Proportional Reasoning-1 - NCERT Solutions Class 8 Maths (Ganita Prakash)

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Proportional Reasoning-1 – NCERT Solutions Class 8 Maths (Ganita Prakash) includes all the questions with solutions given in the NCERT Class 8 Maths (Ganita Prakash).

NCERT Solutions Class 8

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Proportional Reasoning – 1 – NCERT Solutions

Q.1:

Which images look similar and which ones look different?
Do images B and E look like the other three images?
Why?

Solution:

Images A, C, and D look similar, even though they have different sizes. Images B and E look different.
No, images B and E do not look like the other three images (A, C, D). Image B appears elongated, and image E appears compressed and fatter.
Images A, C, and D are similar because they all are rectangles. Image E is a square (width equals height), unlike the rectangular shapes of A, C, and D. Image B, although rectangular, does not look like images A, C, and D. It appears to be more elongated.

Q.2:

Can we observe any pattern to answer this question? Perhaps by measuring the rectangles? What makes images A, C, and D appear similar, and B and E different?

Solution:

Image	Width (in mm)	Height (in mm)
Image A	60	40
Image B	40	20
Image C	30	20
Image D	90	60
Image E	60	60

Comparison between Image A and Image C:

Image C is exactly half the width and half the height of Image A.
This means both the width and height are reduced by the same factor – they are divided by 2.
When both dimensions (width and height) are changed by the same factor using multiplication or division, the shape and look of the image remain the same.
Therefore, even though Image C is smaller, it looks similar to Image A.

Similarly compare Image A and Image D yourself

Comparison between Image A and Image B:

Image B is 20 millimetres less in both width and height compared to Image A.
In this case, the change is made by subtracting a fixed amount, not by dividing or multiplying by the same number.
Also, even though the height of B is half the height of A, the width is not half – so the change is not by the same factor.
When the width and height do not change by the same factor, the shape of the image changes.
That is why Image B looks different from Image A, even though both have the same difference in size.

Similarly compare Image A and Image E yourself

Q.3:

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:

To compare Image A (60 mm width, 40 mm height) with Image D (90 mm width, 60 mm height):

Width factor: 90 {tex}\div{/tex} 60 = 1.5
Height factor: 60 {tex}\div{/tex} 40 = 1.5

The factors are the same (1.5), confirming that the width and height of Image D change by the same factor compared to Image A, making them proportional.
Images A, C, and D look similar because their widths and heights have changed by the same factor. We say that the changes to their widths and heights are proportional.

Q.4: Circle the proportion 4 : 7 :: 12 : 21, if it is true.

Solution:

The given proportion is {tex}4: 7:: 12: 21{/tex}. This can be written as a statement of equality between two ratios: {tex}\frac{4}{7}=\frac{12}{21}{/tex}.
Step-by-Step Verification
Cross-multiplication:
The product of the extremes should be equal to the product of the means for a true…
Calculate the product of the extremes:
The product is {tex}4 \times 21=84{/tex}.
Calculate the product of the means:
The product is {tex}7 \times 12=84{/tex}.
Compare the products:
Since {tex}84=84{/tex}, the products are equal.
The proportion {tex}4: 7:: 12: 21{/tex} is true.

Q.5: Circle the proportion 8 : 3 :: 24 : 6, if it is true.

Solution: Identify the extremes and means:
In the proportion {tex}a: b:: c: d, a{/tex} and {tex}d{/tex} are the extremes, and {tex}b{/tex} and {tex}c{/tex} are the means.
Calculate the product of the extremes:
The product of the extremes is calculated as {tex}8 \times 6=48{/tex}.
Calculate the product of the means:
The product of the means is calculated as {tex}3 \times 24=72{/tex}.
The product of the extremes (48) is compared to the product of the means (72). Since {tex}48 \neq 72{/tex}, the proportion is not true.

Q.6: Circle the proportion 7 : 12 :: 12 : 7, if it is true.

