Proportional Reasoning 2 Class 8 Extra Questions Maths Part 2 Chapter 3

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Class 8 Proportional Reasoning 2 Extra Questions

Class 8 Maths Chapter 3 Proportional Reasoning 2 Extra Questions

Proportional Reasoning 2 Extra Questions Class 8

Question 1.
If the cost of 8 notebooks is ₹ 280, find the cost of 15 such notebooks.
Solution:
Let the cost of 15 notebooks be ₹ x.

The more notebooks, the higher the cost.
This is the case of direct proportion.
So, \(\frac{8}{280}=\frac{15}{x}\)
⇒ 8 × x = 15 × 280
⇒ x = \(\frac{15 \times 280}{8}\)
x = 525
Thus, the cost of 15 notebooks is ₹ 525.

Question 2.
Check whether x andy are inversely proportional to each other in each of the following tables:


Solution:
We find the product xy for the corresponding values of x and y and compare them.
(a) We have 4 × 60 = 240
8 × 30 = 240
15 × 20 = 300
10 × 24 = 240
30 × 12 = 360
36 × 5 = 180
Since the products of the values of x and the corresponding values of y are not the same or fixed.
So, x and y do not vary inversely.

(b) We have 8 × 18 = 144
12 × 12 = 144
9 × 16 = 144
24 × 6 = 144
36 × 4 = 144
3 × 48 = 144
Clearly, xy = 144
Here, the products of the values of x and the corresponding values of y are the same or fixed, i.e., 144.
So, x and y vary inversely.

Question 3.
Shabnam takes 20 minutes to reach her school if she goes at a speed of 6 km/h. If she wants to reach school in 24 minutes, what should be her speed?
Solution:
Given, speed = 6 km/h = \(\frac{6 \times 1000}{60}\) m/min = 100 m/min.
Let the required speed be x m/min.
Thus, we have the following table:

For a constant or fixed distance, speed and time are inversely proportional.
Therefore, 100 × 20 = x × 24
⇒ x = \(\frac{100 \times 20}{24}=\frac{250}{3}\)
Thus, Shabnam’s speed should be \(\frac {250}{3}\) m/min = \(\frac{250 \times 60}{3 \times 1000}\) = 5 km/h

Question 4.
In a scout camp, there is food provision for 350 cadets for 36 days. If 50 cadets leave the camp, for how many days will the provision last?
Solution:
Let the required number of days be x.
Thus, we have the following table:

The fewer cadets, the longer the provision will last.
So, this is the case of inverse proportion.
Therefore, 350 × 36 = 300 × x
⇒ x = \(\frac{350 \times 36}{300}\) = 42
Thus, the provision will last 42 days when 50 cadets leave the camp.

Question 5.
If 24 workers can build a wall in 80 hours, how many extra workers will be required to finish the same work in 60 hours?
Solution:
To finish the work in fewer hours, more workers will be needed.
Therefore, this is the case of inverse proportion.
Let the required number of extra workers be x.
Thus, we have the following table:

Then, 24 × 80 = (24 + x) × 60
⇒ 24 + x = \(\frac{24 \times 80}{60}\) = 32
⇒ x = 32 – 24 = 8
Thus, the number of extra workers required is 8.

Question 6.
A cricket coach schedules practice sessions that include different activities in the following ratio:
time for warm-up/cool-down : time for batting : time for bowling : time for fielding :: 3 : 4 : 3 : 5.
If each session is 150 minutes long, how much time is spent on each activity?
Solution:
The session is 150 minutes long, so the time spent on each activity is:
time for warm-up/cool-down = 150 × \(\frac{3}{3+4+3+5}\)
= 150 × \(\frac {3}{15}\)
= 30 minutes
time for batting = 150 × \(\frac{4}{3+4+3+5}\)
= 150 × \(\frac {4}{15}\)
= 40 minutes
time for bowling = 150 × \(\frac{3}{3+4+3+5}\)
= 150 × \(\frac {3}{15}\)
= 30 minutes
time for fielding = 150 × \(\frac{5}{3+4+3+5}\)
= 150 × \(\frac {5}{15}\)
= 50 minutes

Question 7.
Show that the numbers 22, 33, 42, and 63 are in proportion.
Solution:
Numbers are 22, 33, 42, 63

We have product of extremes = 22 × 63 = 1386
and product of means = 33 × 42 = 1386
∵ Product of extremes = Product of means.
∴ Hence, 22, 33, 42, and 63 are in proportion.

