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Class 8 Maths Chapter 7 Proportional Reasoning 1 Solutions
Ganita Prakash Class 8 Chapter 7 Solutions
Class 8 Maths Ganita Prakash Chapter 7 Solutions Proportional Reasoning 1
Intext (NCERT Page 161)
Question 1.
By what factor should we multiply the ratio 60:40 (image A) to get 90 : 60 (image D)?
Solution:
Factor of change in the first term
= \(\frac{90}{60}=\frac{90 \div 30}{60 \div 30}=\frac{3}{2}\)
Factor of change in the second term
= \(\frac{60}{40}\)
= \(=\frac{60 \div 20}{40 \div 20}\) [∵ HCF (60, 40) = 20]
= \(\frac{3}{2}\)
Required factor is \(\frac{3}{2}\).
Intext (NCERT Page 163)
Question 2.
In my school, there are 5 teachers and 170 students. The ratio of teachers to students in my school is 5 :170. Count the number of teachers and students in your school. What is the ratio of teachers to students in your school? Write it below.
Is the teacher-to-student ratio in your school proportional to the one in my school?
Solution:
Do yourself.
Question 3.
Measure the width and height (to the nearest cm) of the blackboard in your classroom. What is the ratio of width to height of the blackboard?
Solution:
Do yourself.
Question 4.
Can you draw a rectangle in your notebook whose width and height are proportional to the ratio of the blackboard?
Solution:
Do yourself.
Question 5.
Compare the rectangle you have drawn to those drawn by your classmates. Do they all look the same?
Solution:
Do yourself.
Intext (NCERT Page 164-165)
Question 1.
(i) If customers want ‘stronger’ filter coffee, Manjunath mixes 20 ml of decoction with 30 ml of milk. The ratio here is 20 : 30. Why is this coffee stronger?
Solution:
Given, ratio of coffee decoction to milk = 20 : 30
= \(\frac{20 \div 10}{30 \div 10}\) [∵ HCF (20, 30) = 10]
= 2 : 3
Therefore, every 2 parts of decoction, there are 3 parts of milk, indicating a significant amount of decoction. So, the coffee is stronger.
(ii) And when they want ‘lighter’ filter coffee, he mixes 10 ml of Coffee and 40 ml of milk, making the ratio 10 : 40. Why is this coffee lighter?
Solution:
Given, ratio of coffee to milk
= 10 : 40
= \(=\frac{10 \div 10}{40 \div 10}\) [∵ HCF (10, 40) = 10]
= 1 : 4
Therefore, every part of coffee, there are four parts of milk.
So, the higher dilution with milk results in a less concentrated coffee flavor.
Hence, the coffee is lighter.
Question 2.
The following table shows the different ratios in which Manjunath mixes coffee decoction with milk. Write in the last column if the coffee is stronger or lighter than the regular coffee.
Solution:
Let us assume that if the ratio of coffee decoction to milk is 0.5 then it is regular coffee, if its ratio is less than 0.5 then it is lighter coffee and if its ratio is greater than 0.5 then it is stronger coffee.
Ratio of 300 ml to 600 ml = \(\frac{300}{600}=\frac{300 \div 300}{600 \div 300}\)
[∵ HCF (300, 600) = 300]
= \(\frac{1}{2}\) = 0.5 2
So, it is regular coffee.
Ratio of 150 ml to 500 ml
= \(\frac{150}{500}=\frac{150 \div 50}{500 \div 50}=\frac{3}{10}\)
[∵ HCF (150,500) = 50]
= 03 < 0.5
So, it is lighter coffee.
Ratio of 200 ml to 400 ml
= \(\frac{200}{400}=\frac{200 \div 200}{400 \div 200}=\frac{1}{2}\) = 0.5
[∵ HCF (200, 400) = 200]
So, it is regular coffee.
Ratio of 24 ml to 56 ml
= \(\frac{24}{56}=\frac{24 \div 8}{56 \div 8}=\frac{3}{7}\) = 0.4285 < 0.5
[∵ HCF (24,56) = 8]
So, it is lighter coffee.
