Proportional Reasoning 2 Class 8 Solutions Maths Ganita Prakash Part 2 Chapter 3

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Class 8 Maths Ganita Prakash Part 2 Chapter 3 Solutions

Ganita Prakash Class 8 Chapter 3 Solutions Proportional Reasoning 2

Class 8 Maths Ganita Prakash Part 2 Chapter 3 Proportional Reasoning 2 Solutions Question Answer

3.1 Proportionality—A Quick Recap, 3.2 Ratios in Maps

NCERT In-Text Questions (Pages 56-57)

Question 1.
Convert 60,00,000 cm to kilometres.
It is 60 km. Verify this.
Solution:
60,00,000 cm = 60,00,000 ÷ 100
= 60,000 m (∵ 1 m = 100 cm)
= 60,000 ÷ 1000
= 60 km (∵ 1 km = 1000 m)

Question 2.
Using the map above, can you find the geographical distance between Bengaluru and Chennai? Also, find the geographical distance between Mangaluru and Chennai.
[Hint: Use a ruler to find the distance between the cities on the map. Then, use the ratio given on the map to find the actual geographical distance.]
Solution:
Since the distance between Bengaluru and Chennai is 2.5 cm on the map.
So, the geographical distance = 2.5 × 60,00,000 cm
= 1,50,00,000 cm (∵ RF = 1 : 60,00,000)
Now the distance between Mangaluru and Chennai on the map is 5 cm.
So, the geographical distance = 5 × 60,00,000 cm
= 3,00,00,000 cm
= 300 km

Question 3.
Try to find the distances between the same two pairs of cities with different maps that have different scales (ratios). Do they all give the same geographical distance, approximately?
Solution:
Do it yourself.

3.3 Ratios with more than 2 Terms, 3.4 Dividing a Whole in a Given Ratio

Figure it Out (Page 60)

Question 1.
A cricket coach schedules practice sessions that include different activities in a specific 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:
Total time duration of each session = 150 minutes
Total ratio units = 3 + 4 + 3 + 5 = 15 units
So, Time spent on warmup/cool-down = \(\frac {3}{15}\) × 150 = 30 minutes
Time spent on batting = \(\frac {4}{15}\) × 150 = 60 minutes
Time spent on bowling = \(\frac {3}{15}\) × 150 = 30 minutes
Time spent on fielding = \(\frac {5}{15}\) × 150 = 50 minutes

Question 2.
A school library has books in different languages in the following ratio: no. of Odiya books: no. of Hindi books: no. of English books :: 3 : 2 : 1. If the library has 288 Odiya books, how many Hindi and English books does it have?
Solution:
In the given ratio 3 : 2 : 1
The Odiya books correspond to 3 parts.
If 3 parts are 288 ⇒ 1 part is 288 ÷ 3 = 96 books.
Number of Hindi books = 2 × 96 = 192 books.
Number of English books = 1 × 96 = 96 books.

Question 3.
I have 100 coins in the ratio — no. of ₹ 10 coins: no. of ₹ 5 coins: no. of ₹ 2 coins: no. of ₹ 1 coins :: 4 : 3 : 2 : 1. How much money do I have in coins?
Solution:
Given, total number of coins = 100
Ratios of number of coins = ₹ 10 : ₹ 5 : ₹ 2 : ₹ 1 = 4 : 3 : 2 : 1
Total parts = 4 + 3 + 2 + 1 = 10
∵ 10 parts equals 100, 1 part is 100 ÷ 10 = 10
∴ Number of ₹ 10 coins = 4 × 10 = 40
Number of ₹ 5 coins = 3 × 10 = 30
Number of ₹ 2 coins = 2 × 10 = 20
Number of ₹ 1 coins = 1 × 10 = 10
So, total amount = ₹ (40 × 10 + 30 × 5 + 20 × 2 + 10 × 1)
= ₹ (400 + 150 + 40 + 10)
= ₹ 600

Question 4.
Construct a triangle with sidelengths in the ratio 3 : 4 : 5. Will all the triangles drawn with this ratio of sidelengths be congruent to each other? Why or why not?
Solution:
To construct a triangle with sidelengths in the ratio 3 : 4 : 5, choose a scale factor (multiplier) to determine the actual lengths.
Let the scale factor be 2 cm, then the sidelengths of the triangle are 3 × 2 cm = 6 cm, 4 × 2 = 8 cm, and 5 × 2 = 10 cm.

All the triangles drawn with this ratio of sidelengths are not necessarily congruent to each other.
For example, we can draw another triangle with sidelengths 9 cm, 12 cm, and 15 cm, which is not congruent to the previous triangle.
As a ratio only fixes the shape, not the size.

Question 5.
Can you construct a triangle with sidelengths in the ratio 1 : 3 : 5? Why or why not?
Solution:
For any triangle, the sum of the lengths of any two sides must be greater than the third side.
Given the ratio 1 : 3 : 5, take the two smaller sides: 1 + 3 = 4.
But the third side is 5.
Here, 4 < 5, which breaks the triangle inequality rule.
Therefore we can not construct a triangle with sidelengths in the ratio 1 : 3 : 5.

3.5 A Slice of the Pie

Figure it Out (Pages 62 – 63)

Question 1.
A group of 360 people was asked to vote for their favourite season from the three seasons: rainy, winter, and summer. 90 liked the summer season, 120 liked the rainy season, and the rest liked the winter. Draw a pie chart to show this information.
Solution:
Total number of people = 360
Number of people who like:
Rainy season = 120
Summer season = 90
and Winter season = 360 – (120 + 90) = 150
Ratio = 120 : 90 : 150 = 4 : 3 : 5

Now, Central angle for Rainy season = \(\frac{4}{(4+3+5)} \times 360^{\circ}\) = 120°
Central angle for Summer season = \(\frac{3}{(4+3+5)} \times 360^{\circ}\) = 90°
Central angle for Winter season = \(\frac{5}{(4+3+5)} \times 360^{\circ}\) = 150°
Now construct a pie chart shown alongside using these angles.

Question 2.
Draw a pie chart based on the following information about viewers’ favourite type of TV channel: Entertainment — 50%, Sports — 25%, News — 15%, Information — 10%.
Solution:
Given the percentage of favourite TV channels watched by viewers.
Entertainment: 50%, Sports: 25%, News: 15%, Information: 10%
and ratio of these percentage = 50 : 25 : 15 : 10 = 10 : 5 : 3 : 2
Total parts = 10 + 5 + 3 + 2 = 20
∵ The total angle of a pie chart is 360°.
So, value of 1 part = \(\frac {360}{20}\) = 18°
∴ Central angles for each TV channel are as follows:
Entertainment = 10 × 18° = 180°
Sports = 5 × 18° = 90°
News = 3 × 18° = 54°
Information = 2 × 18° = 36°
Now, construct the pie chart using these angles.

Question 3.
Prepare a pie chart that shows the favourite subjects of the students in your class. You can collect the data on the number of students for each subject shown in the table (each student should choose only one subject). Then write these numbers in the table and construct a pie chart:

Solution:
Do it yourself.

3.6 Inverse Proportions

Figure it Out (Page 65)

Question 1.
Which of these are in inverse proportion?

Solution:
For inverse proportion: x₁y₁ = x₂y₂ = x₃y₃ = k (constant)
(i) x₁y₁ = 40 × 20 = 800
x₂y₂ = 80 × 10 = 800
x₃y₃ = 25 × 32 = 800
x₄y₄ = 16 × 50 = 800
∵ x₁y₁ = x₂y₂ = x₃y₃ = x₄y₄ = 800 (constant)
So, x and y are in inverse proportion.

(ii) x₁y₁ = 40 × 20 = 800
x₂y₂ = 80 × 10 = 800
x₃y₃ = 25 × 12.5 = 312.5
x₄y₄ = 16 × 8 = 128
Since, x₁y₁ = x₂y₂ ≠ x₃y₃ ≠ x₄y₄
So, x and y are not in inverse proportion.

(iii) x₁y₁ = 30 × 15 = 450
x₂y₂ = 90 × 5 = 450
x₃y₃ = 150 × 3 = 450
x₄y₄ = 10 × 45 = 450
∵ x₁y₁ = x₂y₂ = x₃y₃ = x₄y₄ = 450 (constant)
So, x and y are in inverse proportion.

Question 2.
Fill in the empty cells if x and y are in inverse proportion.

