Fraction Word Problems on Quadratic Equations | Class 10

Quick answer: Let the denominator be $x$, so a fraction whose numerator is 3 less is $\frac{x - 3}{x}$. If adding 1 to the denominator decreases it by $\frac{1}{15}$, then $\frac{x - 3}{x} - \frac{x - 3}{x + 1} = \frac{1}{15}$, which simplifies to $x^2 - 14x + 45 = 0$. So $x = 9$ or $x = 5$, and the fraction is 6/9 (= 2/3) or 2/5; both values check.

How to set up a fraction word problem

Step 1: Let the denominator (or numerator) be $x$, and write the other part in terms of $x$.

Step 2: Write the new fraction after the change described in the question.

Step 3: Form the equation from the condition given (sum, difference, "decreased by", and so on).

Step 4: Clear the denominators to get a quadratic equation $ax^2 + bx + c = 0$.

Step 5: Solve it, check each root, and write the final fraction, not just $x$.

Problem 1: numerator 3 less than the denominator

The numerator of a fraction is 3 less than its denominator. If 1 is added to the denominator, the fraction is decreased by $\frac{1}{15}$. Find the fraction.

Step 1: Set up the fraction.

Let the denominator be $x$. Then the numerator is $x - 3$ and the fraction is $\frac{x - 3}{x}$.

Step 2: The new fraction.

$\frac{x - 3}{x + 1}$

Step 3: Form the equation. The fraction is decreased by $\frac{1}{15}$, so original − new = $\frac{1}{15}$:

$\frac{x - 3}{x} - \frac{x - 3}{x + 1} = \frac{1}{15}$

Step 4: Take $(x - 3)$ common.

$(x - 3)\left[\frac{1}{x} - \frac{1}{x + 1}\right] = \frac{1}{15}$

$(x - 3) \cdot \frac{1}{x(x + 1)} = \frac{1}{15}$

Step 5: Cross-multiply.

$15(x - 3) = x^2 + x$

$x^2 - 14x + 45 = 0$

Step 6: Factorise. Two numbers with sum −14 and product 45 are −9 and −5.

$(x - 9)(x - 5) = 0$

So $x = 9$ or $x = 5$.

Step 7: Check both values.

Answer: The fraction is $\frac{6}{9}$ (that is, $\frac{2}{3}$) or $\frac{2}{5}$. Both satisfy the conditions.

Note that if you reduce 6/9 to 2/3 first, the numerator is no longer 3 less than the denominator, so write the answer as 6/9 when you state it against the condition.

Problem 2: a number and its reciprocal

The sum of a number and its reciprocal is $\frac{10}{3}$. Find the number.

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

$x + \frac{1}{x} = \frac{10}{3}$

Step 2: Clear the fractions.

$3(x^2 + 1) = 10x$

$3x^2 - 10x + 3 = 0$

Step 3: Factorise. Split the middle term using −9 and −1 (product 9, sum −10):

$3x^2 - 9x - x + 3 = 0$

$(3x - 1)(x - 3) = 0$

So $x = 3$ or $x = \frac{1}{3}$.

Answer: The number is 3 or $\frac{1}{3}$. Check: $3 + \frac{1}{3} = \frac{10}{3}$. ✓

Summary table

Exam tips

• Write "Let the denominator be $x$" clearly at the start; it earns the setting-up mark.
• "Decreased by" means original − new; "increased by" means new − original.
• Look for a common factor (like $x - 3$) before taking the LCM; it saves expansion.
• Check every root in the original condition, and reject only those that are impossible (such as a zero or negative length or count).

Common mistakes

• Subtracting in the wrong order (new − original) when the fraction is "decreased".
• Stopping at $x = 9$ without writing the fraction.
• Discarding a valid root. Here both $x = 9$ and $x = 5$ work.
• Errors when cross-multiplying, such as forgetting to multiply 15 by both terms of $x - 3$.

Frequently Asked Questions

How do you solve fraction word problems using quadratic equations?

Let the denominator be x, write the numerator and the changed fraction in terms of x, form an equation from the condition, clear the fractions to get a quadratic, then solve and check.

What is the fraction if the numerator is 3 less than the denominator and adding 1 to the denominator decreases it by 1/15?

The fraction is 6/9 (which equals 2/3) or 2/5. Both values of the denominator, 9 and 5, satisfy the condition.

What does "the fraction is decreased by 1/15" mean?

It means the new fraction is 1/15 less than the original, so original fraction − new fraction = 1/15.

Which number added to its reciprocal gives 10/3?

The number is 3 or 1/3, because 3 + 1/3 = 10/3.

Should both roots of the quadratic be accepted?

Accept every root that fits the conditions of the problem. Reject a root only if it gives an impossible answer, such as a zero denominator or a negative count.