Mathematics — CBSE Class 10 board question

Let the numerator = x.

Then denominator = 2x + 1, so the fraction = $\dfrac{x}{2x+1}$.

Setting up the equation:

Sum of fraction and its reciprocal = $2\dfrac{16}{21} = \dfrac{58}{21}$

$$\frac{x}{2x+1} + \frac{2x+1}{x} = \frac{58}{21}$$

$$\frac{x^2 + (2x+1)^2}{x(2x+1)} = \frac{58}{21}$$

$$\frac{x^2 + 4x^2 + 4x + 1}{2x^2 + x} = \frac{58}{21}$$

$$\frac{5x^2 + 4x + 1}{2x^2 + x} = \frac{58}{21}$$

Cross-multiplying:

$$21(5x^2 + 4x + 1) = 58(2x^2 + x)$$

$$105x^2 + 84x + 21 = 116x^2 + 58x$$

$$11x^2 - 26x - 21 = 0$$

Factorising:

$$11x^2 - 33x + 7x - 21 = 0$$

$$11x(x - 3) + 7(x - 3) = 0$$

$$(11x + 7)(x - 3) = 0$$

$$x = 3 \quad \text{or} \quad x = -\frac{7}{11}$$

Since x must be a positive integer, x = 3.

The fraction = $\dfrac{3}{7}$.

Source: Chapter 4, Quadratic Equations

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• Examiners expect you to clearly define the variable, form the equation, simplify, factorize, and state the answer — each step earns marks.
• Convert the mixed number $2\tfrac{16}{21}$ to $\tfrac{58}{21}$ immediately to avoid errors.
• Reject $x = -\tfrac{7}{11}$ because the numerator of a fraction described in this context must be a positive integer. Always justify rejection of a root.
• The factorisation step: split $-26x$ as $-33x + 7x$ (since $-33 \times 7 = -231 = 11 \times (-21)$).