Question 8.
Show that the numbers 36, 49, 6, and 7 are not in proportion.
Solution:
Numbers are 36, 49, 6, 7
We have product of extremes = 36 × 7 = 252
and product of means = 49 × 6 = 294
∵ Product of extremes ≠ Product of means.
∴ The numbers 36, 49, 6, and 7 are not in proportion.

Question 9.
Find the fourth proportional of 4, 8, and 12.
Solution:
Let the fourth proportional of 4, 8, and 12 be x.
∴ By definition, the ratios 4 : 8 and 12 : x are in proportion.
∴ Product of extremes = Product of means
⇒ 4 × x = 8 × 12
⇒ x = \(\frac{8 \times 12}{4}\)
⇒ x = 2 × 12
⇒ x = 24
∴ The required fourth proportional is 24.

Question 10.
Find the mean proportional of 6 and 96.
Solution:
We know that mean proportional b of a = 6 and c = 96 is \(\sqrt{a c}\)
⇒ b = \(\sqrt{6 \times 96}=\sqrt{576}\) …(1)
∴ From (1), the mean proportional
b = \(\sqrt{2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3}\)
= 2 × 2 × 2 × 3
= 24

Question 11.
A person spends his salary in the following manner:
Total salary = ₹ 12000
Food = ₹ 1800
Clothing = ₹ 1200
House rent = ₹ 3000
Education of children = ₹ 1200
Misc. expenditure = ₹ 2400
Savings = ₹ 2400
Make a pie chart exhibiting the above information.
Solution:
Let us prepare a table to find the sector angles corresponding to different components (of expenditure) by the formula:

Steps of construction:
1. Draw a circle of any convenient radius (a little larger).
2. Draw a horizontal radius.
3. Starting with this horizontal radius, draw sectors whose central (sector) angles are 54°, 36°, 90°, 36°, 72°, 12°, respectively, as shown in the figure below.

4. Shade the six sectors with different patterns and designs and label (name) each of them.
Also, give a heading to the pie chart. Thus, we obtain the required pie chart as shown in the figure.

Question 12.
Megha’s report card states her marks as follows:

Make a pie chart (circle graph) exhibiting her marks in various subjects.
Solution:
Let us prepare a table to find the sector angles corresponding to different components (or marks) by the formula:

Steps of construction:
1. Draw a circle of any convenient radius (a little larger).
2. Draw any line as a radius (see figure).
3. Starting with this horizontal radius, draw sectors whose central (sector) angles are 72°, 66°, 102°, 66°, 54° as shown in the figure.
(We can take the sector, clockwise or anti-clockwise, as per your convenience.)
4. Shade the five sectors with different patterns and designs and label (name) each of them.
Also, give a heading to the circle graph. Thus, we obtain the required circle as shown in the figure.

Question 13.
Observe the following tables and find if x and y are directly proportional (i.e., are in direct proportion), and if so, also write the constant of direct proportion.

Solution:
(i) The ratios of the corresponding values of x and y in the given table are:
\(\frac{x}{y}: \frac{20}{40}=\frac{1}{2}, \frac{17}{34}=\frac{1}{2}, \frac{14}{28}=\frac{1}{2}, \frac{11}{22}=\frac{1}{2},\) \(\frac{8}{16}=\frac{1}{2}, \frac{5}{10}=\frac{1}{2}, \frac{2}{4}=\frac{1}{2}\)
Since all these ratios are equal (\(\frac {1}{2}\))
∴ (Variables) x and y are in direct proportion, i.e., are directly proportional, and the constant of (direct) proportion is \(\frac {1}{2}\)

(ii) The ratios of the corresponding values of x and y in the given table are
\(\frac{x}{y}: \frac{6}{4}=\frac{3}{2}, \frac{10}{8}=\frac{5}{4}, \frac{14}{12}=\frac{7}{6}, \frac{18}{16}=\frac{9}{8},\) \(\frac{22}{20}=\frac{11}{10}, \frac{26}{24}=\frac{13}{12} \text { and } \frac{30}{28}=\frac{15}{14}\)
Since these ratios are not all equal, variables x and y are not in direct proportion, i.e., are not directly proportional.

