Q1 (i): 4: 7:: 12: 21

To Find: Which of the given statements of proportion are true. Method: A proportion a:b:: c:d is true if the product of the extremes ( a × d ) is equal to the product of the means ( b × c ). Solution: (i) 4: 7:: 12: 21 Product of extremes = 4 × 21 = 84 Product of means = 7 × 12 = 84 Since the…

Q1 (ii): 8: 3:: 24: 6

To Find: Which of the given statements of proportion are true. Method: A proportion a:b:: c:d is true if the product of the extremes ( a × d ) is equal to the product of the means ( b × c ). Solution: (i) 4: 7:: 12: 21 Product of extremes = 4 × 21 = 84 Product of means = 7 × 12 = 84 Since the…

Q1 (iii): 7: 12:: 12: 7

To Find: Which of the given statements of proportion are true. Method: A proportion a:b:: c:d is true if the product of the extremes ( a × d ) is equal to the product of the means ( b × c ). Solution: (i) 4: 7:: 12: 21 Product of extremes = 4 × 21 = 84 Product of means = 7 × 12 = 84 Since the…

Q1 (iv): 21: 6:: 35: 10

To Find: Which of the given statements of proportion are true. Method: A proportion a:b:: c:d is true if the product of the extremes ( a × d ) is equal to the product of the means ( b × c ). Solution: (i) 4: 7:: 12: 21 Product of extremes = 4 × 21 = 84 Product of means = 7 × 12 = 84 Since the…

Q1 (v): 12: 18:: 28: 12

To Find: Which of the given statements of proportion are true. Method: A proportion a:b:: c:d is true if the product of the extremes ( a × d ) is equal to the product of the means ( b × c ). Solution: (i) 4: 7:: 12: 21 Product of extremes = 4 × 21 = 84 Product of means = 7 × 12 = 84 Since the…

Q1 (vi): 24: 8:: 9: 3

To Find: Which of the given statements of proportion are true. Method: A proportion a:b:: c:d is true if the product of the extremes ( a × d ) is equal to the product of the means ( b × c ). Solution: (i) 4: 7:: 12: 21 Product of extremes = 4 × 21 = 84 Product of means = 7 × 12 = 84 Since the…

Q1: Divide ₹4,500 into two parts in the ratio 2: 3.

Given: Total amount = ₹4,500. Ratio = 2:3. To Find: The two parts of the amount. Solution: 1. Find the sum of the terms in the ratio: Sum of parts = 2 + 3 = 5. 2. Calculate the value of one part by dividing the total amount by the sum of the parts: Value of one part = ₹4,500/5 = ₹900. 3. Calculate…

Proportional Reasoning-1 Class 8 Mathematics NCERT Solutions

Q1Figure it Out 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)

$24: 8:: 9: 3$

Solution

To Find: Which of the given statements of proportion are true.

Method: A proportion

a:b :: c:d

is true if the product of the extremes (

a \times d

) is equal to the product of the means (

b \times c

).

Solution:

(i)

$4: 7:: 12: 21$ Product of extremes =

4 \times 21 = 84

Product of means =

7 \times 12 = 84

Since the products are equal, the statement is True.

(ii)

$8: 3:: 24: 6$ Product of extremes =

8 \times 6 = 48

Product of means =

3 \times 24 = 72

Since the products are not equal (

48 \neq 72

), the statement is False.

(iii)

$7: 12:: 12: 7$ Product of extremes =

7 \times 7 = 49

Product of means =

12 \times 12 = 144

Since the products are not equal (

49 \neq 144

), the statement is False.

(iv)

$21: 6:: 35: 10$ Product of extremes =

21 \times 10 = 210

Product of means =

6 \times 35 = 210

Since the products are equal, the statement is True.

(v)

$12: 18:: 28: 12$ Product of extremes =

12 \times 12 = 144

Product of means =

18 \times 28 = 504

Since the products are not equal (

144 \neq 504

), the statement is False.

(vi)

$24: 8:: 9: 3$ Product of extremes =

24 \times 3 = 72

Product of means =

8 \times 9 = 72

Since the products are equal, the statement is True.

Final Answer: The true statements are (i), (iv), and (vi).

Q2Figure it Out 1

Give 3 ratios that are proportional to $4: 9$ .

Solution

Given: The ratio

4:9

.

To Find: Three ratios that are proportional to

4:9

.

Solution: To find proportional ratios, we can multiply both terms of the given ratio by the same non-zero number.

Multiply by 2: $4 \times 2 : 9 \times 2 = 8:18$
Multiply by 3: $4 \times 3 : 9 \times 3 = 12:27$
Multiply by 10: $4 \times 10 : 9 \times 10 = 40:90$

Final Answer: Three ratios proportional to

4:9

are

8:18

,

12:27

, and

40:90

. (Other answers are also possible).

Q3Figure it Out 1

Fill in the missing numbers for these ratios that are proportional to 18 : 24 . 3 : 12 : 20 : 27 :

Solution

Given: The base ratio is

18:24

.

To Find: The missing numbers in the proportional ratios.

