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The Discreet Charm of the DS [#permalink]
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I'm posting the next set of medium/hard DS questions. I'll post OA's with detailed explanations after some discussion. Please, post your solutions along with the answers. Good luck!1. Bonnie can paint a stolen car in x hours, and Clyde can paint the same car in y hours. They start working simultaneously and independently at their respective constant rates at 9:45am. If both x and y are odd integers, is x=y?(1) x^2+y^2<12 (2) Bonnie and Clyde complete the painting of the car at 10:30am Solution: thediscreetcharmoftheds12696220.html#p10396332. Is xy<=1/2? (1) x^2+y^2=1 (2) x^2y^2=0 Solution: thediscreetcharmoftheds12696220.html#p10396343. If a, b and c are integers, is abc an even integer?(1) b is halfway between a and c (2) a = b  c Solution: thediscreetcharmoftheds12696240.html#p10396374. How many numbers of 5 consecutive positive integers is divisible by 4?(1) The median of these numbers is odd (2) The average (arithmetic mean) of these numbers is a prime number Solution: thediscreetcharmoftheds12696240.html#p10396455. What is the value of integer x?(1) 2x^2+9<9x (2) x+10=2x+8 Solution: thediscreetcharmoftheds12696240.html#p10396506. If a and b are integers and ab=2, is a=2?(1) b+3 is not a prime number (2) a>b Solution: thediscreetcharmoftheds12696240.html#p10396517. A certain fruit stand sold total of 76 oranges to 19 customers. How many of them bought only one orange?(1) None of the customers bought more than 4 oranges (2) The difference between the number of oranges bought by any two customers is even Solution: thediscreetcharmoftheds12696240.html#p10396558. If x=0.abcd, where a, b, c and d are digits from 0 to 9, inclusive, is x>7/9?(1) a+b>14 (2) ac>6 Solution: thediscreetcharmoftheds12696240.html#p10396629. If x and y are negative numbers, is x<y?(1) 3x + 4 < 2y + 3 (2) 2x  3 < 3y  4 Solution: thediscreetcharmoftheds12696240.html#p103966510. The function f is defined for all positive integers a and b by the following rule: f(a,b)=(a+b)/GCF(a,b), where GCF(a,b) is the greatest common factor of a and b. If f(10,x)=11, what is the value of x?(1) x is a square of an integer (2) The sum of the distinct prime factors of x is a prime number. Solution: thediscreetcharmoftheds12696240.html#p103967111. If x and y are integers, is x a positive integer?(1) x*y is a prime number. (2) x*y is nonnegative integer. Solution: thediscreetcharmoftheds12696240.html#p103967812. If 6a=3b=7c, what is the value of a+b+c?(1) ac=6b (2) 5b=8a+4c Solution: thediscreetcharmoftheds12696240.html#p1039680
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Re: The Discreet Charm of the DS [#permalink]
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06 Feb 2012, 07:42
Bunuel wrote: nhemdani wrote: Bunuel wrote: 6. If a and b are integers and ab=2, is a=2?
Notice that we are not told that a and b are positive.
There are following integer pairs of (a, b) possible: (1, 2), (1, 2), (2, 1) and (2, 1). Basically we are asked whether we have the third case.
(1) b+3 is not a prime number > rules out 1st and 4th options. Not sufficient. (2) a>b > again rules out 1st and 4th options. Not sufficient.
(1)+(2) Still two options are left: (1, 2) and (2, 1). Not sufficient.
Answer: E. Hi, I was looking at this solution and had a question. If b+3 is not a prime number, only when a number in our set equals 1, in this case the sum os 4 (1+3= 4 (Prime), 2+3, 1+3, 2+3 are all prime numbers), shouldnt the answer be A, as we are only left with option (1,2) so YES a=2 ??? 1 is not a prime number, so the case of (1, 2) is still possible: 2+3=1 not a prime. Hope it's clear. Sorry I got excited Makes sense



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Re: The Discreet Charm of the DS [#permalink]
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06 Feb 2012, 08:43
Hey Bunuel can I ask a question for 12?
We know 6a=3b
And for statement one:
ac =6b. Can't 6b =12a
Then it becomes ac=12a ==> c=12. I know it's wrong since if a is 0 then they will be equal regardless, but can you explain why what I did was wrong?



