Carcass wrote:
\(\lfloor x \rfloor =\) the largest integer less than or equal to x
Quantity A |
Quantity B |
\( \frac{1}{2}\) \(\lfloor x \rfloor \) |
\(\lfloor \frac{1}{2} x \rfloor \) |
First, let's take a moment to get a good idea of what this strange notation means.
A few examples:
\(\lfloor 5.1 \rfloor = 5\)
\(\lfloor 3 \rfloor =3\)
\(\lfloor -1.4 \rfloor =-2\)
\(\lfloor -9.8 \rfloor = -10\)
STRATEGY: Since there's no convenient algebraic way the solve the question, let's start testing some values to see if we can gain some insight into the correct answer.Try x = 0
We get:
QUANTITY A: \( (\frac{1}{2}\))(\(\lfloor x \rfloor \)) \( = (\frac{1}{2})\)\((\lfloor 0 \rfloor)\) = \((\frac{1}{2})(0) = 0\)
QUANTITY B: \(\lfloor \frac{1}{2} x \rfloor \) \(= \lfloor (\frac{1}{2})(0) \rfloor \) \(= \lfloor 0 \rfloor = 0\)
In this case,
the two quantities are equalNow let's try a non-integer value. How about x = 1.2
We get:
QUANTITY A: \( (\frac{1}{2}\))(\(\lfloor x \rfloor \)) \( = (\frac{1}{2})\)\((\lfloor 1.2 \rfloor)\) = \((\frac{1}{2})(1) = \frac{1}{2}\)
QUANTITY B: \(\lfloor \frac{1}{2} x \rfloor \) \(= \lfloor (\frac{1}{2})(1.2) \rfloor \) \(= \lfloor 0.6 \rfloor = 0\)
In this case,
Quantity A is greaterAnswer: D