NCERT Solutions for Exercise 1.3 Class 9 Maths Chapter 1 - Number Systems

NCERT Solutions for Exercise 1.3 Class 9 Maths Chapter 1 - Number Systems

Edited By Vishal kumar | Updated on Sep 29, 2023 04:03 PM IST

NCERT Solutions for Class 9 Maths Chapter 1: Number Systems Exercise 1.3- Download Free PDF

NCERT Solutions for Class 9 Maths Chapter 1: Number Systems Exercise 1.3- Welcome to the updated 9th class maths exercise 1.3 answers created by subject experts of Careers360. NCERT Solutions for Class 9 Maths Exercise 1.3 offers the idea of real numbers and their decimal expansions. Numbers are divided into several categories in Mathematics, including rational numbers, irrational numbers, natural numbers, whole numbers, integers, and real numbers. Exercise 1.3 Class 9 Maths is one of the important exercises of the chapter. 1 Number System.

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  1. NCERT Solutions for Class 9 Maths Chapter 1: Number Systems Exercise 1.3- Download Free PDF
  2. NCERT Solutions for Class 9 Maths Chapter 1 – Number Systems Exercise 1.3
  3. Access Answer of Number Systems Class 9 Chapter 1 Exercise: 1.3
  4. Q1 (i) Write the following in decimal form and say what kind of decimal expansion each has : (i)
  5. More About NCERT Solutions for Class 9 Maths Exercise 1.3
  6. Benefits of NCERT Solutions for Class 9 Maths Exercise 1.3:
  7. Key Features of Class 9 Maths Chapter 1 Exercise 1.3
  8. NCERT Solutions of Class 10 Subject Wise
  9. Subject Wise NCERT Exemplar Solutions
NCERT Solutions for Exercise 1.3 Class 9 Maths Chapter 1 - Number Systems
NCERT Solutions for Exercise 1.3 Class 9 Maths Chapter 1 - Number Systems

NCERT book Class 9 Maths chapter 1 exercise 1.3 consist of 9 questions, 7 of which are brief and 4 of which are lengthy solution kinds primarily based totally on rational and irrational numbers. In this Class 9 Maths chapter 1 exercise 1.3 the ideas linked to the number system are thoroughly discussed. The following activities are included along with Class 9 Maths Chapter 1 Exercise 1.3.

NCERT Solutions for Class 9 Maths Chapter 1 – Number Systems Exercise 1.3

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Access Answer of Number Systems Class 9 Chapter 1 Exercise: 1.3

Q1 (i) Write the following in decimal form and say what kind of decimal expansion each has : (i) \frac{36}{100}

Answer:

We can write \frac{36}{100} as
\Rightarrow \frac{36}{100}= 0.36
Since the decimal expansion ends after a finite number of steps. Hence, it is terminating

Q1 (ii) Write the following in decimal form and say what kind of decimal expansion each has : (ii) \frac{1}{11}

Answer:

We can rewrite \frac{1}{11} as

\Rightarrow \frac{1}{11}= 0.09090909..... = 0.\overline{09}
Since decimal expansion repeats itself so it is a non-terminating recurring decimal expansion.

Q1 (iii) Write the following in decimal form and say what kind of decimal expansion each has : (iii) 4\frac{1}{8}

Answer:

We can rewrite 4\frac{1}{8} as

\Rightarrow 4\frac{1}{8} = \frac{33}{8}= 4.125
Since the decimal expansion ends after a finite number. Therefore, it is terminating

Q1 (iv) Write the following in decimal form and say what kind of decimal expansion each has : (iv) \frac{3}{13}

Answer:

We can rewrite \frac{3}{13} as

\Rightarrow \frac{3}{13} = 0.230769230769 = 0.\overline{230769}
Since decimal expansion repeats itself so it is a non-terminating recurring decimal expansion.

Q1 (v) Write the following in decimal form and say what kind of decimal expansion each has: (v) \frac{2}{11}

Answer:

We can rewrite \frac{2}{11} as

\Rightarrow \frac{2}{11} = 0.181818......= 0.\overline{18}
Since decimal expansion repeats itself so it is a non-terminating recurring decimal expansion.

Q1 (vi) Write the following in decimal form and say what kind of decimal expansion each has : (vi) \frac{329}{400}

Answer:

We can rewrite \frac{329}{400} as

\Rightarrow \frac{329}{400}= 0.8225
Since decimal expansion ends after finite no. of figures. Hence, it is terminating.

Q3 (i) Express the following in the form \frac{p}{q} , where p and q are integers and q ≠ 0. (i) 0.\bar{6}

Answer:

Let x = 0.\overline6= 0.6666.... -(i)

Now, multiply by 10 on both sides

10x= 6.6666...