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Re: The Discreet Charm of the DS [#permalink]
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06 Feb 2012, 09:24
kys123 wrote: Hey Bunuel can I ask a question for 12?
We know 6a=3b
And for statement one:
ac =6b. Can't 6b =12a
Then it becomes ac=12a ==> c=12. I know it's wrong since if a is 0 then they will be equal regardless, but can you explain why what I did was wrong? Also, 6a = 3b = 7c Can we say a/b= 1/2, b/c = 7/3, and a/c = 7/6 a) ac = 6b, therefore c = 6b/a substituting this in b/c => b / (6b/a) = 7/3 => a =14, b=28, c = 12 Isnt A also sufficient? Am I ignoring something?



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Re: The Discreet Charm of the DS [#permalink]
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06 Feb 2012, 09:54
kys123 wrote: Hey Bunuel can I ask a question for 12?
We know 6a=3b
And for statement one:
ac =6b. Can't 6b =12a
Then it becomes ac=12a ==> c=12. I know it's wrong since if a is 0 then they will be equal regardless, but can you explain why what I did was wrong? ac=12a (here you can not reduce by a and write c=12 as you exclude possibility of a=0) > a(c12)=0 > either a=0 OR c=12. So, we get either a=b=c=0 or a=14, b=28 and c=12. nhemdani wrote: Also,
6a = 3b = 7c
Can we say a/b= 1/2, b/c = 7/3, and a/c = 7/6 a) ac = 6b, therefore c = 6b/a substituting this in b/c => b / (6b/a) = 7/3 => a =14, b=28, c = 12
Isnt A also sufficient? Am I ignoring something? Your doubt is partially addressed above, though there is another thing: from 6a = 3b you can not write a/b=1/2 because b can be zero and we can not divide by zero. The same for other ratios you wrote. Hope it's clear.
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Re: The Discreet Charm of the DS [#permalink]
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06 Feb 2012, 20:52
Bunuel wrote: 12. If 6a=3b=7c, what is the value of a+b+c?
Given: \(6a=3b=7c\) > least common multiple of 6, 3, and 7 is 42 hence we ca write: \(6a=3b=7c=42x\), for some number \(x\) > \(a=7x\), \(b=14x\) and \(c=6x\).
(1) ac=6b > \(7x*6x=6*14x\) > \(x^2=2x\) > \(x=0\) or \(x=2\). Not sufficient.
(2) 5b=8a+4c > \(5*14x=8*7x+4*14x\) > \(70x=80x\) > \(10x=0\) > \(x=0\) > \(a=b=c=0\) > \(a+b+c=0\). Sufficient.
Answer: B. I did this for option 1  ac=6b ac=2 * 3 * b since 6a=3b=7c a c = 2 * 6a c = 12 now we can find a and b also, so seems sufficient. so where am i going wrong?



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Re: The Discreet Charm of the DS [#permalink]
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07 Feb 2012, 00:11
Well cause there is 2 options for statement 1.
ac = 3b ==> ac = 12a Now if a was 0 then 0*12 = 0*c ==> 6*0=3*0=7*0 a+b+c= 0 or
12a = ca c=12 7*12(c)=3*28(b)=6*14(a) a+b+c= 54.
There is 2 possible solutions, so you do not know if it's 0 or 54



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Re: The Discreet Charm of the DS [#permalink]
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09 Feb 2012, 02:12
HI Bunuel,
I personally dont feel very comfortable with your solution for Q9. Just not very intuitive for me.
I tried to solve it graphically but failed. Can u please help.
Thanks in advance.
Note: I really enjoyed doing these set of questions. U r taking GMAT club Quant practice questions to a next level alltogether. Thanks



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Re: The Discreet Charm of the DS [#permalink]
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24 Feb 2012, 14:52
I have seen references to use your guides in many threads, Bunuel. Now I can see why. I will be sure to use your challenge sets in the coming weeks to hopefully boost my quant score into the 4850 range. Thank you so much for your invaluable contributions for relative GMAT newbies, like myself.



