Category: Class 8

AP State Syllabus 8th Class Maths Solutions 1st Lesson Rational Numbers InText Questions

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 1 Rational Numbers InText Questions and Answers.

8th Class Maths 1st Lesson Rational Numbers InText Questions and Answers

Do this

Question 1.
Consider the following collection of numbers 1, \(\frac{1}{2}\), -2, 0.5, 4\(\frac{1}{2}\), \(\frac{-33}{7}\), 0, \(\frac{4}{7}\), \(0 . \overline{3}\), 22, -5, \(\frac{2}{19}\), 0.125. Write these numbers under the appropriate category. [A number can be written in more than one group]  (Page No. 2)
Answer:
i) Natural numbers 1, 22
ii) Whole numbers 0, 1, 22
iii) Integers 0, 1, 22, -5, -2
iv) Rational numbers 1, \(\frac{1}{2}\), -2, 0.5, 4\(\frac{1}{2}\), \(\frac{-33}{7}\), 0, \(\frac{4}{7}\), \(0 . \overline{3}\), 22, -5, \(\frac{2}{19}\), 0.125 etc.
Would you leave out any of the given numbers from rational numbers? No
Is every natural number, whole number and integer is a rational number? Yes

Question 2.
Fill the blanks in the table.     (Page No. 6)
Answer:

Question 3.
Complete the following table.     (Page No. 9)
Answer:

Question 4.
Complete the following table.      (Page No. 13)
Answer:

Question 5.
Complete the following table.      (Page No. 16)
Answer:

Question 6.
Complete the following table.      (Page No. 17)
Answer:

Question 7.
Represent – \(\frac{13}{5}\) on the number line.     (Page No. 22)
Answer:
Representing – \(\frac{13}{5}\) on the number line.

Try These

Question 1.
Hamid says \(\frac{5}{3}\) is a rational number and 5 is only a natural number. Shikha says both are rational numbers. With whom do you agree?       (Page No. 3)
Answer:
I would not agree with Hamid’s argument. Since \(\frac{5}{3}\) is a rational number. But ‘5’ is not only
a natural number, it is also a rational number.
Since every natural number is a rational number,
According to Shikha’s opinion \(\frac{5}{3}\), 5 are rational numbers.
∴ I agree with Shikha’s opinion.

Question 2.
Give an example to satisfy the following statements.        (Page No.3)
i) All natural numbers are whole numbers but all whole numbers need not be natural numbers.
ii) All whole numbers are integers but all integers are not whole numbers.
iii) All integers are rational numbers but all rational numbers need not be integers.
Answer:
i) ‘0’ is not a natural number.
∴ Every whole number is not a natural number. (∵ N ⊂ W)
ii) -2, -3, -4 are not whole numbers.
∴ All integers are not whole numbers. (∵ W ⊂ Z)
iii) \(\frac{2}{3}\), \(\frac{7}{4}\) are not integers.
∴ Every rational number is not an integer. (∵ Z ⊂ Q)

Question 3.
If we exclude zero from the set of integers is it closed under division? Check the same for natural numbers.    (Page No. 6)
Answer:
If ‘0’ is subtracted from the set of integers then it becomes Z – {0}.
Closure property under division on integers.
Ex: -4 ÷ 2 = -2 is an integer.
3 ÷ 5 = \(\frac{3}{5}\) is not an integer.
∴ Set of integers doesn’t satisfy closure property under division.
Closure property under division on natural numbers.
Ex: 2 ÷ 4 = \(\frac{1}{2}\) is not a natural number.
∴ Set of natural numbers doesn’t satisfy closure property under division.

Question 4.
Find using distributivity.     (Page No. 16)
A) \(\left\{\frac{7}{5} \times\left(\frac{-3}{10}\right)\right\}+\left\{\frac{7}{5} \times\left(\frac{9}{10}\right)\right\}\)
B) \(\left\{\frac{9}{16} \times 3\right\}+\left\{\frac{9}{16} \times-19\right\}\)
Answer:
Distributive law: a × (b + c) = ab + ac
A)

B)

Question 5.
Write the rational number for the points labelled with letters, on the number line.       (Page No. 22)
i)

ii)

Answer:
i) A = \(\frac{1}{5}\), B = \(\frac{4}{5}\), C = \(\frac{5}{5}\) = 1, D = \(\frac{7}{5}\), E = \(\frac{8}{5}\), F = \(\frac{10}{5}\) = 2.
ii) S = \(\frac{-6}{4}\), R = \(\frac{-6}{4}\), Q = \(\frac{-3}{4}\), P = \(\frac{-1}{4}\)

Think, discuss and write

Question 1.
If a property holds good with respect to addition for rational numbers, whether it holds good for integers? And for whole numbers? Which one holds good and which doesn’t hold good?     (Page No. 15)
Answer:
Under addition the properties which are followed by set of rational numbers are also followed by integers.

Question 2.
Write the numbers whose multiplicative inverses are the numbers themselves.      (Page No. 15)
Answer:
The number T is multiplicative inverse of itself.
∵ 1 × \(\frac{1}{1}\) = 1 ⇒ 1 × 1 = 1
∴ The multiplicative inverse of 1 is 1.

Question 3.
Can you find the reciprocal of ‘0’ (zero)? Is there any rational number such that when it is multiplied by ‘0’ gives ‘1’?
          (Page No. 15)
Answer:
The reciprocal of ‘0’ is \(\frac{1}{0}\).
But the value of \(\frac{1}{0}\) is not defined.
∴ There is no number is found when it is multiplied ‘0’ gives 1.
∵ 0 × (Any number) = 0

∴ No, there is no number is found in place of ‘A’.

Question 4.
Express the following in decimal form.     (Page No. 28)
i) \(\frac{7}{5}\), \(\frac{3}{4}\), \(\frac{23}{10}\), \(\frac{5}{3}\),\(\frac{17}{6}\),\(\frac{22}{7}\)
ii) Which of the above are terminating and which are non-terminating decimals?
iii) Write the denominators of above rational numbers as the product of primes.
iv) If the denominators of the above simplest rational numbers has no prime divisors other than 2 and 5 what do you observe?
Answer:
i) \(\frac{7}{5}\) = 0.4,
\(\frac{3}{4}\) = 0.75,
\(\frac{23}{10}\) = 2.3,
\(\frac{5}{3}\) = 1.66… = \(1 . \overline{6}\),
\(\frac{17}{6}\) = 2.833… = \(2.8 \overline{3}\),
\(\frac{22}{7}\) = 3.142
ii) From the above decimals \(\frac{7}{5}\), \(\frac{3}{4}\), \(\frac{23}{10}\) are terminating decimals.
While \(\frac{5}{3}\),\(\frac{17}{6}\),\(\frac{22}{7}\) are non-terminating decimals
iii) By writing the denominators of above decimals as a product of primes is

iv) If the denominators of integers doesn’t have factors other than 2 or 5 and both are called terminating decimals.

Question 5.
Convert the decimals \(0 . \overline{9}\), \(14 . \overline{5}\) and \(1.2 \overline{4}\) to rational form. Can you find any easy method other than formal method?     (Page No. 31)
Answer:
Let x = \(0 . \overline{9}\)
⇒ x = 0.999 ……. (1)
The periodicity of the above equation is ‘1’. So it is to be multiplied by 10 on both sides.
⇒ 10 × x = 10 × 0.999
10x = 9.999 …….. (2)
From (1) & (2)

∴ x = 1 or \(0 . \overline{9}\) = 1
Second Method:
\(0 . \overline{9}\) = 0 + \(0 . \overline{9}\)
= 0 + \(\frac{9}{9}\)
= 0 + 1
= 1

Let x = \(14 . \overline{5}\)
⇒ x = 14.55 …….. (1)
The periodicity of the equation (1) is 1.
So it should be multiplied by 10 on both sides.
⇒ 10 × x = 10 × 14.55
10x = 145.55 …….. (2)

Second Method:

Let x = \(1.2 \overline{4}\)
⇒ x= 1.244 …….. (1)
Here periodicity of equation (1) is 1. So it should be multiplied by 10 on both sides.
⇒ 10 × x = 10 × 1.244
10 x = 12.44 …….. (2)

Second Method:

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 10 Direct and Inverse Proportions Ex 10.4 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 10th Lesson Direct and Inverse Proportions Exercise 10.4

