Kerala Syllabus 8th Standard Maths Solutions Chapter 9 Negative Numbers

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Kerala State Syllabus 8th Standard Maths Solutions Chapter 9 Negative Numbers

Negative Numbers Text Book Questions and Answers

Textbook Page No 165

Question 1.
Using the principles above, compute the following:
i. 5 – 10
ii. -10 + 5
iii. -5 – 10
iv. -5 – 5
vi. \(-\frac{1}{2}+1 \frac{1}{2}\)
vii. \(-\frac{1}{2}-1 \frac{1}{2}\)
viii. \(-\frac{1}{2}+\frac{1}{2}\)
Solution:
i. 5 – 10 = – (10 – 5) = -5
(since for x – y = -(y – x ))

ii. – 10 + 5 = 5 – 10 = – (10 – 5) = -5
(since for -x + y = y – x and x – y = – (y – x))

iii. -5 – 10 = – (5 + 10 ) = -15
(since for – x – y = -(x + y))

iv. -5 – 5 = – (5 + 5) = -10
(since for -x – y = -(x + y))

v. -5 + 5 = 5 – 5 = 0
(since for -x + y = y – x)
Kerala Syllabus 8th Standard Maths Solutions Chapter 9 Negative Numbers 1

Question 2.
Take as x different positive numbers, negative numbers and zero, and compute x + 1, x – 1, 1 – x. Check whether the equations below hold for all numbers.
i. (1 + x) + (1 – x) = 2
ii. x – (x – 1) = 1
iii. 1 – x = -(x – 1)
Solution:
If x = 1
x + 1 = 1 + 1 = 2
x – 1 = 1 – 1 = 0
1 – x = 1 – 1 = 0
If x = 2
x + 1 = 2 + 1 = 3
x – 1 = 2 – 1 = 1
1 – x = 1 – 2 – 1
If x = o
x + 1 = 0 + 1
x – 1 = 0 – 1 = -1
1 – x = 1 – 0 = 1
If x =-1
x + 1 = -1 + 1 = 1 – 1 = 0
x – 1 = -1 – 1 = -2
1 – x = 1 – (-1) = 1 + 1 = 2
If x = -2
x + 1 = -2 + 1 = -1
x – 1 = -2 – 1 = -3
1 – x = 1 – (-2) = 1 + 2 = 3

i. (1 + x) + (1 – x)
In x = 1, (1 + x) + (1 – x) = 2 + 0 = 2
In x = 2, (1 + x) + (1 – x) = 3 + (-1) = 3 – 1 = 2
In x = o, (1 + x) + (1 – x) = 1 + 1 = 2
In x = -1, (1 + x) + (1 – x) = 0 + 2 = 2
In x – 2, (1 + x) + (1 – x) – 1 + 3 = 3 – 1 = 2
(1 + x) + (1 – x) = 2 , for all values of x

ii. x – (x – 1)
In x = 1, x – (x – 1) = 1 – 0 = 1
In x = 2, x – (x – 1) = 2 – 1 = 1
In x = o, x – (x – 1) = 0 – (-1) = 1
In x = -1, x – (x – 1) = -1 – (-2) = -1 + 2 = 1
In x =-2, x – (x – 1) = -2 – (-3) = -2 + 3 = 1
x – (x – 1) = 1, for all values of x

iii. 1 – x
In x = 1, 1 – x = o = -(x – 1)
In x = 2, 1 – x = -1 = -(1) = -(x – 1)
In x = o, 1 – x = 1 = -(-1) = -(x – 1)
In x = -1, 1 – x = 2 = -(-2) = -(x – 1)
In x = -2, 1 – x = 3 = -(-3) = -(x – 1)
1 – x = -(x – 1), for all values of x

Kerala Syllabus 8th Standard Maths Solutions Chapter 9 Negative Numbers

Question 3.
Taking different numbers as x, y and compute x + y, x – y. Check whether the following hold for all kinds of numbers.
i. (x + y) – x = y
ii. (x + y) – y = x
iii. (x – y) + y = x
Solution:
If we take x = 6 and y = 2, x + y = 6 + 2 = 8 and x – y = 6 – 2 = 4
If x = -6 and y = 2, x + y = -6 + 2 = -4 and x – y = -6 – 2 = -8
If x = 6 and y = -2, x + y = 6 + (-2) = 4
and x – y = 6 – (-2) = 8
If x = -6 and y = -2, x + y = (-6) + (-2)
= -8 and x – y = (-6) – (-2 ) = -4

