Kerala Syllabus Class 9 Maths Chapter 7 Solutions Negative Numbers

Students often refer to Kerala Syllabus 9th Standard Maths Textbook Solutions Chapter 7 Negative Numbers Extra Questions and Answers Notes to clear their doubts.

Kerala SCERT Class 9 Maths Chapter 7 Solutions Negative Numbers

Negative Numbers Class 9 Kerala Syllabus Questions and Answers

Kerala State Syllabus 9th Standard Maths Chapter 7 Negative Numbers Solutions Questions and Answers

Class 9 Maths Chapter 7 Kerala Syllabus – Measures And Numbers

Intext Questions And Answers

Question 1.
Solve the following questions.
(i) 6 – 8
Answer:
6 – 8 = -(8 – 6)
= -2

(ii) -6 + 8
Answer:
-6 + 8 = 8 – 6
= 2

(iii) -6 – 8
Answer:
-6 – 8 = -(6 + 8)
= -14

(iv) 2\(\frac{1}{2}\) – 3\(\frac{1}{2}\)
Answer:
2\(\frac{1}{2}\) – 3\(\frac{1}{2}\)
= \(\frac{5}{2}-\frac{7}{2}=-\left(\frac{7}{2}-\frac{5}{2}\right)\)
= \(-\left(\frac{7-5}{2}\right)=-\left(\frac{2}{2}\right)\)
= -1

(v) -2\(\frac{1}{2}\) + 3\(\frac{1}{2}\)
Answer:
-2\(\frac{1}{2}\) + 3\(\frac{1}{2}\)
= \(-\frac{5}{2}+\frac{7}{2}\)
= \(\frac{7}{2}-\frac{5}{2}=\frac{2}{2}\)
= 1

(vi) -2\(\frac{1}{2}\) – 3\(\frac{1}{2}\)
Answer:
-2\(\frac{1}{2}\) – 3\(\frac{1}{2}\)
= \(-\frac{5}{2}-\frac{7}{2}\)
= \(-\left(\frac{5}{2}+\frac{7}{2}\right)=-\left(\frac{12}{2}\right)\)
= -6

Class 9 Maths Kerala Syllabus Chapter 7 Solutions – Position And Number

Intext Question And Answer

Question 1.
Here is a table giving different initial positions and displacements:

Draw pictures to find the final positions and write in the table. Then make another table with the displacements as just numbers, positive or negative and check if in each row, the last number is the sum of the first two numbers.
Answer:
First row:

Second row:

Third row:

Fourth row:

Fifth row:

Sixth row:

Seventh row:

Eighth row:

The given table can be completed as follows;

Initial Position	Displacement	Final Position
7	3 right	10
3	7 right	10
-7	3 right	-4
-3	7 right	4
7	3 left	4
3	7 left	-4
-7	3 left	-10
-3	7 left	-10

Table with the displacements as just numbers, positive or negative;

Initial Position	Displacement	Final Position
7	3	10
3	7	10
-7	3	-4
-3	7	4
7	-3	4
3	-7	-4
-7	. -3	-10
-3	-7	-10

Consider the first row; 7 + 3 = 10
Consider second row; 3 + 7 = 10
Consider third row; -7 + 3 = 3 – 7 = -(7 – 3) = -4
Therefore, in each row, the last number is the sum of the first two numbers.

Textual Questions And Answers

Question 1.
Complete the table below:

Answer:
6 + (-10) = 6 – 10
= -(10 – 6)
= -4

-6 + 10 = 10 – 6
= 4

-6 + (-10) = -6 – 10
= -(6 + 10)
= -16

-6 + 6 = 6 – 6
= 0

6 + (-6) = 6 – 6
= 0

X	y	x + y
6	-10	-4
-6	10	4
-6	-10	-16
-6	6	0
6	-6	0

Question 2.
Find two pairs of numbers x and y satisfying each of the following conditions:
i) x positive, y negative with, x + y = 1
ii) x negative, y positive with, x + y = 1
iii) x positive, y negative with x + y = – 1
iv) negative, y positive with x + y = – 1
Answer:
i) x = 2, y =-1 and x = 5, y = -4
ii) x = -3, y = 4 and x = -6, y = 7
iii) x = 1, y = -2 and x = 3, y = -4
iv) x = -3, y = 2 and x = -5, y = 4

Question 3.
Complete the table below:

Answer:
First row:
(x + y) + z = (2.+ 4) + (-5)
= 6 – 5
= 1
x + (y + z) = 2 + (4 + (-5))
= 2 + (4 – 5)
= 2 + (-1)
= 2 – 1
= 1

