Since the values (a, b and c) are 3 consecutive integers, then the range must be 2.
We know this, because b is 1 greater than a, which means c is 2 greater than a
Range = c - a = 2

As far as the average is concerned, we know very little. So, let's TEST some possible cases:

CASE A: a, b and c COULD equal 1, 2 and 3 (respectively), in which case the average = (1 + 2 + 3)/3 = 2
So, we get:
Quantity A: 2
Quantity B: 2
In this case, the two quantities are EQUAL

CASE B: a, b and c COULD equal 0, 1 and 2 (respectively), in which case the average = (0 + 1 + 2)/3 = 1
So, we get:
Quantity A: 2
Quantity B: 1
In this case, Quantity A IS GREATER

Answer: D

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Let a, b, c be 1,2,3 respectively

The range would be 3 - 1 = 2

The average = 6/3 = 2 we get once case where they are equal.

Try a = - 5 b = -4 and c = -3

The range is -3 - (-5) = -3 + 5 = 2

The average is -5 -4 -3 = -12/3 = -4

Range is larger than average. This leads to answer choice D

Posted from my mobile device

Case B ain't right.

Oops, I totally ignored the fact that a < b < c < 4
I've edited my solution.

Thanks for the heads up!!

Cheers,
Brent
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0 is an integer! That's the most important fact to know here

For example, if given integers are 0<1<2<4 then option A will be greater
But it given integers are 1<2<3<4 then option B will be greater

Thus, as we don't know the exact values of integers option D is the correct answer

QA:
Given that a b and c are consecutive integers, with a being the smallest we can express:
a=x
b=x+1
c=x+2
range is (max-min)=x+2-x=2
so QA is always 2.

QB:
Knowing that they are consecutive integers the average can be expressed as:
\(\frac{(x+x+1+x+2)}{3}\)
\(\frac{(3x+3)}{3}\)
\(avg=x+1\)
Seeing that this value depends on x, we need to find the highest possible value of a
This occurs when: c=3, b=2, a=1
so x<=1
Notice that for the case of x=1, QB and QA are the same
For x<1, QA is larger than QB.

Two outcomes, D