(iii) The ratios of the corresponding values of x and y in the given table are:
\(\frac{5}{15}=\frac{1}{3}, \frac{8}{24}=\frac{1}{3}, \frac{12}{36}=\frac{1}{3}, \frac{15}{60}=\frac{1}{4},\) \(\frac{18}{72}=\frac{1}{4} \text { and } \frac{20}{100}=\frac{1}{5}\)
Since these ratios are not all equal, variables x and y are not in direct proportion, i.e., are not directly proportional.

Question 14.
The following are the car parking charges near a railway station upto
4 hours – ₹ 60
8 hours – ₹ 100
12 hours – ₹ 140
24 hours – ₹ 180
Check if the parking charges are in direct proportion to the parking time.
Solution:
Let x denote the number of parking hours, and y denote the parking charges in ₹.
The ratios of the corresponding values of x and y in the given table are:
\(\frac{x}{y}: \frac{4}{60}=\frac{1}{15}, \frac{8}{100}=\frac{2}{25}, \frac{12}{140}=\frac{3}{35},\) \(\frac{24}{180}=\frac{12 \times 2}{12 \times 15}=\frac{2}{15}\)
Since these ratios are not all equal, variables x and y are not in direct proportion, i.e., are not directly proportional.

Question 15.
(a) A mixture of paint is prepared by mixing I part of red pigments with 8 parts of base.
In the following table, find the parts of the base that need to be added.

(b) In the above question, if 1 part of a red pigment requires 75 ml of base, how much red pigment should we mix with 1800 ml of base?
Solution:
(a) Given: 1 part of red pigment is mixed with 8 parts of base to prepare the mixture.
∴ The variables x (parts of red pigments) and y (parts of base) are in direct proportion. …(1)
Let the missing entries be a, b, c, and d.
Given table becomes:

Because it is a case of direct proportion [By (1)], the ratios of corresponding entries of x and y are equal.
\(\frac{1}{8}=\frac{4}{a}, \frac{1}{8}=\frac{7}{b}, \frac{1}{8}=\frac{12}{c}, \frac{1}{8}=\frac{20}{d}\)
Cross-multiplying
a = 8 × 4 = 32
b = 8 × 7 = 56
c = 8 × 12 = 96
and d = 8 × 20 = 160
∴ Parts of the base that need to be added are as follows:

(b) Given: x = 1, y = 75ml
x = ?, y = 1800 ml
Because x and y are in direct proportion [By (1)],
therefore \(\frac{1}{x}=\frac{75}{1800}\) (Ratio of corresponding values is same)
Cross-multiplying,
75x = 1800
⇒ x = \(\frac{1800}{75}\)
⇒ x = 24
∴ 24 parts of red pigments should be mixed with 1800 ml of base.

Question 16.
In which of the following tables x and y vary inversely, i.e., x and y are inversely proportional:

Solution:
(i) x₁y₁ = 6 × 24 = 144
x₂y₂ = 3 × 48 = 144
x₃y₃ = 36 × 4 = 144
x₄y₄ = 72 × 2 = 144
x₅y₅ = 16 × 9 = 144
∴ x₁y₁ = x₂y₂ = x₃y₃ = x₄y₄ = x₅y₅
∴ xy = k = 144 for all corresponding values of x and y.
∴ x andy are inversely proportional.

(ii) x₁y₁ = 9 × 8 = 72
x₂y₂ = 18 × 4 = 72
x₃y₃ = 2 × 30 = 60
x₄y₄ = 12 × 6 = 72
We can see that all values of xy are not equal.
∴ x and y are not inversely proportional.

Question 17.
If x and y vary inversely, fill in the blanks:

Solution:
(i) Because x and y vary inversely (given), therefore xy remains constant and
8 × 10 (from the first column of the given table)
= 80
= k
As xy = k = 80
∴ y = \(\frac {k}{x}\) and x = \(\frac {k}{y}\)
∴ First missing entry in row of y = \(\frac {k}{x}\)
= \(\frac {k}{2}\)
= \(\frac {80}{2}\)
= 40
Again first missing entry in row of x = \(\frac {k}{y}\)
= \(\frac {80}{20}\)
= 4
Now second missing entry in row of y = \(\frac {k}{x}\)
= \(\frac {80}{5}\)
= 16

(ii) Because x and y vary inversely (given), therefore xy remains constant = 16 × 4 = 64
and 128 × 0.5 = 64
∴ First missing in row of y = \(\frac {64}{32}\) = 2
and second missing entry in row of y = \(\frac {64}{8}\) = 8

Question 18.
If x varies inversely as y and x = 6 when y = 6, find x when y = 9.
Solution:
Because x and y vary inversely (given), xy is constant
i.e., x₁y₁ = x₂y₂
⇒ 6 × 6 = x × 9
⇒ 36 = 9x
⇒ 9x = 36 [∵ If a = b, then b = a]
Dividing both sides by 9,
⇒ \(\frac{9 x}{9}=\frac{36}{9}\)
⇒ x = 4

Question 19.
In a television game show, the prize money of ₹ 1,00,000 is to be divided equally amongst the winners. Complete the following table and find whether the prize money given to an individual winner is directly or inversely proportional to the number of winners.

