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12 Jan 2023, 01:57
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
(hard)
Question Stats:
58% (03:04) correct
42%
(02:36)
wrong
based on 155
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P is a set of all integers between x and y, inclusive, and Q is a set of all integers between x and z, inclusive, where x, y and z are integers. The median of set P is (5/6)*y. The median of set Q is (2/3)*z. If R is a set of all integers between y and z, inclusive, what fraction of z is the median of set R?
A. ½
B. 7/12
C. 2/3
D. ¾
E. 3/2
A. ½
B. 7/12
C. 2/3
D. ¾
E. 3/2
12 Jan 2023, 13:47
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Aabhash777
We can assume numbers to solve the question
Assume y = 6
Median of set P = 5/6 * 6 = 5
P = {4,5,6}
Median of Q is (2/3)*z
Assume z = 12
Median = 2*4 = 8
Q = { 4, 5, 6, 7, 8, 9, 10, 11, 12}
R = {6, 7, 8, 9, 10, 11, 12}
Median of R = 9
Ratio =\( \frac{9 }{ 12} = \frac{3}{4}\)
Option D
12 Jan 2023, 22:01
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Aabhash777
x, y and z are integers and P, Q and R are sets of consecutive integers.
Take one statement at a time:
The median of set P is (5/6)*y
Median is the middle value.
Median of x...y is (5/6)y. The easiest case of this would be when median is 5 and y = 6
Then x...y must be 4, 5, 6.
The median of set Q is (2/3)*z.
Set Q is x...z so as per our case, x = 4. So Q is 4...??
Since median is (2/3)z, z is a multiple of 3.
If z is 9, median of 4, 5, 6, 7,8 9 is 6.5 which is not valid.
If z is 12, median of 4, 5, 6 ,7, 8, 9, 10, 11, 12 is 8 which is (2/3)z. Hence, z = 12
what fraction of z is the median of set R?
Since y = 6 and z = 12, R is 6, 7, 8, 9, 10, 11, 12 for which median is 9.
Fraction = 9/12 = 3/4
Answer (D)
