Chapter 3 - The world of numbers

NCERT Solutions for Class 9 Maths Chapter 3 The World of Numbers

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1.0Download NCERT Class 9 Maths Chapter 3 The World of Numbers Solutions

The World of Numbers chapter helps students understand different types of numbers, their properties, and fundamental concepts through simple explanations. Build a strong conceptual foundation with textbook-based practice and download the FREE CBSE-aligned NCERT Solutions PDF, featuring step-by-step answers based on the latest NCERT syllabus.

NCERT Solutions Class 9 Maths Chapter 3

2.0Learning Outcomes - NCERT Class 9 Maths Chapter 3 The World of Numbers

Identify and classify different types of numbers, including natural, whole, integers, rational, and irrational numbers.
Represent rational and irrational numbers accurately on the number line.
Understand the properties and relationships of different number systems.
Perform operations involving rational numbers with confidence.
Distinguish between terminating and non-terminating decimal expansions.
Apply the laws of exponents to simplify numerical expressions.
Use number concepts to solve real-life and textbook-based problems effectively.
Develop logical reasoning and problem-solving skills using the number system.

3.0Detailed Class 9 Maths Chapter 3 The World of Numbers Solutions

Exercise : 3.1

1. A merchant in the port city of Lothal is exchanging bags of spices for copper ingots. He receives 15 ingots for every 2 bags of spices. If he brings 12 bags of spices to the market, how many copper ingots will he leave with?

Sol. Given that he receives 15 ingots for every 2 bags of spices

Thus, Cost of 2 bags of spices = 15 ingots And, we can write Cost of 1 bag of spices $= \frac{15}{2}$ ingots Cost of 12 bag of spices $= 12 \times$ Cost of 1 bag of spices $12 \times \frac{15}{2}$ ingots $= 6 \times 15$ $= 90$ ingots

2. Look at the sequence of numbers on one column of the Ishango bone: 11, 13, 17, 19. What do these numbers have in common? List the next three numbers that fit this pattern.

Sol. The Ishango bone has numbers 11, 13, 17, 19 All these are prime numbers.

Next 3 numbers in the pattern $= 23, 29, 31$ Note: Prime numbers are those numbers which have only two factors-1 and the number themselves.

3. We know that Natural Numbers are closed under addition (the sum of any two natural numbers is always a natural number). Are they closed under subtraction? Provide a couple of examples to justify your answer.

Sol. Subtracting Natural numbers Natural numbers are numbers starting from 1 , like $1, 2, 3, 4, 5, \dots$

Subtracting any two natural numbers $100 - 5 = 95$ Here, 95 is a natural number And, $3 - 6 = - 3$ Here, -3 is not a natural number Similarly, $1 - 10 = - 9$ Here, -9 is not a natural number Since on subtracting Natural numbers, the result is not always a Natural Number, we can say that

Natural Numbers are NOT closed under Subtraction.

4. Ancient Indians used the joints of their fingers to count, a practice still seen today. Each finger has 3 joints, and the thumb is used to count them. How many can you count on one hand? How does this relate to the ancient base-12 counting systems?

Sol. We can count till 12 in one hand This is because humans could easily count to 12 on a single hand without putting down a tool they were holding. Hence, 12 became a very natural "grouping" number. This is why many ancient cultures used base-12 (dozenal) systems, which is the exact reason we still have 12 months in a year, 12 hours on a clock face, and 12 inches in a foot.

Exercise : 3.2

1. The temperature in the high-altitude desert of Ladakh is recorded as $4^{\circ} C$ at noon. By midnight, it drops by $1 5^{\circ} C$ . What is the midnight temperature?

Sol. Initial temperature $= 4^{\circ} C$ Temperature drop at midnight $= 1 5^{\circ} C$ Thus, Midnight temperature = Initial Temperature - Temperature drop at midnight $= 4 - 15$ $= - 15 + 4$ $= - 1 1^{\circ} C$

2. A spice trader takes a loan (debt) of ₹850. The next day, he makes a profit (fortune) of ₹1,200. The following week, he incurs a loss of ₹450. Write this sequence as an equation using integers and calculate his final financial standing. Sol. Debt is negative Profit is positive Loss is negative Final financial standing $= (- 850) + 1200 + (- 450)$ $= - 850 - 450 + 1200$ $= - 1300 + 1200$ = -100

3. Calculate the following using Brahmagupta's laws: (i) $(- 12) \times 5$ (ii) $(- 8) \times (- 7)$ (iii) $0 - (- 14)$ (iv) $(- 20) \div 4$

Sol. (i) Since Positive × Negative $=$ Negative We can write

$- 12 \times 5 = - (12 \times 5) = - 60$

(ii) Since Negative × Negative $=$ Positive We can write $(- 8) \times (- 7) = 8 \times 7 = 56$ (iii) So, we can write

$0 - (- 14) = 0 + 14 = 14$

(iv) So, we can write

$(- 20) \div 4 = \frac{- 20}{4} = - 5$

Explain, using a real-world example of debt, why subtracting a negative number is the same as adding a positive number (e.g., $10 - (- 5) = 15$ ).

