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
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
\[ P \# Q = \frac{P+PQ}{Q} \] \[ P \, \$ \, Q = \frac{P-PQ}{Q} \] \[ P \; ? \; Q = \frac{Q+PQ}{P} \] \[ P \, @ \, Q = \frac{Q-PQ}{P} \]
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
Show 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.
Updated On: Jul 15, 2026
- 0
- \( -2 \)
- \( \dfrac{4}{3} \)
- Infinity
Show Solution
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The Correct Option is A
Solution and Explanation
Step 1: Write the four operator values for \(P=1\), \(Q=2\).
Here \(PQ = 1 \times 2 = 2\).
\[ P \# Q = \frac{1+2}{2} = \frac{3}{2} \] \[ P \, \$ \, Q = \frac{1-2}{2} = -\frac{1}{2} \] \[ P \; ? \; Q = \frac{2+2}{1} = 4 \] \[ P \, @ \, Q = \frac{2-2}{1} = 0 \]
Step 2: Rewrite the division of two fractions as a single fraction, before substituting.
For any numbers, \( \dfrac{x}{y} \div \dfrac{u}{v} = \dfrac{x}{y} \times \dfrac{v}{u} = \dfrac{xv}{yu} \). Applying this with \(x = P \# Q\), \(y = P \, \$ \, Q\), \(u = P \; ? \; Q\) and \(v = P \, @ \, Q\) gives:
\[ \left( \frac{P \# Q}{P \, \$ \, Q} \right) \div \left( \frac{P \; ? \; Q}{P \, @ \, Q} \right) = \frac{(P \# Q)(P \, @ \, Q)}{(P \, \$ \, Q)(P \; ? \; Q)} \]
Writing it this way avoids ever dividing by \(P \, @ \, Q = 0\) directly, since \(P \, @ \, Q\) only appears multiplied on the top, never as a divisor.
Step 3: Substitute the numbers.
\[ \frac{(P \# Q)(P \, @ \, Q)}{(P \, \$ \, Q)(P \; ? \; Q)} = \frac{\left(\frac{3}{2}\right)(0)}{\left(-\frac{1}{2}\right)(4)} = \frac{0}{-2} = 0 \]
Step 4: Why the other options are wrong.
Option (d), Infinity, is what a student gets if they wrongly compute \(P \; ? \; Q \div P \, @ \, Q\) as \(4 \div 0\) on its own, without first combining it properly with the rest of the expression, since \(P \, @ \, Q\) actually ends up multiplied in the numerator once the full division is rewritten as a single fraction. Option (b), \(-2\), comes from only computing \( \dfrac{P \, \$ \, Q}{P \; ? \; Q}\) or a similar partial ratio and stopping early. Option (c), \(4/3\), comes from mixing up which term sits on top and which sits at the bottom while combining the two ratios.
Final Answer:
\[ \boxed{0} \]
Here \(PQ = 1 \times 2 = 2\).
\[ P \# Q = \frac{1+2}{2} = \frac{3}{2} \] \[ P \, \$ \, Q = \frac{1-2}{2} = -\frac{1}{2} \] \[ P \; ? \; Q = \frac{2+2}{1} = 4 \] \[ P \, @ \, Q = \frac{2-2}{1} = 0 \]
Step 2: Rewrite the division of two fractions as a single fraction, before substituting.
For any numbers, \( \dfrac{x}{y} \div \dfrac{u}{v} = \dfrac{x}{y} \times \dfrac{v}{u} = \dfrac{xv}{yu} \). Applying this with \(x = P \# Q\), \(y = P \, \$ \, Q\), \(u = P \; ? \; Q\) and \(v = P \, @ \, Q\) gives:
\[ \left( \frac{P \# Q}{P \, \$ \, Q} \right) \div \left( \frac{P \; ? \; Q}{P \, @ \, Q} \right) = \frac{(P \# Q)(P \, @ \, Q)}{(P \, \$ \, Q)(P \; ? \; Q)} \]
Writing it this way avoids ever dividing by \(P \, @ \, Q = 0\) directly, since \(P \, @ \, Q\) only appears multiplied on the top, never as a divisor.
Step 3: Substitute the numbers.
\[ \frac{(P \# Q)(P \, @ \, Q)}{(P \, \$ \, Q)(P \; ? \; Q)} = \frac{\left(\frac{3}{2}\right)(0)}{\left(-\frac{1}{2}\right)(4)} = \frac{0}{-2} = 0 \]
Step 4: Why the other options are wrong.
Option (d), Infinity, is what a student gets if they wrongly compute \(P \; ? \; Q \div P \, @ \, Q\) as \(4 \div 0\) on its own, without first combining it properly with the rest of the expression, since \(P \, @ \, Q\) actually ends up multiplied in the numerator once the full division is rewritten as a single fraction. Option (b), \(-2\), comes from only computing \( \dfrac{P \, \$ \, Q}{P \; ? \; Q}\) or a similar partial ratio and stopping early. Option (c), \(4/3\), comes from mixing up which term sits on top and which sits at the bottom while combining the two ratios.
Final Answer:
\[ \boxed{0} \]
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