Quadratic Equation Class 10 ICSE — Bansal Chapter 6

Written by Greya LakshmiPublished on 18 September 2026Updated on 28 September 2026

This chapter of the Bansal ICSE Class 10 Mathematics book teaches you to solve word problems with quadratic equations. Chapter 6, Solving Simple Problems Based on Quadratic Equations, lays out a three-step method for turning each statement into an equation. The quadratic equation class 10 ICSE problems here follow the patterns that appear in board exams. Each section handles one type: numbers, time and work, geometry, speed, and cost price.

ICSE Class 10 Mathematics Bansal Chapter 6 Solving Simple Problems Based on Quadratic Equations PDF

What is inside Chapter 6

13 pages · 7 sections · 13 worked examples · 50 exercise questions

Section	Title
6.1	Introduction
6.2	Problems Based on Numbers
6.3	Problems Based on Time and Work
6.4	Problems Based on Geometrical Figures
6.5	Problems Based on Distance, Speed and Time
6.6	Problems on C.P. and S.P.
6.7	Miscellaneous Problems

EXERCISE 6(B): 13 questions
EXERCISE 6(C): 10 questions
EXERCISE 6(D): 11 questions
EXERCISE 6(E): 16 questions

6.1 Introduction

This section sets the working method for the whole chapter. It begins with one fact: you already know how to solve a quadratic equation. The previous chapter on quadratic equations and earlier classes covered factorising and the formula. Here the task shifts. You must read a word problem, turn it into an equation, and then solve it.

The book gives three steps to follow:

Represent the unknown quantity by the variable $x$ .
Translate the given statement to form an equation in terms of $x$ .
Solve the equation.

Why does the order matter? Naming the unknown first keeps the translation clean. The equation you build is a quadratic, so solving uses techniques you already know. Every later section applies these same three steps. Practise them now and the rest of the chapter becomes a set of familiar patterns.

A simple demonstration: “a number and its square add to 12.” Step one sets $x$ . Step two gives $x + x^2 = 12$ , so $x^2 + x - 12 = 0$ . Step three solves it as $(x + 4)(x - 3) = 0$ , giving $x = 3$ when the number must be positive. That is the whole method in action.

6.2 Problems Based on Numbers

This section works with natural numbers, whole numbers, and consecutive numbers. A natural number is a positive integer: 1, 2, 3, and so on. A whole number includes zero. The key move is to name one number $x$ and express the others in terms of it.

The book shows three patterns. First, two natural numbers differing by 3, with squares summing to 117. Second, a whole number where five times it is three less than twice its square. Third, two natural numbers summing to 8, where the difference of their reciprocals is $\frac{2}{15}$ .

For the reciprocal pattern, the equation comes from writing $\frac{1}{x} - \frac{1}{8-x} = \frac{2}{15}$ . Solving gives $x = 20$ or $x = 3$ . Since the sum is 8, reject $x = 20$ . The numbers are 3 and 5.

Why reject negative or oversized roots? The word “natural” or “whole” restricts the answer set. A natural number cannot be negative, and a part cannot exceed the stated sum.

Original worked example: find two natural numbers that differ by 4 and whose squares sum to 40. Let the numbers be $x$ and $x + 4$ . Then:

$x^2 + (x+4)^2 = 40$

Expanding gives $2x^2 + 8x + 16 = 40$ , so $x^2 + 4x - 12 = 0$ . Factorising gives $(x + 6)(x - 2) = 0$ . So $x = -6$ or $x = 2$ . Reject $-6$ because it is not natural. The numbers are 2 and 6.

6.3 Problems Based on Time and Work

This section uses the reciprocal of time as a rate. If one worker completes a job in $x$ hours, that worker’s rate is $\frac{1}{x}$ of the job per hour. Two workers together add their rates.

The book’s pattern: A takes 6 hours less than B, and together they finish in 13 hours 20 minutes. The time converts to $13 + \frac{20}{60} = \frac{40}{3}$ hours. If B takes $x$ hours, then A takes $(x - 6)$ hours, and:

$\frac{1}{x-6} + \frac{1}{x} = \frac{3}{40}$

Solving gives $x = 30$ or $x = \frac{8}{3}$ . Reject the smaller root because $x - 6$ would be negative. B alone takes 30 hours.

