Students often refer to Kerala Syllabus 9th Standard Maths Textbook Solutions Chapter 9 Real Numbers Extra Questions and Answers Notes to clear their doubts.
Kerala SCERT Class 9 Maths Chapter 9 Solutions Real Numbers
Real Numbers Class 9 Kerala Syllabus Questions and Answers
Kerala State Syllabus 9th Standard Maths Chapter 9 Real Numbers Solutions Questions and Answers
Class 9 Maths Chapter 9 Kerala Syllabus – Absolute Value
Textual Questions And Answers
Question 1.
Complete the table below:
i) Expand the table by taking some more pairs x, y of numbers. Do you see any relation between \xy\ and |x||y|?
ii) Prove that |xy| = |x||y| for any two numbers x and y.
Answer:
i) If the table is expanded,
From the table, we get that the values of |xy| and |x||y| are equal,
ii) If both the numbers are positive.
Considering the values as, x = 6 and y = 4
Then, |xy| = xy = |x||y|
|(6) (4)| = 24 = |6||4|
If both the numbers are negative.
Considering the values as, x = -6 and y = -4
Then, |xy| – xy = |x||y|
|(-6)(-4)| = 24= |-6||-4|
If one of the number is positive and the another one is negative.
Considering the values as, x = 6 and y = -4
Then, |xy| = xy = |x||y|
|(6)(-4)| = 24= |6||-4|
Hence it is clear that |xy| = |x||y| for any two numbers x and y
Question 2.
Complete the table below:
Expand the table by taking some more pairs x, y of numbers. Do you see any relation between |x| + |y| and |x + y|?
Answer:
If the table is expanded,
From the table it is clear that the values of |x| + |y| is not equal to |x + y|
Class 9 Maths Kerala Syllabus Chapter 9 Solutions – Distances
Textual Questions And Answers
Question 1.
Find x satisfying each of the equations below
i) |x| = 5
ii) |x – 3 | = 2
iii) |x – 2 | = 3
iv) |x + 2 | = 3
Answer:
i) If | x | = 5
Then the values of x will be 5 and -5
That is, x = 5 or x = -5
ii) |x – 3| = 2
If x > 3
Then |x – 3| = x – 3
That means, x – 3 = 2
x = 2 + 3 = 5
x = 5
If x < 3
Then |x – 3| = 3 – x
That means, 3 – x = 2
-x = 2 – 3 = -1
x = 1
If |x – 3| = 2 then the values of x will be x = 1 or x = 5
iii) |x – 2| = 3
If x > 2
Then, |x — 2 | = x – 2
That means, x – 2 = 3
x = 3 + 2 = 5
x = 5
If x < 2
Then |x – 2| = 2 – x
That means, 2 – x = 3
-x = 3 – 2 = 1
x = – 1
If |x – 2 | = 3 then the values of x will be x
iv) |x + 2 | = 3
If x > 2
Then, |x + 2 | = x – (-2)
That means, x – (-2) = 3
x + 2 = 3
x = 3 – 2 = 1
x = 1
If x < 2
Then, |x + 2 | = |x – (-2)| = (-2) – x
That means, (-2) – x = 3
– x = 3 + 2= 5
x = -5
If |x + 2| = 3 then the values of x will be x = 1 or x = -5
Question 2.
Find between which numbers x should lie to satisfy each of the equations below:
i) |x| < 3
ii) |x – 2 | < 1
iii) |x – 11 < 2
iv) |x + 11 < 2
Answer:
i) If |x| = 3
Then the possible values of x will be 3 and -3
But if |x| < 3, then the possible values of x will be in between 3 and -3
That means, —3 < x < 3
ii) |x – 2| < 1
The point x can either be in the left or right of the point 2, (distance from 2 is less than 1).
If x is in the right side of the point 2, whose distance is 1.
Then x – 2 = 1
x = 1 + 2 = 3
x = 3
If x is in the left side of the point 2, whose distance is 1.