Sol. You have a room at $1 0^{\circ} C$ . It's a bit chilly because there are some open windows in the room. You close a window that was letting in $- 5^{\circ}$ worth of cold. You are subtracting the cold. $10 - (- 5) = 15$ Subtracting a drop in temperature is identical to adding a rise in temperature.

Exercise : 3.3

1. Prove that the following rational numbers are equal: (i) $\frac{2}{3}$ and $\frac{4}{6}$ (ii) $\frac{5}{4}$ and $\frac{10}{8}$ (iii) $- \frac{3}{5}$ and $- \frac{6}{10}$ (iv) $\frac{9}{3}$ and 3

Sol. (i) First Fraction: $\frac{2}{3} = \frac{2}{3}$ Second Fraction: $\frac{4}{6} = \frac{2}{3}$ (dividing numerator and denominator by 2 )

Therefore, $\frac{2}{3}$ and $\frac{4}{6}$ are equal. (ii) First Fraction: $\frac{5}{4} = \frac{5}{4}$

Second Fraction: $\frac{10}{8} = \frac{5}{4}$ (dividing numerator and denominator by 2 )

Therefore, $\frac{5}{4}$ and $\frac{10}{8}$ are equal. (iii) First Fraction: $\frac{- 3}{5} = \frac{- 3}{5}$

Second Fraction: $\frac{- 6}{10} = \frac{- 3}{5}$ (dividing numerator and denominator by 2 ) Therefore, $\frac{- 3}{5}$ and $\frac{- 6}{10}$ are equal. (iv) First Fraction: $\frac{9}{3} = 3$ Hence, $\frac{9}{3}$ and 3 are equal.

2. Find the sum: (i) $\frac{2}{5} + \frac{3}{10}$ (ii) $\frac{7}{12} + \frac{5}{8}$ (iii) $- \frac{4}{7} + \frac{3}{14}$

Sol. (i) LCM of 5 and $10 = 10$ Simplifying the first number: $\frac{2}{5} = \frac{4}{10}$ [Making the same denominator] Now, the sum $= \frac{4}{10} + \frac{3}{10}$ $= \frac{7}{10}$ (ii) $LCM$ of 12 and $8 = 24$

Simplifying the first number: $\frac{7}{12} = \frac{14}{24}$ [Making the same denominator] Simplifying the second number: $\frac{5}{8} = \frac{15}{24}$ [Making the same denominator] Now, the sum $= \frac{14}{24} + \frac{15}{24}$ $= \frac{29}{24}$ (iii) LCM of 7 and $14 = 14$

Simplifying the first number: $\frac{- 4}{7} = \frac{- 8}{14}$ [Making the same denominator] So, the sum $= \frac{- 8}{14} + \frac{3}{14}$ $= \frac{- 5}{14}$

3. Find the difference: (i) $\frac{5}{6} - \frac{1}{4}$ (ii) $\frac{11}{8} - \frac{3}{4}$ (iii) $- \frac{7}{9} - (- \frac{2}{3})$

Sol. (i) LCM of 6 and $4 = 12$ Simplifying the first number: $\frac{5}{6} = \frac{10}{12}$ [Making the same denominator] Simplifying the second number: $\frac{1}{4} = \frac{3}{12}$ [Making the same denominator] So, the difference $= \frac{10}{12} - \frac{3}{12} = \frac{7}{12}$ (ii) LCM of 8 and $4 = 8$

Simplifying the second number: $\frac{3}{4} = \frac{6}{8}$ [Making the same denominator] So, the difference $= \frac{11}{8} - \frac{6}{8} = \frac{5}{8}$ (iii) $\frac{- 7}{9} - (\frac{- 2}{3}) = \frac{- 7}{9} + \frac{2}{3}$

LCM of 9 and $3 = 9$ Simplifying the second number: $\frac{2}{3} = \frac{6}{9}$ [Making the same denominator] So, the difference $= \frac{- 7}{9} + \frac{6}{9} = \frac{- 1}{9}$

4. Find the product: (i) $\frac{2}{3} \times \frac{3}{10}$ (ii) $\frac{7}{11} \times \frac{5}{8}$ (iii) $- \frac{4}{7} \times \frac{5}{14}$

Sol. (i) $\frac{2}{3} \times \frac{3}{10}$

$= \frac{( 2 \times 3 )}{( 3 \times 10 )} = \frac{6}{30} = \frac{1}{5} [After simplification]$

(ii) $\frac{7}{11} \times \frac{5}{8}$

$\frac{( 7 \times 5 )}{( 11 \times 8 )} = \frac{35}{88}$

(iii) $\frac{- 4}{7} \times \frac{5}{14}$

$= \frac{( - 4 \times 5 )}{( 7 \times 14 )} = \frac{- 20}{98} = \frac{- 10}{49}$

[After simplification]