Solution:
Since the total prize money is fixed to be ₹ 1,00,000 and all winners are to get equal prize, the more the number (x) of winners, the less the prize (y) for each winner.
∴ x and y are in inverse proportion.
Hence, xy = constant
= 1 × 1,00,000 (from the first row of the given table)
= ₹ 100,000
= k
Also, 2 × 50,000 = ₹ 1,00,000
∴ First missing entry in column of y = \(\frac {k}{x}\)
= \(\frac {100000}{4}\)
= ₹ 25,000
Second missing entry in column of y = \(\frac {100000}{5}\) = ₹ 20,000
Third missing entry in column of y = \(\frac {100000}{8}\) = ₹ 12,500
Fourth missing entry in column of y = \(\frac {100000}{10}\) = ₹ 10,000
Fifth missing entry in column of y = \(\frac {100000}{20}\) = ₹ 5,000

Question 20.
Find the ratio of the following.
(i) Speed of bike 25 km/h to the speed of car 75 km/h.
Solution:
Here, units are same.
Ratio of speed of bike to the speed of car
= \(\frac{25 \mathrm{~km} / \mathrm{h}}{75 \mathrm{~km} / \mathrm{h}}=\frac{1}{3}\) or 1 : 3

(ii) 25 paise to ₹ 50.
Solution:
Here, units are different, so first we convert both into same unit.
We know that ₹ 1 = 100 paise
Ratio of 25 paise to ₹ 50 = \(\frac{25 \text { paise }}{50 \times 100 \text { paise }}\)
[∵ ₹ 1 = 100 paise]
= \(\frac{1}{200}\) or 1:200

(iii) Length of a rope 20 m to the length of a pole 50 m.
Solution:
Here, units are same.
Ratio of the length of rope to the length of pole
= \(\frac{20 \mathrm{~m}}{50 \mathrm{~m}}=\frac{2}{5}\) or 2 :5

(iv) Height of a man 1.6 m to his weight 55 kg.
Solution:
Here, the quantities are of different kinds and have different units. Therefore, we cannot find the ratio of height of a man 1.6 m to his weight 55 kg.

Question 21.
A map has scale 1:50,00,000. Two cities are 4 cm apart on the map. What is the actual distance?
Solution:
Given, scale = 1 : 50,00,000
So, 1 cm = 50,00,000 cm then, 4 cm = 4 × 50,00,000
= 2,00,00,000 cm
Now, converting cm to km, we get
2,00,00,000 cm = 200 km [∵ 1 cm = \(\frac{1}{100000}\)m]
Hence, the actual distance is 200 km.

Question 22.
If the angles of triangle are in the ratio 1 : 2 : 3, what are the measures of these three angles?
Solution:
We know that the sum of all angles of the triangle is 180°. So angles are

Question 23.
A coach schedules 150 minutes for activities in the ratio warm-up : batting : bowling : fielding :: 3 : 4 : 3 : 5. Calculate the exact minutes for each.
Solution:
The given ratio is 3 :4 : 3 : 5.
The total number of ratio parts = 3 + 4 + 3 + 5 = 15
If the total session time is 150 minutes
∴ The value of one ratio part = \(\frac{150}{15}\) = 10 min/part
Warm-up time = 3 × 10 = 30 minutes
Batting time = 4 × 10 = 40 minutes
Bowling time = 3 × 10 = 30 minutes
Fielding time = 5 × 10 = 50 minutes

Question 24.
Observe the following table and find x and y are directly proportional or not?

Solution:
We have,

When x = 3 and y = 9 then
\(\frac{x}{y}=\frac{3}{9}=\frac{1}{3}\)

When x = 18 and y = 54 then
\(\frac{x}{y}=\frac{18}{54}=\frac{1}{3}\)

When x = 17 and y = 51 then
\(\frac{x}{y}=\frac{17}{51}=\frac{1}{3}\)

When x = 19 and y = 57 then
\(\frac{x}{y}=\frac{19}{57}=\frac{1}{3}\)

When x = 16 and v = 48 then
\(\frac{x}{y}=\frac{16}{48}=\frac{1}{3}\)

Here, we see that value of \(\frac{x}{y}\) remains constant in all conditions, so x and y are directly proportional.

