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## Kerala State Syllabus 9th Standard Maths Solutions Chapter 10 Real Numbers

### Real Numbers Textual Questions and Answers

Textbook Page No. 160

**Hss Live Guru 9th Maths Kerala Syllabus Question 1.**

Find the distance between the two points on the number line, denoted by each pair of numbers given below: .

Answer:

**Real Numbers Class 9 Kerala State Board Question 2.**

Find the midpoint of each pair of points in the first problem.

Answer:

**Real Numbers Class 9 State Board Kerala Syllabus Question 3.**

The part of the number line between the points denoted by the numbers 1/3 and 1/2 is divided into four equal parts. Find the numbers de-noting the ends of each such part.

Answer:

So the portion between 1/3 and 1/2 is divided into 4 equal parts the points are

Textbook Page No. 164

**Hsslive Guru 9th Maths Kerala Syllabus Question 1.**

Find those x satisfying each of the equations below:

i. |x – 1| = |x – 3|

ii. |x – 3| = |x – 4|

iii. |x + 2| = |x – 5|

iv. |x| = |x + 1|

Answer:

i. \(|x-1|=|x-3|\)

\(|x-1|\) means distance between x and 1

\(|x-3|\) means distance between x and 3. Therefore the distance from x to 1 and 3 are equal.

x is in between 1 and 3 , that is x is the midpoint of 1 and 3.

\(x=\frac{1}{2} \times(1+3)=\frac{1}{2} \times 4=2\)

ii. \(|x-3|=|x-4|\)

The distance from x to 3 and 4 are equal.

∴ x is in between 3 and 4 , that is x is the midpoint of 3 and 4.

iii. \(\begin{array}{l}{|x+2|=|x-5|} \\ {|x+2|=|x-(-2)|}\end{array}\)

The distance from x to -2 and 5 are equal.

∴ x is in between -2 and 5 , that is x is the midpoint of -2 and 5.

iv. |x| = |x + 1|

|x + l] = ]x – (-l)|

The distance from x to -1 and 0 are equal.

∴ x is in between -1 and 0, that is x is the midpoint of -1 and 0.

**9th Std Kerala Syllabus Maths Solutions Question 2.**

Prove that if 1 < x < 4 and 1 < y < 4, then |x – y| < 3.

Answer:

1 < x < 4 , possible values of x = 2, 3 1 possible values of y = 2, 3

|x – y| = |2 – 2| = 0 < 3

|x – y| = |3 – 2| – 1 < 3

|x – y| = |2 – 3| = |-l| = l < 3

|x – y| = |3 – 3| = 0 < 3

From this if 1 < x < 4, 1 < y < 4, then |x – y| < 3.

**Hss Live Guru Maths 9 Kerala Syllabus Question 3.**

Prove that if x < 3 and y > 7, then |x – y) > 4 .

Answer:

Since x < 3, the maximum value of x is less than 3. Since y > 7 the minimum value of y is greater than 7.

Then the difference between the minimum values of y and the maximum value of x is greater than 7 – 3 = 4.

i.e., |x – y| > 4

**9th Standard Maths Notes Kerala Syllabus Question 4.**

Find two numbers x, y such that

\(|x+y|=|x|+|y|\)

Answer:

1. If x = 3, y =7, then

| x + y | = |13 + 71| = 3 + 7 = 10

|x| + |y| = |3| + |7| = 3 + 7 = 10

|x + y| = |x| + |y|

2. If x = -6, y =-9 , then

|x + y| = |-6 + -9|

= l-15| = 15

**Class 9 Maths Chapter 10 Kerala Syllabus Question 5.**

Are there numbers x,y such that |x + y| < |x| + |y| ?

Answer:

x = 5, y = -3

|x+y| = |5 + -3| = |5 – 3| = |2| = 2

|x| = |5| = 5

|y| = |-3| = 3 |x| + |y| = 5 + 3 = 8

Among x, y, if one is a positive number and the other is a negative number, then

| x + y | < |x| + |y|. Question 6. Are there numbers x, y such that |x + y| > |x| + |y| ?

Answer:

No

**Real Numbers 9th Standard Kerala Syllabus Question 7.**

What are the numbers x, for which

|x – 2| + |x – 8| = 6

Answer:

\(|x-2|\) means distance between x and 2.

x can be to the right or left side of the number 2.

\(|x-8|\) means distance between x and 8

x can be to the right or left side of the number 8.

If we add the distance between x and 2 with x and 8 we get 6.

When x is to the left side of 2, when we add the distance of x to 2 and 8 we get a number which is greater than 6.

Distance between 2 and 8

= |2 – 8| = | -6| = 6

The difference between 2 and 8 is 6, so the position of x is between 2 and 8.