5. Find the quotient: (i) $\frac{2}{3} \div \frac{3}{10}$ (ii) $\frac{7}{11} \div \frac{5}{8}$ (iii) $- \frac{4}{7} \div \frac{5}{14}$

Sol. (i) $\frac{2}{3} \div \frac{3}{10}$

$= \frac{2}{3} \times \frac{10}{3} = \frac{( 2 \times 10 )}{( 3 \times 3 )} = \frac{20}{9}$

(ii) $\frac{7}{11} \div \frac{5}{8}$

$= \frac{7}{11} \times \frac{8}{5} = \frac{( 7 \times 8 )}{( 11 \times 5 )} = \frac{56}{55}$

(iii) $\frac{- 4}{7} \div \frac{5}{14}$

$= \frac{- 4}{7} \times \frac{14}{5} = \frac{( - 4 \times 14 )}{( 7 \times 5 )} = \frac{- 56}{35} = \frac{- 8}{5}$

[After simplification]

6. Show that : $(\frac{1}{2} + \frac{3}{4}) \times \frac{8}{3} = \frac{1}{2} \times \frac{8}{3} + \frac{3}{4} \times \frac{8}{3}$ .

Sol. LHS $= (\frac{1}{2} + \frac{3}{4}) \times \frac{8}{3}$

$= (\frac{2}{4} + \frac{3}{4}) \times \frac{8}{3}$

[First add inside the bracket: $\frac{1}{2} = \frac{2}{4}$ ]

$= (\frac{5}{4}) \times \frac{8}{3} = \frac{( 5 \times 8 )}{( 4 \times 3 )} = \frac{40}{12} = \frac{10}{3} [After simplification]$

RHS $= \frac{1}{2} \times \frac{8}{3} + \frac{3}{4} \times \frac{8}{3}$ $= \frac{( 1 \times 8 )}{( 2 \times 3 )} + \frac{( 3 \times 8 )}{( 4 \times 3 )}$ $= \frac{8}{6} + \frac{24}{12}$ $= \frac{4}{3} + \frac{6}{3}$ [After simplification: $\frac{8}{6} = \frac{4}{3}$ and $\frac{24}{12} = \frac{6}{3}$ ] $= \frac{10}{3}$ Since LHS = RHS, Therefore, $(\frac{1}{2} + \frac{3}{4}) \times \frac{8}{3} = \frac{1}{2} \times \frac{8}{3} + \frac{3}{4} \times \frac{8}{3}$ Hence proved.

7. Simplify the following using the distributive property: $\frac{7}{9} (\frac{6}{7} - \frac{3}{4})$ .

Sol. Using distributive property: $\frac{7}{9} \times (\frac{6}{7} - \frac{3}{4})$ $= \frac{7}{9} \times \frac{6}{7} - \frac{7}{9} \times \frac{3}{4}$ $= \frac{6}{9} - \frac{21}{36}$ $= \frac{2}{3} - \frac{7}{12}$ $= \frac{8}{12} - \frac{7}{12}$ [LCM of 3 and $12 = 12$ and $\frac{2}{3} = \frac{8}{12}$ ] $= \frac{1}{12}$

8. Find the rational number $x$ such that: $\frac{5}{6} (x + \frac{3}{5}) = \frac{5}{6} x + \frac{1}{2}$ .

Sol. Given: $(\frac{5}{6}) (x + \frac{3}{5}) = (\frac{5}{6}) \times x + \frac{1}{2}$ $\Rightarrow (\frac{5 x}{6}) + (\frac{5}{6} \times \frac{3}{5}) = (\frac{5}{6}) x + \frac{1}{2}$ $\Rightarrow (\frac{5 x}{6}) + \frac{15}{30} = (\frac{5}{6}) x + \frac{1}{2}$ $\Rightarrow \frac{15}{30} = \frac{1}{2}$ , which is universal truth. So, $(\frac{5}{6}) (x + \frac{3}{5}) = (\frac{5}{6} x + \frac{1}{2})$ is true for every value of x . Therefore, x can be any rational number.

Exercise : 3.4

1. Represent the rational number $\frac{2}{3}, \frac{- 5}{4}$ and $1 \frac{1}{2}$ on a single number line.

Sol.

On the number line: $\frac{- 5}{4}$ lies between -2 and -1 $\frac{2}{3}$ lies between 0 and 1 $\frac{3}{2}$ lies between 1 and 2

2. Find three distinct rational numbers that lie strictly between $\frac{- 1}{2}$ and $\frac{1}{4}$

Sol. Given numbers: $\frac{- 1}{2}$ and $\frac{1}{4}$ LCM of 2 and $4 = 4$ Converting $\frac{- 1}{2}$ to common denominator: $\frac{- 1}{2} = \frac{- 2}{4}$ Now the given numbers: $\frac{- 2}{4}$ and $\frac{1}{4}$ Now apply ( $n + 1$ ) rule, where n is equal to no. of distinct rational you have to find, just multiply ( $n + 1$ ) to numerator and denominator of both numbers

So, numbers are $\frac{- 2}{4} \times \frac{4}{4}$ and $\frac{1}{4} \times \frac{4}{4}$ $\frac{- 8}{16}$ and $\frac{4}{16}$ Therefore, three rational numbers between them are $\frac{- 6}{16}, 0, \frac{1}{16}$ .