Question 1.
Rice costing ₹480 is needed for 8 members for 20 days. What is the cost of rice required for 12 members for 15 days?
Solution:
Method – 1: Number of men and rice required to them are in inverse proportion.
Number of men ∝ \(\frac{1}{\text { No. of days }}\)

⇒ Compound ratio of 8:12 and 20: 15
= \(\frac{8}{12}=\frac{20}{15}\) = \(\frac{8}{9}\) …………….. (2)
From (1), (2)
480 : x = 8 : 9
⇒ \(\frac{480}{x}=\frac{8}{9}\)
⇒ x = \(\frac{480 \times 9}{8}\) = ₹540
∴ The cost of required rice is ₹ 540

Method – II :
\(\frac{M_{1} D_{1}}{W_{1}}=\frac{M_{2} D_{2}}{W_{2}}\)
M₁ = No. of men
D₁ = No .of days
W₁ = Cost of rice
∴ M₁ = 8
D₁ = 20
W₁ = ₹ 480
M₂ = 12
D₂ = 15
W₂ = ? (x)

⇒ x = 45 x 12 = ₹ 540
The cost of required rice = ₹ 540/-

Question 2.
10 men can lay a road 75 km. long in 5 days. In how many days can 15 men lay a road 45 km. long?
Solution:
\(\frac{M_{1} D_{1}}{W_{1}}=\frac{M_{2} D_{2}}{W_{2}}\)
∴ M₁ = 10
D₁ = 5
W₁ = 75
M₂ = 15
D₂ = ?
W₂ = 45

∴ x = 2
∴ No. of days are required = 2

Question 3.
24 men working at 8 hours per day can do a piece of work in 15 days. In how many days can 20 men working at 9 hours per day do the same work?
Solution:
M₁D₁H₁ = M₂D₂H₂
∴ M₁ = 24
D₁ = 15 days
H₁ = 8 hrs
M₂ = 20
D₂ = ?
H₂ = 9 hrs
⇒ 24 × 15 × 8 = 20 × x × 9

∴ No. of days are required = 16
[ ∵ No. of men and working hours are in inverse]

Question 4.
175 men can dig a canal 3150 m long in 36 days. How many men are required to dig a canal 3900 m. long in 24 days?
Solution:
\(\frac{M_{1} D_{1}}{W_{1}}=\frac{M_{2} D_{2}}{W_{2}}\)
M₁ = 175
D₁ = 36
W₁ = 3150
M₂ = ?
D₂ = 24
W₂ = 3900

∴ No. of workers are required = 325

Question 5.
If 14 typists typing 6 hours a day can take 12 days to complete the manuscript of a book, then how many days will 4 typists, working 7 hours a day, can take to do the same job?
Solution:
M₁D₁H₁ = M₂D₂H₂
M₁ = 14
D₁ = 12 days
H₁ = 6
M₂ = 4
D₂ = ?
H₂ = 7
⇒ 14 × 12 × 6 = 4 × x × 7

⇒ x = 36
∴ No. of days are required = 36
[ ∵ No of men and working hours are in inverse proportion]

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 10 Direct and Inverse Proportions Ex 10.3 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 10th Lesson Direct and Inverse Proportions Exercise 10.3

Question 1.
Siri has enough money to buy 5 kg of potatoes at the price of ₹ 8 per kg. How much can she buy for the same amount if the price is increased to ₹ 10 per kg?
Solution:
Number of kgs of potatoes to their price are in inverse proportion.
∴ x₁y₁ = x₂ y₂
⇒ 8 × 5 = 10 × x
⇒ x = \(\frac{8 \times 5}{10}\) = 4 kgs
∴ 4 kgs of potatoes will be purchased at the rate of ₹ 10 per kg.

Question 2.
A camp has food stock for 500 people for 70 days. ¡f200 more people join the camp, how long will the stock last?
Solution:
Number of persons and their food stock are in inverse proportion.
⇒ x₁y₁ = x₂ y₂ (Let y₂ = x say)
⇒ 500 × 70 = (500 + 200) × x
⇒  x = \(\frac{500 \times 70}{700}\) = 5 × 10
∴ x = 50
∴ The food will be stock for (200 + 500) 700 men = 50 days

Question 3.
36 men can do a piece of work in 12 days. ¡n how many days 9 men can do the same work?
Solution:
Number of workers and number of days are in inverse proportion
∴ x₁y₁ = x₂ y₂ let y₂ = x (say)
= 36 × 12 = 9 × x
x = \(\frac{36 \times 12}{9}\) = 48
∴ x = 48 days

Question 4.
A cyclist covers a distance of28 km in 2 hours. Find the time taken by him to cover a distance of 56 km with the same speed.
Solution:
Time and distance are in direct proportion.
∴ \(\frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}}\) , Let y₂ = x (say)
⇒ \(\frac{28}{2}\) = \(\frac{56}{x}\)
⇒ x = \(\frac{56}{14}\)
∴ x = 4 hours

Question 5.
A ship can cover a certain distance in 10 hours at a speed of 16 nautical miles per hour. By how much should its speed be increased so that it takes only 8 hours to cover the same distance? (A nautical mile in a unit of measurement used at sea distance or sea water i.e. 1852 metres).
Solution:
Speed and distance are in inverse proportion.
⇒ x₁y₁ = x₂ y_{2 ,} Let x₂ = x (say)
⇒ 16 × 10 = x × 8
⇒ x = \(\frac{16 \times 10}{8}\)= 20
∴ x = 20
∴ The speed to be increased
= 20 – 16 = 4 nautical miles

Question 6.
5 pumps are required to fill a tank in 1\(\frac { 1 }{ 2 }\) hours. How many pumps of the same type are used to fill the tank in half an hour.
Solution:
Number of pumps and time to fill the tanks are in inverse proportion.
⇒ x₁y₁ = x₂ y₂
⇒ 5 × 1\(\frac { 1 }{ 2 }\) = x x 1\(\frac { 1 }{ 2 }\)
⇒ 5 × \(\frac { 3 }{ 2 }\) = x x \(\frac { 1 }{ 2 }\)
⇒ x = 5 × 3 = 15
∴ Number of pumps required = 15

Question 7.
If 15 workers can build a wall in 48 hours, how many workers will be required to do the same work in 30 hours?
Solution:
Number of workers and time are in inverse proportion.
⇒ x₁y₁ = x₂ y₂
⇒ 15 × 48 = x × 30
⇒ x = \(\frac{15 \times 48}{30}\) = 24
∴ Number of workers required = 24

Question 8.
A School has 8 periods a day each of45 minutes duration. How long would each period become ,if the school has 6 periods a day? ( assuming the number of school hours to be the same)
Solution:
Time and number of periods are in inverse proportion.
⇒ x₁y₁ = x₂ y₂
⇒ 45 × 8 = x × 6
⇒ \(\frac{45 \times 8}{6}\)
⇒ 60 minutes

Question 9.
If z varies directly as xand inversely as y. Find the percentage increase in z due to an increase of 12% in x and a decrease of 20% in y.
Solution:
Given that
z varies directly as x and inversely as y So, z ∝ x (1); z ∝ 1/y ……………… (2)
From (1) & (2), z ∝ \(\frac{\mathrm{x}}{\mathrm{y}}\)

Let x₁ = 100x, x₂ = 112x
(∵ It increases 12%)
y₁ = 100y, y₂ = 80y
(∵ It decreases 20%)
From (3),

∴ z is increased in 40%

Question 10.
If x + 1 men will do the work in x + 1 days, find the number of days that (x + 2) men can finish the same work.
Solution:
Number of workers and number of days are in inverse proportion.
⇒ x₁y₁ = x₂ y₂
⇒ (x + 1) (x + 1) = (x + 2) x k
⇒ k = \(\frac{(x+1)(x+1)}{(x+2)}\)
∴ k = \(\frac{(x+1)^{2}}{(x+2)}\)

Question 11.
Given a rectangle with a fixed perimeter of 24 meters, if we increase the length by 1 m the width and area will vary accordingly. Use the following table of values to look at how the width and area vary as the length varies.
What do you observe? Write your observations in your note books

Solution:

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 10 Direct and Inverse Proportions Ex 10.2 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 10th Lesson Direct and Inverse Proportions Exercise 10.2

Question 1.
Observe the following tables and fmd which pair of variables (x and y) are in inverse proportion

Solution:
i) From the given table if the value of x is decreases then the value of ‘y’ is increases.
∴ x, y are in inverse proportion.
ii) From the given table if the value of x is increases then the value of y is decreases.
∴ x, y are in inverse proportion.
iii) From the given table the value of x is decreases then the value of y is increases.
∴ x, y are in inverse proportion.