\Rightarrow 10x = 6 + x \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \(using \ (i))

\Rightarrow 9x = 6

\Rightarrow x = \frac{6}{9} = \frac{2}{3}

Therefore, \frac{p}{q} form of 0.\bar{6} is \frac{2}{3}

Q3 (ii) Express the following in the form \frac{p}{q} , where p and q are integers and q ≠ 0. (ii) 0.4\bar{7}

Answer:

We can write 0.4\overline7 as

\Rightarrow 0.4\overline7 = \frac{4}{10}+ \frac{0.777..}{10} -(i)

Now,

Let x = 0.\overline7= 0.7777.... -(ii)

Now, multiply by 10 on both sides

10x= 7.7777...

\Rightarrow 10x = 7 + x \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \(using \ (ii))

\Rightarrow 9x = 7

\Rightarrow x = \frac{7}{9}
Now, put the value of x in equation (i). we will get

\Rightarrow 0.4\overline7 = \frac{4}{10}+ \frac{7}{10\times 9}= \frac{4}{10}+ \frac{7}{90} = \frac{36+7}{90} = \frac{43}{90}

Therefore, \frac{p}{q} form of 0.4\overline7 is \frac{43}{90}

Q3 (iii) Express the following in the form \frac{p}{q} , where p and q are integers and q ≠ 0. (iii) 0.\overline{001}

Answer:

Let x = 0.\overline{001}= 0.001001.... -(i)

Now, multiply by 1000 on both sides

1000x= 1.001001...

\Rightarrow 1000x = 1 + x \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \(using \ (i))

\Rightarrow 999x = 1

\Rightarrow x = \frac{1}{999}

Therefore, \frac{p}{q} form of 0.\overline{001} is \frac{1}{999}

Q4 Express 0.99999 .... in the form \frac{p}{q} . Are you surprised by your answer?

Answer:

Let x = 0.\overline{9}= 0.9999.... -(i)

Now, multiply by 10 on both sides

10x= 9.999....

\Rightarrow 10x = 9 + x \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \(using \ (i))

\Rightarrow 9x = 9

\Rightarrow x = \frac{9}{9} = 1

Therefore, \frac{p}{q} form of 0.999.... is 1

The difference between 1 and 0.999999 is o.000001 which is almost negligible.

Therefore, 0.999 is too much closer to 1. Hence, we can write 0.999999.... as 1

Q5 What can the maximum number of digits be in the repeating block of digits in the decimal expansion of \frac{1}{17} ? Perform the division to check your answer.

Answer:

We can rewrite \frac{1}{17} as

\Rightarrow \frac{1}{17} = 0.05882352941176470588235294117647= 0.\overline{0588235294117647}
Therefore, there are total 16 number of digits be in the repeating block of digits in the decimal expansion of \frac{1}{17}

Q6 Look at several examples of rational numbers in the form \frac{p}{q} (q ≠ 0), where p and q are integers with no common factors other than 1 and having terminating decimal representations (expansions). Can you guess what property q must satisfy?

Answer:

We can observe that when q is 2, 4, 5, 8, 10… then the decimal expansion is terminating. For example:

\frac{3}{2}= 1.5 , denominator q = 2^1

\frac{8}{5}= 1.6 , denominator q = 5^1

\frac{15}{10} = 1.5 , denominator q =10=2\times 5= 2^1 , 5^1

Therefore,

It can be observed that the terminating decimal can be obtained in a condition where prime factorization of the denominator of the given fractions has the power of 2 only or 5 only or both.

Q7 Write three numbers whose decimal expansions are non-terminating non-recurring.

Answer:

Three numbers whose decimal expansions are non-terminating non-recurring are
1) 0.02002000200002......
2) 0.15115111511115.......
3) 0.27227222722227.......

Q8 Find three different irrational numbers between the rational numbers \frac{5}{7} and \frac{9}{11} .

Answer:

We can write \frac{5}{7} as

\Rightarrow \frac{5}{7} = 0.714285714285.... = 0.\overline{714285}

And \frac{9}{11} as

\Rightarrow \frac{9}{11} = 0.818181.... = 0.\overline{81}
Therefore, three different irrational numbers between the rational numbers \frac{5}{7} and \frac{9}{11} are

1) 0.72737475....
2) 0.750760770780...
3) 0.790780770760....

Q9 (i) Classify the following numbers as rational or irrational : \sqrt{23}

Answer:

We can rewrite \sqrt{23} in decimal form as

\Rightarrow \sqrt{23} = 4.7958152....