Original worked example: A takes 3 hours less than B, and together they finish in 2 hours. Let B take $x$ hours, so A takes $(x - 3)$ . Then:

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

This becomes $\frac{2x-3}{x(x-3)} = \frac{1}{2}$ . Cross-multiplying gives $x^2 - 7x + 6 = 0$ , so $(x-6)(x-1)=0$ . Reject $x = 1$ because $x - 3$ is negative. B takes 6 hours, and A takes 3 hours.

Check: $\frac{1}{3} + \frac{1}{6} = \frac{1}{2}$ . The rate equation balances.

6.4 Problems Based on Geometrical Figures

Geometry problems use two tools: Pythagoras’ theorem and the area-perimeter relation. For a right triangle, Pythagoras gives $a^2 + b^2 = c^2$ . For a rectangle, area equals length times breadth, and perimeter equals twice the sum of the two sides.

The book’s hypotenuse pattern: the shorter side is $x$ , the longer is $(x + 7)$ , and the hypotenuse is 13. Pythagoras gives $x^2 + (x+7)^2 = 13^2$ . Expanding and simplifying gives $x^2 + 7x - 60 = 0$ , which factors to $(x+12)(x-5)=0$ . Reject $x = -12$ because a side cannot be negative. The sides are 5 cm and 12 cm.

Original worked example: a right triangle has hypotenuse 10 cm, and the other two sides differ by 2 cm. Let the shorter side be $x$ , so the longer is $x + 2$ . Then $x^2 + (x+2)^2 = 10^2$ . Expanding gives $2x^2 + 4x + 4 = 100$ , so $x^2 + 2x - 48 = 0$ , which factors to $(x+8)(x-6)=0$ . Reject $-8$ . The sides are 6 cm and 8 cm.

Check: $6^2 + 8^2 = 36 + 64 = 100 = 10^2$ . The triangle satisfies Pythagoras.

6.5 Problems Based on Distance, Speed and Time

This section builds equations from the relation $\text{Time} = \frac{\text{Distance}}{\text{Speed}}$ . The difference in times becomes the right-hand side of the equation.

The book’s pattern: a car’s speed increases by 10 km/hr, and the 72 km journey time drops by 36 minutes. The 36 minutes converts to $\frac{36}{60}$ hours. If the original speed is $x$ , then:

$\frac{72}{x} - \frac{72}{x+10} = \frac{36}{60}$

Solving gives $x = -40$ or $x = 30$ . Reject the negative speed. The original speed is 30 km/hr.

Original worked example: a cyclist raises speed by 4 km/hr and covers 60 km in 30 minutes less. Let the original speed be $x$ . Then $\frac{60}{x} - \frac{60}{x+4} = \frac{30}{60} = \frac{1}{2}$ . This becomes $\frac{240}{x(x+4)} = \frac{1}{2}$ , so $x^2 + 4x - 480 = 0$ , which factors to $(x+24)(x-20)=0$ . Reject the negative root. The original speed is 20 km/hr.

Check: at 20 km/hr, the trip takes 3 hours; at 24 km/hr, it takes 2.5 hours. The difference is 30 minutes.

6.6 Problems on C.P. and S.P.

This section turns profit and loss statements into quadratic equations. The core relation is $\text{C.P.} - \text{Loss} = \text{S.P.}$ , and loss percent is always calculated on the cost price.

The book’s pattern: an article sells for ₹24, and the loss percent equals the cost price. If the C.P. is ₹ $x$ , then the loss is $x\%$ of $x$ , which is $\frac{x^2}{100}$ . The equation is $x - \frac{x^2}{100} = 24$ , giving $x^2 - 100x + 2400 = 0$ .

Solving gives $x = 60$ or $x = 40$ . Both roots are kept because both are valid cost prices. A trader could have paid either amount and still lost the stated percent.

Original worked example: an article sells for ₹21, and the loss percent equals the cost price. Let the C.P. be ₹ $x$ . Then $x - \frac{x^2}{100} = 21$ , giving $x^2 - 100x + 2100 = 0$ , which factors to $(x - 30)(x - 70) = 0$ . Both 30 and 70 are valid, so the C.P. is ₹30 or ₹70.

Why keep both roots here when number problems reject one? Because cost price has no upper constraint from the statement. Only a negative root would be rejected, and neither root here is negative.