Solution: A proportion is a statement that two ratios are equal. The given proportion is {tex}7: 12:: 12: 7{/tex}. This can be written as {tex}\frac{7}{12}=\frac{12}{7}{/tex}.
Step-by-Step Solution
Identify the ratios:
The first ratio is {tex}\frac{7}{12}{/tex}, and the second ratio is {tex}\frac{12}{7}{/tex}.
Check for equality:
For the proportion to be true, the two ratios must be equal.
This means {tex}\frac{7}{12}{/tex} must be equal to {tex}\frac{12}{7}{/tex}.
Perform cross-multiplication:
To check for equality, the cross-products can be compared.
Calculate the cross-products:
The first cross-product is {tex}7 \times 7=49{/tex}. The second cross-product is {tex}12 \times 12=144{/tex}.
Compare the cross-products:
Since {tex}49 \neq 144{/tex}, the two ratios are not equal.
The proportion {tex}7: 12:: 12: 7{/tex} is not true. Therefore, it should not be circled.

Q.7: Circle the proportion 21 : 6 :: 35 : 10, if it is true.

Solution: The first ratio is {tex}21: 6{/tex}. This can be written as a fraction {tex}\frac{21}{6}{/tex}.
The greatest common divisor (GCD) of 21 and 6 is 3 .
Dividing both the numerator and the denominator by 3 , the simplified ratio is {tex}\frac{21 \div 3}{6 \div 3}=\frac{7}{2}{/tex}.
Step 2: Simplify the second ratio
The second ratio is {tex}35: 10{/tex}. This can be written as a fraction {tex}\frac{35}{10}{/tex}.
The greatest common divisor (GCD) of 35 and 10 is 5 .
Dividing both the numerator and the denominator by 5 , the simplified ratio is {tex}\frac{35 \div 5}{10 \div 5}=\frac{7}{2}{/tex}.
Step 3: Compare the simplified ratios
The simplified first ratio is {tex}\frac{7}{2}{/tex}.
The simplified second ratio is {tex}\frac{7}{2}{/tex}.
Since {tex}\frac{7}{2}=\frac{7}{2}{/tex}, the two ratios are equal.
The proportion {tex}21: 6:: 35: 10{/tex} is true.

Q.8: Circle the proportion 12 : 18 :: 28 : 12, if it is true.

Solution: Express the ratios as fractions:
The proportion {tex}12: 18:: 28: 12{/tex} can be written as {tex}\frac{12}{18}=\frac{28}{12}{/tex}.
Simplify the first ratio:
The fraction {tex}\frac{12}{18}{/tex} is simplified by dividing both the numerator and the denominator by their greatest common divisor, which is 6. This results in {tex}\frac{12 \div 6}{18 \div 6}=\frac{2}{3}{/tex}.
Simplify the second ratio:
The fraction {tex}\frac{28}{12}{/tex} is simplified by dividing both the numerator and the denominator by their greatest common divisor, which is 4. This results in {tex}\frac{28 \div 4}{12 \div 4}=\frac{7}{3}{/tex}.
Compare the simplified ratios:
The simplified ratios are {tex}\frac{2}{3}{/tex} and {tex}\frac{7}{3}{/tex}.
The proportion {tex}12: 18:: 28: 12{/tex} is not true and therefore should not be circled.

Q.9: Circle the proportion {tex}24: 8:: 9: 3{/tex}, if it is true.

Solution: Simplify the first ratio:
The ratio {tex}24: 8{/tex} can be written as the fraction {tex}\frac{24}{8}{/tex}.
Simplify the second ratio:
The ratio {tex}9: 3{/tex} can be written as the fraction {tex}\frac{9}{3}{/tex}.
Compare the simplified ratios:
Since both simplified ratios are equal to {tex}\frac{3}{1}{/tex}, the proportion is true.
The proportion {tex}24: 8:: 9: 3{/tex} is true.

Q.10: Give 3 ratios that are proportional to 4 : 9.

Solution: To find ratios proportional to {tex}4: 9{/tex}, we multiply both terms by the same number:
{tex} 4 \times 2: 9 \times 2=8: 18 {/tex}
{tex} 4 \times 3: 9 \times 3=12: 27 {/tex}
{tex} 4 \times 5: 9 \times 5=20: 45 {/tex}
So, three ratios proportional to {tex}4: 9{/tex} are {tex}8: 18 ; 12: 27{/tex}; and {tex}20: 45{/tex}.

Q.11: Fill in the missing numbers for these ratios that are proportional to 18 : 24.
3 : ________, 12 : ________, 20 : ________, 27 : ________.