Ratio of 100 ml to 300 ml
= \(\frac{1}{2}\)
[∵ HCF (100, 300) = 100]
So, it is lighter coffee.
| Coffee Decoction (in ml) | Milk (in ml) | Regular/Strong/ Light |
| 300 | 600 | Regular |
| 150 | 500 | Light |
| 200 | 400 | Regular |
| 24 | 56 | Light |
| 100 | 300 | Light |
Figure it Out (NCERT Page 165-167)
Question 1.
Circle the following statements of proportion that are true.
(i) 4 : 7 :: 12 : 21
Solution:
We have,4:7:: 12:21
⇒ \(\frac{4}{7}=\frac{12}{21}\)
⇒ \(\frac{4}{7}=\frac{12 \div 3}{21 \div 3}\) [∵ HCF (12, 21) = 3]
⇒ \(\frac{4}{7}=\frac{4}{7}\), which is true.
(ii) 8 : 3 :: 24 : 6
Solution:
We have, 8 : 3 :: 24 : 6
⇒ \(\frac{8}{3}=\frac{24}{6}\)
⇒ \(\frac{8}{3}=\frac{24 \div 6}{6 \div 6}\) [∵ HCF (24, 6) = 3]
⇒ \(\frac{8}{3}=\frac{4}{1}\), which is false.
(iii) 7 : 12 :: 12 : 7
Solution:
We have, 7 : 12 :: 12 : 7
⇒ \(\frac{7}{12}=\frac{12}{7}\), which is false.
(iv) 21: 6 :: 35 : 10
Solution:
⇒ \(\frac{21}{6}=\frac{35}{10}\)
⇒ \(\frac{21 \div 3}{6 \div 3}=\frac{35 \div 5}{10 \div 5}\)
[∵ HCF (21, 6) = 3 and HCF (35, 10) = 5]
⇒ \(\frac{7}{2}=\frac{7}{2}\), which is true.
(v) 12 : 18 :: 28 : 12
Solution:
We have, 12 : 18 :: 28 : 12
⇒ \(\frac{21}{6}=\frac{35}{10}\)
⇒ \(\frac{21 \div 3}{6 \div 3}=\frac{35 \div 5}{10 \div 5}\)
[∵ HCF (12, 18) = 6 and HCF (28, 12) = 4]
⇒ \(\frac{2}{3}=\frac{7}{3}\), which is false.
(vi) 24 : 8 :: 9 : 3
Solution:
We have,
⇒ \(\frac{24}{8}=\frac{9}{3}\)
⇒ \(\frac{24 \div 8}{8 \div 8}=\frac{9 \div 3}{3 \div 3}\)
[∵ HCF (24,8) = 8 and (9,3) =3]
⇒ \(\frac{3}{1}=\frac{3}{1}\), which is true.
Question 2.
Give 3 ratios that are proportional to 4 : 9.
Solution:
We have, 4 : 9 = 4 × 2 : 9 × 2 = 8 : 18
[changing by the same factor]
4 : 9 = 4 × 3 : 9 × 3 = 12 : 27 and 4 : 9 = 4 × 4 : 9 × 4 = 16 : 36
So, the three ratios proportional to 4 : 9 are 8 : 18, 12 : 27 and 16: 36.
Question 3.
Fill in the missing numbers for these ratios that are proportional to 18 : 24.
Solution:
We have, 18:24 = \(\frac{1}{2}\) [∵ HCF (18,24) = 6]
For first ratio, 3 : ………………
Let missing term be x.
∴ \(\frac{1}{2}\) ⇒ 3x = 3 × 4
⇒ x = \(\frac{1}{2}\) ⇒ x = 4
∴ Ratio = 3:4
For second ratio, 12
∴ Let missing term be x.
∴ \(\frac{12}{x}=\frac{3}{4}\)
⇒ x = \(\frac{12 \times 4}{3}\)
⇒ x = 4 × 4
⇒ x = 16
∴ Ratio = 12 : 16
For third ratio, 20
Let missing term be x.
