Students often refer to Ganita Prakash Class 8 Solutions and Class 8 Maths Chapter 7 Proportional Reasoning 1 NCERT Solutions Question Answer to verify their answers.
NCERT Class 8 Maths Chapter 7 Proportional Reasoning 1 Solutions Question Answer
Ganita Prakash Class 8 Chapter 7 Solutions Proportional Reasoning 1
NCERT Class 8 Maths Ganita Prakash Chapter 7 Proportional Reasoning 1 Solutions Question Answer
7.1 Observing Similarity in Change
NCERT In-Text Questions (Page 160)
Question 1.
Can you check by what factors the width and height of image D change as compared to image A? Are the factors the same?
Solution:
Width of image D = 90 mm and width of image A = 60 mm.
∵ 90 = \(\frac {3}{2}\) × 60
∴ Width of image D is \(\frac {3}{2}\) times the width of image A.
Height of image D = 60 mm and height of image A = 40 mm.
∵ 60 = \(\frac {3}{2}\) × 40
∴ Height of image D is \(\frac {3}{2}\) times the height of image A.
Therefore, the width and height of image D have changed by the same factor as compared to image A.
7.2 Ratios
NCERT In-Text Questions (Page 161)
Question 1.
By what factor should we multiply the ratio 60 : 40 (image A) to get 90 : 60 (image D)?
Solution:
Image A – 60 : 40
Multiplying both the terms by \(\frac {3}{2}\), we get
60 × \(\frac {3}{2}\) : 40 × \(\frac {3}{2}\), which is 90 : 60, the ratio of width to height in image D.
So, we need to multiply the ratio 60 : 40 (image A) by the factor \(\frac {3}{2}\) to get 90 : 60 (image D).
7.4 Problem Solving with Proportional Reasoning
NCERT In-Text Questions (Page 163)
Question 1.
Example 4: 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.
Solution:
Do it yourself.
Question 2.
Is the teacher-to-student ratio in your school proportional to the one in my school?
Solution:
Do it yourself.
Question 3.
Example 5: 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 it 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 it yourself.
Question 5.
Compare the rectangle you have drawn to those drawn by your classmates. Do they all look the same?
Solution:
Do it yourself.
Filter Coffee!
NCERT In-Text Questions (Page 165)
Question 1.
Why is this coffee stronger?
Solution:
For regular coffee, the ratio of decoction to milk is 15 : 35.
For stronger coffee, the ratio of decoction to milk is 20 : 30.
To compare, we need to make either the first terms of both ratios equal or the second terms. Therefore,
15 : 35 = 15 × 4 : 35 × 4 = 60 : 140
20 : 30 = 20 × 3 : 30 × 3 = 60 : 90
Thus, for a 60 mL decoction of regular coffee, he uses 140 mL of milk, while for the stronger coffee, he uses 90 mL of milk.
Since in stronger coffee there is less milk for the same amount of decoction compared to regular coffee.
Therefore, the coffee is stronger.
Question 2.
Why is this coffee lighter?
Solution:
For regular coffee, the ratio of decoction to milk is 15 : 35.
For lighter coffee, the ratio is 10 : 40.
Now, 15 : 35 = 15 × 2 : 35 × 2 = 30 : 70
10 : 40 = 10 × 3 : 40 × 3 = 30 : 120
We see that for 30 mL of decoction of regular coffee, he uses 70 mL of milk, while for lighter coffee, he uses 120 mL of milk.
Since in lighter coffee, there is more milk for the same amount of decoction compared to regular coffee.
That’s why it is lighter.
Question 3.
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:
Figure it Out (Pages 165-167)
Question 1.
Circle the following statements of proportion that are true.
(i) 4 : 7 :: 12 : 21
(ii) 8 : 3 :: 24 : 6
(iii) 7 : 12 :: 12 : 7
(iv) 21 : 6 :: 35 : 10
(v) 12 : 18 :: 28 : 12
(vi) 24 : 8 :: 9 : 3
Soluti0n:
(i) The ratio 4 : 7 is already in its simplest form.
To find the simplest form of 12:21, we need to divide both terms by their HCF.
The HCF of 12 and 21 is 3.
Dividing both terms by 3, we get 4 : 7.
Since both ratios (in their simplest forms) are the same, they are proportional.
(ii) The ratio 8 : 3 is already in its simplest form.
Now, 24 ÷ 6 : 6 ÷ 6 = 4 : 1 (∵ HCF of 24 and 6 = 6)
∴ 8 : 3 ≠ 4 : 1
Since both ratios are not the same, they are not proportional.