Solution: First, let's simplify the base ratio

18:24

. The Highest Common Factor (HCF) of 18 and 24 is 6.

18 \div 6 : 24 \div 6 = 3:4

. So, any proportional ratio must be equivalent to

3:4

.

3 : ____ This is the simplest form of the ratio. So, the missing number is 4. The ratio is $3:4$ .
12 : ____ Let the missing number be $x$ . So, $3:4 :: 12:x$ . To get 12 from 3, we multiply by 4 ( $3 \times 4 = 12$ ). So, we must multiply 4 by the same factor: $x = 4 \times 4 = 16$ . The ratio is $12:16$ .
20 : ____ Let the missing number be $y$ . So, $3:4 :: 20:y$ . To get 20 from 3, the factor is $\frac{20}{3}$ . So, we must multiply 4 by the same factor: $y = 4 \times \frac{20}{3} = \frac{80}{3}$ . The ratio is $20:\frac{80}{3}$ .
27 : ____ Let the missing number be $z$ . So, $3:4 :: 27:z$ . To get 27 from 3, we multiply by 9 ( $3 \times 9 = 27$ ). So, we must multiply 4 by the same factor: $z = 4 \times 9 = 36$ . The ratio is $27:36$ .

Final Answer: The completed ratios are: 3 : 4 12 : 16 20 : $\frac{80}{3}$ 27 : 36

Q4Figure it Out 1

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

To Find: Which of the given rectangles are similar to each other.

Method: This question requires measuring the dimensions of rectangles from a figure in the textbook, which is not provided. The general method to solve this problem is as follows:

Use a ruler (scale) to measure the width (w) and the height (h) of each rectangle shown in the figure.
For each rectangle, write down the ratio of its width to its height, i.e., $w:h$ .
Reduce each ratio to its simplest form by dividing both terms by their Highest Common Factor (HCF).
Compare the simplest forms of the ratios for all the rectangles.
Rectangles that have the same ratio in their simplest form are similar to each other.

Example: If Rectangle A has width 6 cm and height 4 cm, its ratio is

6:4

, which simplifies to

3:2

. If Rectangle B has width 9 cm and height 6 cm, its ratio is

9:6

, which simplifies to

3:2

. Since both rectangles have the same simplified ratio (

3:2

), they are similar.

Q5Figure it Out 1

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

Task: To draw a smaller and a bigger rectangle that are similar to a given rectangle.

Method: This question is an activity based on a figure which is not provided. Here is the general method to complete the task:

First, measure the width (w) and height (h) of the rectangle shown in the textbook.
Calculate the ratio of its width to height, $w:h$ , and simplify it if possible.
To draw a smaller similar rectangle: Choose a scaling factor that is less than 1 (e.g., $\frac{1}{2}$ ). Multiply both the original width and height by this factor. For example, the new width would be $w \times \frac{1}{2}$ and the new height would be $h \times \frac{1}{2}$ . Draw a rectangle with these new dimensions.
To draw a bigger similar rectangle: Choose a scaling factor that is greater than 1 (e.g., 2). Multiply both the original width and height by this factor. For example, the new width would be $w \times 2$ and the new height would be $h \times 2$ . Draw a rectangle with these new dimensions.

Comparison with Classmates' Drawings:

Are they all the same? No, they will likely not be the same size.
Why are they different? They are different because each student might have chosen a different scaling factor to create their smaller and bigger rectangles. For example, one student might use a factor of $\frac{1}{2}$ for the smaller rectangle, while another might use $\frac{1}{3}$ . Both resulting rectangles would be smaller and similar to the original, but they would be different sizes from each other.
Are they wrong? No, as long as the ratio of width to height in their new rectangles is the same as the original rectangle's ratio, their drawings are correct. There are infinitely many similar rectangles that can be drawn.

Q6Figure it Out 1

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.

(a)

(b)

Solution

To Find: The ratio of grey bricks to coloured bricks in the given patterns.

Method: This question requires analyzing patterns from a figure which is not provided. The general method to solve this problem for each pattern (a) and (b) is:

Identify a repeating unit of the pattern in the brick wall.
Within one repeating unit, count the number of grey bricks.
Within the same repeating unit, count the number of coloured bricks.
Write the ratio as: (Number of grey bricks) : (Number of coloured bricks).
Simplify this ratio by dividing both numbers by their Highest Common Factor (HCF).

Example for a hypothetical pattern (a):

Suppose one repeating unit of the pattern has 6 grey bricks and 3 coloured bricks.
The ratio is $6:3$ .
The HCF of 6 and 3 is 3.
The simplified ratio is $(6 \div 3) : (3 \div 3) = 2:1$ .

Example for a hypothetical pattern (b):

Suppose one repeating unit of the pattern has 8 grey bricks and 12 coloured bricks.
The ratio is $8:12$ .
The HCF of 8 and 12 is 4.
The simplified ratio is $(8 \div 4) : (12 \div 4) = 2:3$ .

Q7Figure it Out 1

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- head : torso torso : arms torso : legs Now, draw a figure with head, torso, arms, and legs with equivalent ratios as above. Does the drawing look more realistic if the ratios are proportional? Why? Why not?