Solution: Simplifying {tex}18: 24{/tex}
H.C.F {tex}=6 \rightarrow 18 / 6: 24 / 6=3: 4{/tex}.
So any proportional ratio must equal {tex}3: 4{/tex}.
(i) 3 : ________ {tex}=3: 4{/tex}
(ii) 12 : ________ {tex}=3 \times 4: 4 \times 4=12: 16{/tex}.
(iii) 20 : ________ {tex}=3 \times 20 / 3: 4 \times 20 / 3=20: 80 / 3{/tex}.
(iv) 27 : ________ {tex}=3 \times 9: 4 \times 9=27: 36{/tex}.

Q.12: 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:

Rectangle	Width	Height	Ratio
A	0.5 cm	1.5 cm	{tex}0.5: 1.5=1: 3{/tex}
B	1.5 cm	1 cm	{tex}1.5: 1=3: 2{/tex}
C	4.5 cm	2 cm	{tex}4.5: 2=9: 4{/tex}
D	3.5 cm	1 cm	{tex}3.5: 1=7: 2{/tex}
E	0.5 cm	1.5 cm	{tex}0.5: 1.5=1: 3{/tex}

Since rectangles A and E have the same simplified ratio 1 : 3. So, they are similar to each other.

Q.13: 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 it yourself.

Q.14: 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:

Grey bricks : Coloured bricks = 18 : 33 = 6 : 11.
Gey bricks : Coloured bricks = 48 : 71.

Q.15: 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-

Solution: My friend’s body measurements:

Head = 22 cm
Torso (neck to hip) = 50 cm
Arms (shoulder to fingertip) = 60 cm
Legs (hip to foot) = 80 cm

Head : Torso = 22 : 50
Simplify by dividing both by 2 {tex}\rightarrow{/tex} 11 : 25.
Torso : Arms = 50 : 60
Simplify by dividing both by 10 {tex}\rightarrow{/tex} 5 : 6.
Torso : Legs = 50 : 80
Simplify by dividing both by 10 {tex}\rightarrow{/tex} 5 : 8.

So the ratios are:

Head : Torso = 11 : 25
Torso : Arms = 5 : 6
Torso : Legs = 5 : 8

Q.16: The Earth travels approximately 940 million kilometres around the Sun in a year. How many kilometres will it travel in a week?

Solution: We are told that the Earth travels 940 million kilometres around the Sun in one year.
Now we want to find out how far it travels in one week.
Step 1: How many weeks are there in a year?
There are 52 weeks in a year.
Step 2: Divide the total distance by the number of weeks
We divide 940 million kilometres by 52:
{tex} \frac{940, { 000,000} \ {km}}{52}{/tex} {tex}=18,076,923.08 \text { million } {km} \text { (approx) } {/tex}
The Earth travels about 18,076,923.08 million kilometres in one week.

Q.17: 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:

Perimeter of the house {tex}=(9+15+12+9+9+6+9+9+12)=90 {ft}{/tex}
Length of inner wall {tex}=(12+6) {ft}=18 {ft}{/tex}
{tex}\therefore{/tex} Total length of wall required to build {tex}=(90+2 \times 18) {ft}=126 {ft}{/tex}
To build a wall of 10 -feet, he requires approximately 1450 bricks.
Let be the total number of bricks required {tex}={x}{/tex}
Therefore, {tex}10: 126:: 1450: x{/tex}
Or, {tex}10 / 126=1450 / x{/tex}
Or, {tex}x=145 \times 126{/tex}
Or, {tex}x=18270{/tex}
Hence, 18270 bricks would he need to build the house.

Q.18: Divide ₹4,500 into two parts in the ratio 2 : 3.