Then, 2 – x = 1
– x = 1 – 2 = – 1 .
x = 1
Therefore, |x – 2| < 1, then the possible values of x will be in between 1 and 3
That is, 1 < x < 3.
iii) |x – 1| < 2
The point x can either be in the left or right of the point 1, (distance from 1 is less than 2).
If x is in the right side of the point 1, whose distance is 2.
Then, x – 1 = 2
x = 2 + 1 = 3
x = 3
If x is in the left side of the point 1, whose distance is 2.
Then, 1 – x — 2
-x = 2 – 1 = 1
x = – 1
Therefore, |x – 1| < 2, then the possible values of x will be in between -1 and 3
That is, – 1 < x < 3.
iv) |x + 1| < 2
The point x can either be in the left or right of the point (-1), (distance from (-1) is less than 2).
If x is in the right side of the point (-1), whose distance is 2.
Then, x – (- 1) = 2 – x + 1 = 2
x = 2 – 1 = 1
x = 1
If x is in the left side of the point (-1), whose distance is 2.
Then, – 1 – x = 2
– x = 2 + 1 = 3
x = – 3
Therefore, |x + 1| < 2, then the possible values of x will be in between -3 and 1
That is, – 3 < x < 1.
Question 3.
Find the integers satisfying each of the equations in problem (2).
Answer:
i) |x| < 3
The possible values of x will be in between 3 and -3
Therefore the integer values of x = -2, -1, 0, 1, 2
ii) |x – 2| < 1
The possible values of x will be in between 1 and 3
Therefore the integer values of x – 2
iii) |x – 1| < 2
The possible values of x will be in between -1 and 3
Therefore the integer values of x = 0, 1, 2
iv) |x + 1| < 2
The possible values of x will be in between -3 and 1
Therefore the integer values of x = -2, -1, 0
SCERT Class 9 Maths Chapter 9 Solutions – Midpoint
Intext Questions and Answers
Question 1.
What number x satisfies the equation |x – 2| = |x – 5|
Answer:
Geometrical meaning of the equation |x – 2| = |x – 5| is that the point marked by x on the number line is at the same distance from the points marked by the numbers 1 and 4
That means, the point marked by x is the midpoint of the points marked by 2 and 5
So, x = \(\frac{1}{2}\) × (2 + 5) = \(\frac{7}{2}\)
Textual Questions And Answers
Question 1.
Find the number which mark the midpoint of the points marked by each pair of numbers given below on the number line:
i) 1, -5
ii) -1, -5
v) -2.5, 3.5
vi) 1.3, 8.7
vii) -√2, -√3
viii) -√3, √7
Answer:
Question 2.
Find the number which mark the points dividing the distance between the points marked by 1 and 2 into four equal parts, on the number line.
Answer:
There should be three points which divides the distance between the points marked by 1 and 2 into four equal parts.
To get the three points we have to find the midpoint of
Midpoint of 1 and 2 = \(\frac{1}{2}\)(1 + 2) = \(\frac{3}{2}\)
Midpoint of 1 and \(\frac{3}{2}\) = \(\frac{1}{2}\)(1 + \(\frac{3}{2}\)) = \(\frac{5}{4}\)
Midpoint of \(\frac{3}{2}\) and 2 = \(\frac{1}{2}\)(\(\frac{3}{2}\) + 2) = \(\frac{7}{4}\)
Question 3.
Find 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 = \(\frac{1}{2}\)(1 + 3) = \(\frac{4}{2}\) = 2
ii) x = \(\frac{1}{2}\)(3 + 4) = \(\frac{7}{2}\)
iii) x = \(\frac{1}{2}\)(-2 + 5) = \(\frac{3}{2}\)
iv) x = \(\frac{1}{2}\)(0 + -1) = \(\frac{-1}{2}\)
Real Numbers Class 9 Extra Questions and Answers Kerala Syllabus
Question 1.
