NCERT Solutions for Class 9 Maths Chapter 3 The World of Numbers – Ganita Manjari (2026-27)

Think and Reflect 

1. Can you explain why we need q ≠ 0 in the definition of a rational number?

Solution:

• A rational number is a number that can be written in the form p/q, where p and q are integers, and q is not equal to 0.

• Here, q is the denominator. The denominator tells us into how many equal parts a quantity is divided. If q = 0, then the fraction p/q would mean that we are trying to divide a number by zero.

• Division by zero is not possible in mathematics because it does not give a meaningful or fixed value. For example, if we try to divide something into zero equal parts, there is no way to understand how large each part should be.

• Therefore, q must not be equal to 0 in the definition of a rational number.

Think and Reflect

Question 1: While adding or subtracting two rational numbers having different denominators, how will you make the denominators equal?

Solution:

• When two rational numbers have different denominators, we cannot directly add or subtract their numerators. First, we need to make their denominators the same.

• To do this, we find a common denominator. Usually, we take the LCM of the given denominators because it gives the smallest common denominator.

• After finding the LCM, we convert each rational number into an equivalent rational number with that common denominator. This is done by multiplying both the numerator and denominator by a suitable number.

• Once both rational numbers have the same denominator, we can add or subtract only the numerators and keep the denominator the same.

For example, to add 1/3 and 1/4:

The denominators are 3 and 4.

LCM of 3 and 4 is 12.

Now convert both fractions:

• 1/3 = 4/12

• 1/4 = 3/12

Now add:

4/12 + 3/12 = 7/12

So, by making the denominators equal, addition or subtraction becomes easier.

Question 2: Verify the distributive law for rational numbers.

Solution:

The distributive law for rational numbers says that if a, b, and c are rational numbers, then:

a × (b + c) = a × b + a × c

Let us verify this using rational numbers.

Take:

• a = 2/3

• b = 1/4

• c = 3/8

First, solve the left-hand side.

Left-hand side = a × (b + c)

So,

• a × (b + c) = 2/3 × (1/4 + 3/8)

Now add 1/4 and 3/8.

The LCM of 4 and 8 is 8.

• 1/4 = 2/8

So,

• 1/4 + 3/8 = 2/8 + 3/8 = 5/8

Now multiply:

• 2/3 × 5/8 = 10/24

Simplify 10/24:

• 10/24 = 5/12

So, left-hand side = 5/12

Now solve the right-hand side.

Right-hand side = a × b + a × c

So,

• a × b + a × c = 2/3 × 1/4 + 2/3 × 3/8

Now multiply separately.

• 2/3 × 1/4 = 2/12 = 1/6

• 2/3 × 3/8 = 6/24 = 1/4

Now add:

• 1/6 + 1/4

The LCM of 6 and 4 is 12.

• 1/6 = 2/12

• 1/4 = 3/12

So,

• 1/6 + 1/4 = 2/12 + 3/12 = 5/12

So, right-hand side = 5/12

We got:

Left-hand side = 5/12

Right-hand side = 5/12

Since both sides are equal, the distributive law is verified for rational numbers.

Think and Reflect

1. Try and represent 8/5 and -7/4 on a number line.

Solution:

Image 1

To represent 8/5 and -7/4 on a number line, first understand their positions.

8/5 is a positive rational number.

• 8/5 = 1 3/5

So, 8/5 lies between 1 and 2.

To mark 8/5, divide the interval between 1 and 2 into 5 equal parts. Then count 3 parts to the right of 1. That point represents 8/5.

Now consider -7/4.

• -7/4 = -1 3/4

So, -7/4 lies between -2 and -1.

To mark -7/4, divide the interval between -2 and -1 into 4 equal parts. Since -7/4 is equal to -1 3/4, it is closer to -2 than to -1. It can be marked as the first division to the right of -2, or the third division to the left of -1.

Thus, on the number line:

• 8/5 is between 1 and 2.

• -7/4 is between -2 and -1.
  
  

Think and Reflect

Can √2 be written as a rational number p/q?

Solution: Students can try this question themselves as a practice activity.

Think and Reflect

Question 1: Try to prove the irrationality of √3 using the approach of proof by contradiction. Will the same approach work for √5, √7 or √10?

Solution:

To prove that √3 is irrational, we use the method of contradiction.

First, assume the opposite of what we want to prove.

Suppose √3 is rational.

Then √3 can be written in the form p/q, where p and q are integers, q is not equal to 0, and p and q have no common factor other than 1.

So,

• √3 = p/q

Now square both sides.

This gives:

• 3 = p squared / q squared

So,

• p squared = 3q squared

This means p squared is divisible by 3.

Since 3 is a prime number, if p squared is divisible by 3, then p must also be divisible by 3.

So, let:

• p = 3k

where k is an integer.

Now substitute p = 3k in p squared = 3q squared.

Then:

• (3k) squared = 3q squared

So,

• 9k squared = 3q squared

Divide both sides by 3.

We get:

• 3k squared = q squared

This means q squared is divisible by 3.

Therefore, q is also divisible by 3.

Now we have found that both p and q are divisible by 3.

But this is not possible because we assumed that p and q have no common factor other than 1.

So, our assumption that √3 is rational is wrong.

Therefore, √3 is irrational.

Now, will the same method work for √5, √7 and √10?

• Yes, the same contradiction method works for √5 and √7 because 5 and 7 are prime numbers. If p squared is divisible by 5 or 7, then p must also be divisible by 5 or 7.

• For √10, the method can also be used, but with a small adjustment because 10 is not prime. Since 10 = 2 × 5, if p squared is divisible by 10, then p must have factors 2 and 5. This again leads to a contradiction that p and q have a common factor.

• So, the contradiction method works for √3, √5, √7, and √10.

• In fact, this method can be used to prove the irrationality of the square root of any number that is not a perfect square.

Question 2: We have seen how to obtain a line whose length is a rational number. How do we obtain lines whose lengths are irrational?

Solution: Students can try this question themselves as a practice-based activity.

Think and Reflect

1. Try to extend this method for constructing line segments of lengths √3 and √5 using a ruler and a compass. Generalize this method to construct a line segment of any length of the form √n, where n is a positive integer.

Solution: The construction can be extended by repeatedly using right-angled triangles and the Baudhayana-Pythagoras Theorem.

Image 2

To construct √3:

First construct √2 using a right triangle with both perpendicular sides of length 1 unit.

The hypotenuse of this triangle will be √2 because:

1 square + 1 square = 2

So, the hypotenuse is √2.

Now, from the endpoint of the segment representing √2, draw a perpendicular line of length 1 unit.

Join the new endpoint to the starting point O.

This creates a new right triangle whose sides are √2 and 1.

Using the Baudhayana-Pythagoras Theorem:

Hypotenuse square = (√2) square + 1 square
= 2 + 1
= 3

Therefore, the new hypotenuse is √3.

Now, using a compass, take this length as radius and mark it on the number line from O. The point obtained represents √3.

To construct √5:

A simple method is to first mark the length 2 units on the number line, because 2 is the same as √4.

