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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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16 Apr 2016, 09:25
bikographer wrote: 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. Considering (2), and rearranging it, 2x  3y < 1 .......... (i) substituting x=2 and y=1 in (i), we get 4 + 3 = 1 which is not <1 This means that the condition doesn't hold good, and hence x < y doesn't hold good. Could you please point out where exactly I am making a mistake in my process? \(3x + 4 < 2y + 3\) means that \(3x<2y1\), not 2x  3y < 1.
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The Discreet Charm of the DS [#permalink]
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16 Apr 2016, 18:14
Bunuel wrote: bikographer wrote: 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. Considering (2), and rearranging it, 2x  3y < 1 .......... (i) substituting x=2 and y=1 in (i), we get 4 + 3 = 1 which is not <1 This means that the condition doesn't hold good, and hence x < y doesn't hold good. Could you please point out where exactly I am making a mistake in my process? \(3x + 4 < 2y + 3\) means that \(3x<2y1\), not 2x  3y < 1. Bunuel, I was talking only about the second statement, and in your reply you have talked about the first statement. My doubt, which I have highlighted by taking x = 2 and y = 1 and applying these values to statement (2), still remains.



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Re: The Discreet Charm of the DS [#permalink]
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17 Apr 2016, 08:05
bikographer wrote: 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. Considering (2), and rearranging it, 2x  3y < 1 .......... (i) substituting x=2 and y=1 in (i), we get 4 + 3 = 1 which is not <1 This means that the condition doesn't hold good, and hence x < y doesn't hold good. Could you please point out where exactly I am making a mistake in my process? The statements in DS questions are true. So, if you pick numbers you should pick such that they satisfy the statement. x and y cannot be 2 and 1 respectively because they do not satisfy 2x  3 < 3y  4.
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The Discreet Charm of the DS [#permalink]
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06 Nov 2016, 04:18
Bunuel wrote: 11. If x and y are integers, is x a positive integer?
(1) x*y is a prime number > since only positive numbers can be primes, then: x*y=positive > x=positive. Sufficient
(2) x*y is nonnegative integer. Notice that we are told that x*y is nonnegative, not that it's positive, so x can be positive as well as zero. Not sufficient.
Answer: A. I think the option should be E Option 1 x * y will result in positive, where x is negative. Can you explain further..??



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06 Nov 2016, 05:06
hardik0491 wrote: Bunuel wrote: 11. If x and y are integers, is x a positive integer?
(1) x*y is a prime number > since only positive numbers can be primes, then: x*y=positive > x=positive. Sufficient
(2) x*y is nonnegative integer. Notice that we are told that x*y is nonnegative, not that it's positive, so x can be positive as well as zero. Not sufficient.
Answer: A. I think the option should be E Option 1 x * y will result in positive, where x is negative. Can you explain further..?? If x is negative, then x*y = negative*nonnegative = nonpositive, which cannot be prime, since only positive numbers are primes.
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15 Dec 2016, 10:16
Hello Bunuel,
For question 9, option B  What if we assign the 100 for x and 99 for y? x<y, but the equation (rearranged) 3y2x>1 does not hold true. Please help!



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10 Jun 2017, 10:30
Bunuel wrote: 4. How many numbers of 5 consecutive positive integers is divisible by 4?
(1) The median of these numbers is odd > the median of the set with odd number of terms is just a middle term, thus our set of 5 consecutive numbers is: {Odd, Even, Odd, Even, Odd}. Out of 2 consecutive even integers only one is a multiple of 4. Sufficient.
(2) The average (arithmetic mean) of these numbers is a prime number > in any evenly spaced set the arithmetic mean (average) is equal to the median > mean=median=prime. Since it's not possible that median=2=even, (in this case not all 5 numbers will be positive), then median=odd prime, and we have the same case as above. Sufficient.
Answer: D. Bunuel Can you pls help in understanding this : "Out of 2 consecutive even integers only one is a multiple of 4. Sufficient." from statement 1.Thanks



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10 Jun 2017, 10:50
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.
Hi Bunuel, a small doubt instead of \(2x+8\geq{0}\) > \(x\geq{4}\) this logic if we take x+10 to be both positive and negative we get 2 vale by solving equation x=2 and x=6. however by putting values back in eqtn we can see that only for x=2 equation is satisfying. Is this correct approach?



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10 Jun 2017, 11:47
KARISHMA315 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.
Hi Bunuel, a small doubt instead of \(2x+8\geq{0}\) > \(x\geq{4}\) this logic if we take x+10 to be both positive and negative we get 2 vale by solving equation x=2 and x=6. however by putting values back in eqtn we can see that only for x=2 equation is satisfying. Is this correct approach? Yes, that's also a correct way of solving. Good thing you did is that you did not forget to test both values after you got them.
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26 Jul 2017, 04:52
Is this valid? Statement 1 2x²+9<9x 2x²<9x9 2x²<9(x1) 2x²÷(x1)<9 plugin 1 2(1)²÷(11)=2÷0=undefined plugin 2 2(2)²÷(21)=8÷1<9 valid any no.greater than 2 will give no greater than 9 if we take a negative no than the sign will change and inequality will not be valid. Statement 2 x+10=2x+8 x+10=2(x+4) x+10w÷2=X+4 Only X=2 satisfies the equation 2+10÷2=12÷2=6 2+4=6 [qote="Bunuel"] 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: http://gmatclub.com/forum/x24x94661.html#p731476 http://gmatclub.com/forum/inequalitiestrick91482.htmlhttp://gmatclub.com/forum/everythingis ... me#p868863http://gmatclub.com/forum/xyplane7149 ... ic#p841486Hope it helps.[/quote] Posted from my mobile device



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05 Sep 2017, 23:43
Bunuel wrote: 11. If x and y are integers, is x a positive integer?
(1) x*y is a prime number > since only positive numbers can be primes, then: x*y=positive > x=positive. Sufficient
(2) x*y is nonnegative integer. Notice that we are told that x*y is nonnegative, not that it's positive, so x can be positive as well as zero. Not sufficient.
Answer: A. Hi Bunuel, I understood that x*y=positive; but if y<=0 then y=y and in this case x can be negative. Am I missing something? Thanks.



