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Re: The Discreet Charm of the DS
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03 Feb 2012, 14:56
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
out of 5 consecutive intergers there are only 2 option either there will 1 or 2 numbers which will divide by 4. There will 2 numbers if the first number fo sequence will be divible by 4. So we have to find out if first number is divisible by 4.
Stmt 1  Median is odd means first number is odd ( odd , even , odd, even, odd) So sufficuent to tell that there will only 1 number which be divible by 4. Stmt 2  In case of consecutive numbers median is always equal to average. Numbers are positive so first numbers cannot be 0 to make 2 as median/average. Which means average/mean is odd, Again going by logic in stmt 1 first number should be odd which is also sufficient
Answer D.



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Re: The Discreet Charm of the DS
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03 Feb 2012, 14:56
5. What is the value of integer x? (1) 2x^2+9<9x (2) x+10=2x+8
Stmt 1  2x^2+9<9x Put the values 5, 4, 3, 2, 1, 0, 1, 2, 3, 4, 5 So the only value that satifies the equation is 2. Sufficient
Stmt 2 x+10=2x+8 In number line take one number X =  10. Condition one X >  10 which will result in positive equation. x+10=2x+8 x = 2 x = 2 greater than 10 so satifies the equation.
Condition two X <  10 which will result in negative equation. x+10 =  2x  8 x = 6 which is is not less than 10 which cannot the solution of inequality.
So we get only one value X = 2 Sufficient
Answer D



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Re: The Discreet Charm of the DS
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03 Feb 2012, 14:56
6. If a and b are integers and ab=2, is a=2? (1) b+3 is not a prime number (2) a>b
Possible values of a and b are : (1,2), (2,1), (1,2), (2,1) we have to find out if a = 2 or get if b is 1.
Stmt 1  b+3 is prime number. Possible values of b 2 and 1 Not sufficient.
Stmt 2 a > b Pssible values of b are 1, 2. Not sufficient
Combining stmt 1 and stmt 2 no common answers. So insufficient.
Answer E.



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Re: The Discreet Charm of the DS
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03 Feb 2012, 14:56
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
76 oranges / 19 customers.
Stmt 1  Let's say everyone got maximum 4 oranges to 76 distributed. And if someone got less than 4 and someone would have more than 4 which is not possible. So sufficient to tell that noone got 1 orange.
Stmt 2  Difference will be even if, even  even OR odd  odd.
Say suppose they got odd, the nall of them have to recieve odd number else the difference will not even. Which is not possible as there 76 oranges which wont divide oddly. But if its even if someone gets 2 then someone can get 6 which will make the difference even. All evens will evenly divide 79 between 19 of them.
Sufficient
Answer D



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Re: The Discreet Charm of the DS
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03 Feb 2012, 14:56
8. 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
7/9 = .7777777 so question is asking if a > 7, if a = 7 then is b > 7 , if b = 7 then is c > 7 and so forth.
stmt 1  a+b > 14 so possible values of (a,b) = (8,7) , (7,8), (9,6), (6,9) So a can be = 7 or even > 7 so insufficient.
stmt 2  a  c > 6 possible values of a,c = (7,0), (8,1), (9,2) so a can take multiple values so insufficient.
Combining stmt 1 and stmt 2. a can taken values as 7,8,9 you will see that if a = 7 then b is 8 which is greater than 7/9 All other values will result yes X < 7/9 which is sufficient
Answer C



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Re: The Discreet Charm of the DS
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03 Feb 2012, 14:56
9. If x and y are negative numbers, is x<y? (1) 3x + 4 < 2y + 3 (2) 2x  3 < 3y  4
Given x is ve and y is ve
Stmt 1  3x + 4 < 2y + 3 which can be written as 3x < 2y  1 suppose x < y then 3x < 3y You cannot dedude from here. So insufficient.
Stmt 2  2x  3 < 3y  4 which cane be written as 2x + 1 < 3y which can be furhter written as 2y > 3y
therefore 2x + 1 < 2y even after 1 2x is smaller than 2y that proves x < y.
So answer B



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Re: The Discreet Charm of the DS
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03 Feb 2012, 14:56
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? (1) x is a square of an integer (2) The sum of the distinct prime factors of x is a prime number.
Given 10 + x / GCF (10,x) = 11. Possible values of GCF(10,x) = 1,2,5,10. So 10+x can be 11, 22, 55, 110 ; which will result in x = 1,12, 45, 100
Stmt 1  x can be 1 and 10 so not sufficient. Stmt 2  x can be 12 , 100 so not sufficient.
Combining stmt 1 and stmt 2 we get x = 100.
So answer is C



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Re: The Discreet Charm of the DS
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03 Feb 2012, 14:57
11. If x and y are integers, is x a positive integer? (1) x*y is a prime number. (2) x*y is nonnegative integer.
prime # are always positive
Stmt 1  y will be always positive to x*y to be prime z has to be positve. Sufficient Stmt 2  Same concept x will be positive number Sufficient
Answer is D



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Re: The Discreet Charm of the DS
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03 Feb 2012, 14:57
12. If 6a=3b=7c, what is the value of a+b+c? (1) ac=6b (2) 5b=8a+4c
Stmt 1 = I get the values of a,b and c but a,b,c can be 0 also. So insufficient. Stmt 2  Insufficient because of same reason.



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Re: The Discreet Charm of the DS
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03 Feb 2012, 15:00
sourabhsoni wrote: 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
The catch is that they are working independently.
Answer B Just to clarify more, working independently does not mean they are painting different cars. They are still painting the same car. Only that, the events are mutually exclusive, as you say in probabilistic terms, so that rates are not affected when they work simultaneously. Another way : 1) more than one solution possible. 2) let's say 1)x=y=1. Work will be completed in 1/2 hour i.e. 10.15 am. 2)x=y=3. Work will be completed in 3/2 hour i.e. 11.15 am. Not possible. B.
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Re: The Discreet Charm of the DS
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03 Feb 2012, 16:53
Solutions for: 4, 5, 7, 8, 9, and 10 are correct. +1 for each. sourabhsoni wrote: 6. If a and b are integers and ab=2, is a=2? (1) b+3 is not a prime number (2) a>b
Possible values of a and b are : (1,2), (2,1), (1,2), (2,1) we have to find out if a = 2 or get if b is 1.
Stmt 1  b+3 is prime number. Possible values of b 2 and 1 Not sufficient.
Stmt 2 a > b Pssible values of b are 1, 2. Not sufficient
Combining stmt 1 and stmt 2 no common answers. So insufficient.
Answer E. As for this one: first of all (1) says that b+3 is NOT a prime number. Next, on the GMAT, two data sufficiency statements always provide TRUE information and these statements never contradict each other. So, we can not have the case when (1) says that b is either 2 or 1 and (2) says that b is either 1 or 2, because in this case statements would contradict each other. Hope it's clear. P.S. Funny thing: answer is still E.
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Re: The Discreet Charm of the DS
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03 Feb 2012, 17:37
Bunuel wrote: sourabhsoni wrote: 2. Is xy<=1/2? (1) x^2+y^2=1 (2) x^2y^2=0
My funda  Area of square is largest among all the quadilateral with same perimeter. Stmt 1  Only possible values of x and y are 1/Sqrt(2). So sufficient as xy = 1/2 Stmt 2  Only says x and y are equal. Not sufficient Answer A You are close to correct reasoning for (1), though from it you can not say that xy =1/2 and the only possible value for x and y are 1/Sqrt(2). Consider the following example: 0^1+1^2=1. As for (2): x^2y^2=0 doesn't mean that x=y. 1) x2 + y2 = 1 is a circle with radius 1, and origin as center. We are only concerned with 1st and 3rd quad (think why? ), Draw a line y=x which intersects circle at x=y=1/sqrt(2). We can observe, as x becomes greater than 1/sqrt2, y gets lesser than 1/sqrt2 moving on the circle (below y=x line). Hence xy<= 1/2. Similarly, as y becomes greater than 1/sqrt2, x gets lesser than 1/sqrt2 moving on the circle (above y=x line). Hence xy<= 1/2. Hence sufficient. For more mathematically inclined, draw graph of xy=1/2. It has only one point of contact with the circle, tangential at x=y=1/sqrt(2).
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Re: The Discreet Charm of the DS
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03 Feb 2012, 17:47
vailad wrote: Bunuel wrote: sourabhsoni wrote: 2. Is xy<=1/2? (1) x^2+y^2=1 (2) x^2y^2=0
My funda  Area of square is largest among all the quadilateral with same perimeter. Stmt 1  Only possible values of x and y are 1/Sqrt(2). So sufficient as xy = 1/2 Stmt 2  Only says x and y are equal. Not sufficient Answer A You are close to correct reasoning for (1), though from it you can not say that xy =1/2 and the only possible value for x and y are 1/Sqrt(2). Consider the following example: 0^1+1^2=1. As for (2): x^2y^2=0 doesn't mean that x=y. 1) x2 + y2 = 1 is a circle with radius 1, and origin as center. We are only concerned with 1st and 3rd quad (think why? ), Draw a line y=x which intersects circle at x=y=1/sqrt(2). We can observe, as x becomes greater than 1/sqrt2, y gets lesser than 1/sqrt2 moving on the circle (below y=x line). Hence xy<= 1/2. Similarly, as y becomes greater than 1/sqrt2, x gets lesser than 1/sqrt2 moving on the circle (above y=x line). Hence xy<= 1/2. Hence sufficient. For more mathematically inclined, draw graph of xy=1/2. It has only one point of contact with the circle, tangential at x=y=1/sqrt(2). Yes, the answer to the question is indeed A. +1. Though let me point out that there are at least two other solutions that are shorter and easier. I'll post them in couple of days, after some more discussion.
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Re: The Discreet Charm of the DS
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03 Feb 2012, 17:55
Bunuel wrote: vailad wrote: Bunuel wrote: 2. Is xy<=1/2? (1) x^2+y^2=1 (2) x^2y^2=0 1) x2 + y2 = 1 is a circle with radius 1, and origin as center. We are only concerned with 1st and 3rd quad (think why? ), Draw a line y=x which intersects circle at x=y=1/sqrt(2). We can observe, as x becomes greater than 1/sqrt2, y gets lesser than 1/sqrt2 moving on the circle (below y=x line). Hence xy<= 1/2. Similarly, as y becomes greater than 1/sqrt2, x gets lesser than 1/sqrt2 moving on the circle (above y=x line). Hence xy<= 1/2. Hence sufficient. For more mathematically inclined, draw graph of xy=1/2. It has only one point of contact with the circle, tangential at x=y=1/sqrt(2). Yes, the answer to the question is indeed A. +1. Though let me point out that there are at least two other solutions that are shorter and easier. I'll post them in couple of days, after some more discussion. IMO, if you can imagine drawing the circle and y=x line, it takes less than 30 sec. to figure it out. It only takes so much longer to explain in text. And ofcourse, no need to draw the xy graph. Will wait though for the even faster method, if any.
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Re: The Discreet Charm of the DS
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03 Feb 2012, 18:08
vailad wrote: IMO, if you can imagine drawing the circle and y=x line, it takes less than 30 sec. to figure it out. It only takes so much longer to explain in text. And ofcourse, no need to draw the xy graph. Will wait though for the even faster method, if any. If you answered this question in less than 30 sec using this approach then all I can say is great job! As for the easier/faster solution, little hint: it involves simplest algebraic manipulation.
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Re: The Discreet Charm of the DS
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03 Feb 2012, 18:20
Bunuel wrote: If you answered this question in less than 30 sec using this approach then all I can say is great job!
As for the easier/faster solution, little hint: it involves simplest algebraic manipulation. F***ing good ! (xy)2 = 12xy >=0 => xy>=1/2. Now this should take 15sec !
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Re: The Discreet Charm of the DS
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03 Feb 2012, 18:22
vailad wrote: Bunuel wrote: If you answered this question in less than 30 sec using this approach then all I can say is great job!
As for the easier/faster solution, little hint: it involves simplest algebraic manipulation. F***ing good ! (xy)2 = 12xy >=0 => xy>=1/2. Now this should take 15sec ! That's it. +1 again. Now tell me which one is easier/faster? Just a little typo there: xy<=1/2.
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Re: The Discreet Charm of the DS
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04 Feb 2012, 02:03
sourabhsoni wrote: 11. If x and y are integers, is x a positive integer? (1) x*y is a prime number. (2) x*y is nonnegative integer.
prime # are always positive
Stmt 1  y will be always positive to x*y to be prime z has to be positve. Sufficient Stmt 2  Same concept x will be positive number Sufficient
Answer is D Statement 2: x can be zero or +. ( x*y is a non negative integer. i.eit can be 0 as well.. NS)



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Re: The Discreet Charm of the DS
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04 Feb 2012, 04:59
Bunuel wrote: vailad wrote: Bunuel wrote: If you answered this question in less than 30 sec using this approach then all I can say is great job!
As for the easier/faster solution, little hint: it involves simplest algebraic manipulation. F***ing good ! (xy)2 = 12xy >=0 => xy>=1/2. Now this should take 15sec ! That's it. +1 again. Now tell me which one is easier/faster? Just a little typo there: xy<=1/2. Haha. I said, then and there, this should take 15 sec. i.e. double the time I took. P.S. : As we see, the coordinate geometry method might not be the best way to solve this particular problem. Having said that I must say one should keep this approach at the back of the mind. Solving complex inequalities can be monumental at times, and certainly more prone to errors while coordinate geometry is graphical, clean and quick.
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Re: The Discreet Charm of the DS
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05 Feb 2012, 04:31
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.
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