At the point representing 2, draw a perpendicular of length 1 unit.

Join the endpoint of this perpendicular to the starting point O.

Now, a right triangle is formed with sides 2 and 1.

Using the Baudhayana-Pythagoras Theorem:

Hypotenuse square = 2 square + 1 square
= 4 + 1
= 5

Therefore, the hypotenuse is √5.

Using a compass, transfer this length to the number line from O. The marked point represents √5.

Generalisation for √n:

This method can be continued to construct √2, √3, √4, √5, and so on.

Each time, take the previously constructed length √(n - 1).

From its endpoint, draw a perpendicular of length 1 unit.

Join the new endpoint to the starting point O.

By the Baudhayana-Pythagoras Theorem:

Hypotenuse square = (√(n - 1)) square + 1 square
= n - 1 + 1
= n

Therefore, the new hypotenuse becomes √n.

Hence, by repeating this process, we can construct a line segment of length √n for any positive integer n.

Think and Reflect

Try to find the decimal expansions of 10/3 and 11/12. What do you observe about the repetition of the digits after the decimal point?

Solution: Students can try this themselves by dividing 10 by 3 and 11 by 12, then observing whether the digits after the decimal point terminate or repeat.

Think and Reflect

The decimal expansion of p/q will be terminating precisely when the prime factors of q are only 2, only 5 or both 2 and 5. Can you explain why?

Solution: Students can think about this by checking how denominators made only of 2s and 5s can be converted into powers of 10, which gives terminating decimals.

Think and Reflect 

1. Consider this puzzle: What is the square root of -1? We know that 1 × 1 = 1. We also know that (-1) × (-1) = 1. There is no Real Number that, when multiplied by itself, results in a negative number. Thus, √-1 cannot exist on the number line.

Solution: Students can try to reason this themselves by checking why the square of every real number is always either positive or zero, but never negative.

Exercise Set 3.1

Question 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?

Solution:

The merchant gets 15 copper ingots for every 2 bags of spices.

So, the exchange rate is:

2 bags of spices = 15 ingots

Now, he brings 12 bags of spices.

We need to find how many groups of 2 bags are present in 12 bags.

12 ÷ 2 = 6

This means the merchant can make 6 complete exchanges.

For each exchange, he receives 15 copper ingots.

So, total copper ingots received:

6 × 15 = 90

Therefore, the merchant will leave with 90 copper ingots.

Final Answer:

The merchant will receive 90 copper ingots.

Question 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.

Solution:

The given numbers are:

11, 13, 17, 19

Each of these numbers is a prime number.

A prime number is a number that has only two factors: 1 and the number itself.

For example:

• 11 has only 1 and 11 as factors.

• 13 has only 1 and 13 as factors.

• 17 has only 1 and 17 as factors.

• 19 has only 1 and 19 as factors.

So, the common feature is that all the given numbers are prime numbers.

Now, we need to write the next three prime numbers after 19.

The numbers after 19 are:

20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31

Among these, the next prime numbers are:

23, 29 and 31

Final Answer:

The given numbers are prime numbers.

The next three numbers are 23, 29 and 31.

Question 3: We know that Natural Numbers are closed under addition, that is, 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.

Solution:

Natural numbers are:

1, 2, 3, 4, 5, and so on.

A set is said to be closed under an operation if applying that operation to any two numbers of the set always gives a number from the same set.

Natural numbers are closed under addition.

For example:

3 + 5 = 8

Here, 3 and 5 are natural numbers, and their sum 8, is also a natural number.

Now, let us check subtraction.

Example 1:

5 - 3 = 2

Here, the answer is 2, which is a natural number.

Example 2:

3 - 5 = -2

Here, the answer is -2, which is not a natural number.

Example 3:

4 - 7 = -3

Here also, -3 is not a natural number.

So, subtraction of two natural numbers does not always give a natural number.

Therefore, natural numbers are not closed under subtraction.

Final Answer:

No, natural numbers are not closed under subtraction.

For example:

5 - 3 = 2, which is a natural number.

But 3 - 5 = -2, which is not a natural number.

Question 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?

Solution:

On one hand, we use four fingers for counting.

The thumb is not counted as a finger in this method. Instead, the thumb is used to touch and count the joints of the other four fingers.

Each of the four fingers has 3 joints.

So, the total number of joints is:

4 × 3 = 12

This means we can count up to 12 using one hand.

This method is connected to the base-12 counting system.

In a base-12 system, numbers are grouped in sets of 12 instead of sets of 10.

Since one hand can naturally count 12 joints, ancient people may have found it useful to count in groups of 12.

This is why the base-12 system was practical and widely used in many older counting traditions.

Final Answer:

On one hand, we can count 12 joints.

This relates to the base-12 system because the method naturally groups numbers in sets of 12.

Exercise Set 3.2 Class 9 Maths Solutions

Question 1: The temperature in the high-altitude desert of Ladakh is recorded as 4 °C at noon. By midnight, it drops by 15 °C. What is the midnight temperature?

Solution:

The temperature at noon is 4 °C.

By midnight, the temperature drops by 15 °C.

A drop in temperature means we subtract.

So, midnight temperature:

4 - 15 = -11

Therefore, the temperature at midnight is -11 °C.

Final Answer:

The midnight temperature is -11 °C.

Question 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.

Solution:

A loan is treated as debt, so it is represented by a negative integer.

Loan of ₹850 = -850

A profit is represented by a positive integer.

Profit of ₹1,200 = +1200

A loss is represented by a negative integer.

Loss of ₹450 = -450

So, the sequence can be written as:

-850 + 1200 - 450

Now, calculate step by step.

First:

-850 + 1200 = 350

Now subtract the loss:

350 - 450 = -100

The final result is -100.

This means the trader is still in debt of ₹100.

Final Answer:

The equation is:

-850 + 1200 - 450 = -100

The trader’s final financial standing is -₹100, which means he has a debt of ₹100.

Question 3: Calculate the following using Brahmagupta’s laws:

(i) (-12) × 5

Solution:

We have:

(-12) × 5

Here, -12 represents a debt, and 5 represents a fortune.

According to Brahmagupta’s laws, the product of a debt and a fortune is a debt.

So, the answer will be negative.

12 × 5 = 60

Therefore:

(-12) × 5 = -60

Final Answer: -60

(ii) (-8) × (-7)

Solution:

We have:

(-8) × (-7)

Here, both numbers are negative.

According to Brahmagupta’s laws, the product of two debts is a fortune.

This means the product of two negative numbers is positive.

8 × 7 = 56

Therefore:

(-8) × (-7) = 56

Final Answer: 56

(iii) 0 - (-14)

Solution:

We have:

0 - (-14)

Subtracting a negative number is the same as adding the corresponding positive number.

So:

0 - (-14) = 0 + 14

0 + 14 = 14

Therefore, the answer is 14.

Final Answer: 14

(iv) (-20) ÷ 4

Solution:

We have:

(-20) ÷ 4

Here, -20 is negative, and 4 is positive.

According to Brahmagupta’s laws, a debt divided by a fortune gives a debt.