Question 2.
A school wants to spend ₹6000 to purchase books. Using this data, fill the following table.

Solution:

Question 3.
Take a squared paper and arrange 48 squares in different number of rows as shown below.

What do you observe? As R increases, C decreases
(i) Is R₁:R₂ = C₂:C₁?
(ii) Is R₃:R₄ = C₄:C₃?
(iii) Is R and C inversely proportional to each other?
(iv) Do this activity with 36 squares.
Solution:
(i) Is R₁:R₂ = C₂:C₁
⇒ 2 : 3 = 16 : 24

(ii) Is R₃:R₄ = C₄:C₃

R₁:R₂ = C₂:C₁

(iii) R₃:R₄ = C₄:C₃
⇒ 4 : 6 = 8 : 12
\(\frac{4}{6}=\frac{8}{12}=\frac{4 \times 2}{6 \times 2}=\frac{4}{6} \Rightarrow \frac{4}{6}=\frac{4}{6}\) =
∴ R₃:R₄ = C₄:C₃

(iv) Do this activity with 36 squares.

From the above table we can conclude that if number of rows are increases then number of columns are decreases.

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 10 Direct and Inverse Proportions Ex 10.1 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 10th Lesson Direct and Inverse Proportions Exercise 10.1

Question 1.
The cost of 5 meters of a particular quality of cloth is ₹ 210. Find the cost of(i) 2 (ii) 4
(iii) 10 (iv) 13 meters of cloth of the same quality.
Solution:
The cost of 5 m of a cloth = ₹ 210
The length of a cloth and its price are in direct proportion.
i) \(\frac{\mathrm{x}_{1}}{\mathrm{y}_{1}}=\frac{\mathrm{x}_{2}}{\mathrm{y}_{2}}\)
Here x₁ = 5, y₁ = 210
x₂ = 2, y₂ = ?

Question 2.
Fill the table.

Solution:

Question 3.
48 bags of paddy costs ₹ 16, 800 then find the cost of 36 bags of paddy.
Solution:
Number of bags of paddy and their cost are in direct proportion.
⇒ \(\frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}}\) , x₁ = 16,800
x₂ = 36 y₂ = ?

= 3 × 4200
y₂ = ₹ 12600
∴ The cost of 36 bags of paddy = ₹ 12600

Question 4.
The monthly average expenditure of a family with 4 members is 2,800. Find the
monthly average expenditure ofa family with only 3 members.
Solution:
Number of family members and their expenditure are in direct proportion.
⇒ \(\frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}}\) , x₁ = 4
y₁ = 2,800
x₂ = 3 y₂ = ?

= 3 × 700 = 2100
y₂ = ₹ 2100
The expenditure for 3 members = ₹ 2100

Question 5.
In a ship of length 28 m, height of its mast is 12 m. If the height of the mast in its model is
9 cm what is the length of the model ship?
Solution:
The length of ship and the height of its mast are in direct proportion.
⇒ \(\frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}}\) , x₁ = 28
y₁ = 12
x₂ = ? y₂ = 9

x₂ = 7 × 3 = 21
∴ The length of model ship = 21 m

Question 6.
A vertical pole of 5.6 m height casts a shadow 3.2 m long. At the same time find (j) the
length of the shadow cast by another pole 10.5 m high (ii) the height of a pole which casts
a shadow 5m long.
Solution:
length of a vertical pole and length of its shadow are in direct proportion.
⇒ \(\frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}}\)

i) x₁ = 5.6
y₁ = 3.2
x₂ = 10.5 y₂ = ?

∴The length of the shadow = 6 cm

ii) x₁ = 5.6 m x₂ = ?
y₁ = 3.2 m y₂ = 5
⇒ \(\frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}}\)

∴ x₂ = 8.75

Question 7.
A loaded truck travels 14 km in 25 minutes. If the speed remains the same, how far can it travel in 5 hours?
Solution:
Time and distance are in direct proportion
⇒ \(\frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}}\)
x₁ = 14 km , x₂ = ?
y₁ = 25min = \(\frac{25}{60} \mathrm{hr}=\frac{5}{12} \mathrm{hr}\) = y₂ = 5hrs
⇒ \(x_{2}=\frac{x_{1} \times y_{2}}{y_{1}}=\frac{14 \times 5}{5}=\frac{14 \times \not 5 \times 12}{\not 5}\)
= 168 km
∴ Lorry travelled in 5 hrs = 168km

Question 8.
If the weight of 12 sheets of thick paper is 40 grams, how many sheets of the same paper would weigh 16 \(\frac { 2 }{ 3 }\) kilograms?
Solution:
Number of pages and their weight are in direct proportion.
⇒ \(\frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}}\)
x₁ = 12 km , x₂ = ?
y₁ = 40 gm
y₂ = 16 \(\frac { 2 }{ 3 }\) gm = \(\frac { 50 }{ 3 }\) x 1000 gm
= \(\frac { 50000 }{ 3 }\) gm

From (1)

∴ Number of pages = 5000

Question 9.
A train moves at a constant speed of 75 km/hr.
(i) How far will it travel in 20 minutes?
(ii) Find the time required to cover a distance of 250 km.
Solution:
Speed of the train = 75 km/hr
i) The distance travelled in 20 min.
d = s x t = 75 x 20 min
= 75 x = 25 km
= \(75 \times \frac{20}{60}=\frac{75}{3}\) = 25 km

ii) Time taken to travel 250 km
t = \(\frac{d}{s}=\frac{250}{75}\)
t = \(\frac{10}{3}\) hrs

Question 10.
The design of a microchip has the scale 40:1. The length of the design is 18cm, find the actual length of the micro chip?
Solution:
The scale of the design of a microchip
= 40 : 1
The length of the design = 18 cm
The actual length of microchip = ?
The length of the design and actual length of the microchip are in direct proportion.
⇒ \(\frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}}\)
x₁ = 40 km , x₂ = 18
y₁ = 1
y₂ = ?
⇒ \(\frac{40}{1}=\frac{18}{y_{2}}\)
⇒ \(\frac{18}{40}=\frac{9}{20}\) cm
∴ The original (actual) length of the microchip = [latexs]\frac{9}{20}[/latex]cm

Question 11.
The average age of consisting doctors and lawyers is 40. If the doctors average age is 35 and the lawyers average age is 50, fmd the ratio of the number of doctors to the number of lawyers.
Solution:
Let the number of doctors = x
Number of lawyers = y
The average age of doctors = 35
The total age of doctors = 35 × x
= 35 x years
The average age of lawyers = 50
∴ The total age of lawyers = 50 x y
= 50y
According to the sum
\(\frac{35 x+50 y}{x+y}\) = 40
⇒ 35x + 50y = 40x + 40y
⇒ 40x – 35x = 50y – 40y
⇒ 5x = lOy
⇒ \(\frac{x}{y}=\frac{10}{5}\) (or)
x : y = 2 : 1
∴ The ratio of number of doctors to lawyers = 2:1

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 9 Area of Plane Figures Ex 9.2 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 9th Lesson Area of Plane Figures Exercise 9.2

Question 1.
A rectangular acrylic sheet is 36 cm by 25 cm. From it, 56 circular buttons, each of diameter 3.5 cm have been cut out. Find the area of the remaining sheet.
Solution:
The dimensions of a rectangular acrylic sheet = 36 cm × 25 cm.
∴ Its area = l × b
= 36 × 25 = 900 sq.cm
The diameter of a circular button = 3.5 cm
radius = r = \(\frac{d}{2}=\frac{3.5}{2}\) = 1.75 cm.
∴ Area of each button = πr²
= \(\frac{22}{7}\) × (1.75)²
= \(\frac{22}{7}\) × 1.75 × 1.75
= 9.625 sq.cm.
The area of 56 such buttons = 56 × 9.625 = 539 sq.cm.
∴ The area of remaining sheet
= Area of a rectangular sheet – Area of 56 buttons
= 900 – 539
= 361 sq.cm

Question 2.
Find the area of a circle inscribed in a square of side 28 cm.
[Hint. Diameter of the circle is equal to the side of the square]

Solution:
Diameter of a circle = d = side of a square = 28 cm.
∴ d = 28 cm
⇒ r = \(\frac{d}{2}=\frac{28}{2}\) = 14 cm
∴ Area of a circle = \(\frac{22}{7}\) × 14 × 14
= 22 × 28
= 616 sq.cm.

