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Re: a, b, and c are consecutive integers such that a < b < c < 4 [#permalink]
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GreenlightTestPrep wrote:
sandy wrote:
a, b, and c are consecutive integers such that a < b < c < 4.

Quantity A
Quantity B
The range of a, b, and c
The average of a, b, and c


A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.


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 8, 9 and 10 (respectively), in which case the average = (8 + 9 + 10)/3 = 9
So, we get:
Quantity A: 2
Quantity B: 9
In this case, Quantity B IS GREATER

Answer: D

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Case B ain't right.
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Re: a, b, and c are consecutive integers such that a < b < c < 4 [#permalink]
indiragre18 wrote:
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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Re: a, b, and c are consecutive integers such that a < b < c < 4 [#permalink]
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
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Re: a, b, and c are consecutive integers such that a < b < c < 4 [#permalink]
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
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