i. (x + y) – x
If x = 6 and y = -2, (x + y) – x
= 8 – 6 = 2 = y
If x = -6 and y = 2, (x + y) – x
= -4 – (-6) = -4 + 6 = 2 = y
If x = 6 and y = -2, (x + y) – x
= 4 – 6 = – 2 = y
If x = -6 and y = -2, (x + y) – x
= -8 – (-6) = -8 + 6 = -2 = y
(x + y) – x = y, for all values of x and y

ii. (x + y) – y
If x = 6 and y = 2, (x + y) – y
= 8 – 2 = 6 = x
If x = -6 and y = 2, (x + y) – y
= -4 – 2 = -6 = x
If x = 6 and y = -2, (x + y) – y
= 4 – (-2) = 4 + 2 = 6 = x
If x = -6 and y = -2, (x + y) – y
= -8 – (-2) = -8 + 2 = -6 = x
(x + y) – y = x, for all values of x and y

iii. (x – y) + y
If x = 6 and y = 2, (x – y) + y
= 4 + 2 = 6 = x
If x = -6 and y = 2, (x – y) + y
= -8 + 2 = -6 = x
If x = 6 and y = -2, (x – y) + y
= 8 + (-2) = 8 – 2 = 6 = x
If x = -6 and y =- 2, (x – y) + y
= -4 + (-2) = -6 = x
(x – y) + y = x, for all values of x and y

Question 4.
Check whether the equation are identities. Write the patterns got from each, on taking x = 1, 2, 3, 4, 5 and x = -1, -2, -3, -4, -5.
i. -x + (x + 1) = 1
ii. -x + (x + 1) + (x + 2) – (x + 3) = o
iii. -x – (x + 1) + (x + 2) + (x + 3) = 4
Solution:
i) -x + (x + 1) = 1
If x = 1, -x + (x + 1)
= -1 + (1 + 1) = -1 + 2 = 1
If x = 2, -x + (x + 1)
= -2 + (2 + 1) = -2 + 3 = 1
If x = 3, -x + (x + 1)
= -3 + (3 + 1) = -3 + 4 = 1
If x = 4 -x + (x + 1)
= 4 + (4 + 1) = -4 + 5 = 1
If x = 5, -x + (x + 1)
= -5 + (5 + 1) = -5 + 6 = 1
If x = -1, -x + (x + 1)
= -(-1) + (-1 + 1) = 1 +0 = 1
If x = -2, -x + (x + 1)
= -(-2) + (-2 + 1) = 2 + (-1) = 1
If x = -3, -x + (x + 1)
= -(-3) + (-3 + 1) = 3 + (-2) = 1
If x = -4, -x + (x + 1)
= -(-4) + (-4 + 1) = 4 + (-3) = 1
If x = -5, -x + (x + 1)
= -(-5) + (-5 + 1) = 5 + (-4) = 1
It is an identity.

ii. -x + (x + 1) + (x + 2) – (x + 3) = 0
If x = 1, -x + (x + 1) + (x + 2) – (x + 3)
= -1 + (1 + 1) + (1 + 2) – (1 + 3)
= -1 + 2 + 3 – 4 = 0
If x = 2, -x + (x + 1) + (x + 2) – (x + 3)
= -2 + (2 + 1) + (2 + 2) – (2 + 3)
= -2 + 3 + 4 – 5 = 0
If x = 3, -x + (x + 1) + (x + 2) – (x + 3)
= -3 + (3 + 1) + (3 + 2) – (3 + 3)
= -3 + 4 + 5 – 6 = o
If x = 4 -x + (x + 1) + (x + 2) – (x + 3)
= -4 + (4 + 1) + (4 + 2) – (4 + 3)
= -4 + 5 + 6 – 7 = 0
If x = 5, -x + (x + 1) + (x + 2) – (x + 3)
= -5 + (5 + 1) + (5 + 2) – (5 + 3)
= -5 +6 +7 -8= 0
If x = -1, -x + (x + 1) + (x + 2) – (x + 3)
= -(-1) + (-1 + 1) + (-1 + 2) – (-1 + 3)
= 1 + 0 + 1 – 2 = 0
If x = -2, -x + (x + 1) + (x + 2) – (x + 3)
= 2 + (-2 + 1) + (-2 + 2) – (-2 + 3)
= 2 + -1 + 0 – 1 = 0
If x = -3, -x + (x + 1) + (x + 2) – (x + 3)
= 3 + (-3 + 1) + (-3 + 2) – (-3 + 3)
= 3 + -2 + -1 – 0 = 0
If x = -4, -x + (x + 1) + (x + 2) – (x + 3)
= 4 + (-4 + 1) + (-4 + 2) – (-4 + 3)
= 4 + -3 + -2 – (-1) = 0
If x = -5, -x + (x + 1) + (x + 2) – (x + 3)
= 5 + (-5 + 1) +(-5 + 2) – (-5 + 3)
= 5 + -4 + -3 – (-2) = o
It is an identity.

