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, 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 hatting: 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:
Given, time for warm-up/cool-down : time for batting : Time for bowling : time for fielding :: 3 : 4 : 3 : 5
Total number of ratio parts = 3 + 4 + 3 + 5 = 15
Total time of each session = 150 minutes
So, time for warm-up/cool-down = \(\frac {3}{15}\) × 150 = 30 minutes
Time for batting = \(\frac {4}{15}\) × 150 = 40 minutes
Time for bowling = \(\frac {3}{15}\) × 150 = 30 minutes
Time for fielding = \(\frac {5}{15}\) × 150 = 50 minutes
Verification: 30 + 40 + 30 + 50 = 150 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:
Given, No. of Odiya books : No. of Hindi books : No. of English books :: 3 : 2 : 1
Let x be the total number of books.
No. of Odiya books = \(\frac {3}{6}\) × x
⇒ 288 = \(\frac {3}{6}\) × x
⇒ x = 576
∴ No. of Hindi books = \(\frac {2}{6}\) × 576 = 192
and no. of English books = \(\frac {1}{6}\) × 576 = 96

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 no. of ₹ 10 coins : no. of ₹ 5 coins : no. of ₹ 2 coins : no. of ₹ 1 coins :: 4 : 3 : 2 : 1.
Total number of coins = 100
Total number of ratio parts = 4 + 3 + 2 + 1 = 10
no. of ₹ 10 coins = \(\frac {4}{10}\) × 100 = 40
no. of ₹ 5 coins = \(\frac {3}{10}\) × 100 = 30
no. of ₹ 2 coins = \(\frac {3}{10}\) × 100 = 30
no. of ₹ 1 coins = \(\frac {1}{10}\) × 100 = 10
Total money = 40 × 10 + 30 × 5 + 2 × 20 + 1 × 10
= 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:
We can construct triangles with sides in the ratio 3 : 4 : 5.
They will not be congruent to each other.
Reason:
Triangle 1: Let sides 3 cm, 4 cm, 5 cm
Triangle 2: Let sides 6 cm, 8 cm, 10 cm
Triangle 3: Let sides = 9 cm, 12 cm, 15 cm
Though all these triangles have the same ratio (3 : 4 : 5), their actual sizes are different.
Congruent triangles must have the same shape and size.
These triangles have the same shape (they are similar) but different sizes.
Hence, they are not congruent.

Question 5.
Can you construct a triangle with sidelengths in the ratio 1 : 3 : 5? Why or why not?
Solution:
For a triangle to exist, it must satisfy the triangle inequality theorem, which states:
The sum of any two sides of a triangle must be greater than the third side.
Let’s take
Side 1 = 1 cm
Side 2 = 3 cm
Side 3 = 5 cm
1. 1 + 3 < 5 ⇒ 4 > 5 (No)
2. 1 + 5 > 3 ⇒ 6 > 3 (Yes)
3. 3 + 5 > 1 ⇒ 8 > 1 (Yes)
Since the first condition fails (1 + 3 = 4 < 5)
We can’t construct a triangle with these sidelengths.

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:
Given, total people = 360
90 people liked the summer season.
120 people liked the rainy season.

∴ People liked winter season = 360 – (120 + 90) = 150
So, angle for summer season = \(\frac {90}{360}\) × 360° = 90°
Angle for rainy season = \(\frac {120}{360}\) × 360° = 120°
Angle for winter season = \(\frac {150}{360}\) × 360° = 150°
Verification: 90 + 120 + 150 = 360°

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, Entertainment = 50%
Sports = 25%
News = 15%
Information = 10%

Angle for entertainment = 50% of 360°
= \(\frac {50}{100}\) × 360°
= 180°
Angle for sports = 25% of 360°
= \(\frac {25}{100}\) × 360°
= 90°
Angle for news = 15% of 360°
= \(\frac {15}{100}\) × 360°
= 54°
Angle for information = 10% of 360°
= \(\frac {10}{100}\) × 360°
= 36°
Verification: 180° + 90° + 54° + 36° = 360°

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:

Total number of students = 4 + 6 + 9 + 3 + 10 + 12 + 16 = 60

Angle for Language = \(\frac {4}{60}\) × 360° = 24°
Angle for Arts Education = \(\frac {6}{60}\) × 360° = 36°
Angle for Vocational Education = \(\frac {9}{60}\) × 360° = 54°
Angle for Social Science = \(\frac {3}{60}\) × 360° = 18°
Angle for Physical Education = \(\frac {10}{60}\) × 360° = 60°
Angle for Maths = \(\frac {12}{60}\) × 360° = 72°
Angle for Science = \(\frac {16}{60}\) × 360° = 96°

3.6 Inverse Proportions

Figure It Out (Page 65)

Question 1.
Which of these are in inverse proportion?

Solution:
(i) x₁ = 40, x₂ = 80, x₃ = 25, x₄ = 16
y₁ = 20, y₂ = 10, y₃ = 32, y₄ = 50
x₁y₁ = 40 × 20 = 800
x₂y₂ = 80 × 10 = 800
x₃y₃ = 25 × 32 = 800
x₄y₄ = 16 × 50 = 800
So, x₁y₁ = x₂y₂ = x₃y₃ = x₄y₄ = 800
∴ x and y are in inverse proportion.

