📘 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
- Consecutive integer problems
- Geometry applications with radicals
- Age-related word problems
- Profit and percentage calculations
- Tax calculations
- Compound interest/markup
- Temperature conversion
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
The sum of three consecutive integers is forty-two. Find the three integers.
Let first integer = $x$
Second integer = $x + 1$
Third integer = $x + 2$
$x + (x + 1) + (x + 2) = 42$
$3x + 3 = 42$
$3x = 39$
$x = 13$
First integer = $13$
Second integer = $14$
Third integer = $15$
Final Answer: The three consecutive integers are 13, 14, and 15
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 $AB$
Area of triangle = $\frac{1}{2} \times base \times height$
$(1 + \sqrt{15}) = \frac{1}{2} \times (\sqrt{3} + \sqrt{5}) \times AB$
$AB = \frac{2(1 + \sqrt{15})}{\sqrt{3} + \sqrt{5}}$
$AB = \frac{2 + 2\sqrt{15}}{\sqrt{3} + \sqrt{5}} \times \frac{\sqrt{3} - \sqrt{5}}{\sqrt{3} - \sqrt{5}}$
Final Answer: $AB = 4\sqrt{3} - 2\sqrt{5}$ cm
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.
$^\circ F = \frac{9}{5} \times 48 + 32$
$^\circ F = 86.4 + 32 = 118.4$
Final Answer: $118.4^\circ F$
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%?
Taxable Income = Annual Income - Exempted Amount
= 960,000 - 130,000 = 830,000 Rs
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