Solution

Task: This is an activity involving measuring a friend's body parts, calculating ratios, and drawing a proportional figure.

Steps for the Activity:

Measure: With a measuring tape, carefully measure the following lengths on your friend:
- Length of the head (from top of the head to the chin).
- Length of the torso (from the base of the neck to the waist).
- Length of an arm (from shoulder to fingertips).
- Length of a leg (from the hip to the sole of the foot).
Calculate Ratios: Write down the measurements and form the following ratios. For example, if head=20cm, torso=50cm, arm=60cm, leg=90cm:
- head : torso = $20:50$ , which simplifies to $2:5$ .
- torso : arms = $50:60$ , which simplifies to $5:6$ .
- torso : legs = $50:90$ , which simplifies to $5:9$ .
Draw a Proportional Figure: Now, draw a new figure where the parts have the same ratios. You can start by choosing a new length for one part, for example, the head. Let's say you draw a head that is 2 cm long.
- Since head : torso = $2:5$ , the torso in your drawing should be 5 cm long.
- Since torso : legs = $5:9$ , the legs should be $5 \times \frac{9}{5} = 9$ cm long.
- Since torso : arms = $5:6$ , the arms should be $5 \times \frac{6}{5} = 6$ cm long.
- Draw the figure using these new proportional lengths.

Discussion:

Does the drawing look more realistic if the ratios are proportional? Yes, the drawing will generally look more realistic and well-proportioned.
Why? The human body has certain standard proportions. When you draw a figure using the same proportions as a real person, even if it's scaled down, it maintains the natural look of a human form. If the ratios were not proportional (e.g., making the head too large for the torso, or the legs too short for the body), the figure would look distorted and unrealistic. Proportionality is a key principle in art for creating lifelike figures.

Q1Figure it Out 2

The Earth travels approximately 940 million kilometres around the Sun in a year. How many kilometres will it travel in a week?

Solution

Given: Distance travelled by Earth in 1 year = 940 million km =

940,000,000

km. Number of weeks in 1 year

\approx 52

weeks.

To Find: Distance travelled by Earth in 1 week.

Solution: We can find the distance travelled per week by dividing the total annual distance by the number of weeks in a year.

Distance per week =

\frac{\text{Total distance in a year}}{\text{Number of weeks in a year}}

Distance per week =

\frac{940,000,000 \text{ km}}{52 \text{ weeks}}

940,000,000 \div 52 \approx 18,076,923.0769 \text{ km/week}

Rounding to the nearest kilometre, the Earth travels approximately 18,076,923 km in a week.

Final Answer: The Earth will travel approximately 18,076,923 kilometres in a week.

Q2Figure it Out 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 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

To Find: Total number of bricks needed to build the house.

Method: This question depends on a diagram showing the layout and dimensions of the house walls, which is not provided. The general method to solve this problem is as follows:

Find the total length of the walls: From the diagram, add the lengths of all the walls that need to be built. This includes all outer walls and the inner partition wall. Total Length = (Length of Wall 1) + (Length of Wall 2) + ...
Calculate the number of bricks per foot: We are given that a 10-foot wall requires 1450 bricks. Bricks per foot = $\frac{1450 \text{ bricks}}{10 \text{ feet}} = 145$ bricks/foot.
Calculate the total bricks needed: Multiply the total length of the walls (from step 1) by the number of bricks per foot (from step 2). Total Bricks = (Total Length in feet) $\times$ (145 bricks/foot).

Example: Suppose the diagram shows outer walls of 40 ft, 30 ft, 40 ft, and 30 ft, and an inner wall of 30 ft.

Total Length = $40 + 30 + 40 + 30 + 30 = 170$ feet.
Bricks per foot = 145 bricks/foot.
Total Bricks = $170 \text{ feet} \times 145 \text{ bricks/foot} = 24,650$ bricks.

Without the specific dimensions from the diagram, a final numerical answer cannot be provided.

Q3Figure it Out 2

Puneeth’s father went from Lucknow to Kanpur in 2 hours by riding his motorcycle at a speed of $50 \mathrm{~km} / \mathrm{h}$ . If he drives at $75 \mathrm{~km} / \mathrm{h}$ , how long will it take him to reach Kanpur? Can we form this problem as a proportion - $50: 2:: 75: \_\_\_\_$ Would it take Puneeth's father more time or less time to reach Kanpur? Think about it.

Solution

Given: Initial speed (

S_1

) =

50 \text{ km/h}

. Initial time (

T_1

) = 2 hours. New speed (

S_2

) =

75 \text{ km/h}

.

To Find: New time (

T_2

) to reach Kanpur.

Solution: First, we need to find the distance between Lucknow and Kanpur. The distance remains constant. Distance = Speed

\times

Time Distance =

50 \text{ km/h} \times 2 \text{ h} = 100 \text{ km}

.

Now, we can find the new time taken to cover this distance at the new speed. Time =

\frac{\text{Distance}}{\text{Speed}}

T_2 = \frac{100 \text{ km}}{75 \text{ km/h}} = \frac{4}{3}

hours.