∴ \(\frac{20}{x}=\frac{3}{4}\)
⇒ 3x = 20 × 4
⇒ x =\(\frac{20 \times 4}{3}\)
⇒ x =\(\frac{80}{3}\)
∴ Ratio = \(\frac{20}{1}: \frac{80}{3}\)
= 20 × 3, \(\frac{80}{3}\) × 3 = 60 : 80
For fourth ratio, 27 : ….
Let missing term be x.
∴ \(\frac{27}{x}=\frac{3}{4}\)
⇒ 3x = 27 × 4
⇒ x = \(\frac{27 \times 4}{3}\)
⇒ x = 9 × 4
⇒ x = 36
Ratio = 27:36
Question 4.
Look at the following rectangles. Which rectangles are similar to each other? You can verify this by measuring the width and height using a scale and comparing their ratios.
Solution:
Do yourself
Question 5.
Look at the following rectangle. Can you draw a smaller rectangle and a bigger rectangle with the same width to height ratio in your notebooks? Compare your rectangles with your classmates’ drawings.
Are all of them the same? If they are different from yours, can you think why? Are they wrong?
Solution:
Do yourself
Question 6.
The following figure shows a small portion of a long brick wall with patterns made using coloured bricks. Each wall continues this pattern throughout the wall. What is the ratio of grey bricks to coloured bricks? Try to give the ratios in their simplest form.
Solution:
(a) Number of grey brick = 33
Number of coloured bricks = 18
∴ Required ratio = 33 : 18
= \(=\frac{33 \div 3}{18 \div 3}\) [∵ HCF (33, 18) = 3]
= 11 : 6
(b) Number of grey bricks = 71
Number of coloured bricks = 48
∴ Required ratio = 71: 48
Question 7.
Let us draw some human figures. Measure your friend’s body the lengths of their head, torso, arms and legs. Write the ratios as mentioned below
Now, draw a figure with head, torso, arms, and legs with equivalent ratios as above.
Solution:
Do yourself
Question 8.
Does the drawing look more realistic if the ratios are proportional? Why? Why not?
Solution:
Yes,
When the ratios of different parts within a drawing are proportional, it means they maintain the same relative sizes and relationships as they would in reality.
Figure it Out (NCERT Page 170-171)
Question 9.
The Earth travels approximately 940 million kilometres around the Sun in a year. How many kilometres will it travel in a week?
Solution:
Given, number of days in a year = 365
and the Earth travels distance = 940 million km
= 940 × 1000000 km = 940000000 km
Let x km be the distance travelled by earth in one week (or 7 days).
The ratio of distance travelled by earth to number of days needs to be proportional.
∴ 940000000 : 365 :: x : 7
⇒ \(\frac{940000000 \times 7}{365}\) = x
⇒ x = 18027397.26
Hence, the Earth will travel 18027397.26 kilometres in a week.
Question 10.
A mason is building a house in the shape shown in the diagram. He needs to construct both the outer walls and the inner wall that separates two rooms. To build a wall of 10-feet, he requires approximately 1450 bricks. How many bricks would he need to build the house? Assume all walls are of the same height and thickness.
Solution:
Total length of wall
= 12 + 9 + 15 + 12 + 9 + 6 + 9 + 9 + 6 + 9 + 12
= 108 ft
Let the required number of bricks be x.
The ratio of length of wall to number of bricks needs to be proportional.
∴ 10 : 1450 :: 108 : x
⇒ \(\frac{10}{1450}=\frac{108}{x}\)
⇒ x = \(\frac{108 \times 1450}{10}\)
⇒ x = 108 × 145
⇒ x = 15660
So, mason would need 15660 bricks to build the house.
Intext (NCERT Page 172)
Question 1.
The ratio of the volume of a sachet to a small bottle is 6 :180. The ratio of their prices is 2 :154. Are these ratios proportional?
