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8 MATHS CHAPTER 10

 

Kerala Syllabus 8th Standard Maths Solutions Chapter 10 Statistics

May 5, 2021 by Prasanna

You can Download Equal Triangles Questions and Answers, Activity, Notes, Kerala Syllabus 8th Standard Maths Solutions Chapter 10 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 8th Standard Maths Solutions Chapter 10 Statistics

Statistics Text Book Questions and Answers

Textbook Page No 184

Class 8 Maths Chapter Statistics Kerala Syllabus Question 1.
The number of members in 50 households of a village are listed below.

Make a frequency table and answer these questions:
i. How many households have just two members?
ii. How many households have four or less?
iii. How many households have ten or more?
iv. Households of what size occurs the most?
Solution:

i. 5
ii. 5 + 11 + 9 = 25
iii. 1 + 1 = 2
iv. 3

8th Standard Maths Statistics Kerala Syllabus Question 2.
There are 44 children in class 8B. The list shows how far they come from, in kilometres.

Make a frequency table and answer these questions:
i. How many children are from exactly 1 kilometre away?
ii. How many are from more than 5 kilometres?
iii. How many are from between 5 and 10 kilometres?
iv. How many are from more than 10 kilometres?
Solution:

i. 3
ii. 21
iii. 23
iv. 4

8th Standard Maths Guide Kerala Syllabus  Question 3.
The scores of 35 children in a test are given below:

Make a frequency table and answers these questions:
i. How many children scored 20?
ii. How many children got scores between 10 and 20?
iii. How many scored less than 10?
iv. What is the score most number of children got?
Solution:
Score Tally Number of children

i. 1
ii. 32
iii. 0
iv. 15

Textbook Page No 188

Statistics Class 8 Notes Kerala Syllabus Question 4.
Given below are the highest temperatures (in degree Celsius) one day in 40 towns. Make a frequency table.

Solution:

8th Class Maths Statistics Kerala Syllabus Question 5.
The heights (in centimeters) of 45 people who look part in a physical fitness test are given below? Make a frequency table.

Solution:

Textbook Page No 190

Hss Live Guru Maths 8th Kerala Syllabus Question 6.
The table shows the times 30 children took to complete a long distance race. Draw a histogram of this.

Solution:

Hsslive Guru Maths 8th Standard Kerala Syllabus Question 7.
The table shows the daily incomes of 6o households in a locality.

Draw a histogram.
Solution:

Hss Live Std 8 Maths Kerala Syllabus Question 8.
Detail of rainfall in June and July are given in the table below. Draw a histogram.

Solution:

Hsslive Guru 8th Class Maths Kerala Syllabus Question 9.
The time taken by 25 women and 23 men to complete a race are given in the table below. Draw separate histograms for men and women.

Solution:

Kerala Syllabus 8th Standard Maths Notes Question 10.
The weights of 45 children in a class are listed below.


Make a frequency table and draw a histogram.
Solution:

Additional Questions And Answers

8th Standard Maths Kerala Syllabus Question 1.
Complete the table below on the basis of the histogram.

Solution:

8th Class Maths Notes Kerala Syllabus Question 2.
Teacher conducted a test in her class of 45 students. Their score out of a total of 10 are given below.

a. Construct a frequency table representing these details?
b. Construct a bar graph?
Solution:

Hss Live Guru Class 8 Maths Kerala Syllabus Question 3.
The list below gives the daily wages earned by 30 laboures. Prepare a frequency table of these.

a. What is the daily wages of most labourers?
b. How many get 250 rupees a day?
c. How many get the least amount of wages?
Solution:

a. 225
b. 6
c. 5

Class 8 Maths Chapter 10 Kerala Syllabus Question 4.
The table shows daily expenditure of 60 household in a locality. Draw histrogram.

Solution:

8th Standard Maths Notes Kerala Syllabus Question 5.
Given below are the amount of rainfall (in mm) one day in 61 towns. Make a frequency table.


Solution:

8th Std Maths Solution Guide Kerala Syllabus Question 6.
The runs that a batsman got in 40 one – day cricket matches are given below. Make a frequency table and answer these questions.

a. How many centuries did he get?
b. How many half centimes?
c. In how many games did he score less than 50?
Solution:

Question 7.
The height of 30 childrens in a class are listed below. Mark a frequency table and draw a histogram.

Solution:

8 MATHS CHAPTER 9

 

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

Negative Numbers Text Book Questions and Answers

Textbook Page No 165

Negative Numbers Class 8 Kerala Syllabus Question 1.
Using the principles above, compute the following:
i. 5 – 10
ii. -10 + 5
iii. -5 – 10
iv. -5 – 5
vi. −12+112
vii. −12−112
viii. −12+12
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 Notes Pdf 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

Hsslive Guru Maths 8th Standard Kerala Syllabus 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

8th Standard Maths Notes Kerala Syllabus 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.

Kerala Syllabus 8th Standard Maths Notes 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:

Textbook Page No 178

Class 8 Maths Kerala Syllabus 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 Guide 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

Kerala Syllabus 8th Standard Maths 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

8th Std Maths Guide Kerala Syllabus 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

Textbook Page No 180

Kerala Syllabus 8th Standard Maths Notes Malayalam Medium Question 10.

Solution:

Class 8 Kerala Syllabus Maths Question 11.

Solution:

Class 8 Maths Scert Solutions Kerala Syllabus Question 12.
In the equation z=xy−yx, 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:

Additional Questions and Answers

Maths Class 8 Kerala Syllabus  Question 1.
Match the following

Solution:

8th Class Maths Notes Kerala Syllabus Question 2.

Solution:

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.

Solution:

Question 5.
Complete the following table

Solution:

Question 6.
Write whether the answer got on doing the following operations are positive number or negative number.

Solution:

Question 7.

Solution:

Question 8.

Solution:

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 × −12 ]
Solution:
-4 × -5 = 20

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

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

Solution: