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28 Mar 2013, 20:31
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8] Since we have an even number of elements in the set, the median will be the average of the two central numbers. Therefore, the median will be of type: (1/p + 1/q)/2 = (q+p)/(2.p.q), where p and q are primes.
(1) The reciprocal of the median, X = 2pq/(p+q) , is prime. So, pq/(p+q) = 1 (let me know if this passage was not trivial for you). Therefore: pq = p + q , which means p = q = 2. Since 1/2 is also the largest possibility of an element for S, we must have all numbers of the set equal to 1/2. Therefore, Sufficient
(2) The only terminating decimals of type 1/n, where n is a prime number are: 1/2 and 1/5. Set S has to have only elements equal to 1/2 and 1/5, which means the median can be equal to 1/2, 1/5 or something between 1/2 and 1/5. Therefore, the median is always greater than 1/5. Sufficient
Answer: D
(1) The reciprocal of the median, X = 2pq/(p+q) , is prime. So, pq/(p+q) = 1 (let me know if this passage was not trivial for you). Therefore: pq = p + q , which means p = q = 2. Since 1/2 is also the largest possibility of an element for S, we must have all numbers of the set equal to 1/2. Therefore, Sufficient
(2) The only terminating decimals of type 1/n, where n is a prime number are: 1/2 and 1/5. Set S has to have only elements equal to 1/2 and 1/5, which means the median can be equal to 1/2, 1/5 or something between 1/2 and 1/5. Therefore, the median is always greater than 1/5. Sufficient
Answer: D
28 Mar 2013, 20:54
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10] (x+1).(y+1) = 12, results in the possible (x,y) pairs: (1,5), (2,3), (3,2), (5,1)
(1) If 9 is not a factor of N, then x < 2. Therefore N = 3.5^5. Sufficient
(2) If 125 is a factor of N, then y >=3. Therefore (x,y) can be either (1,5) or (2,3). Insufficient
Answer: A
(1) If 9 is not a factor of N, then x < 2. Therefore N = 3.5^5. Sufficient
(2) If 125 is a factor of N, then y >=3. Therefore (x,y) can be either (1,5) or (2,3). Insufficient
Answer: A
28 Mar 2013, 21:02
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11] (BONUS QUESTION)
(1) |x-2| < 2-y
=> y-2 < x - 2 < 2 - y
=> y < x < 4 -y
=> y = 1 (try, plugging y >1, to see what happens to the inequality)
=> 1 < x < 3
=> x = 2
=> x is prime
Sufficient
(2) x + y - 3 = |1-y|
As y is a positive integer, y>=1 always, which means |1-y| = y-1. Therefore, x + y - 3 = y -1, which implies x = 2.
Sufficient
Answer: D
(1) |x-2| < 2-y
=> y-2 < x - 2 < 2 - y
=> y < x < 4 -y
=> y = 1 (try, plugging y >1, to see what happens to the inequality)
=> 1 < x < 3
=> x = 2
=> x is prime
Sufficient
(2) x + y - 3 = |1-y|
As y is a positive integer, y>=1 always, which means |1-y| = y-1. Therefore, x + y - 3 = y -1, which implies x = 2.
Sufficient
Answer: D