That is the value of x is in between 2 and 8 including 2 and 8.

i. e., 2 ≤ x ≤ 8

**Std 9 Maths Kerala Syllabus Question 8.**

What are the numbers x, for which

|x – 2| + |x – 8| = 10 ?

Taking x as different numbers, what all numbers do we get as

|x – 2| + |x – 8| ?

Answer:

|x – 2| + |x – 8| = 10, |x – 2| – |x – 8| = 10

x – 2 + x – 8 = 10

x = 10

( x – 2) – (x – 8) = 10

There is no number x which satisfies this.

– ( x- 2) + (x – 8) = 10

There is no number x which satisf¬ies this.

-(x – 2) – (x – 8) =10

-x + 2 – x + 8=10

-2x = 0

= x = 0

i.e., x = 0, 10

If x = 1

|x – 2| + |x – 8| = |1 – 2| + |1 – 8|

= | -1 | + | -7| = 1 + 7 = 8

If x = 2

|x – 2| + |x – 8| = |2 – 2| + |2 – 8|

= |0| + | -6| = 0 + 6 = 6

If x = 3

|x – 2| + |x – 8| = |3 – 2| + |3 – 8|

= | -1| + | -5| = 1 + 5 = 6

If x = 4

|x – 2| + |x – 8| = |4 – 2| + |4 – 8|

= | -2| + | -4| = 2 + 4 = 6

If x = 5

|x – 2| + |x – 8| = [5 – 2| +15 – 8|

= | -3| + | -3| = 3 + 3 = 6

If x=6

|x – 2| + |x – 8| = |6 – 2| + |6 – 8|

= | -4|+ | -2| = 4 + 2 = 6

If x = 7

|x – 2| + |x – 8| = |7 – 2| + |17 – 8|

= |-5| + |-1| = 5 + 1 = 6

i.e., |x – 2| + |x – 8| = 6 when 2 < x < 8

If x=9

|x – 2| + |x – 8| = |9 – 2| + |9 – 8|

= |7| + |1| = 7 + 1 = 8

If x= 11

|x – 2| + |x – 8| = |11 – 2| + |11 – 8|

= |9| + |3|= 9 + 3 = 12

…………. etc

Taking x as different numbers, different numbers get as |x – 2| + |x – 8|.

### Real Numbers Exam Oriented Questions and Answer

**Hss Live Guru Class 9 Maths Kerala Syllabus Question 1.**

a. Which number indicates the point located 7 units to left of 5 on the number line?

b. Write down the numbers located 2 units apart from these points.

Answer:

a. The point which is 7 units to the left of 5 = 5 – 7 = -2

b. There is one point 2 unit apart to the left and right side of -2 .

The point which is 2 units to the left of -2 = -4

The point which is 2 units to the right of -2 = 0

**9th Standard Maths Textbook Kerala Syllabus Question 2.**

How many numbers are on the number line which are 11 units apart from 5. What are they ? Calculate the distance between the numbers.

Answer:

There is one point 11 unit apart to the left and right side of 5.

Point on the left side = 5 + -11 = -6

Point on the right side = 5 + 11 = 16

Distance between them = 16 – (-6) = 16 + 6 = 22 unit

**Chapter 10 Maths Class 9 Kerala Syllabus Question 3.**

Find the distance between the num-bers given below on the number line.

i. 4, 6

ii. 3, -2

iii. -5, -8

iv. 3/5, 5/6

Answer:

i. Distance = |6 – 4| = |2| = 2

ii. Distance = |3 – (-2)| = |3 + 2| = |5| = 5

iii. Distance = |-5 -(-8)| =|-5 + 8|

= |+3| = 3

**9th Std Maths Notes Kerala Syllabus Question 4.**

The endpoints of one side of an equilateral triangle are -2 and 4 on the number line. Calculate the area and perimeter of the triangle.

Answer:

The distance between -2 and 4 = 4 – (-2) = 6

One side of a equilateral triangle = 6 unit Perimeter of an equilateral triangle

= 3 × 6 = 18 unit Area of an equilateral triangle

**Class 9 Maths Kerala Syllabus Question 5.**

The number x is on the number line, then find x if |x – 10| = – |x – 4|.

Answer:

The distance between x to 10 and 4 are same. Therefore x is the midpoint of 4 and 10.

Real Numbers Class 10 State Syllabus Kerala Syllabus Question 6.

The quadrilateral ABCD is a square. A and B the points denote -2 and 3 on the number line.

a. Draw the square ABCD on the number line

b. Calculate the perimeter of the square ABCD.

Answer:

b. AB = |-2 – 3| = 5 unit

Perimeter = 5 × 4 = 20 unit

Question 7.

