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If the average (arithmetic mean) of the four numbers M, 2M − 1, 3M + 5
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24 Aug 2021, 05:26
24 Aug 2021, 05:26
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Question Stats:
94% (00:51) correct
5% (01:04) wrong based on 18 sessions
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If the average (arithmetic mean) of the four numbers M, 2M − 1, 3M + 5, and 5M − 3 is 25, what is the value of M?
(A) 2 2/11
(B) 4
(C) 8
(D) 9
(E) 9 1/11
Kudos for the right answer and explanation
Question part of the project GRE Quant/Math Extreme Challenge Daily (2021) Edition
GRE - Math Book
(A) 2 2/11
(B) 4
(C) 8
(D) 9
(E) 9 1/11
Kudos for the right answer and explanation
Question part of the project GRE Quant/Math Extreme Challenge Daily (2021) Edition
GRE - Math Book
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Re: If the average (arithmetic mean) of the four numbers M, 2M − 1, 3M + 5
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24 Aug 2021, 05:34
24 Aug 2021, 05:34
1
Carcass wrote:
If the average (arithmetic mean) of the four numbers M, 2M − 1, 3M + 5, and 5M − 3 is 25, what is the value of M?
(A) 2 2/11
(B) 4
(C) 8
(D) 9
(E) 9 1/11
Kudos for the right answer and explanation
Question part of the project GRE Quant/Math Extreme Challenge Daily (2021) Edition
GRE - Math Book
(A) 2 2/11
(B) 4
(C) 8
(D) 9
(E) 9 1/11
Kudos for the right answer and explanation
Question part of the project GRE Quant/Math Extreme Challenge Daily (2021) Edition
GRE - Math Book
The average (arithmetic mean) of the four numbers M, 2M − 1, 3M + 5, and 5M − 3 is 25
We can write: \(\frac{(M) + (2M − 1)+ (3M + 5)+(5M − 3)}{4} = 25\)
Simplify numerator: \(\frac{11M+1}{4} = 25\)
Multiply both sides by 4 to get: \(11M+1 = 100\)
Subtract 1 from both sides: \(11M = 99\)
Solve: \(M = 9\)
Answer: D
Re: If the average (arithmetic mean) of the four numbers M, 2M − 1, 3M + 5
[#permalink]
24 Aug 2021, 06:41
24 Aug 2021, 06:41
Pretty straightforward :
M + 2M -1 + 5M -3 + 3M + 5 / 4 = 25
11 M + 1 = 100
M = 99 /11
M = 9
Hence D
M + 2M -1 + 5M -3 + 3M + 5 / 4 = 25
11 M + 1 = 100
M = 99 /11
M = 9
Hence D