(iii) The ratios 7 : 12 and 12 : 7 are already in their simplest forms.
But, 7 : 12 ≠ 12 : 7
Since both ratios are not the same, they are not proportional.
(iv) Here, 21 ÷ 3 : 6 ÷ 3 = 7 : 2 (∵ HCF of 21 and 6 is 3)
and 35 ÷ 5 : 10 ÷ 5 = 7 : 2 (∵ HCF of 35 and 10 is 5)
∴ 7 : 2 = 7 : 2
Since both ratios (in their simplest forms) are the same, they are proportional.
(v) Here 12 ÷ 6 : 18 ÷ 6 = 2 : 3 (∵ HCF of 12 and 18 is 6)
and 28 ÷ 4 : 12 ÷ 4 = 7 : 3 (∵ HCF of 28 and 12 is 4)
∴ 2 : 3 ≠ 7 : 3
Since both ratios are not the same, they are not proportional.
(vi) Here 24 ÷ 8 : 8 ÷ 8 = 3 : 1 (∵ HCF of 24 and 8 is 8)
and 9 ÷ 3 : 3 ÷ 3 = 3 : 1 (∵ HCF of 9 and 3 is 3)
∴ 3 : 1 = 3 : 1
Since both ratios (in their simplest forms) are the same, they are proportional.
Therefore, the correct answer is:
Question 2.
Give 3 ratios that are proportional to 4 : 9.
______ : ______ ______ : ______ ______ : ______
Soluti0n:
4 × 2 : 9 × 2 = 8 : 18
4 × 3 : 9 × 3 = 12 : 27
4 × 4 : 9 × 4 = 16 : 36
∴ 3 ratios that are proportional to 4 : 9 are 8 : 18, 12 : 27, 16 : 36.
Question 3.
Fill in the missing numbers for these ratios that are proportional to 18 : 24.
3 : ______, 12 : ______, 20 : ______, 27 : ______
Soluti0n:
∵ 18 ÷ 6 = 3
∴ 18 : 24 = 18 ÷ 6 : 24 ÷ 6 = 3 : 4
Let 18x = 12, then x = \(\frac{12}{18}=\frac{2}{3}\)
Multiplying both terms of 18 : 24 by \(\frac {3}{2}\), we get
\(18 \times \frac{2}{3}: 24 \times \frac{2}{3}\) = 12 : 16
Let 18x = 20, then x = \(\frac{20}{18}=\frac{10}{9}\)
Multiplying both terms of 18 : 24 by \(\frac {10}{9}\), we get
\(18 \times \frac{10}{9}: 24 \times \frac{10}{9}=20: \frac{80}{3}\)
Let 18x = 27, then x = \(\frac{27}{18}=\frac{3}{2}\)
Multiplying both terms of 18 : 24 by \(\frac {3}{2}\), we get
\(18 \times \frac{3}{2}: 24 \times \frac{3}{2}\) = 27 : 36
Therefore, the correct answers are 3 : 4, 12 : 16, 20 : \(\frac {80}{3}\), 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.
Soluti0n:
Do it 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?
Soluti0n:
Do it 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.
Soluti0n:
(a) In the given pattern of the bricks, there are 6 coloured bricks after 9 grey bricks.
The ratio of grey bricks to coloured bricks is 9 : 6.
We can convert it in simplest form as 9 ÷ 3 : 6 ÷ 3 = 3 : 2
So, the ratio of grey bricks to coloured bricks is 3 : 2.
(b) In the given pattern, there are 12 coloured bricks for every 16 grey bricks.
Therefore, the ratio of grey bricks to coloured bricks is 16 : 12.
In simplest form, 16 ÷ 4 : 12 ÷ 4 = 4 : 3.
Therefore, the required ratio is 4 : 3.
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 a head, torso, arms, and legs with equivalent ratios as above.
Soluti0n:
Do it yourself.
Question 8.
Does the drawing look more realistic if the ratios are proportional? Why? Why not?
Soluti0n:
Do it yourself.
Figure it Out (Pages 170-172)
Question 1.
The Earth travels approximately 940 million kilometres around the Sun in a year. How many kilometres will it travel in a week?
Soluti0n:
Distance travelled by the Earth in 1 year (365 days) = 940 million km = 940,000,000 km
Let the Earth travel x km in a week (7 days).
Therefore, we have 365 : 7 :: 940,000,000 : x
⇒ x = \(\frac{940,000,000 \times 7}{365}\)
⇒ x = 18,027,397
⇒ x ~ 18 million
∴ The Earth travels approximately 18 million kilometres in a week.