Solution:
Given, the ratio of the volume of a sachet to a small bottle
= 6 : 180
= \(\frac{6 \div 6}{180 \div 6}\)
= 1 : 30
[∵ HCF (6, 180) = 6]
Also, ratio of their prices = 2 : 154
= \(\frac{2 \div 2}{154 \div 2}\) [∵ HCF (2, 154) = 2]
= 1 : 77
Since, 1:30 ≠ 1:77
So, the ratios are not proportional.
Figure It Out (NCERT Page 175)
Question 1.
Divide ₹ 4,500 into two parts in the ratio 2 : 3.
Solution:
Given, amount = ₹ 4500 and ratio = 2 : 3
∴ First part = 4500 × \(\frac{2}{2+3}\)
= 4500 × \(\frac{2}{5}\)
= 900 × 2
= ₹ 1800
and second part = 4500 × \(\frac{3}{2+3}\)
= 4500 × \(\frac{3}{5}\)
= 900 × 3
= ₹ 2700
Question 2.
In a science lab, acid and water are mixed in the ratio of 1:5 to make a solution. In a bottle that has 240 ml of the solution, how much acid and water does the solution contain?
Solution:
Given, in the solution of 240 ml, the ratio of acid to water is 1 : 5.
∴ Quantity of acid = 240 × \(\frac{1}{1+5}\)
= 240 × \(\frac{1}{6}\)
= 40 ml
and quantity of water = 240 × \(\frac{5}{1+5}\)
= 240 × \(\frac{5}{6}\)
= 200 ml
Question 3.
Blue and yellow paints are mixed in the ratio of 3 : 5 to produce green paint. To produce 40 ml of green paint, how much of these two colours are needed?
To make the paint a lighter shade of green, I added 20 ml of yellow to the mixture. What is the new ratio of blue and yellow in the paint?
Solution:
Given, quantity of green paint = 40 ml
and ratio of blue paint to yellow paint = 3 : 5
∴ Quantity of blue paint = 40 × \(\frac{3}{3+5}\)
= 40 × \(\frac{3}{8}\)
= 5 × 3
= 15 ml
and quantity of yellow paint = 40 × \(\frac{5}{3+5}\)
= 40 × \(\frac{5}{8}\)
= 25 ml
Now, quantity of yellow paint after adding 20 ml
= 25 + 20 = 45 ml
∴ Required ratio = 15 : 45
= \(\frac{15 \div 15}{45 \div 15}\) [∵ HCF (15, 45) = 15]
= 1 : 3
Question 4.
To make soft idlis, you need to mix rice and urad dal in the ratio of 2 : 1. If you need 6 cups of this mixture to make idlis tomorrow morning, how many cups of rice and urad dal will you need?
Solution:
Given, ratio of rice to urad dal = 2 : 1
and quantity of mixture = 6 cups
∴ Quantity of rice = 6 × \(\frac{2}{2+1}\)
= 6 × \(\frac{2}{3}\)
= 4 cups
and quantity of urad dal = 6 × \(\frac{1}{2+1}\)
= 6 × \(\frac{1}{3}\)
= 2 cups
Question 5.
I have one bucket of orange paint that I made by mixing red and yellow paints in the ratio of 3 : 5.
I added another bucket of yellow paint to this mixture. What is the ratio of red paint to yellow paint in the new mixture?
Solution:
Given, orange paint is made by mixing red and yellow paints in the ratio of 3:5.
This means that if there are 8 parts of orange paint then for every 3 part of red paint, there are 5 parts of yellow paint.
Another bucket of yellow paint is added to this mixture. Since, we are not given the amount of second bucket, so we assume it as a unit of 8 parts similar to the bucket of orange paint.
After adding another bucket of yellow paint, the new quantity of yellow paint = 8 parts + 5 parts = 13 parts
So, the new ratio of red paint to yellow paint
= 3 parts: 13parts
= 3:13
Figure it Out (NCERT Page 176-177)
Question 1.