If |a + 1| = |a + 5 |, |b – 2| = |b – 6|,

|a – x| = |b -x|, then find x.

Answer:

|a + 1| = |a + 5|

Since the point, ‘a’ represented on the num¬ber line is at a the equal distance from -1 and -5.

That is ‘a’ is midpoint of -1 and -5.

|b – 2| = |b – 6|

Since the point, ‘a’ represented on the number line is at a equal distance from 2 and 6.

That is ‘a’ is midpoint of 2 and 6

|x – a| = |x – b| therefore point x is the

midpoint of a = -3 and b = 4

Question 8.

Find the value of x.

a. |x – 2| = |x – 10|

b. |x + 3| = |x – 7|

Answer:

a. The distance between x to 2 and 10 are same. Therefore x is the midpoint of 2 and 10.

b. |x + 3| = |x – (-3)|

The distance between x to -3 and 7 are same. Therefore x is the midpoint of -3 and 7.

Question 9.

If the difference between two numbers is 4. If one number is 8. What values will the other number have?

Answer:

Consider other number as x, then

|x – 8| = 4

i.e., x – 8 = 4 or 8 – x = 4

x = 12 or x = 4

Other number as 12 or 4.

Question 10.

Find the distance between the pairs of numbers given below.

a. 2, 8

b. -2, 8

c. -2, -8

Answer:

a. Distance between 2 and 8

= |x – y| = |2 – 8| |-6| = 6

b. Distance between -2 and 8

= |-2-8| = |-10| = 10

c. Distance between -2 and -8

= |-2 – (-8)| = |-2 + 8| = |6| = 6

Question 11.

Draw the number line. Mark the position of numbers given below.

Answer:

Question 12.

The number x is located to the right side of 3 on the number line, if

|x – 3|= 0, then

a. What is the value of x ?

b. Calculate |x – 3| + |x – 9|.

Answer:

a. |x – 3| = 0

-(x – 3) = 0 or x – 3 = 0

-x = -3 or x = 3

x=3

b. |x – 3| + |x – 9|

= |3 – 3| + |3 – 9|

= |0| + |-6 | = 6

Question 13.

If |x – 5| = 10, then find the value of x .

Answer:

The distance between x and 5 is 10.

If x is to the right side of 5 , then x = 5 + 10 = 10

If x is to the left side of 5 , then

x = 5 – 10 = -5

Question 14.

If |x + 2| = |x – 8|, then find the value of x ?

Answer:

|x + 2| = |x – (-2)|

The distance of x to -2 and 8 are equal.

Therefore position of x is between -2 and 8.

Question 15.

If |x – 1| =|x – 3|, then find the value of x.

Answer:

Distance from 1 and 3 to x are same.

Therefore position of x is between 1 and 3.

Position of x = \(\frac { 1 + 3 }{ 2 }\) = \(\frac { 4 }{ 2 }\) = 2

Question 16.

Using the common form of rational numbers, prove that the sum, difference, product and quotient of any two rational numbers is again a rational number.

Answer:

i. If a, b, c and d are natural numbers,

\(\frac { a }{ b }\), \(\frac { c }{ d }\) are rational numbers.

Sum and product of all-natural numbers are always natural numbers. So here the numerator and denominator are natural numbers.

That is if x and y are natural numbers

then this can be expressed as x/y.

Therefore the sum is rational

ii. Difference:

a, b, c and d are natural numbers

Hence it is rational.

iii. Product:

a, b, c and d are natural numbers

Hence it is rational.

iv. Quotient:

a, b, c and d are natural numbers

Hence it is rational.

Question 17.

Prove that the product of any irrational number and non – zero rational number is an irrational number.

Answer:

Let a be the irrational number and b the rational number and a × b = c, that is to prove that c is irrational.

Consider c is a rational number, a × b = c

From this we get b = c/a

That is b = \(\frac { One rational }{ other rational }\) = One rational

If we take b as irrational here we get b as a rational number. This is a wrong decision. This is due to our wrong assumption that is c is a rational number. The assumption that c is a rational number is a wrong one.

∴ c is an irrational number.

That is the product of any irrational number and non – zero rational number is an irrational number.

Question 18.

Give an example of two different irrational numbers whose product is a rational number.

Answer:

3√2 and 5√2 are irrational numbers

3√2 × 5√2 = 3 × 5 × √2 × √2 = 3 × 5 × 2 = 30 is a rational nummber. More examples

1. √3 × 7√3 = 4 × 7 × 3 = 84

2. 2√5 × 5√5 = 2 × 5 × 5 = 50

3. √20 × √5 = √20 × √5 = √100 = 10