Note: we can find infinite rational number between any two rational numbers.

3. Simplify the expression: $(- \frac{1}{4}) + (\frac{5}{12})$

Sol. LCM of 4 and 12 = 12 Converting $- \frac{1}{4}$ to common denominator: $\frac{- 1}{4} = \frac{- 3}{12}$ So, $\frac{- 3}{12} + \frac{5}{12} = \frac{2}{12} = \frac{1}{6}$

4. A tailor has $15 \frac{3}{4}$ metres of fine silk. If making one kurta requires $2 \frac{1}{4}$ metres of silk, exactly how many kurtas can he make?

Sol. $15 \frac{3}{4} = \frac{63}{4}$ $2 \frac{1}{4} = \frac{9}{4}$ Total amount of silk cloth $= 15 \frac{3}{4}$ metres In $2 \frac{1}{4}$ metres of silk, number of kurta $= 1$ In 1 metres of silk, number of kurta $= \frac{1}{21/4}$ $= \frac{1}{9/4} \times \frac{63}{4}$ $= \frac{4}{9} \times \frac{63}{4}$ $= \frac{63}{9} = 7$ Therefore, he can make exactly 7 kurtas from $15 \frac{3}{4}$ metres of fine silk.

5. Find three rational numbers between 3.1415 and 3.1416.

Sol. We can find infinite rational number between any two rational numbers.

So, three rational numbers are 3.14151, 3.14152, 3.14153.

6. Can you think of other way to find a rational number between any two rational numbers?

Sol. Yes, there are methods: (i) If $a$ and $b$ are two rational numbers, then $\frac{( a + b )}{2}$ lies between them. (ii) By making common denominator: Convert both numbers to same denominator and choose a number in between. (iii) By decimal expansion: Convert into decimals and pick numbers in between.

Exercise : 3.5

1. Perform the long division for $\frac{1}{13}$ . Identify the repeating block of digits. Does is show cyclic properties if you evaluate $\frac{2}{13}$ ? Now compute $\frac{3}{13}, \frac{4}{13}$ etc. What do you notice?

Sol.

\frac{1}{13} = 0. \overline{076923}

Repeating blocks

= 076923

Now,

\frac{2}{13} = 0 \overline{.153846}

\frac{3}{13} = 0. \overline{230769}

\frac{4}{13} = 0. \overline{307692}

Each decimal is repeating and showing cyclic properties.

4. Classify the following numbers as rational or irrational: (i) $81$ (ii) $12$ (iii) $0.33333 \dots$ (iv) $0.123451234512345 \dots$ (v) $1.01001000100001 \dots$ (Notice the pattern: It is repeating a single block?) (vi) 23.560185612239874790120

Find the explicit fractions in case they are rational. Sol. (i) $81 = 9$ Since 9 can be written as $\frac{9}{1}$ , it is a rational number. Therefore, $81$ is rational. (ii) $12$ $12 = 23$ Since $3$ is irrational, $12$ is irrational. Therefore, $12$ is irrational. (iii) 0.33333... $0.33333 \dots \frac{1}{3}$ This is a repeating decimal, so it is rational. Therefore, 0.33333... is rational. Explicit fraction: $\frac{1}{3}$ (iv) $0.123451234512345 \dots$

The block 12345 repeats continuously. A repeating decimal is rational. Let $x = 0.123451234512345 \dots$ Then, $100000 x = 0.123451234512345 \dots$ Subtract: $100000 x - x = 12345$ $99999 x = 12345$ $x = \frac{12345}{99999} \Rightarrow x = \frac{4115}{33333}$ Therefore, 0.123451234512345... is rational Explicit fraction: $\frac{4115}{33333}$ (v) 1.01001000100001...

This decimal does not terminate and does not repeat.

So, it is non-terminating and nonrepeating. Therefore, 1.01001000100001 ... is irrational. (vi) 23.560185612239874790120

This is a terminating decimal. Every terminating decimal is rational.

Convert into fraction: 23.560185612239874790120 $= \frac{23560185612239874790120}{1000000000000000000000}$ Therefore, 23.560185612239874790120 is rational. Explicit fraction: $\frac{23560185612239874790120}{1000000000000000000000}$

4. The number $0. \overline{9}$ (which means $0.99999..$ . $) i s a r a t i o na l n u mb er . U s in g$ algebra (let $x = 0. \overline{9}$ , multiply by 10 , and subtract), explain why $0. \overline{9}$ is exactly equal to 1 .