Solution:
Given that x and y are in inverse proportion, then
x₁y₁ = x₂y₂ = x₃y₃ = x₄y₄ = constant
∵ x₁y₁ = 16 × 9 = 144
∴ x₂y₂ = 144
⇒ y₂ = \(\frac {144}{12}\) = 12
∴ x₃y₃ = 144
⇒ x₃ = \(\frac {144}{48}\) = 3
∴ x₄y₄ = 144
⇒ y₄ = \(\frac {144}{36}\) = 4

Figure it Out (Pages 67-68)

Question 1.
Which of the following pairs of quantities are in inverse proportion?
(i) The number of taps filling a water tank and the time taken to fill it.
(ii) The number of painters hired and the days needed to paint a wall of fixed size.
(iii) The distance a car can travel and the amount of petrol in the tank.
(iv) The speed of a cyclist and the time taken to cover a fixed route.
(v) The length of cloth bought and the price paid at a fixed rate per metre.
(vi) The number of pages in a book and the time required to read it at a fixed reading speed.
Solution:
Two quantities are in inverse proportion if an increase in one leads to a proportional decrease in the other (and vice versa)
(i) More taps mean the tank fills faster (less time), so the quantities are in inverse proportion.
(ii) More painters will finish the wall in fewer days. So, the quantities are in inverse proportion.
(iii) More petrol allows you to travel a greater distance. So, the quantities are in direct proportion, not in inverse proportion.
(iv) Higher speed means you reach your destination in less time, so the quantities are in inverse proportion.
(v) More cloth costs more money, so the quantities are in direct proportion, not in inverse proportion.
(vi) A book with more pages takes more time to finish at a constant speed, so the quantities are in direct proportion, not in inverse proportion.

Question 2.
If 24 pencils cost ₹ 120, how much will 20 such pencils cost?
Solution:
Cost is directly proportional to the number of pencils.
Given, cost of 24 pencils = ₹ 120
Cost of 1 pencil = ₹ 120 ÷ 24 = ₹ 5
Then, cost of 20 such pencils = 20 × ₹ 5 = ₹ 100

Question 3.
A tank on a building has enough water to supply 20 families living there for 6 days. If 10 more families move in there, how long will the water last? What assumptions do you need to make to work out this problem?
Solution:
If the number of families increases, then the number of days the water lasts will decrease.
So, these quantities are inversely proportional.
Therefore, x₁y₁ = x₂y₂
Thus, 20 × 6 = 30 × y₂
y₂ = \(\frac{20 \times 6}{30}\) = 4
Thus, the water will last for 4 days.
Assumption needs to work out this problem: Do it yourself.

Question 4.
Fill in the average number of hours each living being sleeps in a day by looking at the charts.
Select the appropriate hours from this list: 15, 2.5, 20, 8, 3.5, 13, 10.5, 18.

Solution:
Here, each circle represents a 24-hour day, and each circle is divided into 12 equal segments like a clock, where each segment = 2 hours.
So, sleeping hours are as follows:
Giraffe: Covers around 1\(\frac {1}{4}\) segments = 1.25 × 2 = 2.5 hours.
Elephant: Covers around 1\(\frac {3}{4}\) segments = 1.75 × 2 = 3.5 hours.
Human: Covers around 4 segments = 4 × 2 = 8 hours.
Dog: Covers around 5\(\frac {1}{4}\) segments = 5.25 × 2 = 10.5 hours.
Cat: Covers around 6\(\frac {1}{2}\) segments = 6.5 × 2 = 13 hours.
Squirrel: Covers around 7\(\frac {1}{2}\) segments = 1.25 × 2 = 2.5 hours.
Snake: Covers around 9 segments = 9 × 2 = 18 hours.
Bat: Covers around 10 segments = 10 × 2 = 20 hours.

Question 5.
The pie chart on the right shows the result of a survey carried out to find the modes of transport used by children to go to school. Study the pie chart and answer the following questions.

(i) What is the most common mode of transport?
(ii) What fraction of children travel by car?
(iii) If 18 children travel by car, how many children took part in the survey? How many children use taxis to travel to school?
(iv) By which two modes of transport are equal numbers of children travelling?
Solution:
From the given pie chart, the central angles for various modes of transport are:
Bus = 120°
Walk = 90°
Cycle = 60°
Two-wheeler = 60°
and Car = 360° – (120° + 90° + 60° + 60°)
= 360° – 330°
= 30°
(i) The most common mode of transport is the bus (As it has the largest slice, central angle 120°).
(ii) Fraction of children who travel by car:
\(\frac{\text { Centre angle for car }}{\text { Total angle }}=\frac{30^{\circ}}{360^{\circ}}=\frac{1}{12}\)
So, the fraction is \(\frac {1}{12}\).
(iii) If 18 children represent the “Car” segment (\(\frac {1}{12}\) part of the total),
then the total number of children is 18 × 12 = 216.
The pie chart does not list “Taxi” as a category.
Therefore, based strictly on the provided data, 0 children are recorded using a taxi.
(iv) The two modes with equal numbers of children travelling are Cycle and Two-wheeler.
This is because both categories have the same central angle of 60°.

