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# If a, b, c, and d are consecutive integers and a < b < c < d, what is

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Math Expert
Joined: 02 Sep 2009
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If a, b, c, and d are consecutive integers and a < b < c < d, what is  [#permalink]

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22 Jan 2018, 00:54
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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) $$d - \frac{5}{2}$$

(B) $$d - 2$$

(C) $$d - \frac{3}{2}$$

(D) $$d + \frac{3}{2}$$

(E) $$\frac{4d-6}{7}$$

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Re: If a, b, c, and d are consecutive integers and a < b < c < d, what is  [#permalink]

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22 Jan 2018, 01:02
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Since they are consecutive integers, therefore a=d-3; b=d-2; c=d-1;

Taking average,

(d-3+d-2+d-1+d)/4 = (4d-6)/4 = d-3/2

Option C.

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Re: If a, b, c, and d are consecutive integers and a < b < c < d, what is  [#permalink]

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22 Jan 2018, 06:05
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1

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  [#permalink]

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22 Jan 2018, 07:11
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

Since consecutive integers, thus
a = d-3
b = d-2
c = d-1

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  [#permalink]

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22 Jan 2018, 07:36
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 c-2, c-1,c , c+1
A.M = (4c-2)/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  [#permalink]

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22 Jan 2018, 22:21
2
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

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Re: If a, b, c, and d are consecutive integers and a < b < c < d, what is  [#permalink]

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22 Jan 2018, 22:29
1. Av= $$\frac{(A+D)}{2}$$ (as it is consecutive evenly spaced numbers)
2. Range = D-A=3 (as it is consecutive numbers with increment of 1)
3. Substitute A= D-3 in 1. Av=$$\frac{(D-3+D)}{2}$$ ->Av=$$\frac{(2D-3)}{2}$$
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  [#permalink]

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23 Jan 2018, 07:53
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

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Re: If a, b, c, and d are consecutive integers and a < b < c < d, what is  [#permalink]

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19 Aug 2018, 09:45
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) $$d - \frac{5}{2}$$

(B) $$d - 2$$

(C) $$d - \frac{3}{2}$$

(D) $$d + \frac{3}{2}$$

(E) $$\frac{4d-6}{7}$$

since the integers are consecutive .
in terms of d
we have sum = d + (d-1) + (d-2) + (d-3)

average = $$\frac{(4d-6)}{4}$$

=d-3/2
Re: If a, b, c, and d are consecutive integers and a < b < c < d, what is   [#permalink] 19 Aug 2018, 09:45