Question

Answer the question on the basis of the information given below.

\[ P \# Q = \frac{P+PQ}{Q} \] \[ P \, \$ \, Q = \frac{P-PQ}{Q} \] \[ P \; ? \; Q = \frac{Q+PQ}{P} \] \[ P \, @ \, Q = \frac{Q-PQ}{P} \]

The value of \( \left( \dfrac{P \# Q}{P \, \$ \, Q} \right) \div \left( \dfrac{P \; ? \; Q}{P \, @ \, Q} \right) \), where \(P = 1\) and \(Q = 2\), is

• A. 0
• B. \( -2 \)
• C. \( \dfrac{4}{3} \)
• D. Infinity

Hint:

Instead of dividing by P ? Q over P @ Q directly (which involves dividing by zero), rewrite the whole expression as one fraction, (P # Q)(P @ Q) over (P $ Q)(P ? Q), then substitute the numbers.

Answer 1

The Correct Option is A

Understanding the Concept:
The expression combines four custom operators through one division of two ratios. Plugging in numbers before simplifying the algebra can create a zero in an inconvenient place, so this method simplifies the operator values symbolically first, using $P=1$ to make the arithmetic light.

Key Approach:
With $P=1$, each formula becomes simpler: $1 \# Q = \dfrac{1+Q}{Q}$, $1 \, \$ \, Q = \dfrac{1-Q}{Q}$, $1 \; ? \; Q = Q+Q = 2Q$, and $1 \, @ \, Q = Q-Q = 0$ for any value of $Q$, since $P \, @ \, Q = Q-PQ = Q-Q = 0$ whenever $P=1$.

Detailed Explanation:
Since $P \, @ \, Q = 0$ for every $Q$ when $P=1$, the second ratio in the question, $\dfrac{P\;?\;Q}{P\,@\,Q}$, has a zero denominator, so it should never be evaluated as a standalone number. Instead, treat the full statement $\dfrac{P\#Q}{P\,\$\,Q} \div \dfrac{P\;?\;Q}{P\,@\,Q}$ as one combined fraction by flipping the second ratio and multiplying: this gives $\dfrac{(P\#Q)\times(P\,@\,Q)}{(P\,\$\,Q)\times(P\;?\;Q)}$, where $P\,@\,Q=0$ now sits in the numerator, not the denominator.
With $Q=2$: $1\#2 = \dfrac{1+2}{2} = \dfrac{3}{2}$, $1\,\$\,2 = \dfrac{1-2}{2} = -\dfrac{1}{2}$, $1\;?\;2 = 2\times 2 = 4$, and $1\,@\,2 = 0$.
Substituting: numerator $= \dfrac{3}{2}\times 0 = 0$, denominator $= -\dfrac{1}{2}\times 4 = -2$.
So the value is $\dfrac{0}{-2} = 0$.
Since the numerator is zero and the denominator, $-2$, is a normal nonzero number, the whole fraction is safely and exactly $0$, with no division-by-zero problem anywhere in the calculation.

Final Answer:
Rewriting the division as a single fraction before substituting keeps the zero in the numerator, giving a clean value of $0$, so the correct choice is option (a). \[ \boxed{0} \]