So, the answer will be negative.

20 ÷ 4 = 5

Therefore:

(-20) ÷ 4 = -5

Final Answer: -5

Question 4: Explain, using a real-world example of debt, why subtracting a negative number is the same as adding a positive number, for example, 10 - (-5) = 15.

Solution:

Suppose you have ₹10.

Now imagine that you also have a debt of ₹5.

A debt of ₹5 is represented as -5.

If this debt is removed, your financial condition improves.

Removing a debt of ₹5 is the same as gaining ₹5.

So:

10 - (-5)

means you are starting with ₹10 and removing a debt of ₹5.

This becomes:

10 + 5 = 15

Therefore:

10 - (-5) = 15

This shows that subtracting a negative number is the same as adding a positive number.

Final Answer:

Subtracting a negative means removing a debt. Removing a debt increases the value, so 10 - (-5) becomes 10 + 5 = 15.

Exercise Set 3.1 Class 9 Maths Solutions

Question 1: Prove that the following rational numbers are equal:

(i) 2/3 and 4/6

Solution:

To prove that two rational numbers are equal, we can use cross multiplication.

Two rational numbers a/b and c/d are equal if:

a × d = b × c

Here, the rational numbers are 2/3 and 4/6.

Now compare them by cross multiplication.

2 × 6 = 12

3 × 4 = 12

Since both products are equal, the two rational numbers are equal.

Therefore:

2/3 = 4/6

Final Answer:

2/3 and 4/6 are equal rational numbers.

(ii) 5/4 and 10/8

Solution:

We need to check whether 5/4 and 10/8 are equal.

Using cross multiplication:

5 × 8 = 40

4 × 10 = 40

Both products are the same.

So, the given rational numbers are equal.

Therefore:

5/4 = 10/8

Final Answer:

5/4 and 10/8 are equal rational numbers.

(iii) -3/5 and -6/10

Solution:

We need to prove that -3/5 and -6/10 are equal.

Using cross multiplication:

-3 × 10 = -30

5 × -6 = -30

Since both products are equal, the two rational numbers are equal.

Therefore:

-3/5 = -6/10

Final Answer:

-3/5 and -6/10 are equal rational numbers.

(iv) 9/3 and 3

Solution:

The number 3 can be written as 3/1.

So, we have to compare 9/3 and 3/1.

Using cross multiplication:

9 × 1 = 9

3 × 3 = 9

Both products are equal.

Therefore, 9/3 and 3 are equal.

Also, 9 divided by 3 gives 3.

Final Answer:

9/3 and 3 are equal rational numbers.

Question 2: Find the sum:

(i) 2/5 + 3/10

Solution:

To add rational numbers, we first make the denominators the same.

The denominators are 5 and 10.

LCM of 5 and 10 is 10.

Now convert 2/5 into a fraction with a denominator of 10.

2/5 = 4/10

So:

2/5 + 3/10 = 4/10 + 3/10

Now add the numerators:

4 + 3 = 7

Therefore:

4/10 + 3/10 = 7/10

Final Answer:

2/5 + 3/10 = 7/10

(ii) 7/12 + 5/8

Solution:

The denominators are 12 and 8.

LCM of 12 and 8 is 24.

Now convert both fractions into equivalent fractions with a denominator of 24.

7/12 = 14/24

5/8 = 15/24

Now add:

14/24 + 15/24 = 29/24

The fraction 29/24 is already in its simplest form.

Final Answer:

7/12 + 5/8 = 29/24

(iii) -4/7 + 3/14

Solution:

The denominators are 7 and 14.

LCM of 7 and 14 is 14.

Convert -4/7 into a fraction with a denominator of 14.

-4/7 = -8/14

Now add:

-8/14 + 3/14 = -5/14

So, the sum is -5/14.

Final Answer:

-4/7 + 3/14 = -5/14

Question 3: Find the difference:

(i) 5/6 - 1/4

Solution:

To subtract rational numbers, first make the denominators the same.

The denominators are 6 and 4.

LCM of 6 and 4 is 12.

Now convert both fractions into equivalent fractions with a denominator of 12.

5/6 = 10/12

1/4 = 3/12

Now subtract:

10/12 - 3/12 = 7/12

Final Answer:

5/6 - 1/4 = 7/12

(ii) 11/8 - 3/4

Solution:

The denominators are 8 and 4.

LCM of 8 and 4 is 8.

So, keep 11/8 as it is.

Convert 3/4 into a fraction with a denominator of 8.

3/4 = 6/8

Now subtract:

11/8 - 6/8 = 5/8

Final Answer:

11/8 - 3/4 = 5/8

(iii) -7/9 - (-2/3)

Solution:

We have:

-7/9 - (-2/3)

Subtracting a negative number is the same as adding the corresponding positive number.

So:

-7/9 - (-2/3) = -7/9 + 2/3

Now convert 2/3 into a fraction with a denominator of 9.

2/3 = 6/9

Now add:

-7/9 + 6/9 = -1/9

Final Answer:

-7/9 - (-2/3) = -1/9

Question 4: Find the product:

(i) 2/3 × 3/10

Solution:

To multiply rational numbers, multiply the numerators together and the denominators together.

So:

2/3 × 3/10 = (2 × 3) / (3 × 10)

= 6/30

Now simplify 6/30.

Both 6 and 30 are divisible by 6.

6/30 = 1/5

Final Answer:

2/3 × 3/10 = 1/5

(ii) 7/11 × 5/8

Solution:

Multiply the numerators:

7 × 5 = 35

Multiply the denominators:

11 × 8 = 88

So:

7/11 × 5/8 = 35/88

The fraction 35/88 cannot be simplified further.

Final Answer:

7/11 × 5/8 = 35/88

(iii) -4/7 × 5/14

Solution:

Multiply the numerators:

-4 × 5 = -20

Multiply the denominators:

7 × 14 = 98

So:

-4/7 × 5/14 = -20/98

Now simplify -20/98.

Both 20 and 98 are divisible by 2.

-20/98 = -10/49

Final Answer:

-4/7 × 5/14 = -10/49

Question 5: Find the quotient:

(i) 2/5 ÷ 3/10

Solution:

To divide one rational number by another, multiply the first rational number by the reciprocal of the second rational number.

The reciprocal of 3/10 is 10/3.

So:

2/5 ÷ 3/10 = 2/5 × 10/3

Now multiply:

2 × 10 = 20

5 × 3 = 15

So:

2/5 × 10/3 = 20/15

Now simplify 20/15.

Both 20 and 15 are divisible by 5.

20/15 = 4/3

Final Answer:

2/5 ÷ 3/10 = 4/3

(ii) 7/11 ÷ 5/8

Solution:

The reciprocal of 5/8 is 8/5.

So:

7/11 ÷ 5/8 = 7/11 × 8/5

Now multiply:

7 × 8 = 56

11 × 5 = 55

Therefore:

7/11 × 8/5 = 56/55

Final Answer:

7/11 ÷ 5/8 = 56/55

(iii) -4/7 ÷ 5/14

Solution:

The reciprocal of 5/14 is 14/5.