Question 3.
Find the area of the shaded region in each of the following figures.
[Hint:d + \(\frac{d}{2}\) + d/2 = 42]
d = 21
∴ side of the square 21 cm
i)

Solution:

∴ Side of a square, d = 21 cm.
Radius of a semicircle = r = \(\frac{\mathrm{d}}{2}=\frac{21}{2}\) = 10.5 cm
∴ Area of a semi-circle = \(\frac{\pi r^{2}}{2}\)

= 11 × 1.5 × 10.5
= 173.250 sq.cm.

∴ Area of 4 shaded semi-circles
= 4 × Area of each semi-circle
= 4 × 173.25 = 693 sq.cm.

Diameter of a big circle = 21 m

Diameter of a semicircle = 10.5 m.
Radius = \(\frac{10.5}{2}\) = 5.25 m.
∴ Area of semi-circle = \(\frac{\pi r^{2}}{2}\)

= 11 × 0.75 × 5.25
= 43.3125 sq.m.
Area of two semicircles
= 2 × 43.3125 = 86.6250 sq.m.
∴ Area of shaded region
= Area of big circle – Area of 2 semicircles
= 346.5 – 86.6250
= 259.8750 sq.m.

Question 4.
The adjacent figure consists of four small semi-circles of equal radii and two big semi-circles of equal radii (each 42 cm). Find the area of the shaded region
Solution:
Diameter of a bigger semi circle d = 42 cm
Radius, r = \(\frac { 42 }{ 2 }\) = 21 cm
Area of a bigger semi circle = \(\frac{\pi r^{2}}{2}\)

= 11 × 3 × 21
= 693 sq.cm

∴ Area of the shaded region
= Area of two bigger semi circles – Area of two smaller semi circles + Area of two smaller semi circles
= Area of two bigger semi circles
= 2 × Area of bigger semi circles
= 2 × 693 = 1386 sq.cm.

Question 5.
The adjacent figure consists of four half circles and two quarter circles.
If OA=OB=OC=OD= 14cm. Find the area of the shaded region.

Solution:

OA = OB = OC = OD = 14CM
Radius of \(\frac { 1 }{ 4 }\) of circle BXD, r = 14cm

= 154 sq.cm.
∴ Area of the quarter circle AYC
= 154 sq.cm
.’. Area of the shaded region = Area of quarter circle BXD – Area of semi circle OPB + Area of semi circle OQD + Area of quarter circle AYC – Area of semi circle ARO + Area of semi circle OSC
= Area of quarter circle BXD + Area of quarter circle AYC
= 154 + 154
= 308 sq.cm.

Question 6.
In adjacent figure A, B, C and D are centres of equal circles which touch externally is pairs and ABCD is a square of side 7 cm. Find the area of the shaded region.

Solution:

Side of a square,
AB = BC = CD = DA = 7 cm
Area of a square = s × s
= 7 × 7 = 49 sq.cm.
From the above figure,
Radius of a circle, r = \(\frac{\text { Side of a square }}{2}\)
= \(\frac{7}{2}\) = 3.5
Areas of 4 equal sectors are equal.
Angle of a sector APQ, x = 90°
Radius, r = 3.5 cm
Area of sector APQ

= 5.5 × 0.5 × 3.5
= 9.625 sq.cm.
Total area of 4 equal sectors = 4 × 9.625 = 38.5 sq.cm.
∴ Area of shaded region = Area of square ABCD – Area of 4 equal sectors
= 49 – 38.5
= 10.5 sq.cm.

Question 7.
The area of an equilateral triangle is 49√ cm². Taking each angular point as centre, a circle is described with radius equal to half the length of the side of the triangle as shown in the figure. Find the area of the portion in the triangle not included in the circles.

Solution:
From the given figure,
Δ ABC is an equilateral triangle.
∴ Area of an equilateral triangle
= 49√3 sq.em (given)
3 equal circles (whose radii are equal) are touch externally.
Each angle of a sector in the circle = 60°,
Radius r = 7 cm.
∴ The areas of 3 sectors are equal.
∴ Area of a sector APQ

∴ The area of 3 sectors
= 3 × \(\frac { 77 }{ 3 }\) = 77 sq. cm
∴ The area of the portion in the triangle not included in the circles
= area of ΔABC — area of 3 equal sectors
=49√3 – 77
=49 × 1.7321 – 77 (∵√3 = 1.7321)
= 84.8729 – 77
= 7.8729 sq.cm.

Question 8.
(i) Four equal circles, each of radius ‘a’ touch one another. Find the area between them.
(ii) Four equal circles are described about the four corners of a square so that each circle
touches two of the others. Find the area of the space enclosed between the circumferences
of the circles, each side of the square measuring 14 cm.
Solution:

The side of a square ABCD
AB = BC = CD = DA = a + a = 2a units.
∴ Area of a square = s × s
= 2a × 2a = 4a2 sq.units.
4 equal sectors are there.
In each sector,
The angle of a sector APQ, x = 90°
Radius, r = a units
Area of each sector APQ

∴ Area of 4 equal sectors
= 4 × \(\frac{\pi \mathrm{a}^{2}}{4}\) = πa² sq. units
= 2a × 2a = 4a2 sq.units. 4 equal sectors are there.
∴ The area between the circles
= Area of square sectors
= 4a² – πa²
= a² (4 – π) sq. units
= a² (4 – \(\frac { 22 }{ 7 }\) ) = \(\frac{6 a^{2}}{7}\) sq. units

Question 9.
From a piece of cardbord, in the shape ofa trapezium ABCD, and AB||CD and ∠BCD = 90°,= quarter circle is removed. Given AB = BC = 3.5 cm and DE =2 cm. Calcualte the area of the remaining piece of the cardboard. (Take π to be \(\frac{22}{7}\) )

Solution:
ABCD is a trapezium
AB || CD and ZBCD = 90°
AB = BC = 3.5 cm; DE = 2 cm.
Area of a trapezium ABCD :
Length of parallel sides, AB = 3.5 cm
CD = DE + EC
= 2 + 3.5 = 5.5 cm.

Distance between parallel sides,
BC = 3.5 cm.
∴ Area of a trapezium
= \(\frac { 1 }{ 2 }\) h (a + b)
= \(\frac { 1 }{ 2 }\) × BC (AB + CD)
= \(\frac { 1 }{ 2 }\) × 3.5 (3.5 + 5.5)
= \(\frac { 1 }{ 2 }\) × 3.5 × 9
= 3.5 × 4.5
= 15.75 sq.cm.

Area of quarter circle EBC :
Radius, r = 3.5 cm
∴ Area of quarter circle

= 5.5 × 0.5 × 3.5
= 9.625 sq.cm.

Area of the remaining piece of card board = Area of a trapezium – \(\frac { 1 }{ 2 }\) th of area of a circle
= 15.75 – 9.625
= 6.125 sq.cm.

Question 10.
A horse is placed for grazing inside a rectangular field 70 m by 52 m and is tethered to one corner by a rope 21 m long. How much area can it graze?

Solution:
From the above figure,
The sector represents the grazing area of a horse.
The angle of a sector OPQ, x = 90°
Radius of a sector OPQ, r = 21 m .
∴ Area of a sector


= 346.5 sq.m.
∴ The area of grazing = 346.5 sq.m.

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 9 Area of Plane Figures Ex 9.1 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 9th Lesson Area of Plane Figures Exercise 9.1

Question 1.
Divide the given shapes as instructed.