Sol. Let $x = 0.99999$ ..... Multiply both sides by 10 : $10 x = 9.99999$ ..... Now subtract: $10 x - x = 9.99999 \dots - 0.99999 \dots$ $9 x = 9$ $x = 1$

5. We have seen that the repeating block of $\frac{1}{7}$ is a cyclic number. Try to find more numbers (n) whose reciprocals $(\frac{1}{n})$ produce decimals with repeating blocks that are cyclic. Sol. Given $\frac{1}{7} = 0.142857142857142857 \dots$ $\frac{1}{17} = 0.05882352941176470588235294$ 1176470588235294117647... $\frac{1}{19} = 0.05263157894736842105263157$ $8947368421052631578947368421 \dots$ These reciprocals produce repeating blocks that show cyclic behaviour. Therefore, examples of such numbers are 7, 17, 19, ...

These are numbers whose reciprocals can give cyclic repeating decimals.

End of Chapter Exercise

1. Convert the following rational numbers in the form of a terminating decimal or non-terminating and repeating decimal, whichever the case may be, by the process of long division: (i) $\frac{3}{50}$ (ii) $\frac{2}{9}$

Sol. (i) $\frac{3}{50}$

It is a terminating decimal (ii) $\frac{2}{9}$

\frac{2}{9} = 0. \overline{2}

, it is a non-terminating repeating decimal 2. Prove that

5

is an irrational number.

Sol. We will use proof by contradiction. Assume that $5$ is rational. Then it can be written in the form of $\frac{p}{q}$ :

Let $5 = \frac{p}{q}$ where p and q are integers, $q \neq = 0$ , and $\frac{p}{q}$ is in lowest terms (that is, p and q have no common factor except 1). Squaring both sides, we get: $5 = \frac{( p ^{2} )}{( q ^{2} )}$ So, $p^{2} = 5 q^{2}$ This means $p^{2}$ is divisible by 5 . Therefore, p is also divisible by 5 . So, let $p = 5 k$ for some integer k . Substitute into $p^{2} = 5 q^{2}$ we get: $(5 k)^{2} = 5 q^{2}$ $\Rightarrow 25 k^{2} = 5 q^{2}$ $\Rightarrow 5 k^{2} = q^{2}$ This shows that $q^{2}$ is divisible by 5 . Therefore, q is also divisible by 5 . So, both p and q are divisible by 5 . But this contradicts our assumption that $\frac{p}{q}$ was in lowest terms. Hence, our assumption is wrong. Therefore, $5$ is an irrational number.

3. Convert the following decimal numbers in the form of $\frac{p}{q}$ . (i) 12.6 (ii) 0.0120 (iii) $3.0 \overline{52}$ (iv) $1.2 \overline{35}$ (v) $0. \overline{23}$ (vi) $2.0 \overline{5}$ (vii) $2.12 \overline{5}$ (viii) $3.12 \overline{5}$ (ix) $2. \overline{1625}$

Sol. (i) 12.6 $12.6 = \frac{126}{10} = \frac{63}{5}$ Therefore, $12.6 = \frac{63}{5}$ (ii) 0.0120 $0.0120 = \frac{120}{10000} = \frac{3}{250}$ Therefore, $0.0120 = \frac{120}{10000} = \frac{3}{250}$ (iii) $3.0 \overline{52}$

Let $x = 3.0525252 \dots$ Then, $100 x = 305.252 \dots$ Subtract: (2)-(1) $100 x - x = 305.2 \overline{52} - 3.0 \overline{52}$ $99 x = 3022$ $x = \frac{3022}{990}$ (iv) $1.2 \overline{35}$

Let $x = 1.23535 \dots$

Multiply both sides by 100

$100 x = 123.5 \overline{35}$

Subtract: (2)-(1) $99 x = 122.3$ $x = \frac{1223}{990}$ (v) $0. \overline{23}$

Let $x = 0.232323 \dots$ . Then, $100 x = 23.232323 \dots .$ . Subtract: $100 x - x = 23. \overline{23} - 0. \overline{23}$ $\Rightarrow 99 x = 23$ So, $x = \frac{23}{99}$ Therefore, $0.23 = \frac{23}{99}$ (vi) $2.0 \overline{5}$

Let $x = 2.05555$

Then, $10 x = 20.5555$

Subtract: (2) - (1) $10 x - x = 20. \overline{5} - 2.0 \overline{5}$ $9 x = 18.5$ $x = \frac{185}{90}$ (vii) $2.12 \overline{5}$

Let $x = 2.125555$

Multiply by 10 both sides $10 x = 21.2 \overline{5}$

Subtract (1) from (2) $9 x = 19.13$ $x = \frac{1913}{900}$ (viii) $3.12 \overline{5}$

Let $x = 3.12 \overline{5}$

Multiply both sides by 10 $10 x = 31.2 \overline{5}$

Subtract (1) from (2) $9 x = 28.13$ $x = \frac{2813}{900}$ (ix) 2.1625

Let $x = 2.162516251625$

Then, 10000x = 21625.16251625...