So:

-4/7 ÷ 5/14 = -4/7 × 14/5

Now simplify before multiplying.

14 divided by 7 gives 2.

So:

-4 × 2 / 5 = -8/5

Final Answer:

-4/7 ÷ 5/14 = -8/5

Question 6: Show that: (1/2 + 3/4) × 8/3 = 1/2 × 8/3 + 3/4 × 8/3

Solution:

We have to show that both sides are equal.

First, solve the left-hand side.

Left-hand side:

(1/2 + 3/4) × 8/3

First, add 1/2 and 3/4.

LCM of 2 and 4 is 4.

1/2 = 2/4

So:

1/2 + 3/4 = 2/4 + 3/4 = 5/4

Now multiply by 8/3:

5/4 × 8/3

= 40/12

Now simplify:

40/12 = 10/3

So, left-hand side = 10/3

Now solve the right-hand side.

Right-hand side:

1/2 × 8/3 + 3/4 × 8/3

First term:

1/2 × 8/3 = 8/6 = 4/3

Second term:

3/4 × 8/3

Here, 3 cancels with 3 and 8 divided by 4 gives 2.

So:

3/4 × 8/3 = 2

Now add:

4/3 + 2

Write 2 as 6/3.

4/3 + 6/3 = 10/3

So, right-hand side = 10/3

Since both sides are equal:

Left-hand side = Right-hand side = 10/3

Therefore:

(1/2 + 3/4) × 8/3 = 1/2 × 8/3 + 3/4 × 8/3

Hence proved.

Question 7: Simplify the following using the distributive property: 7/9 (6/7 - 3/4)

Solution:

We need to simplify:

7/9 (6/7 - 3/4)

Using the distributive property:

a(b - c) = ab - ac

So:

7/9 (6/7 - 3/4) = 7/9 × 6/7 - 7/9 × 3/4

Now simplify the first product:

7/9 × 6/7

Here, 7 cancels with 7.

So:

7/9 × 6/7 = 6/9 = 2/3

Now simplify the second product:

7/9 × 3/4

3 and 9 can be simplified.

3/9 = 1/3

So:

7/9 × 3/4 = 7/12

Now subtract:

2/3 - 7/12

Convert 2/3 into a fraction with a denominator of 12.

2/3 = 8/12

So:

8/12 - 7/12 = 1/12

Final Answer:

7/9 (6/7 - 3/4) = 1/12

Question 8. Find the rational number x such that:

5/6 (x + 3/5) = 5/6x + 1/2

Solution:

We are given:

5/6 (x + 3/5) = 5/6x + 1/2

Now expand the left-hand side using the distributive property.

5/6 (x + 3/5)

= 5/6 × x + 5/6 × 3/5

= 5x/6 + 15/30

Now simplify 15/30:

15/30 = 1/2

So, the left-hand side becomes:

5x/6 + 1/2

This is the same as the right-hand side:

5/6x + 1/2

Therefore, the equation becomes:

5x/6 + 1/2 = 5x/6 + 1/2

This is true for every value of x.

So, x can be any rational number.

Final Answer:

x can be any rational number.

Exercise Set 3.4 Class 9 Maths Solutions

Question 1: Represent the rational numbers 2/3, -5/4 and 1 1/2 on a single number line.

Solution:

We need to show all three rational numbers on the same number line.

The given numbers are:

2/3, -5/4 and 1 1/2

First, understand the position of each number.

2/3 is positive and lies between 0 and 1.

-5/4 can be written as -1 1/4. So, it lies between -2 and -1.

1 1/2 can be written as 3/2. So, it lies between 1 and 2.

Now draw a number line and mark the integers:

-2, -1, 0, 1 and 2

To mark -5/4:

-5/4 = -1 1/4

So, divide the interval between -2 and -1 into 4 equal parts.

Mark the first part to the left of -1.

That point represents -5/4.

To mark 2/3:

Divide the interval between 0 and 1 into 3 equal parts.

Mark the second part from 0.

That point represents 2/3.

To mark 1 1/2:

1 1/2 lies halfway between 1 and 2.

So, divide the interval between 1 and 2 into 2 equal parts.

The midpoint represents 1 1/2.

Final Answer:

On the number line:

Image 3

-5/4 lies between -2 and -1.

2/3 lies between 0 and 1.

1 1/2 lies between 1 and 2.

Question 2: Find three distinct rational numbers that lie strictly between -1/2 and 1/4.

Solution:

We need to find three rational numbers greater than -1/2 and less than 1/4.

One simple way is to use the average method.

First, find the average of -1/2 and 1/4.

Average = (-1/2 + 1/4) ÷ 2

Now add -1/2 and 1/4.

LCM of 2 and 4 is 4.

-1/2 = -2/4

So:

-1/2 + 1/4 = -2/4 + 1/4 = -1/4

Now divide by 2:

-1/4 ÷ 2 = -1/8

So, -1/8 lies between -1/2 and 1/4.

Now find another rational number between -1/2 and -1/8.

Average = (-1/2 + -1/8) ÷ 2

Convert -1/2 to the denominator 8.

-1/2 = -4/8

So:

-4/8 + -1/8 = -5/8

Now divide by 2:

-5/8 ÷ 2 = -5/16

So, -5/16 lies between -1/2 and -1/8.

Now find a rational number between -1/8 and 1/4.

Average = (-1/8 + 1/4) ÷ 2

Convert 1/4 to a denominator of 8.

1/4 = 2/8

So:

-1/8 + 2/8 = 1/8

Now divide by 2:

1/8 ÷ 2 = 1/16

So, 1/16 lies between -1/8 and 1/4.

Final Answer:

Three rational numbers strictly between -1/2 and 1/4 are:

-5/16, -1/8 and 1/16

Question 3: Simplify the expression -1/4 + 5/12.

Solution:

We need to simplify:

-1/4 + 5/12

The denominators are 4 and 12.

LCM of 4 and 12 is 12.

Now convert -1/4 into a fraction with a denominator of 12.

-1/4 = -3/12

Now add:

-3/12 + 5/12 = 2/12

Simplify 2/12.

Both 2 and 12 are divisible by 2.

2/12 = 1/6

Final Answer:

-1/4 + 5/12 = 1/6

Question 4: A tailor has 15 3/4 metres of fine silk. If making one kurta requires 2 1/4 metres of silk, exactly how many kurtas can he make?

Solution:

Total silk available = 15 3/4 metres

Silk required for one kurta = 2 1/4 metres

To find the number of kurtas, divide the total silk by the silk required for one kurta.

First, convert the mixed fractions into improper fractions.

15 3/4 = (15 × 4 + 3) / 4

= (60 + 3) / 4

= 63/4

Now:

2 1/4 = (2 × 4 + 1) / 4

= (8 + 1) / 4

= 9/4

Now divide:

63/4 ÷ 9/4

To divide by a fraction, multiply by its reciprocal.

So:

63/4 × 4/9

Now cancel 4 from the numerator and denominator.

We get:

63/9 = 7

Therefore, the tailor can make 7 kurtas.