Second row:
(x + y) + z = (2 + (-4)) + 5
= (2 – 4) + 5
= -2 + 5
= 5 – 2
= 3

x + (y + z) = 2 + (-4 + 5)
= 2 + (5 – 4)
= 2 + 1
= 3

Third row:
(x + y) + z = (-2 + 4) + (-5)
= (4 – 2) – 5
= 2 – 5
= -(5 – 2)
= -3

x + (y + z) = -2 + (4 + (-5))
= -2 + (4 – 5)
= -2 + -(5 – 4)
= -2 + (-1)
= -2 – 1
= -(2 + 1)
= -3

Fourth row:
(x + y) + z = (2 + (-4)) + (-5)
= (2 – 4) + (-5)
= -2 – 5
= -(2 + 5)
= -7

x + (y + z) = 2 + (-4 + (-5))
= 2 + (-4 – 5)
= 2 + -(4 + 5)
= 2 + (-9)
= 2 – 9
= -(9 – 2)
= -7

Fifth row:
(x + y) + z = (-2 + 4) + 5
= (4 – 2) + 5
= 2 + 5
= 7

x + (y + z) = -2 + (4 + 5)
= -2 + 9
= 9 – 2
= 7

Sixth row:
(x + y) + z = (-2 + (-4)) + 5
= -(2 + 4) + 5
= -6 + 5
= 5 – 6
= -1

x + (y + z) = -2 + (-4 + 5)
= -2 + (5 – 4)
= -2 + 1
= 1 – 2
= -1

Seventh row:
(x + y) + z = (-2 +(-4)) +(-5)
= (-2 – 4) – 5
= -(2 + 4) – 5
= -6 – 5
= -(6 + 5)
= -11

x + (y + z) = -2+ (-4 +(-5))
= -2 + (-4 – 5)
= -2 – 9
= -(2 + 9)
= -11

SCERT Class 9 Maths Chapter 7 Solutions – Displacement

Textual Questions And Answers

Question 1.
Take different numbers, positive and negative, as x, y, z and calculate x – (y – z) and (x – y) + z. Are both the same number in all cases?
Answer:
Let, x = 1, y = -1, z = 2
x – (y – z) = 1 -(-1 – 2)
= 1 – (-3)
= 1 + 3 = 4

(x – y) + z = (1 – (-1)) + 2
= (1 + 1) + 2
= 2 + 2
= 4
Both are same.

Question 2.
Find two pairs of numbers x and y, satisfying each of the conditions below:
i) x positive, y negative, x – y = 1
ii) x negative, y positive, x – y = -1
Answer:

Question 3.
(i) Take four consecutive natural numbers or their negatives as a, b, c, d and calculate
a – b – c + d.
ii) Explain using algebra why it is zero for all such numbers.
iii) What do we get if we calculate a + b – c – d instead of a – b – c + d?
iv) What about a – b + c – d?
Answer:
i) Let, a = 1, b = 2, c = 3, d = 4
a – b – c + d = 1 – 2 – 3 + 4
= -1 – 3 + 4
= -4 + 4
= 0

ii) Let a = x, b = x + 1, c = x + 2, d = a – b – c + d

iii) a + b – c – d = 1 + 2 – 3 – 4
= 3 – 3 – 4
= 0 – 4
= -4

iv) a – b + c – d = 1 – 2 + 3 – 4
= -1 + 3 – 4
= 2 – 4
= -2

Negative Numbers Class 9 Kerala Syllabus – Time And Speed

Textual Questions And Answers

Question 1.
Take different numbers, positive and negative, as x, y, z and calculate (x + y)z and xz + yz. Check if the equation (x + y)z = xz + yz holds in all cases.
Answer:
Let, x = 1, y = 2, z = -1.
(x + y)z = (1 + 2)(-1) = 3 × (-1) = -3.
xz + yz = 1 × (-1) + 2 × (-1)
= -1 + (-2)
= -1 – 2 = -3.

In this case, (x + y)z = xz + yz.
Let, x = -1, y = 3, z = -4.
(x + y)z = (-1 + 3)(-4)
= 2 × (-4)
= -8.

xz + yz = -1 × (-4) + 3 × (-4)
= 4 + (-12)
= 4 – 12
= -8.

In this case, (x + y)z = xz + yz.
∴ (x + y)z = xz + yz holds in all cases.