To convert this to hours and minutes:

\frac{4}{3}

hours = 1 and

\frac{1}{3}

hours.

\frac{1}{3}

of an hour =

\frac{1}{3} \times 60

minutes = 20 minutes. So, the new time is 1 hour and 20 minutes.

Analysis of the Proportion:

The problem asks if we can model this as $50: 2:: 75: x$ . This is a direct proportion. If we solve it, we get $50 \times x = 2 \times 75$ , so $x = \frac{150}{50} = 3$ hours. This is incorrect because as speed increases, the time taken should decrease.
This is a case of inverse proportion. As one quantity (speed) increases, the other quantity (time) decreases. The Rule of Three for direct proportion does not apply here.
Would it take more time or less time? Since the speed has increased from 50 km/h to 75 km/h, it will take less time to cover the same distance.

Final Answer: It will take Puneeth's father

\frac{4}{3}

hours, or 1 hour and 20 minutes, to reach Kanpur. The problem cannot be formed as a direct proportion

50: 2:: 75: x

because speed and time are inversely proportional. It would take him less time to reach Kanpur.

Q1Figure it Out 3

Divide ₹4,500 into two parts in the ratio $2: 3$ .

Solution

Given: Total amount = ₹4,500. Ratio =

2:3

.

To Find: The two parts of the amount.

Solution:

Find the sum of the terms in the ratio: Sum of parts = $2 + 3 = 5$ .
Calculate the value of one part by dividing the total amount by the sum of the parts: Value of one part = $\frac{₹4,500}{5} = ₹900$ .
Calculate the value of each share: First part (corresponding to 2) = $2 \times ₹900 = ₹1,800$ . Second part (corresponding to 3) = $3 \times ₹900 = ₹2,700$ .

Verification: The sum of the two parts is

₹1,800 + ₹2,700 = ₹4,500

, which is the total amount.

Final Answer: The two parts are ₹1,800 and ₹2,700.

Q2Figure it Out 3

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: Total volume of solution = 240 mL. Ratio of acid to water =

1:5

.

To Find: The volume of acid and water in the solution.

Solution:

Find the sum of the terms in the ratio: Sum of parts = $1 + 5 = 6$ .
Calculate the volume of acid: Volume of acid = $(\frac{\text{Acid's part}}{\text{Sum of parts}}) \times \text{Total volume}$ Volume of acid = $(\frac{1}{6}) \times 240 \text{ mL} = 40 \text{ mL}$ .
Calculate the volume of water: Volume of water = $(\frac{\text{Water's part}}{\text{Sum of parts}}) \times \text{Total volume}$ Volume of water = $(\frac{5}{6}) \times 240 \text{ mL} = 200 \text{ mL}$ .

Verification: The sum of the volumes is

40 \text{ mL} + 200 \text{ mL} = 240 \text{ mL}

, which is the total volume of the solution.

Final Answer: The solution contains 40 mL of acid and 200 mL of water.

Q3Figure it Out 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

Part 1: Initial Mixture

Given: Total volume of green paint = 40 mL. Ratio of blue paint to yellow paint =

3:5

.

To Find: The volume of blue and yellow paint needed.

Solution:

Sum of the ratio parts = $3 + 5 = 8$ .
Volume of blue paint = $(\frac{3}{8}) \times 40 \text{ mL} = 3 \times 5 \text{ mL} = 15 \text{ mL}$ .
Volume of yellow paint = $(\frac{5}{8}) \times 40 \text{ mL} = 5 \times 5 \text{ mL} = 25 \text{ mL}$ .

Answer for Part 1: 15 mL of blue paint and 25 mL of yellow paint are needed.

Part 2: New Mixture

Given: Initial blue paint = 15 mL. Initial yellow paint = 25 mL. Added yellow paint = 20 mL.

To Find: The new ratio of blue and yellow paint.

Solution:

The amount of blue paint remains the same: 15 mL.
Calculate the new total amount of yellow paint: New yellow paint = Initial yellow paint + Added yellow paint New yellow paint = $25 \text{ mL} + 20 \text{ mL} = 45 \text{ mL}$ .
Form the new ratio of blue to yellow paint: New Ratio = $15 : 45$ .
Simplify the new ratio by dividing both terms by their HCF, which is 15: $15 \div 15 : 45 \div 15 = 1:3$ .

Final Answer: To produce 40 mL of green paint, 15 mL of blue paint and 25 mL of yellow paint are needed. After adding 20 mL of yellow paint, the new ratio of blue to yellow is

1:3

.

Q4Figure it Out 3

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: Total mixture required = 6 cups. Ratio of rice to urad dal =

2:1

.

To Find: Number of cups of rice and urad dal needed.

Solution:

Find the sum of the ratio parts: Sum of parts = $2 + 1 = 3$ .
Calculate the amount of rice needed: Cups of rice = $(\frac{\text{Rice's part}}{\text{Sum of parts}}) \times \text{Total mixture}$ Cups of rice = $(\frac{2}{3}) \times 6 \text{ cups} = 4 \text{ cups}$ .
Calculate the amount of urad dal needed: Cups of urad dal = $(\frac{\text{Urad dal's part}}{\text{Sum of parts}}) \times \text{Total mixture}$ Cups of urad dal = $(\frac{1}{3}) \times 6 \text{ cups} = 2 \text{ cups}$ .