Final Answer:

The tailor can make exactly 7 kurtas.

Question 5: Find three rational numbers between 3.1415 and 3.1416.

Solution:

We need to find three rational numbers that lie between 3.1415 and 3.1416.

First, write both numbers with the same number of decimal places.

3.1415 = 3.14150

3.1416 = 3.14160

Now choose any three decimal numbers between 3.14150 and 3.14160.

For example:

• 3.14151

• 3.14152

• 3.14153

These numbers are greater than 3.14150 and smaller than 3.14160.

Also, they are terminating decimals, so they are rational numbers.

Final Answer:

Three rational numbers between 3.1415 and 3.1416 are:

3.14151, 3.14152 and 3.14153

Question 6: Can you think of other ways to find a rational number between any two rational numbers?

Solution:

Yes, there are many ways to find a rational number between two rational numbers.

One useful method is the common denominator method.

Suppose we need to find a rational number between 1/3 and 1/2.

First, write both fractions with a common denominator.

LCM of 3 and 2 is 6.

So:

1/3 = 2/6

1/2 = 3/6

There is no integer numerator between 2 and 3.

So, we can make the denominator bigger.

Multiply both fractions by 2/2.

1/3 = 4/12

1/2 = 6/12

Now 5/12 lies between 4/12 and 6/12.

Therefore, 5/12 lies between 1/3 and 1/2.

Another method is the average method.

The average of any two rational numbers always lies between them.

For example:

Average of 1/3 and 1/2

= (1/3 + 1/2) ÷ 2

= (2/6 + 3/6) ÷ 2

= 5/6 ÷ 2

= 5/12

So, 5/12 lies between 1/3 and 1/2.

We can also use decimal form.

1/3 = 0.333...

1/2 = 0.5

A number like 0.4 lies between them.

Since 0.4 = 4/10 = 2/5, it is also a rational number between 1/3 and 1/2.

Final Answer:

Yes, rational numbers between two rational numbers can be found using the common denominator method, average method, or decimal method.

Exercise Set 3.5 Class 9 Maths Solutions

Question 1: Without performing long division, determine which of the following rational numbers will have terminating decimals and which will be repeating: 7/20, 4/15 and 13/250. Then check your answers by explicitly performing the long divisions and expressing these rational numbers as decimals.

Solution:

• To decide whether a rational number has a terminating or repeating decimal, first check the denominator in its simplest form.

• A rational number has a terminating decimal if the denominator has only 2 and/or 5 as prime factors.

(i) 7/20

The denominator is 20.

20 = 2 × 2 × 5

The denominator contains only the prime factors 2 and 5.

So, 7/20 will have a terminating decimal.

Now divide:

7 ÷ 20 = 0.35

Therefore:

7/20 = 0.35

Final Answer:

7/20 is a terminating decimal, and its decimal form is 0.35.

(ii) 4/15

The denominator is 15.

15 = 3 × 5

Here, the denominator also contains 3.

Since the denominator has a prime factor other than 2 and 5, the decimal expansion will be non-terminating and repeating.

Now divide:

4 ÷ 15 = 0.26666...

So, the digit 6 repeats.

Therefore:

4/15 = 0.26666...

Final Answer:

4/15 is a non-terminating repeating decimal and its decimal form is 0.26666…

(iii) 13/250

The denominator is 250.

250 = 2 × 5 × 5 × 5

The denominator contains only 2 and 5 as prime factors.

So, 13/250 will have a terminating decimal.

Now divide:

13 ÷ 250 = 0.052

Therefore:

13/250 = 0.052

Final Answer:

13/250 is a terminating decimal, and its decimal form is 0.052.

Question 2: Perform the long division for 1/13. Identify the repeating block of digits. Does it show cyclic properties if you evaluate 2/13? Now compute 3/13, 4/13, etc. What do you notice?

Solution:

Let us first find the decimal expansion of 1/13.

By long division:

1 ÷ 13 = 0.076923076923...

So,

1/13 = 0.076923076923...

The repeating block is:

076923

Now, let us check the decimal expansions of other fractions with a denominator of 13.

• 2/13 = 0.153846153846...

• 3/13 = 0.230769230769...

• 4/13 = 0.307692307692...

• 5/13 = 0.384615384615...

• 6/13 = 0.461538461538...

• 7/13 = 0.538461538461...

• 8/13 = 0.615384615384...

• 9/13 = 0.692307692307...

• 10/13 = 0.769230769230...

• 11/13 = 0.846153846153...

• 12/13 = 0.923076923076...

We notice that every decimal has a repeating block of 6 digits.

The digits are connected to the repeating block of 1/13. In many cases, the digits appear as cyclic shifts or related arrangements of the same pattern.

Final Answer:

The repeating block of 1/13 is 076923. The decimal expansions of 2/13, 3/13, 4/13, and so on also repeat in blocks of 6 digits and show a cyclic pattern.

Question 3: Classify the following numbers as rational or irrational:

(i) √81
(ii) √12
(iii) 0.33333...
(iv) 0.123451234512345...
(v) 1.01001000100001...
(vi) 23.560185612239874790120

Find the explicit fractions in case they are rational.

Solution:

(i) √81

We know that:

9 × 9 = 81

So:

√81 = 9

Since 9 can be written as 9/1, it is a rational number.

Final Answer:

√81 is rational.

Explicit fraction: 9/1

(ii) √12

We can write 12 as:

12 = 4 × 3

So:

√12 = √(4 × 3)

√12 = 2√3

Now, √3 is an irrational number.

Multiplying an irrational number by a non-zero rational number also gives an irrational number.

Therefore, 2√3 is irrational.

Final Answer:

√12 is irrational.

(iii) 0.33333...

The decimal 0.33333... is non-terminating but repeating.

Every repeating decimal is rational.

Let:

x = 0.33333...

Multiply both sides by 10:

10x = 3.33333...

Now subtract the first equation from the second:

10x - x = 3.33333... - 0.33333...

9x = 3

x = 3/9

x = 1/3

Final Answer:

0.33333... is rational.

Explicit fraction: 1/3

(iv) 0.123451234512345...

Here, the block 12345 repeats again and again.

So, it is a non-terminating repeating decimal.

Therefore, it is rational.

Let:

x = 0.123451234512345...

Since the repeating block has 5 digits, multiply by 100000.

100000x = 12345.1234512345...

Now subtract:

100000x - x = 12345.1234512345... - 0.1234512345...

99999x = 12345

x = 12345/99999

Now simplify by dividing the numerator and denominator by 3.

12345 ÷ 3 = 4115

99999 ÷ 3 = 33333

So:

x = 4115/33333

Final Answer:

0.123451234512345... is rational.

Explicit fraction: 4115/33333

(v) 1.01001000100001...

In this decimal, the number of zeros keeps increasing.

The digits do not repeat in a fixed block.

So, this decimal is non-terminating and non-repeating.

A non-terminating and non-repeating decimal is irrational.

Final Answer:

1.01001000100001... is irrational.

(vi) 23.560185612239874790120

This decimal ends after a fixed number of digits.