(i)

Solution:
Rectangle ABCD
Rectangle CEFG
Rectangle FHIJ

(ii)

Solution:
Rectangle ABCD
Rectangle EFGH
Rectangle CHIJ

(iii)

Solution:
Trapezium ABEF Trapezium BCDE

(iv)

Solution:
Triangle ABC
Triangle DEF B
Rectangle ACDF

(v)

Solution:
Triangle BCD
Triangle BDE

Question 2.
Find the area enclosed by each of the following figures

Solution:
i) Area of a pentagon ABCDE = Area of square ACDE + area of Δ ABC
= AE × ED + \(\frac{1}{2}\) AC × BF
= 4 × 4 + \(\frac{1}{2}\) × 4 × 2
[∵ AC = ED = 4 cm;
BF = BD-CD = 6-4 = 2 cm]
= 16 + 4 = 20 sq.cm.

ii)

Area of a figure ABCDEF
= Area of square ABCF + area of trapezium CDEF
= AB × BC + \(\frac{1}{2}\) CG × (ED + CF)
= 18 × 18 + \(\frac{1}{2}\) × 8 × (7+ 18)
[∵h = CG = 8 cm, FC = AB = 18 cm]
= 324 + 4 × 25
= 324 + 100
= 424 cm²

iii)

Area of a figure ABCDEF
= Area of rectangle ABCD + area of trapezium ADEF
= AB × BC + \(\frac{1}{2}\) × GA × (EF + AD)
= 20 × 15 + \(\frac{1}{2}\) × 8 × (6 + 15)
[∵ GA = GB-AB = 28-20 = 8cm;
AD = BC – 15 cm]
= 300 + 4 × 21
= 300 + 84
= 384 sq.cm

Question 3.
Calculate the area of a quadrilateral ABCD when length of the diagonal AC = 10 cm and the lengths of perpendiculars from B and D on AC be 5 cm and 6 cm respectively.
Solution:
From the figire given below

h₁ = 6 cm,
h₂ = 5 cm,
d = 10 cm.
Area of a quadrilateral ABCD
= \(\frac{1}{2}\) d(h₁+h₂) . .
= \(\frac{1}{2}\) AC × (DE + BF)
= \(\frac{1}{2}\) × 10(6+5)
= 5 × 11 = 55sq.cm.

Question 4.
Diagram of the adjacent picture frame has outer dimensions 28 cm × 24 cm and inner dimensions 20 cm × 16 cm. Find the area of shaded part of frame, if width of each section is E the same.

Solution:
The outer dimensions of a picture frame = 28 cm × 24 cm

Length of the outer edge = 28 cm.
Breadth of the outer edge = 24 cm.
Dimensions of inner picture frame = 20 cm × 16 cm
Length of inner edge = 20 cm
Breadth of inner edge = 16 cm.
Area of AABC = \(\frac{1}{2}\) × b × h
= \(\frac{1}{2}\) × AC × BC
= \(\frac{1}{2}\) × 4 × 4 = 8 cm²
Area of rectangle CDEB = l × b
= CD × DE
= 20 × 4
= 80 cm²

Area of ADEF = \(\frac{1}{2}\) × b × h
= \(\frac{1}{2}\) × DF × DE
= \(\frac{1}{2}\) × 4 = 8sq.cm
∴ The area of the shaded region
= ar ΔABC + ar ▭ CDEB + ar ADEF
= 8 + 80 + 8
= 96 sq.cm.

Question 5.
Find the area of each of the following fields. All dimensions are in metres
Solution:

(i) Area of a trapezium ABCH
= \(\frac{1}{2}\) × h(a + b)
= \(\frac{1}{2}\) × 80 × (30 + 40)
= 40 (70)
= 2800 sq.m.

Area of Δ HCD =\(\frac{1}{2}\) × HC × HD
= \(\frac{1}{2}\) × 40 × 80
= 1600 sq.m.

Area of Δ EID = \(\frac{1}{2}\) × El × ID
= \(\frac{1}{2}\) × 60 × 40
= 1200 sq.m.

Area of trapezium FGIE
= \(\frac{1}{2}\) × h(a + b)
= \(\frac{1}{2}\) × 70 × (50 + 60)
= 35 (110)
= 3850 sq.m.

Area of Δ FGA = \(\frac{1}{2}\) × FG × GA
= \(\frac{1}{2}\) × 25 × 50
= 1250 sq.m.

∴ Area of the field

= 2800 + 1600 + 1200 + 3850 + 1250
= 10700 sq.m.

ii) Area of ΔABK = \(\frac{1}{2}\) × KB × KA
= \(\frac{1}{2}\) x 30 × 50
= 750 sq.m.

Area of trapezium KBCI
= \(\frac{1}{2}\) × h(a + b)
= \(\frac{1}{2}\) × 60 (30 + 40)
= 30 (70)
= 2100 sq.m.

Area of trapezium ICDE
= \(\frac{1}{2}\) × h(a + b)
= \(\frac{1}{2}\) × 80 × (40 + 50)
= 40 (90)
= 3600 sq.m.

Area of Δ FHE = \(\frac{1}{2}\) × FH × HE
=\(\frac{1}{2}\) × 20 × 40
= 400 sq.m.

Area of trapezium GJHF
= \(\frac{1}{2}\) × h(a + b)
= \(\frac{1}{2}\) × 80 × (40 + 20)
= 40 (60)
= 2400 sq.m.

∴ Area of ΔGJA = \(\frac{1}{2}\) × GJ × JA
= \(\frac{1}{2}\) × 40 × 70
= 1400 sq.m.

∴ Area of the field

= 750 + 2100 + 3600 + 400 + 2400 + 1400
= 10650 sq.m.

Question 6.
The ratio of the length of the parallel sides of a trapezium is 5:3 and the distance between them is 16cm. If the area of the trapezium is 960 cm², find the length of the parallel sides.
Solution:
The ratio of the length of parallel sides of a trapezium = 5:3.
Let the parallel sides be 5x, 3x say.
∴ a = 5x cm, b = 3x cm
The distance between the parallel sides
(h) = 16 cm.
∴ Area of a trapezium
= \(\frac{1}{2}\) × h(a + b)
= \(\frac{1}{2}\) × 16 x (5x + 3x)
= 8(8x)
= 64 x

According to the problem,
Area of the trapezium = 960 sq.cm.
∴ 64x = 960
x = \(\frac{960}{64}\) = 15
∴ The length of parallel sides
a = 5x = 5 × 15 = 75 cm
b = 3x = 3 × 15 = 45 cm

Question 7.
The floor of a building consists of around 3000 tiles which are rhombus shaped and each of its diagonals are 45 cm and 30 cm in length. Find the total cost of flooring if each tile costs rupees 20 per m².
Solution:

The floor of a building is paved with 3000 tiles.
The shape of each tile is a rhombus.
Diagonals of a tile are
d₁ = 45 cm, d₂ = 30 cm.
Area of each tile = \(\frac{1}{2}\) × d₁ d₂
= \(\frac{1}{2}\) × 45 × 30
= 675 sq.cm.
∴ Area of floor
= 3000 × 675
= 2025000 sq.cm
= \(\frac{2025000}{10000}\) sq m
(∵ 1 sq.m. = 10000 sq.cm) 2025
= \(\frac{2025}{10}\) sq.m.
= 202.5 sq.m.
The cost of tile per square meter = ₹ 20
∴ The cost of flooring = ₹ 202.5 × 20
= ₹ 4050

Question 8.
There is a pentagonal shaped parts as shown in figure. For finding its area Jyothi and Rashida divided it in two different ways. Find the area in both ways and what do you
observe?

Solution:

Area of pentagon ABCDE

= \(\frac{1}{2}\) × AF × (AB + CF) + \(\frac{1}{2}\) FE x (ED + CF)
= \(\frac{1}{2}\) × 7.5 × (15+30) + \(\frac{1}{2}\) x 7.5 x (15+30)
= \(\frac{1}{2}\) × 7.5 × 45 + \(\frac{1}{2}\) x 7.5 x 45
= 2 × (\(\frac{1}{2}\) × 7.5 × 45)
= 7.5 × 45 = 337.50 sq.cm.

Rashida’s Method :

Area of square ABDE = side × side
= AE × ED = 15 × 15 = 225 sq.cm.

Area of ABDC =\(\frac{1}{2}\) × b × h
= \(\frac{1}{2}\) × BD × CF
= \(\frac{1}{2}\) × 15 × 15
( ∵ CF = 30 – 15 = 15 )
= \(\frac{225}{2}\) = 112.50 sq.cm
∴ Area of pentagon ABCDE
= ar □ ABDE + ar ΔBDC
= 225 + 112.50 = 337.50 sq.cm
The area of a polygon can’t changed if its procedures are changed.

