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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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22 Nov 2013, 10:11
Bunuel, What i mean is you took x^2+y^2=(xy)^2 so you concluded the middle factor would be 2xy why not (x+y)^2 with middle factor +2xy As in what made you conclude the former?



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01 Jun 2014, 21:18
Bunuel wrote: 8. If x=0.abcd, where a, b, c and d are digits from 0 to 9, inclusive, is x>7/9?First of all 7/9 is a recurring decimal =0.77(7). For more on converting Converting Decimals to Fractions see: mathnumbertheory88376.html(1) a+b>14 > the least value of a is 6 (6+9=15>14), so in this case x=0.69d<0.77(7) but a=7 and b=9 is also possible, and in this case x=0.79d>0.77(7). Not sufficient. (2) ac>6 > the least value of a is 7 (70=7>6), but we don't know the value of b. Not sufficient. (1)+(2) The least value of a is 7 and in this case from (1) least value of b is 8 (7+8=15>14), hence the least value of x=0.78d>0.77(7). Sufficient. Answer: C. Hi Bunnel, One question here. in st2 i can consider 80 also then why we are considering only least value. in st1 we are considering least value as well as other value. why this is different in st2? Thanks



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Re: The Discreet Charm of the DS [#permalink]
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02 Jun 2014, 01:23
PathFinder007 wrote: Bunuel wrote: 8. If x=0.abcd, where a, b, c and d are digits from 0 to 9, inclusive, is x>7/9?First of all 7/9 is a recurring decimal =0.77(7). For more on converting Converting Decimals to Fractions see: mathnumbertheory88376.html(1) a+b>14 > the least value of a is 6 (6+9=15>14), so in this case x=0.69d<0.77(7) but a=7 and b=9 is also possible, and in this case x=0.79d>0.77(7). Not sufficient. (2) ac>6 > the least value of a is 7 (70=7>6), but we don't know the value of b. Not sufficient. (1)+(2) The least value of a is 7 and in this case from (1) least value of b is 8 (7+8=15>14), hence the least value of x=0.78d>0.77(7). Sufficient. Answer: C. Hi Bunnel, One question here. in st2 i can consider 80 also then why we are considering only least value. in st1 we are considering least value as well as other value. why this is different in st2? Thanks Because considering the least value of a (7), while knowing nothing about b is enough to say that the second statement is not sufficient. If b=8, then the answer is yes but if b is 0, then the answer is no.
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Re: The Discreet Charm of the DS [#permalink]
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03 Jun 2014, 00:39
Hi Bunuel, Referring to below: 4. 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
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.
I didn't quite get the explanation on the second statement. Do you mind explaining with some numbers?
Thanks, Mitesh



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Re: The Discreet Charm of the DS [#permalink]
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03 Jun 2014, 01:49
mrvora wrote: Hi Bunuel, Referring to below: 4. 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
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.
I didn't quite get the explanation on the second statement. Do you mind explaining with some numbers?
Thanks, Mitesh If mean=median=prime=2, then the 5 consecutive integers would be { 0, 1, 2, 3, 4}. But this set is not possible since we are told that the set consists of 5 positive integers and 0 is neither positive nor negative integer. Hence, mean=median= odd prime > {Odd, Even, Odd prime, Even, Odd}. So, the set includes 2 consecutive even numbers. Out of 2 consecutive even integers only one is a multiple of 4. Hope it's clear.
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18 Sep 2014, 04:38
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 Bunnel, Please clear my doubt regarding second statement: assumtion 1 : x = 1 y = 2 it means x > y it will give me 2(1)  3 < 3(2)  4 5 < 10 [False] assumtion 2 : x = 2 y = 1 it means x < y it will give me 2(2)  3 < 3(1)  4 7 < 7 [False] How x < y is satisfying through numbers .. pls help