So, it is a terminating decimal.

Every terminating decimal is rational.

It can be written as a fraction by removing the decimal point and placing the number over a suitable power of 10.

Since there are 21 digits after the decimal point:

23.560185612239874790120 = 23560185612239874790120 / 1000000000000000000000

Final Answer:

23.560185612239874790120 is rational.

Explicit fraction:

23560185612239874790120 / 1000000000000000000000

Question 4: The number 0.9, which means 0.99999..., is a rational number. Using algebra, let x = 0.9, multiply by 10, and subtract. Explain why 0.9 is exactly equal to 1.

Solution:

Let:

x = 0.99999...

Now multiply both sides by 10:

10x = 9.99999...

Now subtract the first equation from the second equation.

10x - x = 9.99999... - 0.99999...

9x = 9

Now divide both sides by 9.

x = 1

But we started with:

x = 0.99999...

Therefore:

0.99999... = 1

Final Answer:

0.99999... is exactly equal to 1.

Question 5: We have seen that the repeating block of 1/7 is a cyclic number. Try to find more numbers n whose reciprocals 1/n produce decimals with repeating blocks that are cyclic.

Solution:

A cyclic number is a number whose digits rotate in a pattern when multiplied by certain numbers.

We know that:

1/7 = 0.142857142857...

The repeating block is 142857.

Now multiply this block by 2, 3, 4, 5 and 6.

• 142857 × 2 = 285714

• 142857 × 3 = 428571

• 142857 × 4 = 571428

• 142857 × 5 = 714285

• 142857 × 6 = 857142

We can see that the digits rotate in each product.

This is why 142857 is called a cyclic number.

More such examples can be found by checking reciprocals of some prime numbers.

For example:

1/17 = 0.0588235294117647...

The repeating block has 16 digits.

Another example:

1/19 = 0.052631578947368421...

The repeating block has 18 digits.

• These numbers show cyclic-like behaviour because the repeating block is connected to rotations of digits when multiplied.

• Such prime numbers are often called full reptend primes. For a prime number p, if the decimal expansion of 1/p has the maximum possible repeating length p - 1, then it may produce cyclic behaviour.

Final Answer:

Examples of such numbers are 17 and 19. Their reciprocals have long repeating blocks that show cyclic-like patterns.

End of Chapter Exercises Solutions

Question 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) 3/50

Solution:

We need to convert 3/50 into decimal form.

Divide 3 by 50.

3 ÷ 50 = 0.06

The decimal ends after two decimal places.

So, it is a terminating decimal.

Final Answer:

3/50 = 0.06

It is a terminating decimal.

(ii) 2/9

Solution:

We need to convert 2/9 into decimal form.

Divide 2 by 9.

2 ÷ 9 = 0.22222...

The digit 2 keeps repeating.

So, it is a non-terminating repeating decimal.

Final Answer:

2/9 = 0.22222...

It is a non-terminating repeating decimal.

Question 2: Prove that √5 is an irrational number.

Solution:

We will prove this by contradiction.

Assume that √5 is rational.

If √5 is rational, then it can be written in the form p/q, where p and q are integers, q is not equal to 0, and p and q have no common factor other than 1.

So:

√5 = p/q

Now square both sides.

5 = p square / q square

Multiply both sides by q square.

p square = 5q square

This means p square is divisible by 5.

If p square is divisible by 5, then p must also be divisible by 5.

So, let:

p = 5k

where k is an integer.

Now substitute p = 5k in p square = 5q square.

(5k) square = 5q square

25k square = 5q square

Divide both sides by 5.

5k square = q square

This means q square is divisible by 5.

So, q is also divisible by 5.

Now we have found that both p and q are divisible by 5.

But this contradicts our assumption that p and q have no common factor other than 1.

Therefore, our assumption is wrong.

Hence, √5 is irrational.

Final Answer:

√5 is an irrational number.

Question 3: Convert the following decimal numbers into the form p/q.

(i) 12.6

Solution:

12.6 has one digit after the decimal point.

So, write it over 10.

12.6 = 126/10

Now simplify.

126 and 10 are both divisible by 2.

126 ÷ 2 = 63

10 ÷ 2 = 5

So:

126/10 = 63/5

Final Answer:

12.6 = 63/5

(ii) 0.0120

Solution:

0.0120 has four digits after the decimal point.

So, write it over 10000.

0.0120 = 120/10000

Now simplify.

Divide the numerator and denominator by 40.

120 ÷ 40 = 3

10000 ÷ 40 = 250

So:

0.0120 = 3/250

Final Answer:

0.0120 = 3/250

(iii) 3.0525252...

Solution:

Let:

x = 3.0525252...

Here, the digit 0 is non-repeating and the block 52 repeats.

First, multiply by 10 to move the non-repeating digit before the decimal point.

10x = 30.525252...

Now multiply by 1000 so that the repeating part aligns.

1000x = 3052.525252...

Now subtract:

1000x - 10x = 3052.525252... - 30.525252...

990x = 3022

x = 3022/990

Now simplify by dividing by 2.

3022 ÷ 2 = 1511

990 ÷ 2 = 495

So:

x = 1511/495

Final Answer:

3.0525252... = 1511/495

(iv) 1.2353535...

Solution:

Let:

x = 1.2353535...

Here, the digit 2 is non-repeating and the block 35 repeats.

First multiply by 10:

10x = 12.353535...

Now multiply by 1000:

1000x = 1235.353535...

Now subtract:

1000x - 10x = 1235.353535... - 12.353535...

990x = 1223

x = 1223/990

The fraction cannot be simplified further.

Final Answer:

1.2353535... = 1223/990

(v) 0.232323...

Solution:

Let:

x = 0.232323...

The repeating block is 23, which has 2 digits.

So, multiply by 100.

100x = 23.232323...

Now subtract the original equation.

100x - x = 23.232323... - 0.232323...

99x = 23

x = 23/99

Final Answer:

0.232323... = 23/99

(vi) 2.05555...

Solution:

Let:

x = 2.05555...

Here, 0 is non-repeating and 5 repeats.

First multiply by 10:

10x = 20.5555...

Now multiply by 100:

100x = 205.5555...

Now subtract:

100x - 10x = 205.5555... - 20.5555...

90x = 185

x = 185/90

Now simplify by dividing by 5.

185 ÷ 5 = 37

90 ÷ 5 = 18

So:

x = 37/18

Final Answer:

2.05555... = 37/18

(vii) 2.125555...

Solution:

Let:

x = 2.125555...

Here, 12 is non-repeating and 5 repeats.

First multiply by 100:

100x = 212.5555...

Now multiply by 1000:

1000x = 2125.5555...

Now subtract:

1000x - 100x = 2125.5555... - 212.5555...

900x = 1913

x = 1913/900

Final Answer:

2.125555... = 1913/900

(viii) 3.125555...

Solution:

Let:

x = 3.125555...

Here, 12 is non-repeating and 5 repeats.

First multiply by 100:

100x = 312.5555...

Now multiply by 1000:

1000x = 3125.5555...