Question 2.
Prove that the equation (x + y)(u + v) = xu + xv + yu + yv holds if x, y, u, v are replaced by – x, -y, -u, -v.
Answer:
(-x + (-y))(-u + (-v)) = (-x – y)(-u – v)
= -(x + y) x -(u + v)
= (x + y)(u + v)
= xu + xv + yu + yv
Hence, the proof.

Question 3.
In each of the equations below, find y when x is the given number:
i) y = x², x = -1
ii) y = x² + 3x + 2, x = -1
iii) y = x² + 3x + 2, x = -2
iv) y = (x + 1)(x + 2), x = -1
v) y = (x + 1)(x + 2), x = -2
Answer:
i) y = (-1)²
= -1 × -1
= 1

ii) y = (-1)² + 3 × (-1) + 2
= 1 + (-3) + 2
= 1 – 3 + 2
= -2 + 2
= 0

iii) y = (-2)² + 3 × (-2) + 2
= 4 + (-6) + 2
= 4 – 6 + 2
= -2 + 2
= 0

iv) y = (-1 + 1)(-1 + 2)
= 0 × 1
= 0

v) y = (-2 + 1)(-2 + 2)
= -1 × 0
= 0

Question 4.
In the equation y = x² + 4x + 4 take different numbers, positive and negative, a x and calculate y. Why is y positive or zero in all cases?
Answer:
If x = -1, y = (-1)² + 4 × (-1) + 4 = 1 – 4 + 4 = -3 + 4=l.
If x = 1, y = 1² + 4 × 1 + 4 = 1 + 4 + 4 = 9.
If x = 2, y = 2² + 4 × 2 + 4 = 4 + 8 + 4 = 16.
If x = -2, y = (-2)² + 4 × (-2) + 4 = 4 – 8 + 4 = -4 + 4 = 0.
Here, y = x² + 4x + 4 = (x + 2)², whatever the number x, its square is always greater than or equal to zero. That’s why y positive or zero in all cases.

Question 5.
Natural numbers, their negatives and zero can be together called integers.
i) How many pairs of integers (x, y) can you find, satisfying the equation x² + y² = 25?
ii) How many pairs of integers (x, y) can you find satisfying x² – y² = 25?
Answer:
i) 0² + 5² = 25 → (0, 5)
5² + 0² = 25 → (5,0)
0² + (-5)² = 25 → (0, -5) .
(-5)² + 0² = 25 → (-5, 0)
3² + 4² = 25 → (3, 4)
42 + 3² = 25 → (4, 3)
(-3)² + (-4)² = 25 → (-3,-4)
(-4)² + (-3)² = 25 → (-4,-3) ‘
(-3)² + 4² = 25 → (-3, 4)
4² + (-3)² = 25 → (4, -3)
(-4)² + 3² = 25 → (-4, 3)
3² + (4)² = 25 → (3,-4)
12 pairs of integers.

ii) 5² – 0² = 25
(-5)² – o² = 25
13² – 12² = 25
(-13)² – 12² = 25
13² – (-12)² = 25
(-13)² – (-12)² = 25
6 pairs of integers: (5,0), (-5,0), (13,12), (-13,12), (13,-12), (-13,-12)

Question 6.
1 × 2 × 3 × 4 × 5 = 120. What is (- 1)(- 2)(- 3)(- 4)(- 5)?
Answer:
(-1) × (-2) × (-3) × (-4) × (-5) = 2 × (-3) × (-4) × (-5)
= -6 × (-4) × (-5)
= 24 × (-5)
= -120

Question 7.
What is (1 × 2 × 3 × 4 × 5) + [(- 1)(- 2)(- 3)(- 4)(- 5)]?
Answer:
1 × 2 × 3 × 4 × 5 = ××120
(-1) × (-2) × (-3) × (-4) × (-5) = -120
(1 × 2 × 3 × 4 × 5) + [(-1) × (-2) × (-3) × (-4) × (-5)]
= 120 + (-120)
= 120 – 120
= 0

Class 9 Maths Negative Numbers Kerala Syllabus – Negative Division

Textual Questions And Answers

Question 1.
Calculate y when x is taken as 2, -2, \(\frac{1}{2}\), –\(\frac{1}{2}\) in the equation y = \(\frac{1}{x}\)
Answer:
If x = 2, y = \(\frac{1}{2}\)
If x = -2, y = \(\frac{1}{-2}\) = -(\(\frac{1}{2}\))
If x = \(\frac{1}{2}\), y = \(\frac{1}{\frac{1}{2}} = 2\) = 2
If x = –\(\frac{1}{2}\), y = \(\frac{1}{\frac{-1}{2}} = 2\) = 2