Question 2.
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 the 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.
Soluti0n:
The total length of all the walls of the house is
12 ft + 9 ft + 15 ft + 12 ft + 15 ft + 9 ft + 12 ft + 9 ft + 6 ft + 9 ft = 108 ft.
Let’s say the mason needs x bricks to build 108 ft of wall.
Therefore, we have,
10 : 108 :: 1450 : x
⇒ x = \(\frac{1450 \times 108}{10}\)
⇒ x = 15660
∴ The mason will need approximately 15,660 bricks to build the house.
NCERT In-Text Questions (Page 172)
Question 1.
Why do you think that the ratio of the prices is not proportional to the ratio of the volumes?
Soluti0n:
Do it yourself.
Figure it Out (Page 175)
Question 1.
Divide ₹ 4,500 into two parts in the ratio 2 : 3.
Soluti0n:
Total parts = 2 + 3
First part = 2 × \(\frac{4500}{2+3}\) = ₹ 800
Second part = 3 × \(\frac{4500}{2+3}\) = ₹ 2700
Therefore, ₹ 4500 can be divided into two parts in the ratio 2 : 3 as ₹ 1800 and ₹ 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?
Soluti0n:
The ratio of acid to water = 1 : 5.
Total parts = 1 + 5
Amount of acid in the solution = 1 × \(\frac{240}{1+5}\) = 40 mL
Amount of water in the solution = 5 × \(\frac{240}{1+5}\) = 200 mL
Therefore, the solution contains 40 mL of acid and 200 mL of water.
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?
Soluti0n:
The ratio of blue to yellow paint = 3 : 5
Total parts = 3 + 5
Amount of blue paint = 4 × \(\frac{40}{3+5}\) = 15 mL
Amount of yellow paint = 5 × \(\frac{40}{3+5}\) = 25 mL
After adding 20 mL more yellow paint, the total amount of yellow paint = 25 + 20 = 45 mL.
New ratio of blue to yellow paint = 15 : 45
Simplified form = 1 : 3
Therefore, the new ratio of blue to yellow paint is 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?
Soluti0n:
The ratio of rice to urad dal = 2 : 1
Total parts = 2 + 1
Number of cups of rice needed = 2 × \(\frac{6}{2+1}\) = 4 cups
Number of cups of urad dal needed = 1 × \(\frac{6}{2+1}\) = 2 cups
Therefore, you will need 4 cups of rice and 2 cups of urad dal.
Question 5.
I have one bucket of orange paint that 1 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?
Soluti0n:
The ratio of red to yellow paint in the first bucket = 3 : 5.
Total paint in the first bucket = 3 + 5 = 8
So, in the first bucket, 3 out of 8 parts are red paint and 5 out of 8 parts are yellow paint.
The second bucket contains only yellow paint, and its total is equal to one full bucket (i.e., 8 parts of yellow paint).
So, after mixing both the buckets, the red paint is the same, that is 3 out of 16 parts, and the yellow paint becomes 13 out of 16 parts.
So, the new ratio of red paint to yellow paint is 3 : 13.
Figure it Out (Pages 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.
Soluti0n:
The ratio of orange juice to apple juice = 600 : 900
Now, 600 ÷ 300 : 900 ÷ 300 = 2 : 3 (∵ HCF of 600 and 900 is 300)
Therefore, the required ratio in its simplest form is 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?
Soluti0n:
Let the number of buses needed this year be x.
So, 162 : 204 :: 3 : x
⇒ x = \(\frac{204 \times 3}{162}=3 \frac{126}{162}\) buses
Since the number of buses must be a whole number, we round up to 4 buses.
No, 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?
Soluti0n:
Dividing the population of Delhi by the area of Delhi, we get
\(\frac{30,000,000}{1484}=20,215 \frac{940}{1484}\) people per sq. km.
Dividing the population of Mumbai by the area of Mumbai, we get
\(\frac{20,000,000}{550}=36,363 \frac{350}{550}\) people per sq. km.
Clearly, there are more people per sq. km in Mumbai as compared to 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?
Soluti0n:
Do it 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\(\frac {1}{2}\) palas of saffron costs \(\frac {3}{7}\) niskas, O expert businessman! Tell me quickly what quantity of saffron can be bought for 9 niskas?”
Soluti0n:
Let x palas of saffron be bought for 9 niskas.
Then, by the rule of three, we have
\(\frac{3}{7}: 9:: 2 \frac{1}{2}: x\)
⇒ x = \(\frac{9 \times 2 \frac{1}{2}}{\frac{3}{7}}=\frac{9 \times 5}{2} \times \frac{7}{3}=52 \frac{1}{2}\)
So, 52\(\frac {1}{2}\) palas of saffron can be bought for 9 niskas.
