Important Questions Class 9 Maths Chapter 3 The World of Numbers

Important Questions Class 9 Maths Chapter 3 The World of Numbers

This page gives Important Questions Class 9 Maths Chapter 3 The World of Numbers. The chapter is about rational numbers. You will find rational numbers between two numbers, show numbers on the number line, change a repeating decimal into p/q form and check if a decimal ends or not.

Some questions also cover the square root spiral and simple proofs with rational numbers. Every answer is solved step by step in easy words. Read each step slowly and try the question yourself first.

Q1. Find four rational numbers between −2/5 and 1/5.

Answer: To find four rational numbers between −2/5 and 1/5, first write them with a larger common denominator.

Multiply numerator and denominator by 5:

−2/5 = −10/25

and

1/5 = 5/25

Now choose any four rational numbers between −10/25 and 5/25.

For example: −9/25, −7/25, −3/25, 2/25

Q2. Represent the rational numbers 2/3, −5/4 and 1½ on a single number line.

Answer: Convert mixed number:

1½ = 3/2

Now to compare the number first convert into decimal:

−5/4 = −1.25

2/3 ≈ 0.67

3/2 = 1.5

To locate −5/4, divide the interval between −1 and −2 into 4 equal parts and mark the first point to the left of −1.

To locate 2/3, divide the interval between 0 and 1 into 3 equal parts and mark the second point to the right of 0.

To locate 1½, divide the interval between 1 and 2 into 2 equal parts and mark the midpoint.

So, order is:

−5/4 < 2/3 < 3/2

So, we can easily mark these points accordingly on the number line.

Q3. Express 0.234234234... in the form p/q where p and q are integers and q ≠ 0.

Answer: Let x = 0.234234234... (1)

⇒ 1000x = 234.234234234... (2)

By subtracting (1) from (2), we get

999x = 234

x = 234/999 = 26/111

Q4. Find the lengths of the hypotenuses of all the right triangles in Fig which is referred to as the square root spiral.

Answer: 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 = 1, 1

Hypotenuse = √(1² + 1²) = √2

Second triangle:

Sides = √2, 1

Hypotenuse = √[(√2)² + 1²] = √3

Third triangle:

Sides = √3, 1

Hypotenuse = √[(√3)² + 1²] = √4 = 2

Similarly for fourth triangle:

Sides = 2, 1

Hypotenuse = √5

For fifth triangle:

Sides = √5, 1

Hypotenuse = √6

For sixth triangle:

Hypotenuse = √7

For seventh triangle:

Hypotenuse = √8 = 2√2

For eighth triangle:

Hypotenuse = √9 = 3

For ninth triangle:

Hypotenuse = √10

For 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, ...

That is, √2, √3, √4, √5, √6, √7, √8, √9, √10, √11, ...

Hence, in general, the hypotenuse lengths follow the pattern: √2, √3, √4, √5, √6, ... up to as many triangles as drawn in the figure.

Q5. 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.

Answer: Given:

a = 7/12

b = 5/6

First express both with the same denominator.

Since, 5/6 = 10/12,

we have: a = 7/12 and b = 10/12

Here, k₂ − k₁ = 10 − 7 = 3, which is not greater than 6.

So we multiply both fractions by 3:

a = 7/12 = 21/36

b = 10/12 = 30/36

Now, k₁ = 21, k₂ = 30, m = 36

and k₂ − k₁ = 30 − 21 = 9 > 6

So, five distinct rational numbers between a and b are:

22/36, 23/36, 24/36, 25/36, 26/36

These all lie strictly between 21/36 and 30/36.

Therefore, exactly five rational numbers between a and b are:

22/36, 23/36, 24/36, 25/36, 26/36

Now, if we want to find n rational numbers between k₁/m and k₂/m, then there must be at least n integers between k₁ and k₂.

The integers strictly between k₁ and k₂ are k₁ + 1, k₁ + 2, ..., k₂ − 1

Hence, the number of such integers is (k₂ − k₁ − 1)

To obtain n rational numbers between k₁/m and k₂/m, we must have at least n such integers.

Therefore, k₂ − k₁ − 1 ≥ n

So, k₂ − k₁ ≥ n + 1

For safety in this method, the condition is taken as k₂ − k₁ > n + 1

Thus, this ensures that enough integer numerators are available between k₁ and k₂ to form n distinct rational numbers.

Q6. Show that the rational number (a + b)/2 lies between the rational numbers a and b.

Answer: Let us assume that a < b.

