If a, b, c, and d are consecutive integers and a

Question

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}

GMATBusters · Accepted Answer

Please see my approach

blueviper · Answer

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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anugrahs · Answer

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

rahul16singh28 · Answer

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)

KarishmaB · Answer

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)

seed · Answer

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)

JeffTargetTestPrep · Answer

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

CounterSniper · Answer

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

CrackverbalGMAT · Answer

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

BrushMyQuant · Answer

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

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