Now subtract:

1000x - 100x = 3125.5555... - 312.5555...

900x = 2813

x = 2813/900

Final Answer:

3.125555... = 2813/900

(ix) 2.162516251625...

Solution:

Let:

x = 2.162516251625...

Here, the repeating block is 1625, which has 4 digits.

So, multiply by 10000.

10000x = 21625.16251625...

Now subtract the original equation.

10000x - x = 21625.16251625... - 2.16251625...

9999x = 21623

x = 21623/9999

Final Answer:

2.162516251625... = 21623/9999

Question 4: Locate the following rational numbers on the number line:

(i) 0.532
(ii) 1.15555...

Solution:

(i) 0.532

The number 0.532 lies between 0.53 and 0.54.

To locate it accurately, first mark 0.53 and 0.54 on the number line.

The difference between 0.53 and 0.54 is 0.01.

Now divide this small interval into 10 equal parts.

Each part will represent 0.001.

Starting from 0.530:

First mark = 0.531

Second mark = 0.532

So, the second small division after 0.53 represents 0.532.

Image 4

Final Answer:

0.532 lies between 0.53 and 0.54, at the second division after 0.53 when the interval is divided into 10 equal parts.

(ii) 1.15555...

First convert 1.15555... into a fraction.

Let:

x = 1.15555...

Multiply by 10:

10x = 11.5555...

Multiply by 100:

100x = 115.5555...

Now subtract:

100x - 10x = 115.5555... - 11.5555...

90x = 104

x = 104/90

Simplify:

104/90 = 52/45

So:

1.15555... = 52/45

Now write it as a mixed number.

52/45 = 1 + 7/45

This means the number lies between 1 and 2.

To locate it on the number line, divide the interval from 1 to 2 into 45 equal parts.

Then count 7 parts after 1.

That point represents 52/45, or 1.15555…

Image 5

Final Answer:

1.15555... = 52/45. It lies between 1 and 2, at the 7th division after 1, when the interval is divided into 45 equal parts.

Question 5: Find 6 rational numbers between 3 and 4.

Solution:

To find rational numbers between 3 and 4, write both numbers with a common denominator.

Let us choose a denominator of 10.

3 = 30/10

4 = 40/10

Now choose any 6 fractions with a denominator of 10 between 30/10 and 40/10.

They are:

31/10, 32/10, 33/10, 34/10, 35/10, 36/10

All these numbers are greater than 3 and less than 4.

Final Answer:

Six rational numbers between 3 and 4 are:

31/10, 32/10, 33/10, 34/10, 35/10 and 36/10

Question 6: Find 5 rational numbers between 2/5 and 3/5.

Solution:

We need to find 5 rational numbers between 2/5 and 3/5.

To create enough numbers between them, write both fractions with a larger common denominator.

Multiply the numerator and denominator by 10.

2/5 = 20/50

3/5 = 30/50

Now choose any 5 fractions between 20/50 and 30/50.

They are:

21/50, 22/50, 23/50, 24/50, 25/50

All these numbers lie between 2/5 and 3/5.

Final Answer:

Five rational numbers between 2/5 and 3/5 are:

21/50, 22/50, 23/50, 24/50 and 25/50

Question 7: Find 5 rational numbers between 1/6 and 2/5.

Solution:

We need to find 5 rational numbers between 1/6 and 2/5.

First, take the LCM of the denominators 6 and 5.

LCM of 6 and 5 is 30.

Now convert both fractions into equivalent fractions with a denominator of 30.

1/6 = 5/30

2/5 = 12/30

Now choose 5 fractions between 5/30 and 12/30.

They are:

6/30, 7/30, 8/30, 9/30, 10/30

All these fractions lie strictly between 1/6 and 2/5.

Final Answer:

Five rational numbers between 1/6 and 2/5 are:

6/30, 7/30, 8/30, 9/30 and 10/30

Question 8: If x/3 + x/5 = 16/15, find the rational number x.

Solution:

We are given:

x/3 + x/5 = 16/15

First, take the LCM of 3 and 5.

LCM of 3 and 5 is 15.

Now write both fractions with a denominator of 15.

x/3 = 5x/15

x/5 = 3x/15

So:

5x/15 + 3x/15 = 16/15

Now add the numerators.

8x/15 = 16/15

Since the denominators are the same, compare the numerators.

8x = 16

Now divide both sides by 8.

x = 16/8

x = 2

Final Answer:

x = 2

Question 9: Let a and b be two non-zero rational numbers such that a + 1/b = 0. Without assigning numerical values, determine whether ab is positive or negative. Justify your answer.

Solution:

We are given:

a + 1/b = 0

Move 1/b to the other side.

a = -1/b

Now multiply both sides by b.

ab = -1

Since -1 is a negative number, ab is negative.

Final Answer:

ab is negative.

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

Solution:

Let the rational number be x.

The decimal expansion is terminating, and the last non-zero digit occurs in the 4th decimal place.

This means the number has exactly four decimal places after the decimal point, and the fourth decimal digit is not zero.

So, the number can be written as:

x = p/10000

Since 10000 = 10^4, we can write:

x = p/10^4

Here, p is an integer.

Also, p should not be divisible by 10.

Reason:

If p were divisible by 10, then one zero would cancel from the numerator and denominator.

That would make the decimal terminate before the 4th decimal place.

But the question says the last non-zero digit occurs exactly in the 4th decimal place.

Therefore, p is not divisible by 10.

Now:

10^4 = 10000

10000 = 2^4 × 5^4

When p/10^4 is simplified, some factors of 2 or 5 may get cancelled depending on p.

But since p is not divisible by 10, p cannot contain both 2 and 5 as factors together.

So, after simplification, at least one of the powers 2^4 or 5^4 will remain in the denominator.

However, both 2^4 and 5^4 doesn't need to remain in the denominator.

Final Answer:

The rational number can be written as p/10^4, where p is an integer not divisible by 10. In the lowest form, the denominator will contain only powers of 2 and 5, and at least one of the fourth powers will remain, but both need not remain.

Question 11: Without performing division, determine whether the decimal expansion of 18/125 is terminating or non-terminating. If it terminates, state the number of decimal places.

Solution:

We need to check the denominator of 18/125.

First, check whether the fraction is in its lowest form.

18 and 125 have no common factor other than 1.

So, 18/125 is already in lowest form.

Now factorize the denominator.

• 125 = 5 × 5 × 5

• 125 = 5^3

The denominator has only the prime factor 5.

So, the decimal expansion will be terminating.

Now we need to find the number of decimal places.

To convert 125 into a power of 10, multiply it by 8.

125 × 8 = 1000

So:

18/125 = 18 × 8 / 125 × 8

= 144/1000

= 0.144

The decimal has 3 decimal places.

Final Answer:

18/125 has a terminating decimal expansion with 3 decimal places.

Question 12: A rational number in its lowest form has a denominator of 2^3 × 5. How many decimal places will its decimal expansion have? Explain your answer.

Solution:

The denominator is:

2^3 × 5

This means:

2^3 × 5 = 8 × 5 = 40

A rational number has a terminating decimal if its denominator has only powers of 2 and 5.