We have to show that:

a < (a + b)/2 < b

First, compare a and (a + b)/2:

Since a < b,

adding a to both sides gives:

a + a < a + b

2a < a + b

Dividing both sides by 2:

a < (a + b)/2

Now, compare (a + b)/2 and b:

Since a < b, adding b to both sides gives:

a + b < b + b

a + b < 2b

Dividing both sides by 2:

(a + b)/2 < b

Thus, a < (a + b)/2 < b

Therefore, the rational number (a + b)/2 lies between a and b.

Note: If b < a, then similarly we get:

b < (a + b)/2 < a

So in either case, (a + b)/2 lies between a and b.

Q7. 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.

Answer: Given:

x + y + z = 0 ...(1)

xy + yz + zx = 0 ...(2)

We use the identity:

(x + y + z)² = x² + y² + z² + 2(xy + yz + zx)

From (1), (x + y + z)² = 0² = 0

From (2), xy + yz + zx = 0

So, 0 = x² + y² + z² + 2(0)

Therefore, x² + y² + z² = 0

Now, x², y² and z² are all non-negative rational numbers.

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

Hence, x² = 0, y² = 0, z² = 0

Therefore, x = 0, y = 0, z = 0

So, all the rational numbers x, y and z are simultaneously zero.

Q8. 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/10⁴, 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 2⁴ or 5⁴? Give reasons.

Answer: If the last non-zero digit occurs in the 4th decimal place, then the number has exactly 4 decimal places.

Let the number be a.b₁b₂b₃b₄, where b₄ ≠ 0.

By shifting the decimal point 4 places to the right, we obtain an integer p. Therefore, the number can be written as p/10⁴.

Since the last non-zero digit is in the 4th decimal place, p is not divisible by 10. Otherwise, p would end in 0 and a factor of 10 could be cancelled, which would mean that the decimal expansion actually ends before the 4th decimal place.

Hence, the number can be written in the form p/10⁴, where p is an integer not divisible by 10.

Now, 10⁴ = 10000 = 2⁴ × 5⁴, so the given number can always be written as p/(2⁴ × 5⁴).

When this fraction is reduced to its lowest form, only common factors of 2 and 5 can get cancelled.

But p is not divisible by 10, so p cannot be divisible by both 2 and 5.

If p is not divisible by 2, no factor 2 gets cancelled, so the denominator in the lowest form is still divisible by 2⁴.

If p is not divisible by 5, no factor 5 gets cancelled, so the denominator in the lowest form is still divisible by 5⁴.

Therefore, yes, it is necessary that the denominator in the lowest form is divisible by 2⁴ or 5⁴ (it need not be divisible by both).

For example,

0.9375 = 9375/10000 = 15/16

Here, the denominator in the lowest form is 16 = 2⁴, which is divisible by 2⁴ but not by 5⁴.

Also, 0.0002 = 2/10000 = 1/5000, and 5000 = 2³ × 5⁴, which is divisible by 5⁴ but not by 2⁴.

Hence, a terminating decimal whose last non-zero digit occurs in the 4th decimal place can be written as p/10⁴, and its denominator in the lowest form is always divisible by 2⁴ or 5⁴.

Q9. 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.

Answer: Given: 18/125

The denominator is 125 = 5³

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 × 8 = 1000

So, 18/125 = 144/1000 = 0.144

Thus, it terminates with 3 decimal places.

Therefore, 18/125 is a terminating decimal with 3 decimal places.

Q10. If x/3 + x/5 = 16/15, find the rational number x.

Answer: Given: x/3 + x/5 = 16/15

Taking x common, we have:

x(1/3 + 1/5) = 16/15

Now, 1/3 + 1/5 = 5/15 + 3/15 = 8/15

So, x × 8/15 = 16/15

Therefore, x = (16/15) ÷ (8/15)

= (16/15) × (15/8)

= 16/8

= 2

Therefore, the rational number x is 2.

Q11. Find 5 rational numbers between 1/6 and 2/5.

Answer: Write the fractions with common denominator.

Take, LCM (6, 5) = 30

1/6 = 5/30

2/5 = 12/30

Now the fractions between 5/30 and 12/30 are:

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

Therefore, five rational numbers between 1/6 and 2/5 are:

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

These may also be written in simplest form as:

1/5, 7/30, 4/15, 3/10, 1/3

Q12. Locate the following rational numbers on the number line.

(i) 0.532

(ii) 1.15

Answer: (i) 0.532

0.532 lies between 0 and 1.

More precisely: 0.532 = 532/1000

So, we have to mark a point between 0.53 and 0.54, closer to 0.53.

(ii) 1.15

1.15 = 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 closer to 1.16.

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