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18 Sep 2014, 11:05
ajaym28 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. Hi Bunnel, Please clear my doubt regarding second statement: assumtion 1 : x = 1 y = 2 it means x > y it will give me 2(1)  3 < 3(2)  4 5 < 10 [False] assumtion 2 : x = 2 y = 1 it means x < y it will give me 2(2)  3 < 3(1)  4 7 < 7 [False] How x < y is satisfying through numbers .. pls help When you pick numbers, they must satisfy the statement you are testing and the info given in the stem. So, when picking numbers for (2) numbers must satisfy 2x  3 < 3y  4 and x and y must be negative (stem). Then you should try to get an YES and NO answer to the question to prove insufficiency.
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18 Sep 2014, 11:18
Bunuel wrote: ajaym28 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. Hi Bunnel, Please clear my doubt regarding second statement: assumtion 1 : x = 1 y = 2 it means x > y it will give me 2(1)  3 < 3(2)  4 5 < 10 [False] assumtion 2 : x = 2 y = 1 it means x < y it will give me 2(2)  3 < 3(1)  4 7 < 7 [False] How x < y is satisfying through numbers .. pls help When you pick numbers, they must satisfy the statement you are testing and the info given in the stem. So, when picking numbers for (2) numbers must satisfy 2x  3 < 3y  4 and x and y must be negative (stem). Then you should try to get an YES and NO answer to the question to prove insufficiency. Thanks Bunnel, just one more ques putting number in that kind of equations will be good idea or solving it by algebra(like you did) is a good idea ? i mean how to figure what to use and when ?



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Re: The Discreet Charm of the DS [#permalink]
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18 Sep 2014, 11:29
ajaym28 wrote: Bunuel wrote: ajaym28 wrote: Hi Bunnel,
Please clear my doubt regarding second statement:
assumtion 1 :
x = 1 y = 2 it means x > y
it will give me 2(1)  3 < 3(2)  4 5 < 10 [False]
assumtion 2 :
x = 2 y = 1 it means x < y
it will give me 2(2)  3 < 3(1)  4 7 < 7 [False]
How x < y is satisfying through numbers .. pls help
When you pick numbers, they must satisfy the statement you are testing and the info given in the stem. So, when picking numbers for (2) numbers must satisfy 2x  3 < 3y  4 and x and y must be negative (stem). Then you should try to get an YES and NO answer to the question to prove insufficiency. Thanks Bunnel, just one more ques putting number in that kind of equations will be good idea or solving it by algebra(like you did) is a good idea ? i mean how to figure what to use and when ? When you decide that a statement is sufficient based only on plugin method you should make sure that you tried several different numbers (and saw some pattern maybe), and even in this case you may not be 100% sure that the answer would be correct. Though if several numbers give the same answer and you are able to see some pattern, then you can make an educated guess that a statement is sufficient and moveon. Generally on DS questions when plugging numbers, your goal is to prove that the statement is NOT sufficient. So you should try to get an YES answer with one chosen number(s) and a NO with another. Hope it's clear.
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18 Sep 2014, 11:45
yes, it's clear i will try apply in couple of question and will see how fast and accurate i can be. Thanks Bunnel



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Re: The Discreet Charm of the DS [#permalink]
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25 Sep 2014, 08:17
Bunuel wrote: 10. 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?
Notice that the greatest common factor of 10 and x, GCF(10,x), naturally must be a factor of 10: 1, 2, 5, and 10. Thus from f(10,x)=11 we can get four different values of x:
GCF(10,x)=1 > \(f(10,x)=11=\frac{10+x}{1}\) > \(x=1\); GCF(10,x)=2 > \(f(10,x)=11=\frac{10+x}{2}\) > \(x=12\); GCF(10,x)=5 > \(f(10,x)=11=\frac{10+x}{5}\) > \(x=45\); GCF(10,x)=10 > \(f(10,x)=11=\frac{10+x}{10}\) > \(x=100\).
(1) x is a square of an integer > \(x\) can be 1 or 100. Not sufficient.
(2) The sum of the distinct prime factors of x is a prime number > distinct primes of 12 are 2 and 3: \(2+3=5=prime\), distinct primes of 45 are 3 and 5: \(3+5=8\neq{prime}\) and distinct primes of 100 are 2 and 5: \(2+5=7=prime\). \(x\) can be 12 or 100. Not sufficient.
(1)+(2) \(x\) can only be 100. Sufficient.
Answer: C. Hi Bunuel, My question here is why did we stop at 10? GCF(10,3) = 30. But we only checked up to 10. Why is that?