Subtract(1) from (2) $10000 x - x = 21625.16251625 \dots$ -2.16251625.. $\Rightarrow 9999 x = 21623$ So, $x = \frac{21623}{9999}$ Therefore, $2.1625 = \frac{21623}{9999}$

4. Locate the following rational number on the number line. (i) 0.532 (ii) 1.15

Sol. (i) 0.532

0.532 lies between 0 and 1 .

More precisely: $0.532 = \frac{532}{1000}$ So, we have to mark a point between 0.53 and 0.54 , slightly after 0.53. (ii) $1.1 \overline{5}$

1.15555...

It lies between 1 and 2. More precisely: 1.15 < 1.15555... < 1.16 So, we have to mark a point between 1.15 and 1.16, slightly after 1.15.

5. Find 6 rational numbers between 3 and 4.

Sol. Apply ( $n + 1$ ) rule $n = 6$ [Number of rational number we have to find] $n + 1 = 7$ [Multiply with numerator and denominator of both numbers] $\frac{3 \times 7}{1 \times 7}$ and $\frac{4 \times 7}{1 \times 7}$ $\frac{21}{7}$ and $\frac{28}{7}$ Hence six rational numbers are $\frac{22}{7}, \frac{23}{7}, \frac{24}{7}, \frac{25}{7}, \frac{26}{7}$ and $\frac{27}{7}$ .

6. Find 5 rational numbers between $\frac{2}{5}$ and $\frac{3}{5}$ .

Sol. Apply ( $n + 1$ ) rule $n = 5$ (Number of rational number, we have to find) n + 1 (Multiply with numerator and denominator of both number) $\frac{2 \times 6}{5 \times 6}$ and $\frac{3 \times 6}{5 \times 6}$ $\frac{12}{30}$ and $\frac{18}{30}$ Hence five rational numbers are $\frac{13}{30}, \frac{14}{30}, \frac{15}{30}, \frac{16}{30}$ and $\frac{17}{30}$ .

7. Find 5 rational numbers between $\frac{1}{6}$ and $\frac{2}{5}$

Sol. First of all make denominator same $\frac{1 \times 5}{6 \times 5} = \frac{5}{30}$ and $\frac{2 \times 6}{5 \times 6} = \frac{12}{30}$ Apply ( $n + 1$ ) rule $n = 6$ [Number of rational number we have no find] $n + 1 = 6$ [Multiply with numerator and denominator of both numbers] $\frac{5 \times 6}{30 \times 6}$ and $\frac{12 \times 6}{30 \times 6}$ $\frac{30}{180}$ and $\frac{72}{180}$ Hence five rational number are $\frac{35}{180}, \frac{40}{180}, \frac{45}{180}, \frac{50}{180}$ and $\frac{60}{180}$

8. If $\frac{x}{3} + \frac{x}{5} = \frac{16}{15}$ , find the rational number $x$ .

Sol. Given: $\frac{x}{3} + \frac{x}{5} = \frac{16}{15}$ Taking $x$ common, we have: $x (\frac{1}{3} + \frac{1}{5}) = \frac{16}{15}$ Now, $\frac{1}{3} + \frac{1}{5} = \frac{5}{15} + \frac{3}{15} = \frac{8}{15}$ So, $x \times \frac{8}{15} = \frac{16}{15}$ Therefore, $x = (\frac{16}{15}) \div (\frac{8}{15})$ $= (\frac{16}{15}) \times (\frac{15}{8})$ $= \frac{16}{8} = 2$

9. Let a and b be two non-zero rational numbers such that $a + \frac{1}{b} = 0$ . Without assigning any numerical values, determine whether ab is positive or negative. Justify your answer.

Sol. Given: $a + \frac{1}{b} = 0$ So, $a = \frac{- 1}{b}$ Multiply both sides by b: $ab = - 1$ Since $ab = - 1$ , which is negative, Therefore, ab is negative.

10. A rational number has a terminating decimal expansion whose last non-zero digit occurs in the $4^{th}$ decimal place. Show that such a number can be written in the form $\frac{p}{1 0 ^{4}}$ , where p is an integer not divisible by 10 . Is it necessary that the denominator of this rational number, when written in the lowest form, is divisible by $2^{4}$ or $5^{4}$ ? Give reason.

Sol. If the last non-zero digit occurs in the $4^{th}$ decimal place, then the number has exactly 4 decimal places.

So, it can be written as: $\frac{p}{1 0 ^{4}}$ , where p is an integer.