Anagh mixes 600 ml of orange juice with 900 ml of apple juice to make a fruit drink. Write the ratio of orange juice to apple juice in its simplest form.
Solution:
Given, quantity of orange juice = 600 ml and quantity of apple juice = 900 ml
Required ratio = 600 : 900
= \(\frac{600 \div 300}{900 \div 300}\) [∵ HCF (600, 900) = 300]
= 2 : 3
Question 2.
Last year, we hired 3 buses for the school trip. We had a total of 162 students and teachers who went on that trip and all the buses were full. This year we have 204 students. How many buses will we need? Will all the buses be full?
Solution:
Let the required number of buses be x.
The ratio of number of buses to number of students needs to be proportional.
∴ 3 : 162 :: x : 204
⇒ \(\frac{3}{162}=\frac{x}{204}\)
⇒ x = \(\frac{3 \times 204}{162}\)
⇒ x = \(\frac{204}{54}\)
⇒ x = 3.77 ≈ 4
∴ Required number of buses = 4
Now, capacity per bus = \(\frac{162}{3}\) = 54
Total capacity of the bus 4 × 54 = 216
Since, 216 > 204
So, not all the buses will not be full.
Question 3.
The area of Delhi is 1,484 sq. km and the area of Mumbai is 550 sq km. The population of Delhi is approximately 30 million and that of Mumbai is 20 million people. Which city is more crowded?
Why do you say so?
Solution:
Given, population of Delhi = 30 million
and area of Delhi = 1484 sq km
∴ Population density of Delhi
= \(\frac{30 \times 1000000}{1484}\) = 20215.6 people/sq km
Also, population of Mumbai = 20 million
and area of Mumbai = 550 sq km
∴ Population density of Mumbai
= \(\frac{20 \times 1000000}{550}\)
= 36363.6 people/sq km
Since, population density of Mumbai is greater than population density of Delhi.
So, Mumbai is more crowded.
Question 4.
A crane of height 155 cm has its neck and the rest of its body in the ratio 4 : 6. For your height, if your neck and the rest of the body also had this ratio, how tall would your neck be?
Solution:
Do yourself
Question 5.
Let us try an ancient problem from Lilavati. At that time weights were measured in a unit named palas and niskas was a unit of money. “If 2\(\frac{1}{2}\) pa/as of saffron costs \(\frac{3}{7}\) niskas, 0 expert businessman! tell me quickly what quantity of saffron can be bought for 9 niskas?”
Solution:
Let the required quantity of saffron be x palas.
The ratio of quantity of saffron to its costs needs to be proportional.
∴Required quantity of saffron = 52\(\frac{1}{2}\) palas
Question 6.
Harmain is a 1-year-old girl. Her elder brother is 5 years old. What will be Harmain’s age when the ratio of her age to her brother’s age is 1: 2?
Solution:
Let x be the age of Harmain when the ratio of her age to her brother’s age 1: 2.
Since, there is a difference of 4 years between the age of Harmain and her brother’s age.
∴Her brother’s age = x + 4
According to the question,
\(\frac{x}{x+4}=\frac{1}{2}\)
⇒ 2x = x + 4
⇒ 2x – x = 4
⇒ x = 4
∴Harmain’s age = 4 years
Question 7.
The mass of equal volumes of gold and water are in the ratio 37 : 2. If 1 litre of water is 1 kg in mass, what is the mass of 1 litre of gold?
Solution:
Given, the mass of equal volumes of gold and water are in the ratio 37 : 2.
Also, 1 litre of water = 1kg in mass
Let x kg be the mass of 1 litre of gold.
According to the question,
\(\frac{x}{1}=\frac{37}{2}\)
⇒ x = 18.5
So, the mass of 1 litre of gold is 18.5 kg.
Question 8.
It is good farming practice to apply 10 tonnes of cow manure for 1 acre of land. A farmer is planning to grow tomatoes in a plot of size 200 ft by 500 ft. How much manure should he buy? (Please refer to the section on Unit Conversions earlier in this chapter).
Solution:
Given, length of plot = 500 ft and breadth of plot = 200 ft
∴ Area = Length × Breadth
= 500 × 200
= 100000 sq ft
= \(\left(\frac{100000}{43560}\right)\) acre [∴1 acre = 43,560 sq ft]
= 2295 acre
∴ Required cow manure = 2.295 × 10
= 22.95 tonnes
Question 9.
A tap takes 15 seconds to fill a mug of water. The volume of the mug is 500 ml. How much time does the same tap take to fill a bucket of water if the bucket has a 10-litre capacity?
Solution:
Given, the volume of the mug is 500 ml.
∵ 1 L = 1000 ml
∴ 10 L = 1000 x 10
= 10000 ml
Let the required time be x seconds.
The ratio of time to volume of mug needs to be proportional.
∴ 15 : 500 :: x : 10000
⇒ \(\frac{15}{500}=\frac{x}{10000}\)
⇒ x = \(\frac{15 \times 10000}{500}\)
⇒ x = 15 × 20
⇒ x = 300
∴ Required time = 300 sec.
Question 10.
One acre of land costs ₹ 15,00,000. What is the cost of 2,400 square feet of the same land?
Solution:
We know that, 1 acre = 43560 sq feet
Let the required cost be ₹ x.
The ratio of quantity of land to its cost needs to be proportional.
∴ 43560 : 1500000 :: 2400 : x
⇒ \(\frac{43560}{1500000}=\frac{2400}{x}\)
⇒ x = \(\frac{2400 \times 1500000}{43560}=\frac{3600000000}{43560}\)
⇒ x = 82644.62
So, the cost of 2400 square feet of the same land is ₹ 82644.62.
Question 11.
A tractor can plough the same area of a field 4 times faster than a pair of oxen. A farmer wants to plough his 20-acre field. A pair of oxen takes 6 hours to plough an acre of land. How much time would it take if the farmer used a pair of oxen to plough the field? How much time would it take him if he decides to use a tractor instead?
Solution:
Given, time taken by an oxen to plough 1 acre of land = 6 hours
Let x hours be the time taken to plough 20 acres.
The ratio of the area of the field to ploughing time with oxen needs to be proportional.
∴ 1 : 6 :: 20 : x
⇒ \(\frac{1}{6}=\frac{20}{x}\)
⇒ x = 120
So, the time taken by an oxen to plough 20 acres of land is 120 hours.
∵ A tractor can plough the same area 4 times faster than a pair of oxen.
∴ Time taken by a tractor to plough 20 acres of land
= \(\frac{120}{4}\) = 30 hr
Question 12.
The ₹ 10 coin is an alloy of copper and nickel called ‘cupro-nickel’. Copper and nickel are mixed in a 3 :1 ratio to get this alloy. The mass of the coin is 7.74 grams. If the cost of copper is ₹ 906 per kg and the cost of nickel is ₹ 1,341 per kg, what is the cost of these metals in a ₹ 10 coin?
Solution:
Given, mass of the coin = 7.74 g
and ratio of copper to nickel = 3 : 1
∴ Mass of copper = 7.74 × \(\frac{3}{3+1}\)
= 7.74 × \(\frac{3}{4}\)
= 1.935 × 3
= 5.805 g
= \(\left(\frac{5.805}{1000}\right)\) kg [∵ 1 g = \(\frac{1}{1000}\)kg]
= 0.005805 kg
and mass of nickel = 7.74 × \(\frac{1}{3+1}\)
= 7.74 × \(\frac{1}{4}\)
= 1.935g
= \(\left(\frac{1935}{1000}\right)\) kg
= 0.001935 kg [∵ 1g = \(\frac{1}{1000}\)kg]
Now, cost of copper = 0.005805 × 906
= ₹ 5.25933
and cost of nickel = 0.001935 × 1341
= ₹ 2.594835
∴ Total cost of metal = cost of copper + cost of nickel
= 525933 + 2.594835
= ₹ 7.854165