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Re: The Discreet Charm of the DS [#permalink]
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25 Sep 2014, 08:27
ronr34 wrote: Bunuel wrote: 10. 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?
Notice that the greatest common factor of 10 and x, GCF(10,x), naturally must be a factor of 10: 1, 2, 5, and 10. Thus from f(10,x)=11 we can get four different values of x:
GCF(10,x)=1 > \(f(10,x)=11=\frac{10+x}{1}\) > \(x=1\); GCF(10,x)=2 > \(f(10,x)=11=\frac{10+x}{2}\) > \(x=12\); GCF(10,x)=5 > \(f(10,x)=11=\frac{10+x}{5}\) > \(x=45\); GCF(10,x)=10 > \(f(10,x)=11=\frac{10+x}{10}\) > \(x=100\).
(1) x is a square of an integer > \(x\) can be 1 or 100. Not sufficient.
(2) The sum of the distinct prime factors of x is a prime number > distinct primes of 12 are 2 and 3: \(2+3=5=prime\), distinct primes of 45 are 3 and 5: \(3+5=8\neq{prime}\) and distinct primes of 100 are 2 and 5: \(2+5=7=prime\). \(x\) can be 12 or 100. Not sufficient.
(1)+(2) \(x\) can only be 100. Sufficient.
Answer: C. Hi Bunuel, My question here is why did we stop at 10? GCF(10,3) = 30. But we only checked up to 10. Why is that? The greatest common factor of 10 and 3 is 1, not 30. 30 is the least common multiple of 10 and 3.
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25 Sep 2014, 12:52
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, Usually in questions such as this one, we have an idea as what numbers would be best to test. I can't find such logic in this question. What are the guidelines when searching for numbers to test?



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09 Oct 2014, 08:32
Bunuel wrote: 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 1.B 2.A 3.B 4.D 5.D 6.E
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20 Oct 2014, 13:46
Bunuel wrote: 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 > this basically means that all customers bought exactly 4 oranges (76/19=4), because if even one customer bought less than 4, the sum will be less than 76. Hence, no one bought only one orange. Sufficient.
(2) 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. But the first case is not possible: the sum of 19 odd numbers is odd and not even like 76. Hence, again no one bought only one=odd orange. Sufficient.
Answer: D. Hi Bunuel, I have a question, in option 2, if we assume that every customer bought even number of oranges and the difference between oranges purchased by any two customers is even, it would mean that every customer bought different number of apples. Using the above analogy, we would not have 19 customers.
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20 Oct 2014, 14:03
Thoughtosphere wrote: Bunuel wrote: 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 > this basically means that all customers bought exactly 4 oranges (76/19=4), because if even one customer bought less than 4, the sum will be less than 76. Hence, no one bought only one orange. Sufficient.
(2) 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. But the first case is not possible: the sum of 19 odd numbers is odd and not even like 76. Hence, again no one bought only one=odd orange. Sufficient.
Answer: D. Hi Bunuel, I have a question, in option 2, if we assume that every customer bought even number of oranges and the difference between oranges purchased by any two customers is even, it would mean that every customer bought different number of apples. Using the above analogy, we would not have 19 customers. What exactly is your question? Why cannot we have a case when each customer bought 4 oranges?
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20 Oct 2014, 20:40
Dear Bunuel, If every one will have 4 then the difference will be 0 which is not even. Posted from my mobile device
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22 Oct 2014, 06:02
Bunuel wrote: 2. Is xy<=1/2?
(1) x^2+y^2=1. Recall that \((xy)^2\geq{0}\) (square of any number is more than or equal to zero) > \(x^22xy+y^2\geq{0}\) > since \(x^2+y^2=1\) then: \(12xy\geq{0}\) > \(xy\leq{\frac{1}{2}}\). Sufficient.
(2) x^2y^2=0 > \(x=y\). Clearly insufficient.
Answer: A. Hi Bunuel, I considered \((x+y)^2\) \(>=\) 0 and arrived at \(xy >=\) \(\frac{1}{2}\). And hence concluded that the statement is insufficient. Please correct me.



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