Also, since the last non-zero digit is in the $4^{th}$ decimal place, p is not divisible by 10 . Otherwise, the decimal would end earlier. Hence, the number can be written in the form: $\frac{p}{1 0 ^{4}}$ , where p is an integer not divisible by 10 . Now, $1 0^{4} = 2^{4} \times 5^{4}$

When the fraction is reduced to lowest form, some common factors between p and $1 0^{4}$ may cancel. Therefore, it is NOT necessary that the denominator in lowest form is divisible by $2^{4}$ or $5^{4}$ .

Example: $0.1250 = \frac{1250}{10000} = \frac{1}{8}$ Here, in lowest form the denominator is 8, which is not divisible by 24 or 54.

Therefore, it is not necessary.

11. Without performing division, determine whether the decimal expansion of $\frac{18}{125}$ is terminating or non-terminating. If it terminates, state the number of decimal places.

Sol. Given: $\frac{18}{125}$ The denominator is $125 = 5^{3}$ Since the denominator has only the prime factor 5, the decimal expansion is terminating.

To find the number of decimal places, make the denominator a power of $10 : 125 \times 8 = 1000$ So, $\frac{18}{125} = \frac{144}{1000} = 0.144$ Thus, it terminates with 3 decimal places. Therefore, $\frac{18}{125}$ is a terminating decimal with 3 decimal places.

12. A rational number in its lowest form has denominator $2^{3} \times 5$ . How many decimal places will its decimal expansion have? Explain your answer.

Sol. Given denominator: $2^{3} \times 5 = 8 \times 5 = 40$ A rational number with denominator of the form $2^{m} \times 5^{n}$ has a terminating decimal expansion.

The number of decimal places is equal to the greater of m and n .

Here: $m = 3, n = 1$ So, number of decimal places $= 3$

13. Let $a = \frac{7}{12}$ and $b = \frac{5}{6}$ . Express both a and b in the form $\frac{k _{1}}{m}$ and $\frac{k _{2}}{m}$ where $k_{1}, k_{2}$ and m are integers and $k_{2} - k_{1} > 6$ . Using the same denominator m , write exactly five distinct rational numbers lying between a and b keeping an integer numerator. Explain why the condition $k_{2} - k_{1} > n + 1$ is necessary to find n such rational numbers between the two rational numbers a and b using this method.

Sol. Given: $a = \frac{7}{12}; b = \frac{5}{6}$ First express both with the same denominator.

Since, $\frac{5}{6} = \frac{10}{12}$ , we have: $a = \frac{7}{12}$ and $b = \frac{10}{12}$ So, we multiply both fractions by 3 : $a = \frac{7}{12} = \frac{21}{36}$ $b = \frac{10}{12} = \frac{30}{36}$

Now, $k_{1} = 21, k_{2} = 30, m = 36$ and $k_{2} - k_{1} = 30 - 21 = 9 > 6$ So, five distinct rational numbers between $a$ and $b$ are: $\frac{22}{36}, \frac{23}{36}, \frac{24}{36}, \frac{25}{36}, \frac{26}{36}$ These all lie strictly between $\frac{21}{36}$ and $\frac{30}{36}$ For safety in this method, the condition is taken as $k_{2} - k_{1} > n + 1$

Thus, this ensures that enough integer numerators are available between kı and $k_{2}$ to form n distinct rational numbers.

14. Three rational numbers $x, y, z$ satisfy $x + y + z = 0$ and $xy + yz + zx = 0$ . Show that all the rational numbers $x, y, z$ must be simultaneously zero.

Sol. Given: $x + y + z = 0$ $x y + yz + z x = 0$

We use the identity: $(x + y + z)^{2} = x^{2} + y^{2} + z^{2} + 2 (x y + yz + z x)$ From (1), $(x + y + z)^{2} = 0^{2} = 0$ From (2), $x y + yz + z x = 0$ So, $0 = x^{2} + y^{2} + z^{2} + 2 (0)$ Therefore, $x^{2} + y^{2} + z^{2} = 0$ Now, $x^{2}, y^{2}$ and $z^{2}$ are all non-negative rational numbers.

The sum of three non-negative numbers is 0 only when each one is 0 .

Hence, $x^{2} = 0, y^{2} = 0, z^{2} = 0$ Therefore, $x = 0 y = 0 z = 0$ So, all the rational numbers $x, y$ and z are simultaneously zero.

15. Show that the rational number $\frac{( a + b )}{2}$ lies between the rational numbers a and b .

Sol. Let us assume that $a < b$ We have to show that: $a < \frac{( a + b )}{2} < b$ First, compare a and $\frac{( a + b )}{2}$ . Since $a < b$ adding a to both sides gives: $a + a < a + b$ $2 a < a + b$ Dividing both sides by $2 : a < \frac{( a + b )}{2}$ Now, compare $\frac{( a + b )}{2}$ and b : Since $a < b$ adding b to both sides gives: $a + b < b + b$ $a + b < 2 b$ Dividing both sides by 2 : $\frac{( a + b )}{2} < b$ Thus, $a < \frac{( a + b )}{2} < b$ Therefore, the rational number $\frac{( a + b )}{2}$ lies between a and b . Note: If $b < a$ then similarly we get: $b < \frac{( a + b )}{2} < a$ So, in either case, $\frac{( a + b )}{2}$ lies between a and b .