Solution: To divide ₹4,500 in the ratio {tex}2: 3{/tex}, follow these simple steps:
Step 1: Add the parts of the ratio
{tex} 2+3=5 \text { parts } {/tex}
Step 2: Find the value of one part
{tex} \frac{\text { ₹4, } 500}{5}=\text { ₹ } 900 {/tex}
Step 3: Multiply to find each part
First part:
{tex} 2 \times \text { ₹ } 900=\text { ₹ } 1,800 {/tex}
Second part:
{tex} 3 \times \text { ₹ } 900=\text { ₹ } 2,700 {/tex}
First part = ₹1,800
Second part = ₹2,700

Q.19: 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: Step 1: Add the parts of the ratio
{tex} 1+5=6 \text { parts } {/tex}
Step 2: Find the value of 1 part
{tex} \frac{240 {~mL}}{6}=40 {~mL} {/tex}
Step 3: Find the amount of acid and water
Acid {tex}=1{/tex} part
{tex}= 1 \times 40=40 {~mL} {/tex}
Water {tex}=5{/tex} parts
{tex}= 5 \times 40=200 {~mL} {/tex}
Acid {tex}=40 {~mL}{/tex}
Water {tex}=200 {~mL}{/tex}

Q.20: 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: The volume represented by one part is determined by dividing the total volume of green paint by the total number of parts:
Step 1: Add the parts of the ratio:
{tex} 3+5=8 \text { parts } {/tex}
Step 2: Find the value of one part:
{tex} \frac{40 {~mL}}{8}=5 {~mL} \text { per part } {/tex}
Step 3: Calculate amounts of blue and yellow:
Blue paint {tex}=3 \times 5=15 {~mL}{/tex}
Yellow paint {tex}=5 \times 5=25 {~mL}{/tex}
An additional 20 ml of yellow paint is added, resulting in a new total of yellow paint:
Step 1: Total yellow paint now {tex}={/tex} original {tex}25 {~mL}+20 {~mL}=45 {~mL}{/tex}
Step 2: Blue paint is still 15 mL
Step 3: Find the new ratio of blue to yellow:
{tex} \text { Blue : Yellow }=15: 45 {/tex}
Simplify the ratio by dividing both numbers by 15 :
{tex} 15 \div 15: 45 \div 15=1: 3 {/tex}
To make 40 mL of green paint:
Blue {tex}=15 {~mL}{/tex}
Yellow {tex}=25 {~mL}{/tex}
After adding 20 mL yellow, the new ratio of blue to yellow paint is {tex}1: 3{/tex}.

Q.21: 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: Step 1: Add the parts of the ratio
{tex} 2+1=3 \text { parts } {/tex}
Step 2: Find the value of one part
{tex} \frac{6 {cups}}{3}=2 {cups} \text { per part } {/tex}
Step 3: Calculate how much rice and urad dal are needed
Rice {tex}=2 \times 2=4{/tex} cups
Urad dal {tex}=1 \times 2=2{/tex} cups
Final answer:
You will need 4 cups of rice and 2 cups of urad dal to make 6 cups of the mixture.

Q.22: 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:
Original orange paint: red and yellow mixed in ratio {tex}3: 5{/tex}
Then, another bucket of yellow paint is added (same amount as one bucket, I assume)
Step 1: Let’s assume the bucket size is 1 unit (for easy calculation).
Red paint {tex}=3{/tex} parts
Yellow paint = 5 parts
So total paint in the bucket {tex}=3+5=8{/tex} parts
Step 2: After adding 1 bucket of yellow paint, the yellow paint increases by 8 parts (since one bucket = total original mixture).
New yellow paint = original 5 parts +8 parts {tex}=13{/tex} parts
Red paint remains {tex}=3{/tex} parts
Step 3: New ratio of red to yellow paint is:
{tex} 3: 13 {/tex}
The new ratio of red paint to yellow paint is {tex}3: 13{/tex}.

Q.23: 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:
Orange juice {tex}=600 {~mL}{/tex}
Apple juice {tex}=900 {~mL}{/tex}
Step 1: Write the ratio
{tex} 600: 900 {/tex}
Step 2: Simplify the ratio
Divide both numbers by their greatest common divisor (GCD), which is 300:
{tex} \frac{600}{300}: \frac{900}{300}=2: 3 {/tex}
The ratio of orange juice to apple juice is {tex}2: 3{/tex}.

Q.24: 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: Last year:
3 buses carried 162 people
So, people per bus:
{tex} \frac{162}{3}=54 \text { people per bus } {/tex}
This year:
We have 204 students
If each bus still holds 54 people, then:
{tex} \frac{204}{54}=3.78 {/tex}
This means we need 4 buses (since 3 buses won’t be enough).
Let’s check how many people will be in each of the 4 buses:
{tex}4 \times 54=216{/tex} people can fit, but we only have 204 students.
So:
{tex} 216-204=12 \text { empty seats } {/tex}
You will need 4 buses
No, not all buses will be full – there will be 12 empty seats.

