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If a, b, c, and d are consecutive integers and a < b < c < d, what is
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22 Jan 2018, 01:54
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If a, b, c, and d are consecutive integers and a < b < c < d, what is the average (arithmetic mean) of a, b, c, and d in terms of d? (A) d – 5/2 (B) d – 2 (C) d – 3/2 (D) d + 3/2 (E) (4d – 6)/7
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Re: If a, b, c, and d are consecutive integers and a < b < c < d, what is
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22 Jan 2018, 02:02
Since they are consecutive integers, therefore a=d3; b=d2; c=d1; Taking average, (d3+d2+d1+d)/4 = (4d6)/4 = d3/2 Option C. Sent from my ONE E1003 using GMAT Club Forum mobile app



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Re: If a, b, c, and d are consecutive integers and a < b < c < d, what is
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22 Jan 2018, 07:05
Please see my approach Bunuel wrote: If a, b, c, and d are consecutive integers and a < b < c < d, what is the average (arithmetic mean) of a, b, c, and d in terms of d?
(A) d – 5/2
(B) d – 2
(C) d – 3/2
(D) d + 3/2
(E) (4d – 6)/7
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Re: If a, b, c, and d are consecutive integers and a < b < c < d, what is
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22 Jan 2018, 08:11
Bunuel wrote: If a, b, c, and d are consecutive integers and a < b < c < d, what is the average (arithmetic mean) of a, b, c, and d in terms of d?
(A) d – 5/2
(B) d – 2
(C) d – 3/2
(D) d + 3/2
(E) (4d – 6)/7 Answer: C Since consecutive integers, thus a = d3 b = d2 c = d1 Mean = (a + b +c + d) / 4 Substituting a, b and c in above we get option C



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Re: If a, b, c, and d are consecutive integers and a < b < c < d, what is
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22 Jan 2018, 08:36
Bunuel wrote: If a, b, c, and d are consecutive integers and a < b < c < d, what is the average (arithmetic mean) of a, b, c, and d in terms of d?
(A) d – 5/2
(B) d – 2
(C) d – 3/2
(D) d + 3/2
(E) (4d – 6)/7 As the numbers are consecutive, let the numbers be c2, c1,c , c+1 A.M = (4c2)/4 = c (1/2).. Also, as d = c+1 from above equation, c = d  1 So, AM = d (3/2)
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Re: If a, b, c, and d are consecutive integers and a < b < c < d, what is
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22 Jan 2018, 23:21
The moment I see consecutive integers and a < b< c< d, I think of 1,2 ,3 ,4. Average is 2.5 which is 4  1.5 i.e. d  3/2 Answer (C)
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Re: If a, b, c, and d are consecutive integers and a < b < c < d, what is
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22 Jan 2018, 23:29
1. Av= \(\frac{(A+D)}{2}\) (as it is consecutive evenly spaced numbers) 2. Range = DA=3 (as it is consecutive numbers with increment of 1) 3. Substitute A= D3 in 1. Av=\(\frac{(D3+D)}{2}\) >Av=\(\frac{(2D3)}{2}\) 4. After dividing by 2 it becomes Av=D\(\frac{3}{2}\). Answer (C)
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Re: If a, b, c, and d are consecutive integers and a < b < c < d, what is
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23 Jan 2018, 08:53
Bunuel wrote: If a, b, c, and d are consecutive integers and a < b < c < d, what is the average (arithmetic mean) of a, b, c, and d in terms of d?
(A) d – 5/2
(B) d – 2
(C) d – 3/2
(D) d + 3/2
(E) (4d – 6)/7 We can label our consecutive integers as: d c = d  1 b = d  2 a = d  3 Using the formula for the arithmetic average (mean), we have: average = (d + d  1 + d  2 + d  3)/4 = (4d  6)/4 = 4d/4  6/4 = d  3/2 Answer: C
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