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11 Jan 2024, 02:30
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
70% (01:17) correct
30%
(01:22)
wrong
based on 93
sessions
History
Date
Time
Result
Not Attempted Yet
Is the median of a, b, and c equal to their average (arithmetic mean)?
(1) \(a≤b≤c\)
(2) \(b = \frac{(a + c)}{2}\)
(1) \(a≤b≤c\)
(2) \(b = \frac{(a + c)}{2}\)
11 Jan 2024, 21:55
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Bunuel
OA is B
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05 Jun 2025, 11:04
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Is the median of a, b, and c equal to their average (arithmetic mean)?
(1) \(a≤b≤c\)
(2) \(b = \frac{(a + c)}{2}\)
(1) \(a≤b≤c\)
Putting a=3. b=4. c= 4;
median = 4
mean = 11/3
median of a, b, and c is not equal to their average (arithmetic mean)-----NO is the answer
Putting a=3. b=3. c= 3;
median = 4
mean = 3
median of a, b, and c equal to their average (arithmetic mean) -----YES is the answer
Can't be determined
(2) \(b = \frac{(a + c)}{2}\)
So, 2b = a + c
c-b = b-a
ie, a, b, c are in A.P.
Putting
b = a+d
c = a+2d
a.b,c ===> a, a+d, a+2d
Mean = a+d
Median = a+d
Clearly, median of a, b, and c will always be equal to their average (arithmetic mean) -----YES is the answer
YES is the answer
(B) is the correct option
(1) \(a≤b≤c\)
(2) \(b = \frac{(a + c)}{2}\)
(1) \(a≤b≤c\)
Putting a=3. b=4. c= 4;
median = 4
mean = 11/3
median of a, b, and c is not equal to their average (arithmetic mean)-----NO is the answer
Putting a=3. b=3. c= 3;
median = 4
mean = 3
median of a, b, and c equal to their average (arithmetic mean) -----YES is the answer
Can't be determined
(2) \(b = \frac{(a + c)}{2}\)
So, 2b = a + c
c-b = b-a
ie, a, b, c are in A.P.
Putting
b = a+d
c = a+2d
a.b,c ===> a, a+d, a+2d
Mean = a+d
Median = a+d
Clearly, median of a, b, and c will always be equal to their average (arithmetic mean) -----YES is the answer
YES is the answer
(B) is the correct option
