Printable Learning Resources for Class 9

The Case Study Class 9 Number System PDF provides structured, exam-focused practice. It allows students to revise important concepts offline. Therefore, preparation becomes more flexible.

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PDF resources are easy to print. Moreover, students can practice anytime. As a result, learning continuity improves.

Alignment with CBSE Exams

The Case Study Class 9 Number System PDF follows the latest CBSE assessment pattern. Consequently, students gain familiarity with application-based questions.

Effective Revision Tool

Using the Case Study Class 9 Number System regularly improves accuracy. In addition, it boosts exam confidence and conceptual understanding.

Case Study: Class 9 Number System

In a high-precision chemistry laboratory, a group of researchers is measuring the density of various rare earth elements. The lead scientist, Dr. Mehta, records the weight of a sample as 1.6̅ grams. He explains to his assistants that this weight is a rational number because it is a recurring decimal. To calibrate the digital scales, they use a reference weight that is exactly √3 units, which they represent on a number line using a right-angled triangle construction where the base is √2 and the height is 1 unit.

During the cooling process, the temperature of a solution drops by a factor represented by the expression 3 / (√7 + √2). Dr. Mehta asks the students to rationalize the denominator to find the simplified cooling constant. Furthermore, the volume of a gas in a pressurized container is calculated using the laws of exponents as (272/3 × 81-1/4) / 3. The lab equipment requires these values to be entered in their simplest real number form. One student notices that the ratio of the circumference to the diameter of their circular glass beakers results in a non-terminating, non-recurring decimal, which they identify as an irrational number.

1. Dr. Mehta wants to express the sample weight 1.6̅ in the form p/q. Which of the following is the correct fractional representation?
Solution: Let x = 1.666… (i). Multiply by 10: 10x = 16.666… (ii). Subtract (i) from (ii): 9x = 15 ⇒ x = 15/9. Simplifying gives x = 5/3.
2. To represent √3 on the number line, the students use a right-angled triangle. If the base is √2 and the height is 1, what is the length of the hypotenuse?
Solution: Using Pythagoras Theorem: H² = B² + P². H² = (√2)² + (1)² = 2 + 1 = 3. Therefore, H = √3.
3. Simplify the cooling constant by rationalizing the denominator: 3 / (√7 + √2).
Solution: Multiply numerator and denominator by the conjugate (√7 – √2): [3(√7-√2)] / [(√7)² – (√2)²] = 3(√7-√2) / (7-2) = 3(√7-√2) / 5.
4. Solve the expression for the gas volume using laws of exponents: (272/3 × 81-1/4) / 3.
Solution: 272/3 = (3³)2/3 = 3² = 9. 81-1/4 = (3⁴)-1/4 = 3⁻¹ = 1/3. Expression: (9 × 1/3) / 3 = 3 / 3 = 1.
5. The ratio of the circumference to the diameter of the beaker (Pi) is described as a non-terminating, non-recurring decimal. This means the number is:
Solution: Any number that has a decimal expansion which is non-terminating and non-recurring cannot be written in p/q form and is therefore an Irrational Number.
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