Q.25: 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: Step 1: Use the formula for population density
{tex} \text { Population Density }=\frac{\text { Population }}{\text { Area }} {/tex}
Delhi:
Population {tex}=30{/tex} million
Area {tex}=1,484 {sq} . {km}{/tex}
{tex} \text { Density of Delhi }=\frac{30,000,000}{1,484} \approx 20,215 \text { people per sq. } {km} {/tex}
Mumbai:
Population = 20 million
Area {tex}=550{/tex} sq. km
{tex} \text { Density of Mumbai }=\frac{20,000,000}{550} \approx 36,363 \text { people per sq. } {km} {/tex}
Final Answer:
Mumbai is more crowded because it has more people living in each square kilometre than Delhi.

Q.26: 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: Ratio of neck and the rest of body {tex}=4: 6{/tex}
Height of the crane is {tex}=155 {~cm}{/tex}
{tex}\therefore{/tex} Height of the neck {tex}=4 /(4+6) \times 155=4 / 10 \times 155=62 {~cm}{/tex}

Q.27: 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 {tex}2 \frac{1}{2}{/tex} palas of saffron costs {tex}\frac{3}{7}{/tex} niskas, O expert businessman! tell me quickly what quantity of saffron can be bought for 9 niskas?”

Solution: Given:
{tex}2 \frac{1}{2}=\frac{5}{2}{/tex} palas of saffron costs {tex}\frac{3}{7}{/tex} niskas.
You want to find how much saffron (in palas) can be bought for 9 niskas.
Step 1: Find cost of 1 pala of saffron
If {tex}\frac{5}{2}{/tex} palas cost {tex}\frac{3}{7}{/tex} niskas, then:
Cost of 1 pala {tex}=\frac{\frac{3}{7}}{\frac{5}{2}}=\frac{3}{7} \times \frac{2}{5}=\frac{6}{35}{/tex} niskas
Step 2: Now, find how many palas can be bought for 9 niskas
If 1 pala costs {tex}\frac{6}{35}{/tex} niskas, then:
{tex} \text { Palas for } 9 \text { niskas }=\frac{9}{\frac{6}{35}}=9 \times \frac{35}{6}=\frac{315}{6}=52.5 \text { palas } {/tex}
You can buy 52.5 palas of saffron for 9 niskas.

Q.28: 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: Present age of the old girl = 1 year and her brother’sage = 5 years
Let’s be, the girl’s age after {tex}x{/tex} years {tex}=(1+x){/tex} years and her brother’s age {tex}=(5+x){/tex} years.
Therefore, {tex}(1+x) /(5+x)=1 / 2{/tex}
Or, {tex}2+2 x=5+x{/tex}
Or, {tex}x=3{/tex}
After 3 years the ratio of her age to her brother’s age is {tex}1: 2{/tex}.
In that time the age of the girl is {tex}=(1+3)=4{/tex} years
And the age of her brother {tex}=(5+3)=8{/tex} years.

Q.29: 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:
Mass of equal volumes of gold and water is in the ratio {tex}37: 2{/tex}
Mass of 1 litre of water {tex}=1 {~kg}{/tex}
We are asked to find the mass of 1 litre of gold
Step 1: Use the ratio
{tex} \text { Gold : Water }=37: 2 {/tex}
If 2 parts of mass {tex}=1 {~kg}{/tex} (water), then:
{tex} 1 \text { part }=\frac{1}{2} {~kg} {/tex}
So, 37 parts (gold)
{tex}= 37 \times \frac{1}{2}=\frac{37}{2}=18.5 {~kg} {/tex}
The mass of 1 litre of gold is 18.5 kg.

Q.30: 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?