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Re: The Discreet Charm of the DS [#permalink]
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25 Feb 2012, 05:27
All,
7. A certain fruit stand sold total of 76 oranges to 19 customers. How many of them bought only one orange? (1) None of the customers bought more than 4 oranges (2) The difference between the number of oranges bought by any two customers is even
In respect to the question above, I assumed that any two of those 19 customer might have bought 5 & 3 oranges and hence I, marked the option insufficient. Bunnel have equated and treated the option in totally different way. I ,lack the skill to convert these sort of condition in to equation.
please can some post or point to the list of similar Word translation sentences and how to convert them in to equation. Im very new to GMAT club so please forgie me if this is the repeated posting.
Thanks, Vids



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Re: The Discreet Charm of the DS [#permalink]
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25 Feb 2012, 05:53
vidhya16 wrote: All,
7. A certain fruit stand sold total of 76 oranges to 19 customers. How many of them bought only one orange? (1) None of the customers bought more than 4 oranges (2) The difference between the number of oranges bought by any two customers is even
In respect to the question above, I assumed that any two of those 19 customer might have bought 5 & 3 oranges and hence I, marked the option insufficient. Bunnel have equated and treated the option in totally different way. I ,lack the skill to convert these sort of condition in to equation.
please can some post or point to the list of similar Word translation sentences and how to convert them in to equation. Im very new to GMAT club so please forgie me if this is the repeated posting.
Thanks, Vids I did not use any equation for this question. Statement (2) says: the difference between the number of oranges bought by ANY two customers is even > in order the difference between ANY number of oranges bought to be even, either all customers must have bought odd number of oranges or all customers must have bough even number of oranges. Now, the sum of 19 odd integers is odd and we have that fruit stand sold total of 76, so even number of oranges, which means that the case where all customers buy odd number of oranges is not possible. And since 1 is odd then no one bought only one orange. Sufficient. As for word translation check this: wordproblemsmadeeasy87346.htmlHope it helps.
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Re: The Discreet Charm of the DS [#permalink]
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24 Mar 2012, 09:38
Bunuel wrote: SOLUTIONS:
1. Bonnie can paint a stolen car in x hours, and Clyde can paint the same car in y hours. They start working simultaneously and independently at their respective constant rates at 9:45am. If both x and y are odd integers, is x=y?
(2) Bonnie and Clyde complete the painting of the car at 10:30am > they complete the job in 3/4 of an hour (45 minutes), since it's neither an integer nor integer/2 then \(x\) and \(y\) are not equal. Sufficient.
Answer: B. please consider adding " working together" to the stmt 2 as I deciphered they both worked separately and ended at the same time so cool enough (though same answer but got carried away by the wording) Thanks once again for nice collection!!
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Re: The Discreet Charm of the DS [#permalink]
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24 Mar 2012, 15:08
yogesh1984 wrote: Bunuel wrote: SOLUTIONS:
1. Bonnie can paint a stolen car in x hours, and Clyde can paint the same car in y hours. They start working simultaneously and independently at their respective constant rates at 9:45am. If both x and y are odd integers, is x=y?
(2) Bonnie and Clyde complete the painting of the car at 10:30am > they complete the job in 3/4 of an hour (45 minutes), since it's neither an integer nor integer/2 then \(x\) and \(y\) are not equal. Sufficient.
Answer: B. please consider adding " working together" to the stmt 2 as I deciphered they both worked separately and ended at the same time so cool enough (though same answer but got carried away by the wording) Thanks once again for nice collection!! Thank you for the suggestion.
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14 May 2012, 16:40
Bunuel wrote: 5. What is the value of integer x?(1) 2x^2+9<9x > factor qudratics: \((x\frac{3}{2})(x3)<0\) > roots are \(\frac{3}{2}\) and 3 > "<" sign indicates that the solution lies between the roots: \(1.5<x<3\) > since there only integer in this range is 2 then \(x=2\). Sufficient. (2) x+10=2x+8 > LHS is an absolute value, which is always non negative, hence RHS must also be nonnegative: \(2x+8\geq{0}\) > \(x\geq{4}\), for this range \(x+10\) is positive hence \(x+10=x+10\) > \(x+10=2x+8\) > \(x=2\). Sufficient. Answer: D. Check this for more on solving inequalities like the one in the first statement: x24x94661.html#p731476 inequalitiestrick91482.htmleverythingislessthanzero108884.html?hilit=extreme#p868863xyplane71492.html?hilit=solving%20quadratic#p841486Hope it helps. Hey Bunuel, Two questions. 1) How did you factor 2x^2+9<9x (ie 2x^29x+9<0) so quickly? I always struggle with factoring polynomials in which a coefficient other than 1 is on the x^2. Did you use the quadratic formula? I am interested in knowing if there is a quicker way than the quadratic formula method. 2) Once you determined that 1.5 and 3 were the roots of the equation, how did you figure that the solution was in between 1.5 and 3 from just looking at the sign "<"?? I used the dumb method of just plugging values that lie from (infinity , 1.5), (1.5, 3) and (3, +infinity). How did you know the sign "<" told you the solution was in in between (1.5,3)? Many thanks Bunuel! Your my hero dude!



