find x*C/y*P. Well we know C/P=5. So x*C/y*P=(x/y)*5. All we need to find is the ratio of x/y.
Statement 1) x/y=3/2 in the first half of the year, ie the company sold 3z computers and 2z printers in the 1st half. x/y=2/1 in the second half of the year, ie the company sold 2z computers and 1z printers in the 2nd half. Hence, x/y=5/3 for the entire year. Shouldn't that be sufficient?
Is my error to assume that z is the same in both the first half and second half? The company may have sold computers 9 computers and 6 printers in the first (z=3), maintaing the 3/2 ratio, but in the second half they might have sold 4 computers and 2 printers (ie z=2). By changing the z for the first and second we get different # of computers and printers sold. Is this why I am wrong?
Thank you so much Bunuel!



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Re: Collection of 12 DS questions [#permalink]
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08 Apr 2012, 12:38
alphabeta1234 wrote: Hey Bunuel,
I am a little confused by question 2.
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.
========================================================
CPrice of a computer P Price of a Printer xnumber of computers sold ynumber of printers sold
C=5p
were trying to find x*C/y*P. Well we know C/P=5. So x*C/y*P=(x/y)*5. All we need to find is the ratio of x/y.
Statement 1) x/y=3/2 in the first half of the year, ie the company sold 3z computers and 2z printers in the 1st half. x/y=2/1 in the second half of the year, ie the company sold 2z computers and 1z printers in the 2nd half. Hence, x/y=5/3 for the entire year. Shouldn't that be sufficient?
Is my error to assume that z is the same in both the first half and second half? The company may have sold computers 9 computers and 6 printers in the first (z=3), maintaing the 3/2 ratio, but in the second half they might have sold 4 computers and 2 printers (ie z=2). By changing the z for the first and second we get different # of computers and printers sold. Is this why I am wrong?
Thank you so much Bunuel! Exactly. Company could have sold ANY number of computers and printers in the ratio of 3/2 in the first half of the year (3 and 2, 6 and 4, ..., 300 and 200, ...). Similarly it could have sold ANY number of computers and printers in the ratio of 2/1 in the second half of the year (2 and 1, 4 and 2, ..., 2,000,000 and 1,000,000 ...). So, infinitely many combinations of ratios are possible for the whole year, and 5 to 3 is just one of them.
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Re: Collection of 12 DS questions [#permalink]
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08 Apr 2012, 14:30
9. Is x^2 equal to xy? (1) x^2  y^2 = (x+5)(y5) (2) x=y
Answer: B.
I am confused. Statement one should be sufficient. We get x=y=5 or x=y=5. If x=y the statement should be sufficient. We cannot mix and match these two pairs, ie x=5 and y=5. If that were the case then statement 1 would be insufficient. x^2=x*y X=5=Y=5 > 25=(5)^2=(5)(5)=25 X=5=5> 25=(5)^2=(5)(5)=25
Both pairs satisfy the above statements. What am I doing wrong?



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Re: Collection of 12 DS questions [#permalink]
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08 Apr 2012, 14:48
alphabeta1234 wrote: 9. Is x^2 equal to xy? (1) x^2  y^2 = (x+5)(y5) (2) x=y
Answer: B.
I am confused. Statement one should be sufficient. We get x=y=5 or x=y=5. If x=y the statement should be sufficient. We cannot mix and match these two pairs, ie x=5 and y=5. If that were the case then statement 1 would be insufficient. x^2=x*y X=5=Y=5 > 25=(5)^2=(5)(5)=25 X=5=5> 25=(5)^2=(5)(5)=25
Both pairs satisfy the above statements. What am I doing wrong? Is x^2 equal to xy?(1) x^2 – y^2 = (x + 5)(y  5) (2) x = y Question: is \(x^2=xy\)? > \(x(xy)=0\) > Equation holds true if \(x=0\) or/and \(x=y\). So, basically question asks: Is \(x=0\) or/and \(x=y\) true? Obviously statement (2) is sufficient, as it gives directly that \(x=y\). (1) \(x^2y^2 = (x + 5)(y  5)\) > \((x+y)(xy)=(x + 5)(y5)\) If \(y=5\) > \((x+5)(x5)=(x+5)(55)\) > \((x+5)(x5)=0\) > Either \(x=5=y\) and in this case answer to the question is YES OR \(x=5\), hence \(x\) is not equal to \(y\) (nor to zero) and in this case answer to the question is NO. So two different answers. Not sufficient. Answer: B.
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Re: Collection of 12 DS questions [#permalink]
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26 Oct 2012, 05:23
Bunuel wrote: alphabeta1234 wrote: Hey Bunuel,
I am a little confused by question 2.
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.
========================================================
CPrice of a computer P Price of a Printer xnumber of computers sold ynumber of printers sold
C=5p
were trying to find x*C/y*P. Well we know C/P=5. So x*C/y*P=(x/y)*5. All we need to find is the ratio of x/y.
Statement 1) x/y=3/2 in the first half of the year, ie the company sold 3z computers and 2z printers in the 1st half. x/y=2/1 in the second half of the year, ie the company sold 2z computers and 1z printers in the 2nd half. Hence, x/y=5/3 for the entire year. Shouldn't that be sufficient?
Is my error to assume that z is the same in both the first half and second half? The company may have sold computers 9 computers and 6 printers in the first (z=3), maintaing the 3/2 ratio, but in the second half they might have sold 4 computers and 2 printers (ie z=2). By changing the z for the first and second we get different # of computers and printers sold. Is this why I am wrong?
Thank you so much Bunuel! Exactly. Company could have sold ANY number of computers and printers in the ratio of 3/2 in the first half of the year (3 and 2, 6 and 4, ..., 300 and 200, ...). Similarly it could have sold ANY number of computers and printers in the ratio of 2/1 in the second half of the year (2 and 1, 4 and 2, ..., 2,000,000 and 1,000,000 ...). So, infinitely many combinations of ratios are possible for the whole year, and 5 to 3 is just one of them. In 2003 Acme Computer’s price for each of its computers was five times the price for each of its printers. What was the ratio of its gross revenue from computers to its gross revenue from printers in 2003? (1) In the first half of 2003, Acme sold computers and printers in a ratio of 3:2; in the second half of 2003, Acme sold computers and printers in the ratio of 2:1. (2) Acme’s 2003 price for each of its computers was $1,000. Bunuel..i got that ans is E..but my question is. Assume..in statement 1..if it wud have been given only one ratio for whole year..like 2 ratio 1 ?? will it b sufficient..i tried it..i got the ans sufficent .. Thank u in Advance
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Re: Collection of 12 DS questions [#permalink]
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25 Apr 2013, 11:19
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.
Ans: D
In the first statement, whether it was excluding or including tax is not given.
So this 1 statement is not sufficient.




Re: Collection of 12 DS questions
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