Question 25.
A car travels 432 km on 48 L of petrol. How far would it travel on 20 L of petrol?
Solution:
Suppose the car travels x km on 20 L of petrol.
Then, the above information can be put in the following tabular form.

We observe that the lesser petrol consumed, lesser the number of kilometres travelled.
So, it is a case of direct proportion.
Ratio of petrol consumed = Ratio of distances travelled
⇒ 48 : 20 = 432 : x
⇒ \(\frac{48}{432}=\frac{20}{x}\)
⇒ 48 × x = 20 × 432
⇒ x = \(\frac{20 \times 432}{48}\) = 180
Hence, the car would travel 180 km on 20 L of petrol.

Question 26.
If a 10 feet pole casts a shadow that is 5 feet long. How long would the shadow be, if you had a 20 feet pole in the same sunlight?
Solution:
Let the shadow of the 20 feet pole be x feet. We form a table as shown below.

Height of the pole	10	20
Length of the shadow	5	X

∴ This case is of the direct proportion.
So, \(\frac{10}{5}=\frac{20}{x}\) or x = \(\frac{20 \times 5}{10}\) = 10
Thus, the shadow of 20 feet pole is 10 feet.

Question 27.
On a map with a scale of 1 inch representing 20 miles, how many inches on the map would represent a distance of 80 miles?
Solution:
Let x inches be represent 80 miles.
Now, the above information can be put in the following tabular form.

Inch	1	X
Miles	20	80

This case is of the direct proportion.
So, 1 : 20 = x : 80
⇒ \(\frac{1}{20}=\frac{x}{80}\)
⇒ x = \(\frac{80}{20}\) = 4 inches

Question 28.
If 5 apples weight 500 g then how many grams would 10 apples weight assuming they all of are the same size and types?
Solution:
Let the weight of 10 apples be x g.
Now, the above information can be put in the following tabular form.

Number of apples	5	10
Weight of apples (in g)	500	x

More the number of apples, the more would be their weight. So, the number of apples and their weight are directly proportional to each other.
So, \(\frac{5}{500}=\frac{10}{x}\) or x = \(\frac{10 \times 500}{5}\) = 1000g

Alternate Method
Two quantities x and y which vary in direct proportion have the relation x = ky or \(\frac{x}{y}\) = k.
Here, k = \(\frac{\text { Number of apples }}{\text { Weight of apples in grams }}\)
= \(\frac{5}{500}=\frac{1}{100}\)
Now, y is the weight of 10 apples. Using the relation,
x = ky
10 = \(\left(\frac{1}{100}\right)\)y
⇒ y = 100 × 10= 1000 g
Thus, 10 apples would weight 1000 g.

Question 29.
Observe the following tables and find which pair of variables (x and y) are in inverse proportion?

Solution:
When x₁ = 60,y₁ = 6 then x₁y₁ = 60 × 6 = 360
When x₂ =50,y₂ = 6 then x₂y₂ =50 × 6 = 300
When x₃ =20,y₃ = 5 then x₃y₃ = 20 × 5 = 100
When x₄ = 40, y₄ = 8 then x₄y₄ = 40 × 8 = 320
Here, x₁y₁ ≠ x₂y₂ ≠ x₃y₃ ≠ x₄y₄
x and y are not in inverse proportion.

Solution:
When, x₁ = 75, y₁ = 15 then x₁y₁ = 75 × 15 = 1125
When x₂ = 45, y₂ = 25 then x₂y₂ = 45 × 25 = 1125
When x₃ = 125, y₃ = 9 then x₃y₃ =125 × 9 = 1125
When x₄ = 225, y₄ = 5 then x₄y₄ = 225 × 5 = 1125
Here, x₁y₁ = x₂y₂ = x₃y₃ = x₄y₄ = 1125
∴ x and y are in inverse proportion.