iii. -x – (x + 1) + (x + 2) + (x + 3) = 4
If x = 1, -x – (x +1) + (x + 2) + (x + 3)
= -1 – (1 + 1) + (1 + 2) + (1 + 3)
= -1 – 2 + 3 + 4 = 4
If x = 2, -x – (x + 1) + (x + 2) + (x + 3)
= -2 – (2 + 1) +(2 + 2) + (2 + 3)
= -2 – 3 + 4 + 5 = 4
If x = 3, -x – (x + 1) + (x + 2) + (x + 3)
= -3 – (3 + 1) + (3 + 2) + (2 + 3)
= -3 – 4 + 5 + 6 = 4
If x = 4 -x – (x + 1) + (x + 2) + (x + 3)
= -4 – (4 + 1) + (4 + 2) + (4 + 3)
= -4 – 5 + 6 + 7 = 4
If x = 5, -x – (x + 1) + (x + 2) + (x + 3)
= -5 – (5 + 1) + (5 + 2) +(5 + 3)
= -5 – 6 + 7 + 8 = 4
If x = -1, -x – (x + 1) + (x + 2) + (x + 3)
= -(-1) – (-1 + 1) + (-1 + 2) + (-1 + 3)
= 1 – 0 + 1 + 2 = 4
If x = -2, -x – (x + 1) + (x + 2) + (x + 3)
= 2 – (-2 + 1) + (-2 + 2) + (-2 + 3)
= 2 – (-1) + 0 + 1 = 4
If x = -3, -x – (x + 1) + (x + 2) + (x + 3)
= 3 – (-3 + 1) + (-3 + 2) + (-3 + 3)
= 3 – (-2) + – 1 + 0 = 4
If x = -4, -x – (x + 1) + (x + 2) + (x + 3)
= 4 – (-4 + 1) + (-4 + 2) + (-4 + 3)
= 4 – (-3) + – 2 + (-1) = 4
If x = -5, -x – (x + 1) + (x + 2) + (x + 3)
= 5 – (-5 + 1) + (-5 + 2) + (-5 + 3)
= 5 – (-4) + -3 + (-2) = 4
It is an identity.

Question 5.
Taking different numbers, positive, negative and zero, as x, y, z and compute x + (y + z) and (x + y) + z. Check whether the equation, x + (y + z) = (x + y) + z holds for all these numbers.
Solution:
Kerala Syllabus 8th Standard Maths Solutions Chapter 9 Negative Numbers 5
Textbook Page No 178

Question 6.
Take various positive and negative numbers as x, y, z and compute (x + y) z and xz + yz. Check whether the equation (x + y) z = xz + yz holds for all these.
Solution:
Kerala Syllabus 8th Standard Maths Solutions Chapter 9 Negative Numbers 6

Kerala Syllabus 8th Standard Maths Solutions Chapter 9 Negative Numbers

Question 7.
In each of the following equations, take x as the given numbers and compute the numbers y.
i. y = x2, x = -5, x = 5
ii. y = x2 + 3x + 2, x = -2
iii. y =x2 + 5x + 4, x = -2, x = -3
iv. y = x3 + 1, x = -1
v. y = x3 + x2 + x + 1
Solution:
i. y = x2 , x = -5 , x = 5
If x =-5, y = x2 = (-5 )2 = -5 × -5 = 25
If x = 5, y = x2 = (5)2 = 5 × 5 = 25

ii. y = x2 + 3x + 2, x = -2
If x = -2, y = x2 + 3x + 2 = (-2)2 + 3x(-2) + 2
= 4 – 6 + 2 = -2 + 2 = 0