Verification: Total cups = 4 cups of rice + 2 cups of urad dal = 6 cups.

Final Answer: You will need 4 cups of rice and 2 cups of urad dal.

Q5Figure it Out 3

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: Initial ratio of red paint to yellow paint =

3:5

. One bucket of this mixture is taken. Another bucket of yellow paint is added.

To Find: The new ratio of red paint to yellow paint.

Let: Let the volume of one bucket be

V

.

Solution:

In the first bucket of orange paint, the total number of parts is $3+5=8$ .
- Volume of red paint = $(\frac{3}{8})V$ .
- Volume of yellow paint = $(\frac{5}{8})V$ .
Another bucket of yellow paint is added. The volume of the added yellow paint is $V$ .
Calculate the total volumes of each paint in the new mixture:
- Total red paint = $(\frac{3}{8})V$ (This does not change).
- Total yellow paint = (Initial yellow paint) + (Added yellow paint) = $(\frac{5}{8})V + V = (\frac{5}{8} + 1)V = (\frac{5+8}{8})V = (\frac{13}{8})V$ .
Form the new ratio of red paint to yellow paint: New Ratio = (Total red paint) : (Total yellow paint) New Ratio = $(\frac{3}{8})V : (\frac{13}{8})V$ .
We can cancel the common factors $V$ and $\frac{1}{8}$ from both sides of the ratio. New Ratio = $3:13$ .

Final Answer: The ratio of red paint to yellow paint in the new mixture is

3:13

.

Q1Figure it Out 4

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: Volume of orange juice = 600 mL. Volume of apple juice = 900 mL.

To Find: The ratio of orange juice to apple juice in its simplest form.

Solution:

Write the initial ratio: Ratio = $600 : 900$ .
To simplify, we can divide both terms by their greatest common divisor. First, divide by 100: $600 \div 100 : 900 \div 100 = 6:9$ .
Now, divide both terms by their HCF, which is 3: $6 \div 3 : 9 \div 3 = 2:3$ .

Final Answer: The ratio of orange juice to apple juice in its simplest form is

2:3

.

Q2Figure it Out 4

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

Given: Last year: 3 buses for 162 people. This year: 204 people (assuming 'students' here refers to the total number of people for consistency).

To Find: Number of buses needed this year and if they will be full.

Solution:

First, find the capacity of one bus. Since all 3 buses were full with 162 people: Capacity per bus = $\frac{162 \text{ people}}{3 \text{ buses}} = 54$ people/bus.
Now, calculate how many buses are needed for 204 people: Number of buses = $\frac{\text{Total people}}{\text{Capacity per bus}} = \frac{204}{54}$ . $204 \div 54 \approx 3.77$ . Since we cannot hire a fraction of a bus, we must hire the next whole number of buses, which is 4.
Determine if the buses will be full:
- The first 3 buses can hold $3 \times 54 = 162$ people. These buses will be full.
- The number of people remaining for the fourth bus is $204 - 162 = 42$ people.
- Since the fourth bus has a capacity of 54, and it will only have 42 people, it will not be full.

Final Answer: We will need 4 buses. Not all the buses will be full; three will be full and the fourth will have 42 people.

Q3Figure it Out 4

The area of Delhi is $1,484 \mathrm{sq} . \mathrm{km}$ and the area of Mumbai is $550 \mathrm{sq} . \mathrm{km}$ . 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: Delhi: Area =

1,484 \text{ sq. km}

Population = 30 million =

30,000,000

Mumbai: Area =

550 \text{ sq. km}

Population = 20 million =

20,000,000

To Find: Which city is more crowded and why.

Solution: To determine which city is more crowded, we need to calculate the population density of each city. Population density is the number of people per unit area.

Population Density =

\frac{\text{Total Population}}{\text{Total Area}}

Population Density of Delhi: Density = $\frac{30,000,000}{1,484} \approx 20,215.6$ people per sq. km.
Population Density of Mumbai: Density = $\frac{20,000,000}{550} \approx 36,363.6$ people per sq. km.

Comparison: The population density of Mumbai (approx. 36,364 people/sq. km) is significantly higher than the population density of Delhi (approx. 20,216 people/sq. km).

Final Answer: Mumbai is more crowded. We say so because it has a higher population density, meaning there are more people living in each square kilometre of area compared to Delhi.

Q4Figure it Out 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

Part 1: Crane's Neck Height

Given: Total height of the crane = 155 cm. Ratio of neck to rest of body =

4:6

.

Solution:

Sum of the ratio parts = $4 + 6 = 10$ .
Crane's neck height = $(\frac{\text{Neck's part}}{\text{Sum of parts}}) \times \text{Total height}$ Crane's neck height = $(\frac{4}{10}) \times 155 \text{ cm} = 0.4 \times 155 \text{ cm} = 62 \text{ cm}$ .