16. Find the lengths of the hypotenuses of all the right triangles in which is referred to as the square root spiral.

Sol. In the square root spiral, each new right triangle is formed by taking: one leg = 1 unit the other leg hypotenuse of the previous triangle

Using Pythagoras theorem: First triangle: Sides = underline 1, 1 Hypotenuse $= (1^{2} + 1^{2}) = 2$ Second triangle: Sides $= 2, 1$ Hypotenuse $= [(2)^{2} + 1^{2}] = 3$ Third triangle: Sides $= 3, 1$ Hypotenuse $= [(3)^{2} + 1^{2}] = 4 = 2$

Fourth triangle: Sides $= 2, 1$ Hypotenuse $= (2^{2} + 1^{2}) = 5$ Fifth triangle: Sides $= 5, 1$ Hypotenuse $= 6$ Sixth triangle: Hypotenuse $= 7$ Seventh triangle: Hypotenuse $= 8 = 22$ Eighth triangle: Hypotenuse $= 9 = 3$ Ninth triangle: Hypotenuse $= 10$ Tenth triangle: Hypotenuse $= 11$ Thus, the lengths of the hypotenuses of the triangles in the square root spiral are: $2, 3, 4, 5, 6, 7, 8, 9, 10, 11 \dots$ That is, $2, 3, 2, 5, 6, 7, 22, 3$ , $10, 11 \dots$

4.0Important Key Concepts of NCERT Solutions for Class 9 Maths Chapter 3 The World of Numbers

Topic	What Students Learn
Natural Numbers, Whole Numbers & Integers	A revisit of the basic number groups and how each group extends the previous one.
Rational Numbers	The definition of rational numbers, their expression in p/q form, and their key properties.
Irrational Numbers	Numbers that cannot be expressed as p/q, along with their decimal expansions.
Real Numbers	How rational and irrational numbers together form the complete set of real numbers.
Number Line Representation	Accurate plotting of rational and irrational numbers on the number line using geometric construction methods.
Decimal Expansions	The difference between terminating, non-terminating repeating, and non-terminating non-repeating decimals.
Operations on Real Numbers	Addition, subtraction, multiplication, and division of real numbers, applying relevant number properties.
Laws of Exponents for Real Numbers	Simplification of expressions with rational exponents using standard exponent laws.
Rationalisation of Denominators	Techniques to rationalise denominators containing surds and simplify complex expressions.
Word Problems & Applications	Application of number system concepts to real-life and application-based questions.

5.0 Exercise-wise NCERT Class 9 Maths Chapter 3 The world of Numbers

Access exercise-wise NCERT Solutions for Class 9 Maths Chapter 3 – The World of Numbers, with every exercise explained through accurate, step-by-step solutions based on the latest NCERT syllabus.

Exercise	Topics Covered
Exercise 3.1	Natural numbers, closure property under addition and subtraction, and a real-world activity connecting finger-joint counting to the ancient base-12 system
Exercise 3.2	Integers, Brahmagupta's laws for signed-number operations, and real-life problems involving profit, loss, debt, and temperature change
Exercise 3.3	Equality of rational numbers, and addition, subtraction, multiplication, and division of rational numbers, including the distributive property
Exercise 3.4	Representing rational numbers on the number line and finding rational numbers between two given numbers
Exercise 3.5	Terminating and non-terminating decimal expansions, cyclic numbers, and converting decimals to p/q form
End of Chapter Exercises	Mixed practice covering proof of irrationality (√5), decimal-to-fraction conversion, and reasoning-based questions on the number system

6.0Quick Revision on Class 9 Maths Chapter 3 - The World of Numbers

7.0Related Study Materials Class 9 Maths

Enhance your learning with ALLEN’s Class 9 Maths study material curated as per latest NCERT syllabus. Along with NCERT Solutions, explore the NCERT textbooks, revision notes, sample papers and previous years’ question papers to strengthen your concepts, revise effectively and prepare confidently for school and CBSE examinations.

CBSE Class 9 Maths Syllabus

Class 9 Maths Revision Notes

NCERT Textbook for Class 9 Maths

CBSE Sample Papers for Class 9 Maths

8.0Advantages of Chapter 3 Maths Class 9 NCERT Solutions

Describes different types of numbers: rational, irrational and real numbers.
Enhances Number Line Visualisation: Enables learners to display various types of numbers correctly.
Decimal Expansions Made Easy: Explains terminating and non terminating decimals easily.
Builds Knowledge of Real Numbers Talks about the properties and relationships between different number sets.
Builds Conceptual Clarity Helps to identify and classify different numbers in the right way.
Prepares for Advanced Concepts: Gives a solid foundation for algebra and higher maths.

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Do these NCERT Solutions include answers to all exercises in Chapter 3?

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