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If p is the average (arithmetic mean) of the first 12 positive multipl
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13 Aug 2021, 09:42
13 Aug 2021, 09:42
Expert Reply
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Question Stats:
88% (01:01) correct
11% (02:41) wrong based on 17 sessions
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If p is the average (arithmetic mean) of the first 12 positive multiples of 6 and if P is the median of the first 12 positive multiples of 6, what is the value of P − p ?
(A) 39
(B) 36
(C) 6
(D) 0
(E) −6
Kudos for the right answer and explanation
Question part of the project GRE Quant/Math Extreme Challenge Daily (2021) Edition
GRE - Math Book
(A) 39
(B) 36
(C) 6
(D) 0
(E) −6
Kudos for the right answer and explanation
Question part of the project GRE Quant/Math Extreme Challenge Daily (2021) Edition
GRE - Math Book
If p is the average (arithmetic mean) of the first 12 positive multipl
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13 Aug 2021, 19:40
13 Aug 2021, 19:40
1
First \(12\) +ve multiples of \(6\) = \(6 , 12 , 18 , 24 , 30 , 36 , 42 , 48 , 54 , 60 , 66 , 72\)
\(p\) is the avg of first \(12\) +ve multiples
\(p = \frac{6+12+18+24+30+36+42+48+54+60+66+72}{12} = 39\)
\(P\) is the median
Median \(P\) = \(\frac{6^{th} no + 7^{th} no}{2}\) \(= \frac{36+42}{2} = 39\)
\(P - p = 0\)
Answer D
\(p\) is the avg of first \(12\) +ve multiples
\(p = \frac{6+12+18+24+30+36+42+48+54+60+66+72}{12} = 39\)
\(P\) is the median
Median \(P\) = \(\frac{6^{th} no + 7^{th} no}{2}\) \(= \frac{36+42}{2} = 39\)
\(P - p = 0\)
Answer D
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Re: If p is the average (arithmetic mean) of the first 12 positive multipl
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13 Aug 2021, 21:01
13 Aug 2021, 21:01
1
Also, for evenly spaced consecutive integers, the mean is always equal to the median
SO the mean-median=0
SO the mean-median=0