Find the absolute value of the following.
i) 7
ii) -11
iii) π
iv) \(\frac{-1}{2}\)
Answer:
i) |7| = 7
ii) |-11| = 11
iii) |π| = π
iv) |\(\frac{-1}{2}\)| = \(\frac{1}{2}\)
Question 2.
Complete the table below:
Expand the table by taking some more pairs x, y of numbers. Do you see any relation between |x – y| and |x| – |y|?
Answer:
If the table is expanded,
From the table it is clear that the values of |x| – |y| is not equal to|x – y|
Question 3.
Complete the missing columns of the given table.
Find a value for |x| = – x
Answer:
From the above table, the negative numbers are used.
Therefore, x = -2
Question 4.
What is the number x for which
i) |x – 2| = 5
ii) |x + 3| = 2
Answer:
i) |x – 2| = 5
If x < 2,
Then |x – 2 | = 2 – x
That means, 2 – x = 5
– x = 5 – 2 = 3
x = – 3
If x > 2,
Then |x – 2| = x – 2
That means, x – 2 = 5
x = 5 + 2 = 7
x = 7
Therefore|x – 2| = 5, then x = – 3 or x – 7
ii) |x + 3| = 2
If x < 3 ,
Then |x + 3| = |x – (-3) | = – 3 – x
That means, -3 – x = 2
– x = 2 + 3 = 5
x = – 5
Therefore |x + 3| = 2 , then x = – 5 or x = -1
Question 5.
Find between which numbers x should lie to satisfy each of the equations |x – 3| < 5 . Also find the integer values of the equation.
Answer:
|x – 3| < 5
The point x can either be in the left or right of the point 3, (distance from 3 is less than 5).
If x is in the right side of the point 3, whose distance is 5.
Then x — 3 = 5
x = 5 + 3 = 8
x = 8
If x is in the left side of the point 3, whose distance is 5.
Then, 3 – x = 5
– x = 5 – 3 = -2
x = – 2
Therefore, |x – 3| < 5, then the possible values of x will be in between -2 and 8
That is, —2 < x <8.
Therefore the possible integer values of x is = -1, 0, 1, 2, 3, 4, 5, 6, 7
Question 6.
Write down the numbers located 3 units apart from 0 on a number line.
Answer:
The number can either be in the left or right of the point 0,
If the number is in the right side of the point 0, whose distance is 3 units.
Then ,0 + 3 = 3
If the number is in the left side of the point 0, whose distance is 3 units.
Then, 0 – 3 = -3
That means, the numbers located 3 units apart from 0 on a number line is -3 and 3.
Question 7.
On a number line, calculate the number that represents the midpoint of each pair of numbers given below,
i) -3,-7
ii) √5, √6
iii) 4\(\frac{1}{9}\), -2\(\frac{3}{4}\)
iv) -1.8, 0.25
Answer:
i) \(\frac{1}{2}\)(-3 + (-7)) = \(\frac{-10}{2}\) = -5
ii) \(\frac{1}{2}\)(√5 + √6)
iii) \(\frac{1}{2}\left(\frac{37}{2}+\frac{-11}{4}\right)=\frac{1}{2} \times \frac{63}{4}=\frac{63}{8}\)
iv) \(\frac{1}{2}\)(-1.8 + 0.25) = 0.775
Question 8.
i) Find x,if |x – 2| = |x – 6|
ii) Find y, if |y – 3| = |y + 1|,
iii) Then find the value of |x – y|?
Answer:
i) x = \(\frac{2+6}{2}=\frac{8}{2}\) = 4
n) y = \(\frac{3-1}{2}=\frac{2}{2}\) = 1
iii) |x – y| = |4 – 1| = 3
Question 9.
i) What is the distance between the points representing the numbers 2 and 7 on the number line?
ii) Find the value of x that satisfies the equation |x – 2| = |x – 7|.
Answer:
i) Distance = 7 – 2 = 5
ii) x = \(\frac{7+2}{2}=\frac{9}{2}\) = 4.5