Question 30.
If x and y are in inverse proportion, then find the value of p.

x	30	25
y	50	P

Solution:
Given, x and y are in inverse proportion.
So, x₁y₁ = x₂y₂
⇒ 30 × 50 = 25 × p [v x₁ = 30, y₁ = 50, x₂ = 25 and y₂ = p]
⇒ p = \(\frac{30 \times 50}{25}\) = 60

Question 31.
If 52 men can do a piece of work in 35 days, then in how many days 28 men will do it?
Solution:
Let 28 men will do the piece of work in x days.
Now, the above information can be put in the following tabular form.

Number of men	52	28
Number of days	35	X

Since, the number of men decreases, the number of days to finish the work increases, therefore it is the case of inverse proportion.
∴Ratio of number of men = Inverse ratio of number of days
∴ 52 :28 = 35 : x
So, \(\frac{52}{28}=\frac{x}{35}\)
⇒ 52 × 35 = 28 × x
⇒ x = \(\frac{52 \times 35}{28}\) = 65
Hence, 28 men will do the work in 65 days.

Question 32.
Suppose that 6 workers take 10 h to complete a construction project. If we want to finish the same project in 5 h, how many workers should we hire?
Solution:
Let the number of workers hired to complete the project in 5 h be y.
Now, the above information can be put in the following tabular form.

Number of hours	10	5
Number of workers	6	y

Since, the number of workers increases, the number of hours decreases.
So, the number of hours and number of workers vary in inverse proportion.
∴ Ratio of number of workers = Inverse ratio of number of hours
10 :5 = y : 6
⇒ \(\frac{10}{5}=\frac{y}{6} \Rightarrow \frac{10 \times 6}{5}\) = y
⇒ y = 12
Hence, 12 workers are required to finish the project in 5 h.

Question 33.
8 pipes are required to fill a tank in 90 min. How long will it take if only 6 pipes of the same type are used?
Solution:
Let x min be the time to fill the tank by 6 pipes.
Now, the above information can be put in the following tabular form.

Number of pipes	8	6
Time (in minutes)	90	X

Lesser the number of pipes, more will be the time required by it to fill the tank, so this is a case of inverse proportion.
Ratio of number of pipes = Inverse ratio of time taken by them
∴ 8 : 6 = x : 90
⇒ \(\frac{8}{6}=\frac{x}{90}\)
⇒ \(\frac{8 \times 90}{6}\) = x
⇒ x = 120 min or 2 h
Thus, time taken to fill the tank by 6 pipes is 120 min or 2 h.

Question 34.
The pie chart (as shown below) represents the activity of 8th grade students.

(i) If there are 180 students in your class then how many of them prefer swimming as their activity according to pie chart?
(ii) If there are 720 students in the class then how many of them prefer reading books?
(iii) If there are 1080 students in the class. How many of them prefer drawing and watching TV?
Solution:
The complete central angle is 360°.
(i) Given, total number of students = 180
Central angle for swimming = 80°
∴Number of students who prefer swimming 80
= \(\frac{80}{360}\) × 180 = 40 students

(ii) Given, total number of students = 720
Central angle for reading books = 45°
Number of students who prefer reading books
= \(\frac{45}{360}\) × 720 = 90 students

(iii) Given, total number of students = 1080
Central angle for both activities = 60°, 100°
∴Number of students who prefer drawing
= \(\frac{60}{360}\) × 1080 =180 students
and number of students who prefer watching TV
= \(\frac{100}{360}\) × 1080 = 300 students
∴ Total number of students who prefer drawing and watching TV = 180 + 300 = 480 students

Question 35.
In the month of July 2004, a house holder spent his monthly salary amounting to ₹ 7200 on different items as given below:

Then, find the central angle of each item and also represent the information in the form of a pie chart in step by steps.
Solution:
Here, total amount = ₹ 7200
∴ Central angle for an item
= (\(\frac{\text { Amount spent on the item }}{\text { Total amount }}\) × 360)

The central angles of the sectors representing different items are computed in the following table

Now, to construct the pie chart representing the above data, we follow the following steps.

Step I Firstly, draw a circle of an appropriate radius and then draw a vertical radius of the circle.

Step II Choose the first central angle. Here, first central angle is of 50°. Draw a sector with central angle 50° in such a way that its one radius coincides with the radius drawn in Step I and another radius is in its anti-clockwise direction.

Step III Construct other sectors representing other items in anti-clockwise sense of magnitudes of their central angles except the sector representing miscellaneous expanses.
This sector is to be drawn in the last.

Step IV Shade the sectors obtained by different designs and label them as shown in figure given below to obtain the required pie chart.

Note Corresponding angle or central angle are same.

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