Exercise 1.3

Class 9 Mathematics Notes | Unit 1 – Real Numbers | Class 9 Mathematics (New Syllabus)

📘 Exercise 1.3

Learn and practice Class 9 Mathematics Exercise 1.3 on Real Numbers with step-by-step solutions. This page includes free downloadable notes and solved examples focusing on word problems, geometry, and real-life mathematical applications as per the Punjab PECTAA 2025 syllabus.

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Exercise 1.3 - Real Numbers

This exercise focuses on word problems, geometry applications, and real-world mathematics involving real numbers.

Exercise 1.3 provides practice problems that help students apply mathematical concepts to solve real-world problems, work with geometric applications, and understand practical uses of mathematics in daily life.

The exercise covers consecutive integer problems, geometry with radicals, age problems, financial calculations, and unit conversions.

Key Topics Covered

Important Note:

This exercise bridges the gap between abstract mathematical concepts and real-world applications, showing how mathematics is used in everyday situations from finance to geometry to temperature conversion.

Sample Problems with Solutions

Problem 1: Consecutive Integers

The sum of three consecutive integers is forty-two. Find the three integers.

1
Define the variables

Let first integer = $x$
Second integer = $x + 1$
Third integer = $x + 2$

2
Set up the equation

$x + (x + 1) + (x + 2) = 42$

3
Solve the equation

$3x + 3 = 42$
$3x = 39$
$x = 13$

4
Find all integers

First integer = $13$
Second integer = $14$
Third integer = $15$

Final Answer: The three consecutive integers are 13, 14, and 15

Problem 2: Geometry Application

The diagram shows right angled $\triangle ABC$ in which the length of $AC$ is $(\sqrt{3} + \sqrt{5})$ cm. The area of $\triangle ABC$ is $(1 + \sqrt{15})cm^2$. Find the length of $\overline{AB}$.

Right Triangle Diagram

$\triangle ABC$ with right angle at B

$AC = \sqrt{3} + \sqrt{5}$ cm (hypotenuse)

Area = $1 + \sqrt{15}$ cm²

Find $AB$

1
Use area formula

Area of triangle = $\frac{1}{2} \times base \times height$
$(1 + \sqrt{15}) = \frac{1}{2} \times (\sqrt{3} + \sqrt{5}) \times AB$

2
Solve for AB

$AB = \frac{2(1 + \sqrt{15})}{\sqrt{3} + \sqrt{5}}$

3
Rationalize the denominator

$AB = \frac{2 + 2\sqrt{15}}{\sqrt{3} + \sqrt{5}} \times \frac{\sqrt{3} - \sqrt{5}}{\sqrt{3} - \sqrt{5}}$

Final Answer: $AB = 4\sqrt{3} - 2\sqrt{5}$ cm

Problem 5: Temperature Conversion

The weather in Lahore was usually warm during the summer of 2024. The TV news reported temperature as high as 48°C. Using the formula $^\circ F = \frac{9}{5} C + 32$, find the temperature in Fahrenheit scale.

1
Apply the conversion formula

$^\circ F = \frac{9}{5} \times 48 + 32$

2
Calculate

$^\circ F = 86.4 + 32 = 118.4$

Final Answer: $118.4^\circ F$

Problem 8: Tax Calculation

The annual income of Tayyab is Rs. 9,60,000 while the exempted amount is Rs. 1,30,000. How much tax would he have to pay at the rate of 0.75%?

1
Calculate taxable income

Taxable Income = Annual Income - Exempted Amount
= 960,000 - 130,000 = 830,000 Rs

2
Calculate tax amount

Tax = 0.75% of 830,000
= $\frac{0.75}{100} \times 830,000 = 6,225$ Rs

Final Answer: Rs. 6,225

Created by Hira Science Academy | Aligned with PECTAA 2025 Syllabus

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