Part 2: Your Neck Height

Given: Ratio of neck to rest of body =

4:6

. Let your total height be

H

cm.

To Find: The height of your neck based on this ratio.

Solution: Using the same principle, your neck would be

\frac{4}{10}

of your total height. Your neck height =

(\frac{4}{10}) \times H = 0.4 H \text{ cm}

.

Example: If your height is 160 cm, your neck height would be:

0.4 \times 160 \text{ cm} = 64 \text{ cm}

.

Final Answer: The crane's neck is 62 cm tall. For a person of height

H

cm, if their body had the same ratio, their neck would be

0.4H

cm tall.

Q5Figure it Out 4

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?"

Solution

Given: Quantity of saffron (

Q_1

) =

2 \frac{1}{2}

palas = 2.5 palas. Cost of saffron (

P_1

) =

\frac{3}{7}

niskas. New amount of money (

P_2

) = 9 niskas.

To Find: Quantity of saffron (

Q_2

) that can be bought for 9 niskas.

Solution: The quantity of saffron is directly proportional to its cost. We can set up a proportion:

Q_1 : P_1 :: Q_2 : P_2

2.5 : \frac{3}{7} :: Q_2 : 9

Using the rule of three (cross multiplication), product of means equals product of extremes:

Q_2 \times P_1 = Q_1 \times P_2

Q_2 \times \frac{3}{7} = 2.5 \times 9

Now, solve for

Q_2

:

Q_2 \times \frac{3}{7} = 22.5

Q_2 = 22.5 \div \frac{3}{7}

Q_2 = 22.5 \times \frac{7}{3}

Q_2 = (\frac{22.5}{3}) \times 7

Q_2 = 7.5 \times 7

Q_2 = 52.5

Final Answer: 52.5 palas of saffron can be bought for 9 niskas.

Q6Figure it Out 4

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

Given: Harmain's current age = 1 year. Brother's current age = 5 years. Future ratio of their ages =

1:2

.

To Find: Harmain's age when the ratio is

1:2

.

Let: Let

x

be the number of years that pass until their ages are in the ratio

1:2

.

Solution: In

x

years:

Harmain's age will be $1 + x$ .
Her brother's age will be $5 + x$ .

At that time, the ratio of their ages will be:

\frac{\text{Harmain's age}}{\text{Brother's age}} = \frac{1+x}{5+x}

We are given that this ratio is

1:2

, so we can set up the equation:

\frac{1+x}{5+x} = \frac{1}{2}

Now, cross-multiply to solve for

x

:

2(1+x) = 1(5+x)

2 + 2x = 5 + x

2x - x = 5 - 2

x = 3

So, this will happen in 3 years.

Harmain's age at that time will be her current age +

x

:

1 + 3 = 4

years old.

Verification: In 3 years, Harmain will be 4 and her brother will be

5+3=8

. The ratio of their ages will be

4:8

, which simplifies to

1:2

. This is correct.

Final Answer: Harmain will be 4 years old.

Q7Figure it Out 4

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: For an equal volume: Ratio of Mass of Gold : Mass of Water =

37:2

. Mass of 1 litre of water = 1 kg.

To Find: Mass of 1 litre of gold.

Solution: We are given the ratio of masses for equal volumes. Let's consider the volume to be 1 litre.

Let

M_{gold}

be the mass of 1 litre of gold and

M_{water}

be the mass of 1 litre of water.

The proportion is:

\frac{M_{gold}}{M_{water}} = \frac{37}{2}

We know that

M_{water} = 1

kg. We can substitute this value into the equation:

\frac{M_{gold}}{1 \text{ kg}} = \frac{37}{2}

Now, solve for

M_{gold}

:

M_{gold} = \frac{37}{2} \times 1 \text{ kg}

M_{gold} = 18.5 \text{ kg}

Final Answer: The mass of 1 litre of gold is 18.5 kg.

Q8Figure it Out 4

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: Manure application rate = 10 tonnes per acre. Plot size = 200 ft

\times

500 ft. Unit conversion: 1 acre = 43,560 square feet.

To Find: Total amount of manure needed.

Solution:

Calculate the area of the farmer's plot in square feet: Area = Length $\times$ Width Area = $200 \text{ ft} \times 500 \text{ ft} = 100,000 \text{ sq. ft}$ .
Convert the area from square feet to acres: Area in acres = $\frac{\text{Area in sq. ft}}{\text{sq. ft per acre}}$ Area in acres = $\frac{100,000}{43,560} \approx 2.29568$ acres.
Calculate the total amount of manure needed: Total manure = (Area in acres) $\times$ (Manure rate per acre) Total manure = $2.29568 \text{ acres} \times 10 \text{ tonnes/acre}$ Total manure = $22.9568$ tonnes.

Rounding to two decimal places, the farmer needs 22.96 tonnes of manure.

Final Answer: The farmer should buy approximately 22.96 tonnes of manure.

Q9Figure it Out 4

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: Time to fill mug (

T_1

) = 15 seconds. Volume of mug (

V_1

) = 500 mL. Volume of bucket (

V_2

) = 10 litres.

