NCERT Solutions Class 9 Maths Chapter 1 Number System: Exercise 1.4

The NCERT Solutions for Class 9 Maths Chapter 1 Number System: Exercise 1.5 will help you understand the laws of exponents for real numbers. In this exercise, you will use the laws of exponents to solve problems effectively. Mastering this topic will help you score better in your school exams.

These NCERT Solutions take into account the current CBSE exam pattern and are consistent with the latest NCERT syllabus. The answers provided will give you an opportunity to learn the proper way to solve each question.

Practicing these exercises will also help you develop strong problem-solving skills and also help in preparing you for competitive exams.

1.0Download NCERT Solutions of Class 9 Maths Chapter 1 Number System Exercise 1.4: Free PDF

Exercise 1.4 of the NCERT Solutions for Class 9 Maths Chapter 1 deals with the laws of exponents. Download the free PDF of NCERT Solutions for exercise 1.4 of NCERT Class 9 Maths Chapter 1 from below:

2.0NCERT Exercise Solutions Class 9 Chapter 1- Number Systems: All Exercises 

3.0NCERT Class 9 Maths Chapter 1 Exercise 1.4 : Detailed Solutions

1. Classify the following numbers as rational or irrational :

(i) 2−√5

(ii) (3+√23)−√23

(iii) 7√7 / 2√7

(iv) 1/√2

(v) 2π

Sol. (i) 2 is a rational number and √5 is an irrational number. Therefore, 2−√5 is an irrational number (difference of a rational and an irrational is irrational).

(ii) (3+√23)−√23 = 3+√23−√23 = 3. Since 3 is a rational number, the expression is a rational number.

(iii) 7√7 / 2√7 = 7/2. Since 7/2 is a rational number, the expression is a rational number.

(iv) 1/√2. Since 1 is a rational number and √2 is an irrational number, their quotient (1/√2) is an irrational number.

(v) 2π. Since 2 is a rational number and π is an irrational number, their product (2π) is an irrational number.

2. Simplify each of the following expressions :

(i) (3+√3)(2+√2)

(ii) (3+√3)(3−√3)

(iii) (√5+√2)²

(iv) (√5−√2)(√5+√2)

Sol. (i) (3+√3)(2+√2) = 3(2+√2)+√3(2+√2)

= 6+3√2+2√3+√6

(ii) (3+√3)(3−√3) = (3)²−(√3)² [using (a+b)(a−b)=a²−b²]

= 9−3 = 6

(iii) (√5+√2)² = (√5)²+2(√5)(√2)+(√2)² [using (a+b)²=a²+2ab+b²]

= 5+2√10+2 = 7+2√10

(iv) (√5−√2)(√5+√2) = (√5)²−(√2)² [using (a−b)(a+b)=a²−b²]

= 5−2 = 3

3. Recall, π is defined as the ratio of the circumference (say c) of a circle to its diameter (say d). That is, π=c/d. This seems to contradict the fact that π is irrational. How will you resolve this contradiction?

Sol. There is no contradiction. When we measure a length with a scale or any other device, we only get an approximate rational value. Therefore, we may not realise that c (circumference) or d (diameter) or both are irrational. The ratio c/d is truly irrational, but our measurement provides only a rational approximation.

4. Represent √9.3 on the number line.

Sol.

1. Draw a line segment AB = 9.3 units on the number line.
2. Extend AB to C such that BC = 1 unit.
3. Find the midpoint O of AC.
4. With O as center and OA (or OC) as radius, draw a semicircle.
5. Draw a line BD perpendicular to AC at point B, intersecting the semicircle at D.
6. Then, BD = √9.3 units.
7. With B as center and BD as radius, draw an arc intersecting the number line at P.
8. Point P represents √9.3 on the number line.

5. Rationalise the denominators of the following :

(i) 1/√7

(ii) 1/(√7−√6)

(iii) 1/(√5+√2)

(iv) 1/(√7−2)

Sol. (i) 1/√7 = (1/√7) × (√7/√7) = √7/7

(ii) 1/(√7−√6) = (1/(√7−√6)) × ((√7+√6)/(√7+√6))

= (√7+√6) / ((√7)²−(√6)²)

= (√7+√6) / (7−6)

= (√7+√6) / 1 = √7+√6

(iii) 1/(√5+√2) = (1/(√5+√2)) × ((√5−√2)/(√5−√2))

= (√5−√2) / ((√5)²−(√2)²)

= (√5−√2) / (5−2)

= (√5−√2) / 3

(iv) 1/(√7−2) = (1/(√7−2)) × ((√7+2)/(√7+2))

= (√7+2) / ((√7)²−2²)

= (√7+2) / (7−4)

= (√7+2) / 3

4.0Key Features and Benefits Class 9 Maths Chapter 1 Number System: Exercise 1.4

• This exercise contains the rules of exponents as per the NCERT syllabus
• Every solution is detailed step by step for better understanding of the method used.
• The problems are in alignment with the questions typical to CBSE school exams. 
• Practicing these NCERT Solutions will encourage you to approach exponent related problems more easily.
• It builds your logical thinking skills which is very important to excel in competitive exams.
• Practicing these solutions will improve your confidence to prepare you for tricky exam situations with an exponent.