Solution: Given:
Manure needed {tex}=10{/tex} tonnes per acre
Plot size {tex}=200 {ft}{/tex} by 500 ft
Need to find: How much manure for the plot?
Step 1: Calculate the area of the plot in square feet
{tex} 200 \times 500=100,000 \text { sq. } {ft} {/tex}
Step 2: Convert the area from square feet to acres
1 acre {tex}=43,560{/tex} sq. ft
{tex} \text { Area in acres }=\frac{100,000}{43,560} \approx 2.2957 \text { acres } {/tex}
Step 3: Calculate manure needed
{tex} \text { Manure }=10 \text { tonnes per acre }{/tex} {tex} \times 2.2957 \text { acres }=22.957 \text { tonnes } {/tex}
The farmer should buy approximately 23 tonnes of cow manure for his plot.

Q.31: 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 {tex}=15{/tex} seconds
Volume of mug {tex}=500 {~mL}{/tex}
Volume of bucket {tex}=10{/tex} litres {tex}=10,000 {~mL}{/tex} (since 1 litre {tex}=1000 {~mL}{/tex} )
Step 1: Find the rate of water flow (volume per second)
{tex} \text { Rate }=\frac{500 {~mL}}{15 \text { seconds }}=\frac{500}{15} \approx 33.33 {~mL} / \text { second } {/tex}
Step 2: Find time to fill the bucket
{tex} \text { Time }=\frac{10,000 {~mL}}{33.33 {~mL} / \text { second }}=300 \text { seconds } {/tex}
Step 3: Convert seconds to minutes
{tex} 300 \text { seconds }=\frac{300}{60}=5 \text { minutes } {/tex}
The tap will take 5 minutes to fill the 10-litre bucket.

Q.32: One acre of land costs ₹15,00,000. What is the cost of 2,400 square feet of the same land?

Solution: We know, 1 acre = 43560 sq ft
One acre of land costs ₹15,00,000, i.e, 43560 sq ft of land costs ₹15,00,000
{tex} \therefore 2,400 \text { square feet of land costs }{/tex} {tex}=\frac{1500000}{43560} \times 2400 \approx ₹ 82645 {/tex}

Q.33: 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:
Pair of oxen takes 6 hours to plough 1 acre
Field size {tex}=20{/tex} acres
Tractor is {tex}{4}{/tex} times faster than oxen
Step 1: Time taken by oxen to plough the whole field
{tex} 6 \text { hours per acre } \times 20 \text { acres }=120 \text { hours } {/tex}
Step 2: Since tractor is 4 times faster, time taken by tractor is
{tex} \frac{120 \text { hours }}{4}=30 \text { hours } {/tex}
Time taken by pair of oxen {tex}=120{/tex} hours
Time taken by tractor {tex}=30{/tex} hours

Q.34: 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:
Copper : Nickel ratio = 3:1
Mass of the coin {tex}=7.74{/tex} grams
Cost of copper = ₹906 per kg
Cost of nickel = ₹1,341 per kg
Step 1: Find the total parts in the ratio
{tex} 3+1-4 \text { parts } {/tex}
Step 2: Find the mass of copper and nickel in the coin
Mass of copper {tex}=\frac{3}{4} \times 7.74-5.805{/tex} grams
Mass of nickel {tex}=\frac{1}{4} \times 7.74-1.935{/tex} grams
Step 3: Convert grams to kilograms (since prices are per kg)
{tex} 1 \text { gram }=\frac{1}{1000} {~kg} {/tex}
Mass of copper {tex}=5.805 \times \frac{1}{1000}-0.005805 {~kg}{/tex}
Mass of nickel {tex}=1.935 \times \frac{1}{1000}-0.001935 {~kg}{/tex}
Step 4: Calculate the cost of copper and nickel in the coin
Cost of copper {tex}=0.005805 \times 906-₹ 5.26{/tex} (approx)
Cost of nickel {tex}=0.001935 \times 1341-{/tex} ₹ 2.59 (approx)
Step 5: Calculate total cost of metals in the coin
{tex} \text { ₹5. } 26 \text { + ₹2.59 – ₹7.85 } {/tex}
The cost of the copper and nickel metals in a ₹10 coin is approximately ₹7.85.

Q.35: By what factor should we multiply the ratio 60 : 40 (image A) to get 90 : 60 (image D)?

Solution: Do it yourself.

Q.36: What is the HCF of 72 and 96?