Now, as the decimal expansion of this number is non-terminating non-recurring.

Therefore, it is an irrational number.

Q9 (ii) Classify the following numbers as rational or irrational : \sqrt{225}

Answer:

We can rewrite \sqrt{225} as

\Rightarrow \sqrt{225} = 15
We can clearly see that it is a rational number because we can represent it in \frac{p}{q} form

Q9 (iii) Classify the following numbers as rational or irrational : 0.3796

Answer:

We can rewrite 0.3796 as

\Rightarrow 0.3796 = \frac{3796}{10000}
Now, we can clearly see that it is a rational number as the decimal expansion of this number is terminating and we can also write it in \frac{p}{q} form.

Q9 (iv) Classify the following numbers as rational or irrational : 7.478478....

Answer:

We can rewrite 7.478478.... as

\Rightarrow 7.478478.... = 7.\overline{478}
Now, as the decimal expansion of this number is non-terminating recurring. Therefore, it is a rational number.

Q9 (v) Classify the following numbers as rational or irrational : 1.101001000100001...

Answer:

In the case of number 1.101001000100001...
As the decimal expansion of this number is non-terminating non-repeating. Therefore, it is an irrational number.

More About NCERT Solutions for Class 9 Maths Exercise 1.3

The topic of real numbers and their decimal expansions was the emphasis of the NCERT solutions for Class 9 Maths exercise 1.3. There are other questions based on decimal forms in Exercise 1.3 Class 9 Maths. Decimal numbers are fractional numbers written in a certain format. The decimal point is the dot that appears between the entire number and fraction parts.

Terminating decimal numbers are decimal numbers that end after a certain number of decimal places. With repeating decimal patterns, rational numbers can be both terminating and non-terminating decimals. Irrational numbers should never have terminating decimals and should instead have non-terminating decimals with no repeating decimal patterns.

Also Read| Number Systems Class 9 Notes

Benefits of NCERT Solutions for Class 9 Maths Exercise 1.3:

• NCERT syllabus Class 9 Maths Exercise 1.3 helps us to understand the basics of the number system in-depth, which is beneficial for us in understanding higher math as well.

• By solving the NCERT solution for Class 9 Maths Chapter 1 exercise 1.3 exercises, we can also study the representation of real numbers on the number line with the help of the decimal expansions.

• Exercise 1.3 Class 9 Maths is well-versed with the basic facts of decimals and their representation on the number line which helps us to solve the exercises smoothly.

Key Features of Class 9 Maths Chapter 1 Exercise 1.3

  1. Easy-to-Understand: The 9th class maths exercise 1.3 answers are presented in a straightforward and easily comprehensible manner, making it accessible to students of various levels.

  2. Free Access: These class 9 maths chapter 1 exercise 1.3 solution are freely accessible to all students, ensuring that cost is not a barrier to learning.

  3. Clear Explanations: Each class 9 maths ex 1.3 solution includes clear explanations to help students understand the concepts and problem-solving techniques.

  4. Step-by-Step Format: The ex 1.3 class 9 solutions are structured in a step-by-step format, guiding students through the problem-solving process.

  5. Syllabus Alignment: The class 9 ex 1.3 solutions align with the prescribed syllabus, covering all relevant topics and concepts.

Also see-

NCERT Solutions of Class 10 Subject Wise

Subject Wise NCERT Exemplar Solutions

Frequently Asked Questions (FAQs)

1. What are irrational numbers, according to NCERT solutions for Class 9 Maths chapter 1 exercise 1.3?

Irrational numbers are numbers that cannot be stated in the form of a fraction with an integer in both the numerator and the denominator. 

2. Is root (49) a rational number?

root(49)=7 which can be written as 7/1 

Thus 49 is a rational number. 

3. Is 0 a rational number?

Yes 0 is a rational number (since 0 can be written as 0/1 , 0/2 etc… ) 

4. What are the different types of decimals, According to NCERT solutions for Class 9 Maths chapter 1 exercise 1.3 ?

Decimal numbers are classified into

  • Recurring Decimal Numbers (repeating or Non-Terminating Decimals) 

  • Non-Recurring Decimal Numbers (non Repeating or Terminating Decimals). 

5. Write the expanded form of 64.3?

The expanded form of 64.3 is 60+4+3/10.

6. Is 0.033 a non-recurring decimal number?

Yes, 0.033 is a non-recurring decimal number. Since 0.033 is a non Repeating and Terminating number. 

7. The dot which is present in between the whole number and fractions part is known as ________.

The dot which is present in between the whole number and fractions part is known as the decimal point. 

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

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A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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