Question 2.
Calculate y when x = -2 and x = –\(\frac{1}{2}\) in the equation y = \(\frac{1}{x-1}+\frac{1}{x+1}\)
Answer:

Question 3.
Calculate z when x and y are taken as the numbers below in the equation
z = \(\frac{x}{y}-\frac{y}{x}\)
(i) x = 10, y = -5
Answer:
Z = \(\frac{10}{-5}-\frac{-5}{10}\)
= -2 – \(\frac{-1}{2}\)
= -(2 –\(\frac{1}{2}\))
= \(\frac{-3}{2}\)

(ii) x = -10, y = 5
Answer:
Z = \(\frac{-10}{5}-\frac{5}{-10}\)
= – 2 – \(\frac{1}{-2}\)
= -(2 – \(\frac{1}{2}\))
= \(\frac{-3}{2}\)

(iii) x = -10, y = -5
Answer:
Z = \(\frac{-10}{-5}-\frac{-5}{-10}\)
= 2 – \(\frac{1}{2}\)
= \(\frac{3}{2}\)

Negative Numbers Class 9 Extra Questions and Answers Kerala Syllabus

Question 1.
Find the value of z from the equation z = x – y with the given values.
a) x = 7, y = 2
b) x = -3, y = -6
c) x = -8, y = 3
d) x = -4, y = 9
Answer:
a) x = 7, y = 2
z = x – y = 7 – 2 = 5

b) x = -3 y = -6
z = x- y = -3- (-6) = -3 + 6 = 3

c) x = -8 y = 3
z = x – y = -8 – 3 = -11

d) x = -4 y = 9
z = x – y = -4 – 9 = -13

Question 2.
If x = 4 y = -3 and z = 8 find the value of
a) (x + y) + z
b) x + (y + z)
c) xyz
d) (x + y)z
e) xy + xz
Answer:
a) (x + y) + z
= (4 – 3) + 8
= (4 – 3) + 8
= 1 + 8
= 9

b) x + (y + z)
= 4 + (-3 + 8)
=4 + 5
= 9

c) yz = 4 × (-3) × 8
= -12 × 8
= -96

d) (x + y)z = (4 – 3) × 8
= 1 × 8
= 8

e) xy + xz = (4 × (-3)) + (4 × 8)
= -12 + 32
= 20

Question 3.
Compute y = x² + 9x – 5 by taking x as the given number.
a) x = 1
b) x = -3
c) x = 0
Answer:
a) x = 1
y = x² + 9x – 5
= 1² + (9 × 1) – 5
= 1 + 9 – 5
= 10 – 5
= 5

b) x = -3
y = x² + 9x – 5
= (-3)² + (9 × (-3)) – 5
= 9 – 27 – 5
= -23

c) x = 0
y = x² + 9x – 5
= 0² + (9 × 0) – 5
= -5

Question 4.
Find y = x⁴ + x³ + x² + x + 2 if x = -1.
Answer:
y = (-1)⁴4 + (-1)³ + (-1)² + (-1) + 2
= 1 +(-1) + 1 + (-1) +2
= 1 – 1 + 1 -1 + 2
= 2

Question 5.
Check whether the equations are identities. Write the patterns got from each, on taking x = 1, 2,3 and x = – 1, -2, -3
a) – x + (x + 3) = 3
b) (x + 2) – (x + 3) = – 1
c) – x – (x + 1) + 2x + 1 = 0
Answer:
If x = 1
a) -x + (x + 3) = -1 + (1 + 3)
= -1 + 4
= 3

b) (x + 2) – (x + 3) = (1 + 2) – (1 + 3)
= 3 – 4
= -1

c) -x – (x + 1) + 2x + 1 = -1 – (1 + 1) + 2 + 1
= -1 – 2 + 2 + 1
= 0

Similarly, if we take x = 2, 3 we can see all these identities are true always.
Now, if x = -1
a) -x + (x + 3)
= -(-1) + (-1 + 3)
= 1 + 2 = 3

b) (x + 2) – (x + 3) = (-1 + 2) – (-1 + 3)
= 1 – 2
= -1

c) -x – (x + 1) + 2x + 1
= -(-1) – (-1 + 1) – 2 + 1
= 1 – 0 – 1
= 0
Similarly if we take x = -2, -3 we can see all these identities are true always.