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05 Sep 2017, 23:49
Yashodhan123 wrote: Bunuel wrote: 11. If x and y are integers, is x a positive integer?
(1) x*y is a prime number > since only positive numbers can be primes, then: x*y=positive > x=positive. Sufficient
(2) x*y is nonnegative integer. Notice that we are told that x*y is nonnegative, not that it's positive, so x can be positive as well as zero. Not sufficient.
Answer: A. Hi Bunuel, I understood that x*y=positive; but if y<=0 then y=y and in this case x can be negative. Am I missing something? Thanks. Yes. An absolute value of a number cannot be negative: a is positive or 0, no matter whether a itself is negative or not. If y <= 0, then y = y, yes, but even in this case y = negative = positive.
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18 Oct 2017, 08:10
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?
Bonnie and Clyde when working together complete the painting of the car ins \(\frac{xy}{x+y}\) hours (sum of the rates equal to the combined rate or reciprocal of total time: \(\frac{1}{x}+\frac{1}{y}=\frac{1}{T}\) > \(T=\frac{xy}{x+y}\)). Now, if \(x=y\) then the total time would be: \(\frac{x^2}{2x}=\frac{x}{2}\), since \(x\) is odd then this time would be odd/2: 0.5 hours, 1.5 hours, 2.5 hours, ....
(1) x^2+y^2<12 > it's possible \(x\) and \(y\) to be odd and equal to each other if \(x=y=1\) but it's also possible that \(x=1\) and \(y=3\) (or viseversa). Not sufficient.
(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 not odd/2 then \(x\) and \(y\) are not equal. Sufficient.
Answer: B. Please explain this partsince it's not odd/2 then x and y are not equal.. how can u say its not odd..pls help!! Thanks



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18 Oct 2017, 08:30
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, consider \(a=1\), \(b=3\) and \(c=5\). Of course it's also possible that \(b=even\), for example if \(a=1\) 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. Why are we assuming that a and c could be even..? or could be odd? Its also possible that a is even and c is odd..in that case b can be odd Pls explain thanks



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18 Oct 2017, 09:58
zanaik89 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?
Bonnie and Clyde when working together complete the painting of the car ins \(\frac{xy}{x+y}\) hours (sum of the rates equal to the combined rate or reciprocal of total time: \(\frac{1}{x}+\frac{1}{y}=\frac{1}{T}\) > \(T=\frac{xy}{x+y}\)). Now, if \(x=y\) then the total time would be: \(\frac{x^2}{2x}=\frac{x}{2}\), since \(x\) is odd then this time would be odd/2: 0.5 hours, 1.5 hours, 2.5 hours, ....
(1) x^2+y^2<12 > it's possible \(x\) and \(y\) to be odd and equal to each other if \(x=y=1\) but it's also possible that \(x=1\) and \(y=3\) (or viseversa). Not sufficient.
(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 not odd/2 then \(x\) and \(y\) are not equal. Sufficient.
Answer: B. Please explain this partsince it's not odd/2 then x and y are not equal.. how can u say its not odd..pls help!! Thanks Please reread the highlighted part. We got that if \(x=y\) then the total time would be odd/2. (2) says that it's NOT odd/2, therefore x does not equal to y.
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18 Oct 2017, 10:00
zanaik89 wrote: 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, consider \(a=1\), \(b=3\) and \(c=5\). Of course it's also possible that \(b=even\), for example if \(a=1\) 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. Why are we assuming that a and c could be even..? or could be odd? Its also possible that a is even and c is odd..in that case b can be odd Pls explain thanks It's not clear what are you trying to say here. If a is even, then abc is even.
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Re: The Discreet Charm of the DS [#permalink]
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18 Oct 2017, 17:50
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. Hi Bunuel, I have two questions.. 1) Cant we do 6a=3b simplify further and plug it back into equation and solve? 6a=3b can be written as 2a=b? 2) In your solution u have got x^2=2x how are thee two values of x? how did u get x=0? Thanks



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Re: The Discreet Charm of the DS [#permalink]
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18 Oct 2017, 20:04
zanaik89 wrote: 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. Hi Bunuel, I have two questions.. 1) Cant we do 6a=3b simplify further and plug it back into equation and solve? 6a=3b can be written as 2a=b? 2) In your solution u have got x^2=2x how are thee two values of x? how did u get x=0? Thanks 1. Not clear what you mean at all. Please show your work. 2. x^2=2x; x^2  2x = 0 x(x  2) = 0 x = 0 or x = 2. P.S. You should know such basic staff before attempting questions, especially hard ones as in this set.
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Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
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Re: The Discreet Charm of the DS [#permalink]
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07 Nov 2017, 06:24
Thanks for the compilation guys! It was super helpful Out of curiosity. What difficulty level are these questions?




Re: The Discreet Charm of the DS
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