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 8 Exploring Geometrical Figures Ex 8.2 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 8th Lesson Exploring Geometrical Figures Exercise 8.2

Question 1.
Cut the bold type English alphabets (capital) and paste in your note book. Draw possible number of lines of symmetry for each of the letter.
(i) How many letters have no linear symmetry?
(ii) How many letters have one line of symmetry?
(iii) How many letters have two lines of symmetry?
(iv) How many letters have more than two lines of symmetry?
(v) Which of them have rotational symmetry?
(vi) Which of then have point symmetry?
Solution:

i) letters don’t have line of symmetry
→ F, G, J, L, N, P, Q, R, S, Z.
ii) Liters which are having 1 line of symmetry
→ B, C, D, E, H, I, K, M, 0, T, U, V,
iii) Letirs which are having 2 lines of symmetry
→ H, I, O, X.
iv) Letters which are having more than two lines of symmetry O, X.
v) Letters which are having rotational symmetry
→ B, D, E, H, I, M, 0, S, T, W, X, Z.
vi) Letters which are having point symmetry
→ 0, X, M, W, H, 1, E, D.

Question 2.
Draw lines of symmetry for the following figures. Identify which of them have point symmetry. Is there any implication between lines of symmetry and point symmetry?

Solution:

The given figures has point of symmetry. Which are having line of symmetry are also having point of symmetry.

Question 3.
Name some natural objects with faces which have atleast one line of symmetry.
Solution:
1) Moon (Full)
2) Face of a human being
3) Orange fruit
4) Sunflower
5) Butterfly

Question 4.
Draw three tessellations and name the basic shapes used on your tessellation.
Solution:
I may notice that the basic shapes used to draw tessellations are pentagon, rectangle, squares, equilateral triangles, hexagons etc. Most tessellations can’t be formed with these shapes.

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 8 Exploring Geometrical Figures Ex 8.1 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 8th Lesson Exploring Geometrical Figures Exercise 8.1

Question 1.
Name five pairs of congruent objects, you use daily.
Solution:
Solution:
i) Pair of ear rings.
ii) Wheels of a cycle.
iii) Equal lengths of scales.
iv) Pair of bangles.
v) Mathematics textbooks of same class.

Question 2.
(a) Draw two congruent figures. Are they similar? Explain
(b) Take two similar shapes. If you slide rotate or flip one of them, does the similarity remain?
Solution:
a) ΔABC ≅ ΔPQR

From ΔABC & ΔPQR
AB = PQ
AC = PR
BC = QR
∠A = ∠P
∠B = ∠Q
∠C = ∠R
∴ Congruent triangles are similar to each other.
But similar triangles are not congruent.

b)

From fig. ΔXYZ and ΔSTU are similar triangles. If these triangles are rotated even they are similar to each other.

Question 3.
If Δ ABC ≅ Δ NMO, name the congruent sides and angles.
Solution:

From the congruent triangles
ΔABC ≅ ΔNMO
AB = NM
BC = MO
AC = NO
∠A = ∠N
∠B =∠M
∠C = ∠O

Question 4.
State whether the following statements are true. Explain with reason.
(i) Two squares of side 3 cm each and one of them rotated through 45° are congruent.
(ii) Any two right triangles with hypotenuse 5 cm, are congruent.
(iii) Any two circles of radii 4 cm each are congruent.
(iv) Two equilateral triangles of side 4 cm each but labeled as A ABC and A LHN arc not congruent.
(v) Mirror image of a polygon is congruent to the original.
Solution:
i) True. If the square is rotated 45° then it is remaining congruent.
ii) True. If the hypotenuse of two right triangles are equal, then the two triangles are congruent to each other, [∵ Their corresponding sides and angles will also be equal]

(iii) True
Since they are having equal radii (r₁ = r₂ = 4 cm )
= 4 cm).
∴ The given circles are congruent.

iv) False.
If two sides of one triangle are equal to the corresponding two sides of another triangle, then the third sides are in the proportion.
∴ ΔABC ≅ ΔLHN

But given tXABC,’L\LHN. Then it is false.

v) True.
Always the mirror images of a polygon is congruent to the original.

Question 5.
Draw a polygon on a square dot sheet. Also draw congruent figures in different directions and mirror image of it.
Solution:

Question 6.
Using a square dot sheet or a graph sheet draw a rectangle and construct a similar figure.
Find the perimeter and areas of both and compare their ratios with the ratio of their
corresponding sides.
Solution:

Question 7.
7 pillars are used to hold a slant iron gudder as shown in the figure. If the distance between every two pillars is 1 m and height of the last piller is 10.5 m. Find the height of pillar.

Solution:

The height of the 1st pillar
h₁ = \(\frac{1}{7}\) x 10.5 = 1.5m.

The height of 2nd pillar
h₂ = \(\frac{2}{7}\) x 10.5 = 3m.

The height of 3rd pillar
h₃ = \(\frac{3}{7}\) x 10.5 = 4.5 m.

The height of 4th pillar
h₄ = \(\frac{4}{7}\) x 10.5 = 6 m.

The height of 5th pillar
h₅ = \(\frac{5}{7}\) x 10.5 = 7.5 m.

The height of 6th pillar
h₆ =\(\frac{6}{7}\) x 10.5 = 9 m.

Question 8.
Standing at 5 m apart from a vertical pole of height 3 m, Sudha observed a building at the back of the piller that tip of the pillar is in line with the top of the building. If the distance
between pillar and building is 10 m, estimate the height of the building. [Here height of
Sudha is neglected]
Solution:
ΔAOAB ~ ΔOCD
The corresponding sides of similar triangles are in a same proportion^?

⇒ h = 3 x 3 = 9
∴ h = height of a building = 9 mts.

Question 9.
Draw a quadrilateral of any measurements. Construct a dilation of scale factor 3. Measure their corresponding sides and verify whether they are similar.
Solution:

Observe the following dilation ABCD, it is a square drawn on a graph sheet.
All vertices A, B, C, D are joined from the sign ’O’ and produced to 3 times the length upto A’, B’, C’ and D’ respec¬tively. Then A’, B’, C’, D’ are joined to form a rectangle which 3 times has enlarged sides of ABCD. Here, 0 is called centre of dilation and \(\frac{\mathrm{OA}^{\prime}}{\mathrm{OA}}=\frac{3}{1}\) = 3 is called scale factor.
∴ □ ABCD ~ □ A’B’C’D’
[ ∵ their corresponding sides are equal]

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 7 Frequency Distribution Tables and Graphs Ex 7.3 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 7th Lesson Frequency Distribution Tables and Graphs Exercise 7.3

Question 1.
The following table gives the distribution of45 students across the different levels of intelligent Quotient. Draw the histogram for the data.

Solution:

Steps of construction:
1. Calculate the difference between mid values of two consecutive classes
∴ h = 75 – 65 = 10
∴ Class interval (C.I) = 10

2. Select such a right scale
on X-axis 1 cm = 10 units
on Y-axis 1 cm 1 student

3. Construct a histogram with C.Is as width and frequencies as lengths.

Question 2.
Construct a histogram for the marks obtained by 600 students in the VII class annual
examinations.

Solution:
Classes will be prepared on class marks.
Step – 1: Take the difference between the mid values of two consecutive classes.
∴ h = 400 – 360 = 40

Step – 2: Let the lower and upper boundaries be taken as x – \(\frac{h}{2}\) , x + \(\frac{h}{2}\)
∴ x – \(\frac{h}{2}\) = 360 – \(\frac{40}{2}\) = 360 – 20 = 340
x + \(\frac{h}{2}\) = 360 + \(\frac{40}{2}\) = 360 + 20 = 380

Step – 3 : Select the scale
on X-axis 1 cm = 1 C.I (mid values)
on Y-axis 1 cm = 20 students

Step – 4 : Take C.I’s as width, frequencies as lengths.
Then construct the histogram.

Class Marks	Class Interval	Frequency
360	340- 380	100
400	380- 420	125
440	420-460	140
480	460-500	95
520	500-540	80
560	540-580	60

Scale : On Y – axis no. of students = 20, On X – axis take marks of students.