(ii) x₁ = 40, x₂ = 80, x₃ = 25, x₄ = 16
y₁ = 20, y₂ = 10, y₃ = 12.5, y₄ = 8
x₁y₁ = 40 × 20 = 800
x₂y₂ = 80 × 10 = 800
x₃y₃ = 25 × 12.5 = 312.5
x₄y₄ = 16 × 8 = 128
So, x₁y₁ = x₂y₂ ≠ x₃y₃ ≠ x₄y₄
∴ x and y are not in inverse proportion.

(iii) x₁ = 30, x₂ = 90, x₃ = 150, x₄ = 10
y₁ = 15, y₂ = 5, y₃ = 3, y₄ = 45
x₁y₁ = 30 × 15 = 450
x₂y₂ = 90 × 5 = 450
x₃y₃ = 150 × 3 = 450
x₄y₄ = 10 × 45 = 450
So, x₁y₁ = x₂y₂ = x₃y₃ = x₄y₄ = 450
∴ x and y are in inverse proportion.

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

Solution:

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:
(i) More taps → Less time to fill the tank
Fewer taps → More time to fill the tank
The quantities change in opposite directions by the same factor.
If we double the no. of taps, the time taken becomes half.
Hence, they are in inverse proportion.

(ii) More painters → Fewer days needed
Fewer painters → More days needed.
If we double the no. of painters, the work gets done in half the time.
Hence, they are in inverse proportion.

(iii) Petrol → It decreases
Distance → It increases
When distance increases, then petrol decreases, so they are in inverse proportion.

(iv) Higher speed → Less time taken
Lower speed → More time taken
For a fixed distance, if speed doubles, time becomes half.
Hence, they are in inverse proportion.

(v) More cloth → More price to pay
Less cloth → Less price to pay
Both quantities decrease together and increase together, so they are in direct proportion.

(vi) More pages → More time to read
Fewer pages → Less time to read
Both quantities decrease together and increase together, so they are in direct proportion.

Question 2.
If 24 pencils cost ₹ 120, how much will 20 such pencils cost?
Solution:
The number of pencils and the cost of pencils are in direct proportion.
If x is the required cost, then
\(\frac{24}{20}=\frac{₹ 120}{x}\)
⇒ x × 24 = ₹ 120 × 20
⇒ x = ₹ 100
So, the cost of 20 such pencils is ₹ 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:
The number of families and the number of days are in inverse proportion.
Assumptions needed
(i) All families use the same amount of water.
(ii) Water usage per family per day is constant.
(iii) No additional water is added to the tank.
Let the water last for x days.
So, 20 × 6 = 30 × x
⇒ x = 4
So, the water will last for 4 days.

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:
Common Sleep Patterns:
Average no. of hours a giraffe sleeps = 2.5 hours
Average no. of hours an elephant sleeps = 3.5 hours
Average no. of hours a boy sleeps = 8 hours
Average no. of hours a dog sleeps = 10.5 hours
Average no. of hours a cat sleeps = 13 hours
Average no. of hours a squirrel sleeps = 15 hours
Average no. of hours a snake sleeps = 18 hours
Average no. of hours a bat sleeps = 20 hours

Question 5.
The pie chart given below 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:
(i) The largest angle is 120°, which corresponds to the Bus.

(ii) The fraction of children who travel by car
\(\frac {30}{360}\) = \(\frac {1}{12}\)

(iii) Let x be the total number of children who took part in the survey.
Then 18 = \(\frac {1}{12}\) × 12
⇒ x = 18 × 12 = 216
The number of children using taxis is 0, as taxis are not a category in this chart.

(iv) Cycle and two-wheeler (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:
When the number of workers increases, the number of days needed to paint the fence decreases.
Assumptions Needed
(i) All workers work at the same speed/rate.
(ii) The work is uniformly distributed among all workers.
(iii) All workers work for the same number of hours each day.
So, the number of workers and the number of days are in inverse proportion.
Let x be the no. of days taken.
3 × 4 = 4 × x
⇒ x = 3
So, they will take 3 days to finish the work.

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:
No. of hours and no. of tanks are in direct proportion.
Let 5 such tanks take x hours.
\(\frac{6}{2}=\frac{x}{5}\)
⇒ x = 15
So, 5 tanks will take 15 hours.

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:
No. of rows and no. of chairs in each row are in inverse proportion.
Let the new arrangement have x rows.
25 × 12 = x × 20
⇒ x = 15
So, the new arrangement has 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:
No. of periods and the duration of each period are in inverse proportion.
Let each period be of x minutes.
8 × 45 = 9 × x
⇒ x = 40
So, each period is 40 minutes.

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:
Let x litres be the capacity of the tank.
Then, water filled by small tank in one hour = \(\frac {x}{3}\) litre
Water filled by large tank in one hour = \(\frac {x}{2}\) litre
So, total water filled by both pumps in one hour = \(\left(\frac{x}{3}+\frac{x}{2}\right)\) litre = \(\frac {5x}{6}\) litre
∴ Time taken by both pumps used together to fill the tank = \(\left(1 \div \frac{5 x}{6}\right) \times x\)
= 1 × \(\frac {6}{5x}\) × x
= \(\frac {6}{5}\) hour
= 1\(\frac {1}{5}\) hour

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:
No. of machines and no. of days are in inverse proportion.
42 × 63 = x × 54
⇒ x = 49
So, 49 machines are required.

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:
Let the car take t hours.
The speed of the car and the time taken are in inverse proportion.
So, 2 × 60 = t × 80
⇒ t = 1.5 hour
So, the car will take 1.5 hours.

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