6.7 Miscellaneous Problems

This final section mixes the earlier types with new settings: sums of natural numbers, two-digit numbers, ages, boats and streams, and the cost of articles. Each problem still follows the same three steps from Section 6.1.

The sum of the first $n$ natural numbers is given by $S = \frac{n(n+1)}{2}$ . The book sets $S = 276$ , giving $n^2 + n - 552 = 0$ , which factors to $(n + 24)(n - 23) = 0$ . Reject $-24$ . So $n = 23$ .

Original worked example: find $n$ when the sum of the first $n$ natural numbers is 210. Set $\frac{n(n+1)}{2} = 210$ , so $n^2 + n - 420 = 0$ , which factors to $(n + 21)(n - 20) = 0$ . Reject $-21$ . So $n = 20$ . Check: $\frac{20 \times 21}{2} = 210$ .

The section also covers two-digit numbers written as $10x + y$ , age problems where one age is a square of another, boat problems using downstream speed $(9 + x)$ and upstream speed $(9 - x)$ , and cost problems where a lower rate buys more articles. All of them collapse into a quadratic after translation.

What to practise from this chapter

This chapter’s exercises build in difficulty. Work through them in order, and check the board-exam years printed beside many questions.

EXERCISE 6(B) — 13 questions on geometrical figures, including right triangles, rectangles, and paths.
EXERCISE 6(C) — 10 questions on speed, trains, and cost price.
EXERCISE 6(D) — 11 miscellaneous questions on sums, ages, and boats.
EXERCISE 6(E) — 16 mixed questions that combine every type from the chapter.

Board exam years appear in the chapter: 2005, 2007, 2008, 2010, 2011, 2012, 2013, 2014, and 2015. These include the speed of a car (2005), an age problem (2007), a picnic budget (2008), a cistern (2010), a division of money (2011), a car’s speed (2012), a book purchase (2013), a two-digit number (2014), and reciprocal numbers (2015).

Common mistakes to avoid:

Rejecting the wrong root. Check the word “natural”, “whole”, or “positive” before choosing.
Forgetting to convert minutes to hours. 36 minutes becomes $\frac{36}{60}$ hours.
Writing reciprocals in the wrong order when the problem states a difference.
Keeping a negative root for a length, speed, or count. Only the positive root is valid.

Here is the comparison of the five problem types:

Problem type	Unknown $x$	Typical equation form
Numbers	a natural or whole number	$x^2 + bx + c = 0$
Time and work	time one worker takes	$\frac{1}{x-a} + \frac{1}{x} = \frac{1}{T}$
Geometrical figures	a side or breadth	$x^2 + (x+k)^2 = h^2$
Distance, speed, time	original speed	$\frac{d}{x} - \frac{d}{x+k} = t$
C.P. and S.P.	cost price	$x - \frac{x^2}{100} = \text{S.P.}$

For more chapters from the Bansal book, see the Class 10 Maths chapter list. The official CISCE website carries the full syllabus and examination details.

Frequently asked questions

How do I decide which root to reject in a word problem on quadratic equations?

Read the type of number the question names. If it says natural, whole, positive, or a length or speed, reject any negative root. If it says the sum of two numbers is 8, reject any root larger than 8. In cost price problems, keep both positive roots unless the statement rules one out.

Why do we write 13 hours 20 minutes as 40/3 hours in time and work problems?

Because the rate equation needs time in hours. Convert 20 minutes to $\frac{20}{60} = \frac{1}{3}$ hours, then add to 13 to get $13\frac{1}{3} = \frac{40}{3}$ hours. The reciprocal of that total time is $\frac{3}{40}$ , which appears on the right side of the equation.

How do I form the equation when a problem says the sum of the reciprocals of two numbers is a fraction?

Name the numbers $x$ and $8 - x$ (or whatever the sum is). Write the reciprocals as $\frac{1}{x}$ and $\frac{1}{8-x}$ . The difference or sum of these two fractions equals the given fraction. Then combine over a common denominator and cross-multiply to get the quadratic.

Why are both roots accepted in the cost price and selling price problem?

Because the statement does not limit the cost price. Any positive value that satisfies $x - \frac{x^2}{100} = \text{S.P.}$ is a possible answer. In the book’s example, ₹60 and ₹40 both work because both give a loss percent that equals the cost price.

Reference: ICSE Class 10 Mathematics textbook material, chapter 6: Solving Simple Problems Based on Quadratic Equations — CISCE-aligned.

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