iii. y = x2 + 5x + 4, x = -2 , x = -3
If x = -2, y = x2 + 5x + 4 = (-2 )2 + 5x(-2) + 4
= 4 – 10 + 4 = -6 + 4 = -2
If x = -3, y = x2 + 5x + 4 = (-3)2 + 5x(-3) + 4
= 9 – 15 + 4 = -6 + 4 = -2

iv. y = x3 + 1, x = -1
If x = -1, y = x3 + 1 = (-1 )3 + 1
= (-1)(-1)(-1) + 1 = 1 × (-1) + 1 = 0

v. y = x3 + x2 + x + 1, x = -1
If x = -1, y = x3 + x2 + x + 1 = (-1)3 +(-1)2 + (-1) + 1 = -1 + 1 – 1 + 1 = o

Question 8.
For a point starting at a point P and travelling along a straight line, time of travel is taken as t and the distance from P as s. The relation between s and t is found to be s = 12t – 2t2, where distances to the right are taken as positive numbers and to the left as negative numbers.
i. Is the position of the point to the right or left of P, till 6 seconds?
ii. Where is the position at 6 seconds?
iii. After 6 seconds?
(Here it is convenient to write 12t – 2t2 = 2t(6 – t).
Solution:
i. S = 12t – 2t2
Distance to the point when time is 1 second = 12t – 2t2 = 12 × 1 – 2 × 12 = 12 – 2 = 10 m
Distance to the point when time is 5 second = 12t – 2t2 = 12 × 5 – 2 × 52 = 60 – 50 = 10 m
Since the distance to the point till 6 seconds is positive. So the position of the point is on the right of P.
ii. Distance to the point when time is 6 second = 12t – 2t2 = 12 × 6 – 2 × 62 = 72 – 72 = o
At the 6th second, the point is at P.
iii. Distance to the point when time is 7 second (after 6 sec)= 12t – 2t2 = 12 × 7 – 2 × 72
= 84 – 2 × 49 = 84 – 98 = -14 metres
This is a negative number, So the position of the point is on the left of P.

Textbook Page No 179

Question 9.
Natural numbers, their negatives and zero are together called integers. How many pair of integers are there, satisfying the equation. x2 + y2 = 25?
Solution:
It is convenient to write it as a table
Kerala Syllabus 8th Standard Maths Solutions Chapter 9 Negative Numbers 50
Textbook Page No 180

Question 10.
Kerala Syllabus 8th Standard Maths Solutions Chapter 9 Negative Numbers 51
Solution:
Kerala Syllabus 8th Standard Maths Solutions Chapter 9 Negative Numbers 52
Kerala Syllabus 8th Standard Maths Solutions Chapter 9 Negative Numbers 53

Kerala Syllabus 8th Standard Maths Solutions Chapter 9 Negative Numbers

Question 11.
Kerala Syllabus 8th Standard Maths Solutions Chapter 9 Negative Numbers 54
Solution:
Kerala Syllabus 8th Standard Maths Solutions Chapter 9 Negative Numbers 55

Question 12.
In the equation \(z=\frac{x}{y}-\frac{y}{x}\), take x as the numbers given below and calculate the number z.
i. x = 10, y = -5
ii. x = -10, y = 5
iii. x = -10, y = -5
iv. x = 5, y = -10
v. x = -5, y = 10
Solution:
Kerala Syllabus 8th Standard Maths Solutions Chapter 9 Negative Numbers 56
Kerala Syllabus 8th Standard Maths Solutions Chapter 9 Negative Numbers 57

Additional Questions and Answers

Question 1.
Match the following
Kerala Syllabus 8th Standard Maths Solutions Chapter 9 Negative Numbers 58
Solution:
Kerala Syllabus 8th Standard Maths Solutions Chapter 9 Negative Numbers 59

Question 2.
Kerala Syllabus 8th Standard Maths Solutions Chapter 9 Negative Numbers 60
Solution:
Kerala Syllabus 8th Standard Maths Solutions Chapter 9 Negative Numbers 61

Kerala Syllabus 8th Standard Maths Solutions Chapter 9 Negative Numbers

Question 3.
Which of the following number is the largest (-1)6, (-1)10, (-1)2, (-1)50
Solution:
If the power of (-1) is even then answer will be 1
If the power of (-1) is odd then answer will be -1
(-1)6 = 1
(-1)10 = 1
(1)2 = 1
(-1)50 = 1
All the given numbers are equal.