To Find: Time to fill the bucket (

T_2

).

Solution:

Ensure units are consistent: The volumes are in mL and litres. Let's convert litres to mL. 1 litre = 1000 mL So, 10 litres = $10 \times 1000 = 10,000$ mL. $V_2 = 10,000$ mL.
Set up a proportion: The time taken to fill is directly proportional to the volume being filled (assuming a constant flow rate). $T_1 : V_1 :: T_2 : V_2$ $15 : 500 :: T_2 : 10,000$
Solve for $T_2$ : We can find the factor of change in volume: $\frac{V_2}{V_1} = \frac{10,000}{500} = 20$ . The time will also change by the same factor. $T_2 = T_1 \times 20 = 15 \text{ seconds} \times 20 = 300$ seconds.
Convert seconds to minutes (optional but helpful): 300 seconds = $\frac{300}{60}$ minutes = 5 minutes.

Final Answer: It takes 300 seconds (or 5 minutes) for the tap to fill the bucket.

Q10Figure it Out 4

One acre of land costs ₹15,00,000. What is the cost of 2,400 square feet of the same land?

Solution

Given: Cost of 1 acre = ₹1,500,000. Area to be priced = 2,400 sq. ft. Unit conversion: 1 acre = 43,560 sq. ft.

To Find: Cost of 2,400 sq. ft. of land.

Solution:

Find the cost per square foot: Cost per sq. ft. = $\frac{\text{Total cost of 1 acre}}{\text{Area of 1 acre in sq. ft.}}$ Cost per sq. ft. = $\frac{₹1,500,000}{43,560 \text{ sq. ft.}} \approx ₹34.43528$ per sq. ft.
Calculate the cost for 2,400 square feet: Cost = (Cost per sq. ft.) $\times$ (Area in sq. ft.) Cost = $34.43528 \times 2,400$ Cost $\approx ₹82,644.67$

Alternatively, using proportions: Let

C

be the cost. Let

A

be the area.

C_1 : A_1 :: C_2 : A_2

1,500,000 : 43,560 :: C_2 : 2,400

C_2 = \frac{1,500,000 \times 2,400}{43,560}

C_2 = \frac{3,600,000,000}{43,560} \approx 82,644.67

Final Answer: The cost of 2,400 square feet of the land is approximately ₹82,644.67.

Q11Figure it Out 4

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: Area of field = 20 acres. Oxen's ploughing rate = 6 hours per acre. Tractor is 4 times faster than oxen.

To Find: Time taken by oxen to plough 20 acres. Time taken by tractor to plough 20 acres.

Solution:

Time taken by oxen: Total time for oxen = (Oxen's rate) $\times$ (Total area) Total time for oxen = $6 \text{ hours/acre} \times 20 \text{ acres} = 120$ hours.
Time taken by tractor: Since the tractor is 4 times faster, it will take $\frac{1}{4}$ of the time that the oxen take. Total time for tractor = $\frac{\text{Time taken by oxen}}{4}$ Total time for tractor = $\frac{120 \text{ hours}}{4} = 30$ hours.

Final Answer: It would take the pair of oxen 120 hours to plough the field. It would take the tractor 30 hours to plough the same field.

Q12Figure it Out 4

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 a ₹10 coin = 7.74 grams. Ratio of Copper : Nickel =

3:1

. Cost of copper = ₹906 per kg. Cost of nickel = ₹1,341 per kg.

To Find: The cost of the metals in one ₹10 coin.

Solution:

Calculate the mass of copper and nickel in one coin:
- Sum of ratio parts = $3 + 1 = 4$ .
- Mass of copper = $(\frac{3}{4}) \times 7.74 \text{ g} = 5.805 \text{ g}$ .
- Mass of nickel = $(\frac{1}{4}) \times 7.74 \text{ g} = 1.935 \text{ g}$ .
Convert the cost of metals from per kg to per gram:
- 1 kg = 1000 g.
- Cost of copper per gram = $\frac{₹906}{1000} = ₹0.906$ per gram.
- Cost of nickel per gram = $\frac{₹1,341}{1000} = ₹1.341$ per gram.
Calculate the cost of the metals in the coin:
- Cost of copper in coin = (Mass of copper) $\times$ (Cost per gram) Cost = $5.805 \text{ g} \times ₹0.906/\text{g} = ₹5.25933$ .
- Cost of nickel in coin = (Mass of nickel) $\times$ (Cost per gram) Cost = $1.935 \text{ g} \times ₹1.341/\text{g} = ₹2.594835$ .
Calculate the total cost:
- Total cost = Cost of copper + Cost of nickel Total cost = $₹5.25933 + ₹2.594835 = ₹7.854165$ .

Rounding to two decimal places, the total cost is ₹7.85.

Final Answer: The cost of the metals in a ₹10 coin is approximately ₹7.85.

Q1Filter Coffee!