Question 3.
Weekly wages of 250 workers in a factory are given in the following table. Construct the histogram and frequency polygon on the same graph for the data given.
using the histogram. (Use separate graph sheets)

Solution:

Class Interval (Weekly wages)	Frequency (No. of workers)	Mid values
500-550	30	525
550-600	42	575
600-650	50	625
650-700	55	675
700-750	45	725
750 -800	28	775
	N = 250

Steps of constructIon:

1. C.I. = Difference of two consecutive mid values = h = 575 – 525 = 50
2. Scale : On X-axis 1 cm = ₹ 50
   On Y-axis 1 cm = lo members
3. Take on X-axis width of C.I, on Y-axis frequncies.
4. Keep points A, B, C, D, E, F, G, H on the mid points of rectangles.
5. The area of a histogram is equal to the area of a polygon.

Question 4.
Ages of 60 teachers in primary schools of a Mandai are given in the following frequency distribution table. Construct the Frequency polygon and frequency curve for the data without using the histogram. (Use separate graph sheets)

Solution:
Construction of frequency polygon:

1. Class interval = The difference between two mid values of classes
   = 30 – 26 = 4
2. Scale : On X-axis take the age of teachers
   On Y-axis take number of teachers.
3. Scale: On X-axis 1 cm = 4units
   On Y- axis 1 cm = 2 units
4. Take the widths of classes on X – axis. Frequencies on Y – axis.
5. The points are formed on the graph sheet are joined by a scale then the required frequency polygon and if the points are joined by hand frequency curve will be formed.

Question 5.
Construct class intervals and frequencies for the foliowing distribution tabie. Also draw the ogive curves for the same.

Solution:
Steps of constructIon:
Step – 1: If the given frequency distribution is in inclusive form, then convert it into an
exclusive form.

Step – 2: Calculate the less than cumulative frequency.

Step – 3: Mark the upper boundaries of the class intervals along X-axis and their corresponding cumulative frequencies along Y-axis.

Select the scale:
X – axis 1 cm = 1 class interval
Y – axis 1 cm = 10 students

Step – 4: Also, plot the lower boundary of the first class (upper boundary of the class previous to first class) interval with cumulative frequency 0.

Step – 5: Join these points by a free hand curve to obtain the required ogive.
Similarly we can construct ‘greater than cumulative frequency curve by taking greater than cumulative on Y – axis and corresponding ‘lower boundaries’ on the
X-axis.

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 7 Frequency Distribution Tables and Graphs Ex 7.2 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 7th Lesson Frequency Distribution Tables and Graphs Exercise 7.2

Question 1.
Given below are the ages of 45 people in a colony.
Construct grouped frequency distribution for the given data with 6 class intervals.

Solution:
Number of classes = 6
Range = Maximum value – Minimum value = 63 – 5 = 58

Question 2.
Number of students in 30 class rooms in a school are given below. Construct a frequency
distribution table for the data with a exclusive class interval of 4 (students).

Solution:
Class Interval (C.I.) = 4
Range = Maximum value – Minimum value = 40 – 15 = 25
No. of classes = \(\frac{\text { Range }}{\mathrm{C} . \mathrm{I}}=\frac{25}{4}\) = 6.25 *(approx ‘6’)

Question 3.
Class intervals in a grouped frequency distribution are given as 4 – 11, 12 – 19, 20 – 27, 28 – 35, 36 – 43. Write the next two class intervals. (1) What is the length of each class
interval? (ii) Write the class boundaries of all classes, (iii) What are the class marks of
each class?
Solution:
The given class intervals are 4-11,12 -19, 20 – 27, 28 – 35, 36 – 43
The next two class intervals are 44 – 51, 52 – 59
i) The length of each class interval is 8
ii) Boundaries of classes :
iii) Class = \(\frac{3.5+11.5}{2}=\frac{15}{2}\) = 7.5

Class Interval	Boundaries of classes	Class Marks
4-11	3.5-11.5	7.5
12-19	11.5-19.5	15.5
20- 27	19.5- 27.5	23.5
28-35	27.5-35.5	31.5
36-43	35.5-43.5	39.5
44-51	43.5-51.5	47.5
52-59	51.5-59.5	55.5

Question 4.
In the following grouped frequency distribution table class marks are given.

(i) Construct class intervals of the data. (Exclusive class intervals)
(ii) Construct less than cumulative frequencies and
(iii) Construct greater than cumulative frequencies.
Solution:

To determine the lower boundary of first class :
The difference between two marks of consecutive classes = h = 22-10 = 12
Let the marks of each class be ‘x’ then the boundaries of the classes be
Lower boundary of first class = x – \(\frac{\mathrm{h}}{2}\) = 10 – \(\frac{12}{2}\) = 10 – 6 = 4
Upper boundary = 10 + \(\frac{12}{2}\) = 10 + 6 = 16
We can determine the remaining classes in the same way.

Question 5.
The marks obtained by 35 students in a test in statistics (out of 50) are as below.

Construct a frequency distribution table with equal class intervals, one of them being 10-20(20 is not included).
Solution:
The frequency distribution table is

Class Interval	Frequency
0-10	2
10-20	10
20-30	4
30-40	9
40-50	10

C.I. = 10 (from 10 – 20)
Range = 48 – 1 = 47
No. of classes = = 4.7 = 5 (approx)

Question 6.
Construct the class boundaries of the following frequency distribution table. Also construct less than cumulative and greater than cumulative frequency tables.

Solution:

Question 7.
Cumulative frequency table is given below. Which type of cumulative frequency is given. Try to build the frequencies of respective class intervals.

Solution:
In the given table frequencies are increasing from top to bottom. So, it is less than cumulative frequency distribution

Less than C.F.	Frequency
3	3
8	(8 – 3) 5
19	(19 – 8) 11
25	(25 – 19)6
30	(30 – 25) 5

∴ Thee required less than C.F. distribution table is

Question 8.
Number of readers in a library are given below. Write the frequency of respective classes.
Also write the less than cumulative fequency table.

Solution:

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 7 Frequency Distribution Tables and Graphs Ex 7.1 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 7th Lesson Frequency Distribution Tables and Graphs Exercise 7.1

Question 1.
Find the arithmetic mean of the sales per day in a fair price shop in a week.
₹ 10000, ₹ 10250, ₹ 10790, ₹ 9865, ₹ 15350, ₹ 10110
Solution:

Question 2.
Find the mean ofthe data; 10.25, 9, 4.75, 8, 2.65, 12, 2.35
Solution:

Question 3.
Mean of eight observations is 25. If one observation 11 is excluded, find the mean of the remaining.
Solution:
Given that
The mean of 8 observations = 25
The sum of 8 observations = 8 × 25
= 200
If the observation 11 is subtracted from the sum
⇒ 200 – 11 = 189
∴ The mean of remaining 7 observations = \(\frac { 189 }{ 7 }\) = 27

Question 4.
Arithmetic mean of nine observations is calculated as 38. But in doing so, an observation 27 is mistaken for 72. Find the actual mean of the data.
Solution:
The mean of 9 observations = 38
The sum of 9 observations = 38 × 9
= 342
If the observation 27 is mistaken for 72
then correct observation = 72 – 27
= 45
Sum of correct observations = 342 + 45
= 387
∴ Correct Mean = \(\frac { 387 }{ 9 }\) = 43

Question 5.
Five years ago mean age of a family was 25 years. What is the present mean age of the family ?
Solution:
When all the observations of the data are increased or decreased by a certain number, the mean also increases or decreases by the same number.
Five years ago mean age of a family = 25 years
Present mean age of the family = 25 + 5 = 30 y

Question 6.
Two years ago the mean age of 40 people was 11 years. Now a person left the group and the mean age is changed to 12 years. Find the age of the person who left the group.
Solution:
The present mean age of 40 people = 11 + 2 = 13 years
∴ The total age of 40 people = 40 × 13
= 520 years
If 1 person is left from the group of 40 people then remaining will be = 40 – 1 = 39.
The mean age of 39 people = 12 years
The total age of 39 people = 39 × 12
= 468 years
Age of the person who left from the group = 520 – 468 = 52 years

Question 7.
Find the sum of deviations of all observations of the data 5,8, 10, 15,22 from their mean.
Solution:
A.M. of the observations 5,8,10,15,22 is


∴ Sum of deviations = 0

Question 8.
If sum of the 20 deviations from the mean is 100, then find the mean deviation.
Solution:

Question 9.
Marks of 12 students inaunittestare givenas4,21, 13, 17,5,9, 10,20, 19, 12,20, 14.
Assume a mean and calculate the arithmetic mean of the data. Assume another number as
mean and calculate the arithmetic mean again. Do you get the same result? Comment.
Solution:
Given observations are 4, 21, 13, 17, 5, 9, 10, 20, 19, 12, 20, 14
Estimated mean = 10 say
A.M. Estimated Mean + Average of deviations

= 10 + \(\frac { 44 }{ 12 }\) = 10 + 3.67 = 13.67
∴ A.M = \(\bar{x}\) = 13.67
Another Estimated Mean = 12 then
\(\bar{x}\) = A.M. = Estimated Mean + Average of deviations

= 12 + \(\frac { 20 }{ 12 }\) = 12 + 1.67 = 13.67
\(\bar{x}\) = 13.67
∴ In all cases \(\bar{x}\) = 13.67

Question 10.
Arithmetic mean of marks (out of 25) scored by 10 students was 15. One of the student, named Karishma enquired the other 9 students and find the deviations from her marks are noted as – 8, – 6, – 3, – 1, 0, 2, 3,4, 6. Find Karishma’s marks.
Solution:
Average marks of 10 students = 15
Total marks of 10 students = 10 × 15
= 150
Marks obtained by Karishma = x say
Deviations of 9 students =
-8, -6-3,-1,0, 2, 3,4,6
∴ x-8 + x-6 + x-3 + x-l + x + x + 2 + x + 3 + x + 4 + x + 6 + x = 150
⇒ 10x- 3 = 150
⇒ 10x = 150 + 3 = 153
x = \(\frac { 153 }{ 10 }\) = 15.3

(Or)

The given observations are
-8,-6,-3,-1,0, 2, 3, 4,6
Let the number of marks of Karisima =
Her deviation from x’ are – 8 + x, – 6 + x, – 3 + x, – 1 + x, x + 0, 2 + x, 3 + x, 4 + x, 6 + x

[ ∵ Mean = 15 is given]
10x – 3 = 15 × 10
10x =150 + 3
10x= 153
x = \(\frac{153}{10}\)
x = 15.3
∴ Number of marks obtained by Karishma = 15.3

Question 11.
The sum of deviations of ‘n’ observations from 25 is 25 and sum of deviations of the same ‘n’ observations from 35 is – 25. Find the mean of the observations.
Solution:
From the value ’25’ the sum of deviations of a data = Σx₁ = 25
∴ Mean of the observation = 25 + \(\frac{\Sigma \mathrm{x}_{1}}{\mathrm{n}}\)
= \(25+\frac{25}{n}=\frac{25 n+25}{n}\)
From the value ’35’ the sum of deviations of a data = Σx₁ = -25
∴ Mean of the observation = 35 + \(\frac{\Sigma \mathrm{x}_{1}}{\mathrm{n}}\)
= \(35+\frac{-25}{n}=\frac{35 n+25}{n}\)
Mean of the given data

(Or)

Let x₁, x₂, x₃,…… x_n are the number of
deviations from 25
∴ Number of observations are
x₁ – 25, x₂ – 25, …………. x_n – 25
∴ Sum of the observations = 25
∴ x₁ – 25 + x₂ – 25 +…………. + x_n – 25 = 25
⇒ x₁+ x₂ + + x_n = 25 + 25n …………(1)
= 25 (n + 1)
Deviations from 35 are
x₁ – 35, x₂ – 35y …………., x_n – 35
∴ Sum of the deviations = – 25
∴ (x₁ – 35) + (x₂ – 35) + … (x_n – 35) = – 25
⇒x₁ + x₁ +………………….. + x_n = – 25 + 35 x n …………….(2)
From (1) and (2),
– 25 + 35n = 25 + 25n
⇒ 10n = 50
n = 5
∴ Mean of the observations

Question 12.
Find the median of the data ; 3.3, 3.5, 3.1, 3.7, 3.2, 3.8
Solution:
The ascending order of 3.3, 3.5, 3.1, 3.7, 3.2, 3.8 is 3.1, 3.2, 3.3, 3.5, 3.7, 3.8
∴ n = 6 (is an even)
Median = Mean of \(\frac{n}{2},\left(\frac{n}{2}+1\right)\) terms
= \(\frac{6}{2}, ([latex]\frac{6}{2}\) + 1)[/latex]
= Mean of 3, 4 terms
= \(\frac{3.3+3.5}{2}\)
= \(\frac{6.8}{2}\) = 3.4

Question 13.
The median of the following observations, arranged in ascending order is 15. 10, 12, 14, x – 3, x, x + 2, 25.Then find x.
Solution:
Given observations are
10, 12, 14, x – 3,x ; x + 2, 25
⇒ n = 7 (is an odd)
∴ Median = \(\frac{n+1}{2}=\frac{7+1}{2}\) = 4th term
= x – 3
⇒ x-3 = 15
⇒ x = 15 + 3
∴ x = 18

Question 14.
Find the mode of 10, 12, 11, 10, 15, 20, 19, 21, 11, 9, 10
Solution:
The given observations are
10, 12, 11, 10, 15, 20, 19,21,11,9, 10
∴ The most frequently occurring observation is 10.
∴ Mode = 10

Question 15.
Mode of certain scores is x. 1f each score is decreased by 3, then find the mode of the new series.
Solution:
The mode of some observations is ‘x’. If ‘3’ is subtracted from every observa¬tion then the mode = x – 3

Question 16.
Find the mode of all digits used in Titing the natural numbers from 1 to 100.
Solution:
Natural numbers from 1 to 100 are 1,2,3, …………….99, 100
∴ It has no mode.
[ ∵ Every number appears only once]

Question 17.
Observations of a raw data are 5, 28, 15, 10, 15, 8, 24. Add four more numbers so that mean and median of the data remain the same, but mode increases by 1.
Solution:
The given observations are 5, 28, 15, 10, 15, 8, 24

Ascending order of the observations 5, 8, 10, 15, 15, 24, 28
Median = \(\frac{n+1}{2}=\frac{7+1}{2}\) = 4 th term
= 15 [n = 7 is an odd]
∴ Mode = 15
If 4 observations are inserted for the given data then mode increases by 1.
Let x₁ = x₂ = x₃ = x say .
∴ The mean
= \(\frac{x+x+x+x_{4}+10}{11}\) = 15
⇒ 3x + x₄ = 165 – 105 = 60
∴ 3x + x₄ = 60
If mode will increases by 1 then
x= 15 +1 = 16
∴ 3x + x₄ = 60
⇒ 3 x 16 + x₄ =60
⇒ x₄ = 60 – 48 = 12
∴ The required 4 observations are 12, 16, 16, 16

Question 18.
If the mean of a set of observations x₁, x₁ , x₁₀ is 20. Find the mean of x₁ + 4, x₂ + 8, x₃ + 12 , x₁₀ + 40.
Solution:
Mean of observations x₁, x₂……………….. ,x₁₀ is

= 20 + 2 × 11 = 20 + 22 = 42

Question 19.
Six numbers from a list of nine integers are 7,8,3, 5,9 and 5. Find the largest possible value of the median of all nine numbers in this list.
Solution:
The given 6 integers are 7, 8, 3, 5, 9, 5
Ascending order = 3, 5, 5, 7, 8, 9
Let the remaining 3 observations be x₁, x₂, x₃ say
Median of 3, 5, 5, 7, 8, 9, x₁, x₂, x₃ is
= \(\frac{n+1}{2}\) [ ∵ n is an odd (9)]
= \(\frac{9+1}{2}\) = 5th term = 8
∴ The largest possible median will be 8

Question 20.
The median of a set of 9 distinct observations is 20. If each of the largest 4 observations of the set is increased by 2, find the median of the resulting set.
Solution:
Let the 9 observations be x₁, x₂………..x₉ say
Median = \(\frac{n+1}{2}\) [∵ n = 9 is an odd]
= \(\frac{n+1}{2}\) = 5th term = x₅ = 20
∴ x₅ = 20
If 2 is added to the last 4 largest numbers then x₁, x₂, x₃, x₄, x₅, x₆ + 2 x₇ + 2, x₈ + 2, x₉ +2
∴ Median = \(\frac{n+1}{2}=\frac{9+1}{2}\) = 5th term
∴ x₅ = 20