Here, the denominator has only 2 and 5, so the decimal expansion will terminate.

Now we need to find the number of decimal places.

The denominator is:

2^3 × 5^1

To make it a power of 10, the powers of 2 and 5 should be equal.

Here, the power of 2 is 3, and the power of 5 is 1.

So, multiply by 5^2 to make the power of 5 also equal to 3.

2^3 × 5^3 = 10^3 = 1000

Therefore, the decimal expansion will have 3 decimal places.

Final Answer:

The decimal expansion will have 3 decimal places.

Question 13: Let a = 7/12 and b = 5/6. Express both a and b in the form k1/m and k2/m, where k1, k2 and m are integers and k2 - k1 > 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 k2 - k1 > n + 1 is necessary to find n such rational numbers between the two rational numbers a and b using this method.

Solution:

We are given:

a = 7/12

b = 5/6

First, write both with the same denominator.

5/6 = 10/12

So:

a = 7/12

b = 10/12

Here:

k1 = 7

k2 = 10

m = 12

Now:

k2 - k1 = 10 - 7 = 3

But 3 is not greater than 6.

So, we need to choose a larger common denominator.

Multiply both fractions by 4/4.

7/12 = 28/48

10/12 = 40/48

So:

a = 28/48

b = 40/48

Now:

k1 = 28

k2 = 40

m = 48

k2 - k1 = 40 - 28 = 12

Since 12 > 6, the condition is satisfied.

Now we can choose exactly five rational numbers between 28/48 and 40/48.

They are:

29/48, 30/48, 31/48, 32/48, 33/48

All these fractions have integer numerators and the same denominator 48.

They lie strictly between 7/12 and 5/6.

Now, let us understand the condition.

Between k1/m and k2/m, the possible fractions with denominator m are:

(k1 + 1)/m, (k1 + 2)/m, (k1 + 3)/m, and so on, up to (k2 - 1)/m.

The number of such fractions is:

k2 - k1 - 1

To get n rational numbers between the two given numbers, we need at least n integer numerators between k1 and k2.

So:

k2 - k1 - 1 should be at least n.

That means:

k2 - k1 should be at least n + 1.

The stronger condition k2 - k1 > n + 1 also ensures that enough rational numbers are available.

Final Answer:

a = 28/48 and b = 40/48.

Five rational numbers between them are:

29/48, 30/48, 31/48, 32/48 and 33/48.

Question 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, and z must be simultaneously zero.

Solution:

We are given:

x + y + z = 0

and

xy + yz + zx = 0

Now square the first equation.

Since:

x + y + z = 0

So:

(x + y + z) square = 0 square

Expanding the left side:

x square + y square + z square + 2xy + 2yz + 2zx = 0

Now group the terms.

x square + y square + z square + 2(xy + yz + zx) = 0

But we are already given:

xy + yz + zx = 0

So:

x square + y square + z square + 2(0) = 0

Therefore:

x square + y square + z square = 0

Now, the square of any rational number is always non-negative.

So, x square, y square and z square are all greater than or equal to 0.

The only way their sum can be 0 is when each square is 0.

So:

x square = 0, y square = 0 and z square = 0

Therefore:

x = 0, y = 0 and z = 0

Final Answer:

All three rational numbers must be zero.

So, x = y = z = 0.

Question 15: Show that the rational number (a + b)/2 lies between the rational numbers a and b.

Solution:

Assume that:

a < b

We need to show that:

a < (a + b)/2 < b

First, prove that (a + b)/2 is greater than a.

Since a < b, add a to both sides.

a + a < a + b

2a < a + b

Now divide both sides by 2.

a < (a + b)/2

So, (a + b)/2 is greater than a.

Now prove that (a + b)/2 is less than b.

Since a < b, add b to both sides.

a + b < b + b

a + b < 2b

Now divide both sides by 2.

(a + b)/2 < b

Therefore:

a < (a + b)/2 < b

Final Answer:

The rational number (a + b)/2 lies between a and b.

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

Image 6

Solution:

In the square root spiral, each new right triangle is formed by using the previous hypotenuse as one side and a new side of length 1 unit.

We use the Baudhayana-Pythagoras Theorem:

Hypotenuse square = Base square + Perpendicular square

Triangle 1:

The two perpendicular sides are 1 and 1.

Hypotenuse square = 1 square + 1 square

= 1 + 1

= 2

So, the hypotenuse is √2.

Triangle 2:

The sides are √2 and 1.

Hypotenuse square = (√2) square + 1 square

= 2 + 1

= 3

So, the hypotenuse is √3.

Triangle 3:

The sides are √3 and 1.

Hypotenuse square = (√3) square + 1 square

= 3 + 1

= 4

So, the hypotenuse is √4 = 2.

Triangle 4:

The sides are 2 and 1.

Hypotenuse square = 2 square + 1 square

= 4 + 1

= 5

So, the hypotenuse is √5.

Triangle 5:

The sides are √5 and 1.

Hypotenuse square = 5 + 1 = 6

So, the hypotenuse is √6.

Triangle 6:

The sides are √6 and 1.

Hypotenuse square = 6 + 1 = 7

So, the hypotenuse is √7.

Triangle 7:

The sides are √7 and 1.

Hypotenuse square = 7 + 1 = 8

So, the hypotenuse is √8.

Triangle 8:

The sides are √8 and 1.

Hypotenuse square = 8 + 1 = 9

So, the hypotenuse is √9 = 3.

Triangle 9:

The sides are 3 and 1.

Hypotenuse square = 3 square + 1 square

= 9 + 1

= 10

So, the hypotenuse is √10.

Triangle 10:

The sides are √10 and 1.

Hypotenuse square = 10 + 1 = 11

So, the hypotenuse is √11.

Final Answer:

The hypotenuse lengths are:

√2, √3, 2, √5, √6, √7, √8, 3, √10 and √11.

Image 7

Key Highlights of NCERT Solutions for Class 9 Maths Chapter 3 The World of Numbers – Ganita Manjari (2026-27)

The NCERT Solutions for Class 9 Maths Chapter 3 The World of Numbers – Ganita Manjari are designed to help students understand the core ideas of the Maths subject in a simple and structured way. This chapter covers important topics such as natural numbers, integers, rational numbers, irrational numbers, real numbers, decimal expansions, and number line representation.

These Class 9 Maths Chapter 3 solutions explain each question step by step, making it easier for students to revise concepts, complete homework, and prepare for school exams. Students can also download the free PDF of NCERT Solutions for Class 9 Maths Chapter 3 The World of Numbers for quick revision anytime during the 2026-27 academic session.

By practising these solutions regularly, students can improve their understanding of number systems, identify terminating and non-terminating decimals, and solve rational number-based questions with better accuracy. Focusing on concept clarity will help Class 9 students build a strong base for higher-level Maths topics.

Access Exercise Wise NCERT Solutions for Chapter 3 Maths Class 9

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Chapter-Specific NCERT Solutions for Class 9 Maths

Given below are the chapter-wise NCERT Solutions for Class 9 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

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