Solution: The prime factorization of 72 is determined:
{tex} 72=2 \times 2 \times 2 \times 3 \times 3=2^3 \times 3^2 \text {. } {/tex}
The prime factorization of 96 is determined:
{tex} 96=2 \times 2 \times 2 \times 2 \times 2 \times 3=2^5 \times 3^1 . {/tex}
The common prime factors are identified, taking the lowest power of each common prime factor:

The common prime factor 2 has a lowest power of {tex}2^3{/tex}.
The common prime factor 3 has a lowest power of {tex}3^1{/tex}.

The HCF is calculated by multiplying these common prime factors with their lowest powers: {tex}{HCF}=2^3 \times 3^1=8 \times 3=24{/tex}.
The HCF of 72 and 96 is 24.

Q.37: To make the lemonade with the same sweetness, how many spoons of sugar should she add?

Solution: To maintain the same sweetness, the ratio of the number of glasses of lemonade to the number of spoons of sugar should be proportional. For 6 glasses of lemonade, she added 10 spoons of sugar. The ratio of glasses of lemonade to spoons of sugar is 6 : 10. If she needs to make 18 more glasses of lemonade, how many spoons of sugar should she use? We can model this problem as – 6 : 10 :: 18 : ? We know that each term in the ratio must change by the same factor, for the ratios to be proportional.

Q.38: How can we find the factor of change in the ratio?

Solution: The first term has increased from 6 to 18. To find the factor of change, we can divide 18 by 6 to get 3. Proportional Reasoning-1 163 The second term should also change by the same factor. When 10 increases by a factor of 3, it becomes 30. Thus, 6 : 10 :: 18 : 30. So, she should use 30 spoons of sugar to make 18 glasses of lemonade with the same sweetness as earlier.

Q.39: Compare the rectangle you have drawn to those drawn by your classmates. Do they all look the same?

Solution: Do it yourself.

Q.40: What factor should we multiply 14 by to get 6? Can it be an integer? Or should it be a fraction?

Solution: To find the factor you multiply 14 by to get 6 , you solve:
{tex} 14 \times ?=6 {/tex}
So,
{tex} ?-\frac{6}{14}=\frac{3}{7} {/tex}
Answer:
The factor is {tex}\frac{3}{7}{/tex}, which is a fraction, not an integer.
So, you need to multiply 14 by a fraction {tex}\frac{3}{7}{/tex} to get 6 . It cannot be an integer because 6 is smaller than 14.

Q.41: Why is this coffee stronger?

Solution: And when they want ‘lighter’ filter coffee, he mixes 10 mL of coffee and 40 mL of milk, making the ratio 10 : 40.

Q.42: Why is this coffee lighter?

Solution: Do it yourself.

Q.43: Does the drawing look more realistic if the ratios are proportional? Why? Why not?

Solution: Do it yourself.

Q.44: Puneeth’s father went from Lucknow to Kanpur in 2 hours by riding his motorcycle at a speed of 50 km/h. If he drives at 75 km/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?

Solution: Given:
Speed {tex}=50 {~km} / {h} \rightarrow{/tex} Time {tex}=2{/tex} hours
New Speed {tex}=75 {~km} / {h} \rightarrow{/tex} Time = ?
We are asked:
Can we form a proportion like:
{tex} 50: 2:: 75:{/tex} ________?
And will it take more time or less time?
Step 1: Understand the relationship
Speed and time are inversely proportional (if distance is the same):
More speed {tex}\Rightarrow{/tex} Less time
Step 2: Form the correct proportion
Since speed and time are inversely related, we should form the proportion as:
{tex} 50: 75=x: 2 {/tex}
Now solve for x (the new time):
{tex} \frac{50}{75}-\frac{x}{2} {/tex}
Cross-multiply:
{tex} 50 \times 2=75 \times x \rightarrow 100=75 x \rightarrow x=\frac{100}{75}{/tex} {tex}=\frac{4}{3}=1 \frac{1}{3} \text { hours } {/tex}
Final Answers:
The correct proportion is:
{tex} 50: 75=x: 2 {/tex}
(Not {tex}50: 2:: 75:{/tex} ________ because that assumes direct proportion.)
Time taken at {tex}75 {~km} / {h}=1{/tex} hour 20 minutes ( {tex}11 / 3{/tex} hours)
It would take less time to reach Kanpur at the faster speed.