There are three positive numbers - a, b and c. The average of a and b is less than the average of b and c by 1. I. Value of c II. Value of a

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1. Let the average of a and b be less than the average of b and c by 1, so

$\frac{a+b}{2}=\frac{b+c}{2}-1$.

2. Multiply both sides by 2 and simplify:

$a+b=b+c-2\;\Rightarrow\;a=c-2$.

3. Therefore $c=a+2$, and since the numbers are positive, $c>a$. Hence Quantity I (value of $c$) is greater than Quantity II (value of $a$).

FINAL_ANSWER: b

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