Quadratic Word Problems on Reciprocals and Fractions

Quick answer: For word problems on a number and its reciprocal, let the number be $x$, so its reciprocal is $\frac{1}{x}$. Multiply through by the denominators to get a quadratic equation, then solve it. For example, if a number plus its reciprocal is $\frac{61}{30}$, then $30x^2 - 61x + 30 = 0$, so the number is $\frac{5}{6}$ or $\frac{6}{5}$. Always check the roots against the conditions in the question.

Some word problems link a number to its reciprocal, or describe a fraction whose numerator and denominator are related. Writing these conditions as equations gives expressions with fractions, such as $x + \frac{1}{x}$. Multiplying through by the denominators turns them into quadratic equations of the form $ax^2 + bx + c = 0$.

In CBSE Class 10 Mathematics, Chapter 4 (Quadratic Equations), word problems on numbers, reciprocals and fractions are common board exam questions.

What You Will Learn

• How to write a number and its reciprocal, and a fraction, in terms of $x$
• How to clear fractions to get a quadratic equation
• Problem 1: a number plus its reciprocal equals $2\frac{1}{30}$
• Problem 2: finding a fraction from a condition on its numerator and denominator
• Problem 3: a natural number that, increased by $12$, equals $160$ times its reciprocal
• How to reject roots that do not fit the question

1. Writing the Equations

For a number and its reciprocal with sum $K$:

$x + \frac{1}{x} = K \implies \frac{x^2 + 1}{x} = K \implies x^2 - Kx + 1 = 0$

2. Problem 1: A Number and Its Reciprocal

Question

The sum of a number and its reciprocal is $2\frac{1}{30}$. Find the number.

Solution

Step 1: Form the equation. Let the number be $x$. Since $2\frac{1}{30} = \frac{61}{30}$:

$x + \frac{1}{x} = \frac{61}{30}$

Step 2: Clear the fractions.

$\frac{x^2 + 1}{x} = \frac{61}{30} \implies 30(x^2 + 1) = 61x \implies 30x^2 - 61x + 30 = 0$

Step 3: Factorise. We need two numbers with product $30 \times 30 = 900$ and sum $-61$: these are $-36$ and $-25$.

$30x^2 - 36x - 25x + 30 = 0$

$6x(5x - 6) - 5(5x - 6) = 0$

$(6x - 5)(5x - 6) = 0$

Step 4: Solve. $x = \frac{5}{6}$ or $x = \frac{6}{5}$.

Answer: The number is $\frac{5}{6}$ or $\frac{6}{5}$.

Check: $\frac{5}{6} + \frac{6}{5} = \frac{25 + 36}{30} = \frac{61}{30}$. ✓ The two answers are reciprocals of each other, as expected.

3. Problem 2: Finding a Fraction

Question

The numerator of a fraction is $3$ less than its denominator. If $2$ is added to both the numerator and the denominator, the sum of the new fraction and the original fraction is $\frac{29}{20}$. Find the original fraction.

Solution

Step 1: Write the fractions. Let the denominator be $x$. Then:

$\text{Original fraction} = \frac{x - 3}{x} \qquad \text{New fraction} = \frac{x - 1}{x + 2}$

Step 2: Form the equation.

$\frac{x - 3}{x} + \frac{x - 1}{x + 2} = \frac{29}{20}$

Step 3: Combine the fractions.

$\frac{(x - 3)(x + 2) + x(x - 1)}{x(x + 2)} = \frac{29}{20}$

$\frac{(x^2 - x - 6) + (x^2 - x)}{x^2 + 2x} = \frac{29}{20} \implies \frac{2x^2 - 2x - 6}{x^2 + 2x} = \frac{29}{20}$

Step 4: Cross-multiply.

$20(2x^2 - 2x - 6) = 29(x^2 + 2x)$

$40x^2 - 40x - 120 = 29x^2 + 58x$

$11x^2 - 98x - 120 = 0$

Step 5: Factorise. We need two numbers with product $11 \times (-120) = -1320$ and sum $-98$: these are $-110$ and $12$.

$11x^2 - 110x + 12x - 120 = 0$

$11x(x - 10) + 12(x - 10) = 0$

$(11x + 12)(x - 10) = 0$

Step 6: Choose the valid root. $x = 10$ or $x = -\frac{12}{11}$. The denominator must give a fraction with whole-number parts, so $x = 10$.

Answer: The original fraction is $\frac{7}{10}$.

Check: $\frac{7}{10} + \frac{9}{12} = \frac{7}{10} + \frac{3}{4} = \frac{14}{20} + \frac{15}{20} = \frac{29}{20}$. ✓

4. Problem 3: A Natural Number and Its Reciprocal

Question

A natural number, when increased by $12$, becomes equal to $160$ times its reciprocal. Find the number.

Solution

Step 1: Form the equation. Let the natural number be $x$.

$x + 12 = \frac{160}{x}$

Step 2: Clear the fraction.

$x^2 + 12x = 160 \implies x^2 + 12x - 160 = 0$

Step 3: Factorise. Two numbers with product $-160$ and sum $12$ are $20$ and $-8$:

$(x + 20)(x - 8) = 0$

Step 4: Choose the valid root. $x = -20$ or $x = 8$. A natural number cannot be negative, so $x = -20$ is rejected.

Answer: The number is $8$.

Check: $8 + 12 = 20$ and $160 \times \frac{1}{8} = 20$. ✓

5. Summary Table

6. Exam Tips and Common Mistakes

Exam Tip: Convert mixed numbers to improper fractions first, e.g. $2\frac{1}{30} = \frac{61}{30}$.

Exam Tip: Always check your answer by substituting it back into the words of the question.

Common Mistake: Forgetting to give a reason for rejecting a root, e.g. "a natural number cannot be negative, so $x = -20$ is rejected".

Common Mistake: In fraction problems, adding the number only to the numerator. Read carefully whether it is added to the numerator, the denominator, or both.

Frequently Asked Questions

How do you solve a number and its reciprocal problem?

Let the number be x and its reciprocal 1/x. Write the given condition as an equation, multiply through by x to clear the fraction, and solve the resulting quadratic.

What is the number if its sum with its reciprocal is 2 1/30?

The number is 5/6 or 6/5.

Why do the two answers come out as reciprocals?

The equation x + 1/x = K does not change if x is replaced by 1/x, so if one number works, its reciprocal works too.

What natural number increased by 12 equals 160 times its reciprocal?

The number is 8, since 8 + 12 = 20 and 160 ÷ 8 = 20.