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Question

If $P + \frac{1}{Q} = 1$ and $Q + \frac{1}{R} = 1$, then what is PQR?

The correct answer is
-1

Solving for PQR from Algebraic Equations

We are given the following algebraic equations:

  • Equation 1: $P + \frac{1}{Q} = 1$
  • Equation 2: $Q + \frac{1}{R} = 1$

The objective is to determine the value of the product $PQR$.

Deriving Expressions for P and R

From Equation 1, we can isolate $P$:

$P = 1 - \frac{1}{Q}$

Combine the terms on the right side:

$P = \frac{Q - 1}{Q}$

From Equation 2, we can isolate $\frac{1}{R}$:

$\frac{1}{R} = 1 - Q$

Assuming $Q \neq 1$ (which must be true, otherwise $1/R = 0$, which is impossible), we find $R$:

$R = \frac{1}{1 - Q}$

Calculating the Product PQR

Substitute the derived expressions for $P$ and $R$ into the product $PQR$:

$PQR = \left(\frac{Q - 1}{Q}\right) \times Q \times \left(\frac{1}{1 - Q}\right)$

Cancel the variable $Q$ present in the numerator and denominator:

$PQR = (Q - 1) \times \frac{1}{1 - Q}$

Factor out $-1$ from $(Q - 1)$:

$PQR = -(1 - Q) \times \frac{1}{1 - Q}$

Cancel the $(1 - Q)$ terms:

$PQR = -1 \times 1$

$PQR = -1$

Conclusion

The value of $PQR$ calculated from the given equations is -1.

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Important Questions from Algebra (Notes)

  1. If $(y-12) = 4\sqrt{5}$, then find the value of $\sqrt{y-3} - \frac{1}{\sqrt{y-3}}$.
  2. If $x^2 + \frac{1}{x^2} = 16$ and $x \neq 0$, then what is the value of $x^4 + \frac{1}{x^4}$?
  3. In the expansion of (x + 9)(x - 6)(x + 5), what is the coefficient of x?
  4. Find the value of $\frac{x+3}{x^2-2x} \times \frac{2x-1}{x^2+2x+4} \times \frac{x^4-8x}{2x^2+5x-3}$
  5. The roots of the equation $ax^3-24x^2+188x-480=0$ are three consecutive even natural numbers. The value of a is _____.
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