Question 4.
Kerala Syllabus 8th Standard Maths Solutions Chapter 9 Negative Numbers 62
Solution:
Kerala Syllabus 8th Standard Maths Solutions Chapter 9 Negative Numbers 63
Kerala Syllabus 8th Standard Maths Solutions Chapter 9 Negative Numbers

Question 5.
Complete the following table
Kerala Syllabus 8th Standard Maths Solutions Chapter 9 Negative Numbers 64
Solution:
Kerala Syllabus 8th Standard Maths Solutions Chapter 9 Negative Numbers 65

Question 6.
Write whether the answer got on doing the following operations are positive number or negative number.
Kerala Syllabus 8th Standard Maths Solutions Chapter 9 Negative Numbers 66
Solution:
Kerala Syllabus 8th Standard Maths Solutions Chapter 9 Negative Numbers 67

Question 7.
Kerala Syllabus 8th Standard Maths Solutions Chapter 9 Negative Numbers 68
Solution:
Kerala Syllabus 8th Standard Maths Solutions Chapter 9 Negative Numbers 69

Question 8.
Kerala Syllabus 8th Standard Maths Solutions Chapter 9 Negative Numbers 698
Solution:
Kerala Syllabus 8th Standard Maths Solutions Chapter 9 Negative Numbers 70
Kerala Syllabus 8th Standard Maths Solutions Chapter 9 Negative Numbers

Question 9.
Calculate (-1)10 + (-1)17 + (-1)21 + (-1)26 + (-1)77
Solution:
(-1)10 + (-1)17 + (-1)21 + (-1)26 + (-1)77
= 1 + (-1) + (-1) + 1 + (-1)
= 1 – 1 – 1 + 1 – 1 = 2 – 3 = -1

Question 10.
Simplify [(-4) × (-5)] + [-16 × \(\frac{-1}{2}\) ]
Solution:
-4 × -5 = 20
Kerala Syllabus 8th Standard Maths Solutions Chapter 9 Negative Numbers 71

Question 11.
If x = 8 and y = -3; find the values of x + y, y + x, x – y, y – x, -x – y and – y – x
Solution:
x + y = 8 + -3 = 5
y + x = -3 + 8 = 5
x – y = 8 – (-3) = 8 + 3 = 11
y – x = -3 – 8 = -11
– x – y = -8 – (-3) = -8 + 3 = -5
-y – x = -(-3) – 8 = 3 – 8 = -5

Kerala Syllabus 8th Standard Maths Solutions Chapter 9 Negative Numbers

Question 12.
If x = 7, y = -6 and z = -2, find the value of
i. (x + y) + z
ii. x + (y + z)
iii. xyz
iv. (x + y)z
v. xy + xz
Solution:
i. (x + y) + z = (7 + -6) + -2 = 1 – 2 = -1
ii. x + (y + z) = 7 + (-6 + -2)
= 7 – 8 = -1
iii. xyz = 7 × -6 × -2 = 84
iv. (x + y)z = (7 + -6) × -2
= 1 × -2 = -2
v. xy + xz = (7 × -6) + (7 × -2)
= -42 – 14
= -56

Question 13.
Compute y = x2 + 9x – 5, for take x as the given number,
i. x = 5
ii. x = -2
iii. x = o
iv. x = -3
Solution:
i. x = 5
y = x2 + 9x – 5
= 52 + 9 × 5 – 5 = 25 + 45 – 5
= 20 + 45 = 65

ii. x = -2
y = x2 + 9x – 5
=(2)2 + 9 × (-2) -5
= 4 – 18 – 5 = 4 – 23 = – 19

iii. x = 0
y = x2 + 9x – 5
= 0 + 9 × 0 – 5 = -5

iv. x = -3
y = x2 + 9x – 5
= (-3)2 + 9 × (-3) – 5
= 9 – 27 – 5 = 9 – 32 = -23

Question 14.
Find y = x4 + x3 + x2 + x + 1, If x = -1
Solution:
y = x4 + x3 + x2 + x + 1
= (-1)4 + (-1)3 + (-1)2 + (-1) + 1
= 1 + (-1) + 1 + (-1) + 1
= 1 – 1 + 1 – 1 + 1
= 0 + 0 + 1 = 1

Question 15.
Complete the table
Kerala Syllabus 8th Standard Maths Solutions Chapter 9 Negative Numbers 75
Solution:
Kerala Syllabus 8th Standard Maths Solutions Chapter 9 Negative Numbers 80

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