Quadratic Word Problems on Cost and Quantity | Class 10

Quick answer: In cost and quantity word problems, use total cost = number of items × cost per item. Let the unknown quantity be $x$, so the cost per item is $\frac{\text{total}}{x}$, then turn the given condition into an equation and solve the resulting quadratic. Reject any negative root, because a number of items cannot be negative. For example, a shopkeeper who buys books for ₹$80$, and would pay ₹$1$ less each for $4$ more books, bought $16$ books.

In everyday buying and selling, total cost = quantity × cost per item. If the total amount is fixed and the number of items changes, the cost per item changes too. Writing this relationship as an equation often leads to a quadratic equation.

One of the two roots usually comes out negative or otherwise impossible, and must be rejected with a reason.

In CBSE Class 10 Mathematics, Chapter 4 (Quadratic Equations), word problems on cost, quantity and total expenditure are common board exam questions.

What You Will Learn

• How to write the cost per item when the total amount is fixed
• How to form an equation from a change in quantity or price
• Problem 1: the shopkeeper buying books
• Problem 2: the NCERT cottage industry pottery problem
• Problem 3: the length of a piece of cloth
• How to reject impossible roots correctly

1. Setting Up the Equation

If a fixed total amount ₹$E$ is spent on $x$ items:

$\text{Cost per item} = \frac{E}{x}$

If the same amount buys $x + k$ items, the new cost per item is $\frac{E}{x + k}$. If the price per item drops by ₹$D$:

$\frac{E}{x} - \frac{E}{x + k} = D$

2. Problem 1: Shopkeeper Buying Books

Question

A shopkeeper buys a number of books for ₹$80$. If he had bought $4$ more books for the same amount, each book would have cost ₹$1$ less. How many books did he buy?

Solution

Step 1: Define the variable. Let the number of books be $x$. Then the cost of each book is ₹$\frac{80}{x}$.

Step 2: New cost. With $x + 4$ books for ₹$80$, each book costs ₹$\frac{80}{x + 4}$.

Step 3: Form the equation. Each book costs ₹$1$ less:

$\frac{80}{x} - \frac{80}{x + 4} = 1$

Step 4: Simplify.

$\frac{80(x + 4) - 80x}{x(x + 4)} = 1 \implies \frac{320}{x^2 + 4x} = 1$

$x^2 + 4x - 320 = 0$

Step 5: Factorise. We need two numbers with product $-320$ and sum $4$: these are $20$ and $-16$.

$x^2 + 20x - 16x - 320 = 0$

$x(x + 20) - 16(x + 20) = 0$

$(x + 20)(x - 16) = 0$

Step 6: Choose the valid root. $x = -20$ or $x = 16$. The number of books cannot be negative, so $x = -20$ is rejected and $x = 16$.

Answer: The shopkeeper bought 16 books.

Check: $16$ books for ₹$80$ cost ₹$5$ each; $20$ books for ₹$80$ cost ₹$4$ each. The difference is ₹$1$. ✓

3. Problem 2: Cottage Industry Pottery (NCERT)

Question

A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was $3$ more than twice the number of articles produced on that day. If the total cost of production on that day was ₹$90$, find the number of articles produced and the cost of each article.

Solution

Step 1: Define the variable. Let the number of articles be $x$. Then the cost of each article is ₹$(2x + 3)$.

Step 2: Form the equation.

$x(2x + 3) = 90 \implies 2x^2 + 3x - 90 = 0$

Step 3: Factorise. We need two numbers with product $2 \times (-90) = -180$ and sum $3$: these are $15$ and $-12$.

$2x^2 + 15x - 12x - 90 = 0$

$x(2x + 15) - 6(2x + 15) = 0$

$(2x + 15)(x - 6) = 0$

Step 4: Choose the valid root. $x = -\frac{15}{2}$ or $x = 6$. The number of articles must be a positive whole number, so $x = 6$.

Step 5: Cost of each article.

$2(6) + 3 = 15$

Answer: 6 articles were produced, and each cost ₹15.

Check: $6 \times 15 = 90$. ✓

4. Problem 3: Length of a Piece of Cloth

Question

A piece of cloth costs ₹$200$. If the piece were $5\text{ m}$ longer and each metre of cloth cost ₹$2$ less, the cost of the piece would remain unchanged. How long is the piece, and what is the original rate per metre?

Solution

Step 1: Define the variable. Let the length be $x$ metres. The rate is ₹$\frac{200}{x}$ per metre.

Step 2: Form the equation. With length $x + 5$ and the same total cost, the rate is ₹$\frac{200}{x + 5}$, which is ₹$2$ less:

$\frac{200}{x} - \frac{200}{x + 5} = 2$

Step 3: Simplify.

$\frac{1000}{x^2 + 5x} = 2 \implies x^2 + 5x - 500 = 0$

Step 4: Factorise. Two numbers with product $-500$ and sum $5$ are $25$ and $-20$:

$(x + 25)(x - 20) = 0$

Step 5: Choose the valid root. Length cannot be negative, so $x = 20$.

Answer: The piece is 20 m long, and the original rate is ₹$\frac{200}{20} =$ ₹10 per metre.

Check: $25\text{ m}$ at ₹$8$ per metre also costs ₹$200$. ✓

5. Summary Table

6. Exam Tips and Common Mistakes

Exam Tip: Always write a sentence rejecting the impossible root, for example "the number of books cannot be negative, so $x = -20$ is rejected".

Exam Tip: Check your answer in the original words of the problem, not just in the equation.

Common Mistake: Subtracting the fractions the wrong way round. With fewer items, each costs more, so $\frac{80}{x}$ is larger than $\frac{80}{x + 4}$. The equation is $\frac{80}{x} - \frac{80}{x + 4} = 1$.

Common Mistake: Forgetting to answer every part. In the pottery problem, you must give both the number of articles and the cost of each.

Frequently Asked Questions

How do you solve cost and quantity word problems with quadratic equations?

Let the unknown quantity be x, write the cost per item as total ÷ x, form an equation from the given condition, then solve the quadratic and reject any impossible root.

How many books did the shopkeeper buy for ₹80?

He bought 16 books, at ₹5 each.

What is the answer to the NCERT pottery problem?

6 articles were produced, and each cost ₹15.

Why do we reject the negative root?

A quantity such as a number of items or a length cannot be negative, so a negative root does not make sense in the problem.