Manjunath usually mixes 15 mL of coffee decoction with 35 mL of milk to make one cup of regular filter coffee. 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.
Coffee Decoction (in mL) Milk (in mL) Regular/Strong/ Light
300 600
150 500
200 400
24 56
100 300

Coffee Decoction (in mL)	Milk (in mL)	Regular/Strong/ Light
300	600
150	500
200	400
24	56
100	300

Solution

Given: Regular coffee has a ratio of coffee decoction to milk of

15 \text{ mL} : 35 \text{ mL}

.

To Find: Whether the coffee in each case is regular, stronger, or lighter.

Solution: First, let's find the simplest form of the ratio for regular coffee. Ratio =

15:35

Dividing both terms by their HCF, which is 5, we get:

15 \div 5 : 35 \div 5 = 3:7

. As a fraction (decoction/milk), this is

\frac{3}{7} \approx 0.4286

.

A coffee is 'stronger' if the ratio of decoction to milk is greater than

3:7

. It is 'lighter' if the ratio is less than

3:7

.

Let's analyze each case:

300 mL decoction, 600 mL milk: Ratio = $300:600 = 1:2$ . As a fraction, this is $\frac{1}{2} = 0.5$ . Since $0.5 > 0.4286$ , this coffee is Stronger.
150 mL decoction, 500 mL milk: Ratio = $150:500 = 15:50 = 3:10$ . As a fraction, this is $\frac{3}{10} = 0.3$ . Since $0.3 < 0.4286$ , this coffee is Lighter.
200 mL decoction, 400 mL milk: Ratio = $200:400 = 1:2$ . As a fraction, this is $\frac{1}{2} = 0.5$ . Since $0.5 > 0.4286$ , this coffee is Stronger.
24 mL decoction, 56 mL milk: Ratio = $24:56$ . The HCF of 24 and 56 is 8. $24 \div 8 : 56 \div 8 = 3:7$ . This is the same as regular coffee. This coffee is Regular.
100 mL decoction, 300 mL milk: Ratio = $100:300 = 1:3$ . As a fraction, this is $\frac{1}{3} \approx 0.3333$ . Since $0.3333 < 0.4286$ , this coffee is Lighter.

Final Answer: The completed table is:

Coffee Decoction (in mL)	Milk (in mL)	Regular/Strong/ Light
300	600	Strong
150	500	Lighter
200	400	Strong
24	56	Regular
100	300	Lighter

Proportional Reasoning-1Class 8 Mathematics NCERT Solutions

Circle the following statements of proportion that are true.

4:7::12:214: 7:: 12: 214:7::12:21

8:3::24:68: 3:: 24: 68:3::24:6

7:12::12:77: 12:: 12: 77:12::12:7

21:6::35:1021: 6:: 35: 1021:6::35:10

12:18::28:1212: 18:: 28: 1212:18::28:12

24:8::9:324: 8:: 9: 324:8::9:3

Solution

Give 3 ratios that are proportional to 4:94: 94:9.

Solution

Fill in the missing numbers for these ratios that are proportional to 18 : 24 . 3 : ____ 12 : ____ 20 : ____ 27 : ____

Solution

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

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

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

Solution

The Earth travels approximately 940 million kilometres around the Sun in a year. How many kilometres will it travel in a week?

Solution

Solution

Solution

Divide ₹4,500 into two parts in the ratio 2:32: 32:3.

Solution

In a science lab, acid and water are mixed in the ratio of 1:51: 51:5 to make a solution. In a bottle that has 240 mL of the solution, how much acid and water does the solution contain?

Solution

Blue and yellow paints are mixed in the ratio of 3:53: 53:5 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

To make soft idlis, you need to mix rice and urad dal in the ratio of 2:12: 12:1. 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

I have one bucket of orange paint that I made by mixing red and yellow paints in the ratio of 3:53: 53:5. 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

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

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

The area of Delhi is 1,484sq.km1,484 \mathrm{sq} . \mathrm{km}1,484sq.km and the area of Mumbai is 550sq.km550 \mathrm{sq} . \mathrm{km}550sq.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

A crane of height 155 cm has its neck and the rest of its body in the ratio 4:64: 64:6. For your height, if your neck and the rest of the body also had this ratio, how tall would your neck be?

Solution

Solution

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 1:21: 21:2 ?

Solution

The mass of equal volumes of gold and water are in the ratio 37:237: 237:2. If 1 litre of water is 1 kg in mass, what is the mass of 1 litre of gold?

Solution

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

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

One acre of land costs ₹15,00,000. What is the cost of 2,400 square feet of the same land?

Solution

Solution

Solution

Solution

$4: 7:: 12: 21$

$8: 3:: 24: 6$

$7: 12:: 12: 7$

$21: 6:: 35: 10$

$12: 18:: 28: 12$

$24: 8:: 9: 3$

Give 3 ratios that are proportional to $4: 9$ .

Fill in the missing numbers for these ratios that are proportional to 18 : 24 . 3 : 12 : 20 : 27 :

Divide ₹4,500 into two parts in the ratio $2: 3$ .

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?

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?

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?

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?

The area of Delhi is $1,484 \mathrm{sq} . \mathrm{km}$ and the area of Mumbai is $550 \mathrm{sq} . \mathrm{km}$ . 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?

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?

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 $1: 2$ ?

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?