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?
Soluti0n:
Difference in their ages = 5 – 1 = 4 years
This age difference will always remain the same.
Let Harmain’s age be x years.
Then her brother’s age will be x + 4 years.
Given, \(\frac{\text { Harmains’s age }}{\text { brother’s age }}=\frac{1}{2}\)
So, x = \(\frac{x}{x+4}=\frac{1}{2}\)
⇒ 2x = x + 4
∴ Harmain’s age will be 4 years when the ratio of her age to her brother’s age is 1 : 2.
Question 7.
The mass of equal volumes of gold and water is in the ratio 37 : 2. If 1 litre of water is 1 kg in mass, what is the mass of 1 litre of gold?
Soluti0n:
For equal volumes, the ratio of masses of gold to water is 37 : 2.
Let the mass of 1 litre of gold be x kg.
Then, by the rule of three, we get
37 : 2 :: x : 1
⇒ x = \(\frac{37 \times 1}{2}=18 \frac{1}{2}\) kg
Therefore, the mass of 1 litre of gold is 18\(\frac {1}{2}\) 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).
Soluti0n:
Area of the plot = 200 ft × 500 ft = 100000 sq ft
Also, 1 acre = 43,560 sq ft.
Let’s say, the farmer should buy x tonnes of cow manure.
Therefore, by the rule of three, we have
43560 : 100000 :: 10 : x
⇒ x = \(\frac{100000 \times 10}{43560}=22 \frac{41680}{43560}\)
⇒ x = \(22 \frac{1042}{1089}\) tonnes
⇒ x ~ 22.96 tonnes
Therefore, the farmer should buy approximately 22.96 tonnes of cow manure.
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?
Soluti0n:
1 litre = 1,000 mL
∴ 10 litre = 10,000 mL
Let’s say the tap takes x seconds to fill the bucket.
Then by the rule of three, we have
500 : 10000 :: 15 : x
⇒ x = \(\frac{10000 \times 15}{500}\)
⇒ x = 300
Also, 300 seconds = 5 minutes.
Therefore, the tap will take 5 minutes to fill the 10-litre bucket.
Question 10.
One acre of land costs ₹ 15,00,000. What is the cost of 2,400 square feet of the same land?
Soluti0n:
1 acre = 43,560 square feet
Let the cost of 2,400 square feet of land be ₹ x.
Then, by the rule of three, we have
43560 : 2400 :: 1500000 : x
⇒ x = \(\frac{2400 \times 1500000}{43560}\)
⇒ x = ₹ 82645
Thus, the cost of 2,400 square feet of land is approximately ₹ 82,645.
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?
Soluti0n:
Time taken by a pair of oxen to plough 1 acre = 6 hours.
Let the pair of oxen plough a 20-acre field in x hours.
Then, we have 1 : 20 :: 6 : x
⇒ x = \(\frac{20 \times 6}{1}\) = 120 hours
Since the tractor is 4 times faster, it will take = \(\frac {120}{4}\) = 30 hours
Therefore, the pair of oxen would take 120 hours, and the tractor would take 30 hours to plough the 20-acre field.
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?
Soluti0n:
The ratio of copper to nickel 3 : 1.
Total parts = 3 + 1
Weight of copper in ₹ 10 coin = 3 × \(\frac{7.74}{3+1}\)
= 5.805 g
= \(\frac {5.805}{1000}\) kg
Weight of nickel in ₹ 10 coin = 1 × \(\frac{7.74}{3+1}\)
= 1.935 g
= \(\frac {1.935}{1000}\) kg
∴ Cost of copper in ₹ 10 coin = \(\frac{906 \times 5.805}{1000}\) = ₹ 5
Cost of nickel in ₹ 10 coin = \(\frac{1314 \times 1.935}{1000}\) = ₹ 3
So, the total cost of these metals in a ₹ 10 coin is approximately ₹ 8.
It’s Puzzle Time! (Page 178)
Binairo, also known as Takuzu, is a logic puzzle with simple rules. Binairo is generally played on a square grid with no particular size. Some cells start filled with two symbols: horizontal and vertical lines. The rest of the cells are empty. The task is to fill cells in such a way that:
1. Each row and each column must contain an equal number of horizontal and vertical lines.
2. More than two horizontal or vertical lines can’t be adjacent.
3. Each row is unique. Each column is unique.
Solve the following Binairo Puzzles:
Soluti0n:
