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

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Bunuel
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If a, b, c, and d are consecutive integers and a < b < c < d, what is [#permalink] New post   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) \(d - \frac{5}{2}\)

(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] New post   22 Jan 2018, 07:05
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Please see my approach :cool:

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] New post   22 Jan 2018, 02:02
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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] New post   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 = 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] New post   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 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] New post   22 Jan 2018, 23:21
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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 [#permalink] New post   22 Jan 2018, 23:29
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] New post   23 Jan 2018, 08:53
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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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Re: If a, b, c, and d are consecutive integers and a < b < c < d, what is [#permalink] New post   19 Aug 2018, 10:45
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
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Re: If a, b, c, and d are consecutive integers and a < b < c < d, what is [#permalink] New post   07 Jan 2022, 00:27
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Assume that the consecutive integers a,b,c,d are 1,2,3,4.

Avg = (1+2+3+4)/4 = 10/8 = 2.5

2.5 can be written as 4 - 1.5 which is same as d - 3/2.

Option C is the answer.

Thanks,
Clifin J Francis
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Re: If a, b, c, and d are consecutive integers and a < b < c < d, what is [#permalink] New post   26 Oct 2022, 10:53
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Given that a, b, c, and d are consecutive integers and a < b < c < d, and we need to find what is the average (arithmetic mean) of a, b, c, and d in terms of d

a, b, c, d are consecutive integers
=> b = a + 1
=> c = b + 1 = a + 2
=> d = c + 1 = a + 3
=> a = d - 3

=> Average of a, b, c and d = \(\frac{a + b + c + d}{4}\) = \(\frac{a + a + 1 + a + 2 + a + 3}{4}\) => a + \(\frac{3}{2}\) = d - 3 + \(\frac{3}{2}\) = d - \(\frac{3}{2}\)

So, Answer will be C
Hope it helps!

Watch the following video to learn How to Sequence problems

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Re: If a, b, c, and d are consecutive integers and a < b < c < d, what is [#permalink] New post   25 Feb 2024, 05:18
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