Question 6.
Three workers can paint a fence in 4 days. If one more worker joins the team, how many days will it take them to finish the work? What are the assumptions you need to make?
Solution:
The number of workers and the number of days are inversely proportional (more workers take fewer days).
Let the required number of days be x.
Using the inverse proportionality, we can write the equation as:
3 × 4 = 4 × x
⇒ x = \(\frac{3 \times 4}{4}\) = 4
Therefore, the fence will be painted in 3 days.
The assumptions we need to make are as follows:
All workers work at the same speed.
The rate of work is constant, i.e., no delay or variations in work efficiency.

Question 7.
It takes 6 hours to fill 2 tanks of the same size with a pump. How long will it take to fill 5 such tanks with the same pump?
Solution:
The number of tanks and time are directly proportional (more tanks need more time).
Let time taken to fill 5 tanks be x hours
\(\frac{2}{6}=\frac{5}{x}\)
⇒ x = \(\frac{5 \times 6}{2}\) = 15 hours
Thus, it will take 15 hours to fill 5 tanks.

Question 8.
A given set of chairs is arranged in 25 rows, with 12 chairs in each row. If the chairs are rearranged with 20 chairs in each row, how many rows does this new arrangement have?
Solution:
The total number of chairs remains the same.
Total number of chairs = 25 × 12 = 300
New arrangement: 20 chairs in each row.
Let the number of rows be x.
Then, 20 × x = 300
⇒ x = 15
Thus, the new arrangement will have 15 rows.

Question 9.
A school has 8 periods a day, each of 45 minutes duration. How long is each period, if the school has 9 periods a day, assuming that the number of school hours per day stays the same?
Solution:
The number of periods and the duration of each period are inversely proportional, since the total school time remains the same.
Let the new duration of each period be x minutes.
So, 9 × x = 8 × 45
⇒ x = \(\frac{8 \times 45}{9}\) = 40
Thus, each period will be 40 minutes long.

Question 10.
A small pump can fill a tank in 3 hours, while a large pump can fill the same tank in 2 hours. If both pumps are used together, how long will the tank take to fill?

Solution:
Time taken by the small pump = 3 hours
Time taken by the large pump = 2 hours
Let us figure out the work done by each pump in 1 hour.
Work done by small pump in 1 hour = \(\frac {1}{3}\)
Work done by large pump in 1 hour = \(\frac {1}{2}\)
So, work done by both pump in 1 hour = \(\frac{1}{3}+\frac{1}{2}=\frac{2+3}{6}=\frac{5}{6}\)
Therefore, together the two pumps fill \(\frac {5}{6}\) of the tank in 1 hour.
Since it is a direct proportion, we know that
1 : x :: \(\frac {5}{6}\) : 1
⇒ \(\frac {5}{6}\) × x = 1 × 1
⇒ x = \(\frac {6}{5}\) hours
⇒ x = 1 hour 12 minutes
Thus, the tank will be filled in 1 hour and 12 minutes.

Question 11.
A factory requires 42 machines to produce a given number of toys in 63 days. How many machines are required to produce the same number of toys in 54 days?

Solution:
Let the required number of machines = x
The number of machines and the number of days are inversely proportional, as the amount of work remains constant.
42 : x :: 54 : 63
⇒ x × 54 = 42 × 63
⇒ x = \(\frac{42 \times 63}{54}\)
⇒ x = 49
Therefore, 49 machines are required to produce the toys in 54 days.

Question 12.
A car takes 2 hours to reach a destination, travelling at a speed of 60 km/h. How long will the car take if it travels at a speed of 80 km/h?
Solution:
When the speed of the car increases, the time taken to reach the destination decreases.
Hence, speed and time are inversely proportional for a fixed distance.
So, x₁y₁ = x₂y₂
Let the required time = x hours
Thus, 60 × 2 = 80 × x
⇒ x = 1.5 hours
⇒ x = 1 hour 30 minutes
Therefore, the car will take 1 hour and 30 minutes to reach the destination.