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Re: The Discreet Charm of the DS [#permalink]
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14 May 2012, 23:19
alphabeta1234 wrote: Bunuel wrote: 5. What is the value of integer x?(1) 2x^2+9<9x > factor qudratics: \((x\frac{3}{2})(x3)<0\) > roots are \(\frac{3}{2}\) and 3 > "<" sign indicates that the solution lies between the roots: \(1.5<x<3\) > since there only integer in this range is 2 then \(x=2\). Sufficient. (2) x+10=2x+8 > LHS is an absolute value, which is always non negative, hence RHS must also be nonnegative: \(2x+8\geq{0}\) > \(x\geq{4}\), for this range \(x+10\) is positive hence \(x+10=x+10\) > \(x+10=2x+8\) > \(x=2\). Sufficient. Answer: D. Check this for more on solving inequalities like the one in the first statement: x24x94661.html#p731476 inequalitiestrick91482.htmleverythingislessthanzero108884.html?hilit=extreme#p868863xyplane71492.html?hilit=solving%20quadratic#p841486Hope it helps. Hey Bunuel, Two questions. 1) How did you factor 2x^2+9<9x (ie 2x^29x+9<0) so quickly? I always struggle with factoring polynomials in which a coefficient other than 1 is on the x^2. Did you use the quadratic formula? I am interested in knowing if there is a quicker way than the quadratic formula method. 2) Once you determined that 1.5 and 3 were the roots of the equation, how did you figure that the solution was in between 1.5 and 3 from just looking at the sign "<"?? I used the dumb method of just plugging values that lie from (infinity , 1.5), (1.5, 3) and (3, +infinity). How did you know the sign "<" told you the solution was in in between (1.5,3)? Many thanks Bunuel! Your my hero dude! 1. Solving and Factoring Quadratics: http://www.purplemath.com/modules/solvquad.htmhttp://www.purplemath.com/modules/factquad.htm2. Solving inequalities: x24x94661.html#p731476 (Check this first) inequalitiestrick91482.htmleverythingislessthanzero108884.html?hilit=extreme#p868863xyplane71492.html?hilit=solving%20quadratic#p841486Hope it helps.
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Re: The Discreet Charm of the DS [#permalink]
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15 May 2012, 11:39
Bunuel wrote: 5. What is the value of integer x?
(1) 2x^2+9<9x > factor qudratics: \((x\frac{3}{2})(x3)<0\) > roots are \(\frac{3}{2}\) and 3 > "<" sign indicates that the solution lies between the roots: \(1.5<x<3\) > since there only integer in this range is 2 then \(x=2\). Sufficient.
(2) x+10=2x+8 > LHS is an absolute value, which is always non negative, hence RHS must also be nonnegative: \(2x+8\geq{0}\) > \(x\geq{4}\), for this range \(x+10\) is positive hence \(x+10=x+10\) > \(x+10=2x+8\) > \(x=2\). Sufficient.
Answer: D.
Hope it helps. Hi bunuel, Isn't x+10=2x+8 be written as Either x+10=2x+8 or x+10=(2x+8) ? and then this should be solved? Please help on this one.



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Re: The Discreet Charm of the DS [#permalink]
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15 May 2012, 11:43
piyushksharma wrote: Bunuel wrote: 5. What is the value of integer x?
(1) 2x^2+9<9x > factor qudratics: \((x\frac{3}{2})(x3)<0\) > roots are \(\frac{3}{2}\) and 3 > "<" sign indicates that the solution lies between the roots: \(1.5<x<3\) > since there only integer in this range is 2 then \(x=2\). Sufficient.
(2) x+10=2x+8 > LHS is an absolute value, which is always non negative, hence RHS must also be nonnegative: \(2x+8\geq{0}\) > \(x\geq{4}\), for this range \(x+10\) is positive hence \(x+10=x+10\) > \(x+10=2x+8\) > \(x=2\). Sufficient.
Answer: D.
Hope it helps. Hi bunuel, Isn't x+10=2x+8 be written as Either x+10=2x+8 or x+10=(2x+8) ? and then this should be solved? Please help on this one. We goth that x is more than or equal to 4. Now, for this range x+10>0 so x+10 expands only as x+10 (x+10=x+10).
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15 May 2012, 12:16
Bunuel wrote: 9. If x and y are negative numbers, is x<y?
(1) 3x + 4 < 2y + 3 > \(3x<2y1\). \(x\) can be some very small number for instance 100 and \(y\) some large enough number for instance 3 and the answer would be YES, \(x<y\) BUT if \(x=2\) and \(y=2.1\) then the answer would be NO, \(x>y\). Not sufficient.
(2) 2x  3 < 3y  4 > \(x<1.5y\frac{1}{2}\) > \(x<y+(0.5y\frac{1}{2})=y+negative\) > \(x<y\) (as y+negative is "more negative" than y). Sufficient.
Answer: B. Hi bunuel, Did not got how u solved option 2.Could you please explain in detail. thanks.



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21 May 2012, 06:32
I plugged in number for question 2 statement (1).
Any numbers that I could think of really met the inequasion.
But do you have an algebric way of showing this rule?



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21 May 2012, 06:35



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25 May 2012, 11:58
Bunuel wrote: 3. If a, b and c are integers, is abc an even integer?
In order the product of the integers to be even at leas on of them must be even
(1) b is halfway between a and c > on the GMAT we often see such statement and it can ALWAYS be expressed algebraically as \(b=\frac{a+c}{2}\). Now, does that mean that at leas on of them is be even? Not necessarily: \(a=1\), \(b=5\) and \(c=3\), of course it's also possible that for example \(b=even\), for \(a=1\) and \(b=7\). Not sufficient.
(2) a = b  c > \(a+c=b\). Since it's not possible that the sum of two odd integers to be odd then the case of 3 odd numbers is ruled out, hence at least on of them must be even. Sufficient.
Answer: B. What about the case when all a,b,c are zero. In this case, abc = 0 and 0 is neither odd nor even. Hence 'E'.




Re: The Discreet Charm of the DS
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