Exercise 7.4

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NCERT Solutions for Class 8 Maths Chapter 7 Proportional Reasoning 1 - Exercise 7.4

Mastering complex motion and flow problems is simple with our NCERT Solutions for Class 8 Maths Ch 7: Proportional Reasoning – Exercise 7.4. This exercise focuses on real-world applications of inverse proportion, specifically looking at how speed affects travel time and how the number of pipes affects the time taken to fill a tank.

Class 8 Math Lessons (Ch 7): Learn to Solve Speed and Flow Problems with Ease! The solutions to Exercise 7.4 will be presented in a step-by-step approach to help students navigate the inverse relationship between rate and time. These solutions follow CBSE Guidelines to help you master competitive exam problems involving relative speed and filling rates.

1.0Download NCERT Solutions for Class 8 Maths Chapter 7 Ex 7.4 : Free PDF

Our easy-to-follow NCERT Solutions for Class 8 Maths Chapter 7 Exercise 7.4 are available for download as a PDF. Perfect for offline study and quick reference.

NCERT Solutions Class 8 Maths Chapter 7 Ex 7.4

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2.0Detailed NCERT Class 8 Maths Chapter 7 Solutions of Exercise 7.4

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.

Sol. Quantity of orange juice $= 600 mL$ Quantity of apple juice $= 900 mL$ ∴ Ratio of orange juice to apple juice $= 600 : 900$ ∴ Ratio in the simplest form $= 600 : 900 = 2 : 3$

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?

Sol. Number of buses for 162 students and teachers $= 3$ Since the buses were full, the capacity of 1 bus $= \frac{162}{3} = 54$ ∴ Ratio of number of seats to the number of buses is $54 : 1$ . We have

$54 : 1 = 2 (54) : 2 (1) = 108 : 2 54 : 1 = 3 (54) : 3 (1) = 162 : 3 54 : 1 = 4 (54) : 4 (1) = 216 : 4$

∴ Capacity of 4 buses $= 216$ ∴ For 204 students, we shall need 4 buses. Since $216 - 204 = 12$ , we have 12 vacant seats in the buses. 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?

Sol. Area of Delhi $= 1, 484 sq . km$ Population of Delhi $= 30$ million Area of Mumbai $= 550 sq . km$ Population of Mumbai $= 20$ million $∴$ Ratio of area to population for Delhi $= 1484 : 30$ $∴$ Ratio of area to population for Mumbai $= 550 : 20$ Factor of change of area $= \frac{550}{1484} = 0.371$ (nearly) Factor of change of population $= \frac{20}{30} = 0.667$ (nearly) Since $0.667 > 0.371$ , Mumbai is more crowded than Delhi. Alternative Method: Ratio of area to population for Delhi = 1484 : 30 Let the density of Delhi and Mumbai be the same, and there be x people in Mumbai. ∴ The ratios 1,484:30 and $550 : x$ are in proportion. $∴ \frac{1484}{30} = \frac{550}{x}$ $\Rightarrow 1484 x = 30 \times 550 = 16, 500$ $\Rightarrow x = \frac{16500}{1484} = 11.118$

There should be 11.118 million people in Mumbai. But the population of Mumbai is 20 million. $∴$ Mumbai is more crowded than Delhi. 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?

Sol. The ratio of the height of the neck and the height of the rest of the body of a crane is $4 : 6$ . My height is 65 inches, i.e., 165 cm . Let the ratio of the height of my neck and the height of the rest of my body also be $4 : 6$ . ∴ Height of my neck $= (\frac{4}{4 + 6} \times 165) cm = 66 cm$ 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 0 expert businessman! Tell me quickly, what quantity of saffron can be bought for 9 niskas?"

Sol. Here, the unit of weight in palas and the unit of money are niskas. Cost of $2 \frac{1}{2}$ palas of saffron $= \frac{3}{7}$ niskas Ratio of weight to price is $2 \frac{1}{2} : \frac{3}{7}$ or $\frac{5}{2} : \frac{3}{7}$ or $35 : 6$ Let x palas of saffron be bought for 9 niskas. ∴ Ratio of weight to price is $x : 9$ These ratios are in proportion. ∴ 35:6:: $x : 9$ $\Rightarrow \frac{35}{6} = \frac{x}{9}$ $\Rightarrow 6 x = 35 \times 9$ $\Rightarrow x = 52.5$ $∴ 52.5$ palas of saffron can be bought for 9 niskas. 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 $1 : 2$ ?

Sol. The ages of Harmain and her brother are 1 year and 5 years. Let x years be the time until the ratio of their ages is $1 : 2$ . Age of Harmain after x years $= (1 + x)$ years. Age of her brother after x years $= (5 + x)$ years. ∴ After x years, ratio of their ages $= 1 + x : 5 + x$ The ratios are in proportion. ∴ 1:2::1 + x:5 + x $\Rightarrow \frac{1}{2} = \frac{1 + x}{5 + x}$ $\Rightarrow 5 + x = 2 (1 + x)$ $\Rightarrow 5 + x = 2 + 2 x$ $\Rightarrow 2 x - x = 5 - 2$ $\Rightarrow x = 3$ ∴ After 3 years, the age of Harmain $= 1 + 3 = 4$ years. Verification: After 3 years, age of her brother $= 5 + 3 = 8$ years Also, the ratio of their ages $= 4 : 8 = 1 : 2$ . 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?

Sol. The ratio of masses of gold and water, when their volumes are the same, is $37 : 2$ . Mass of 1 litre of water $= 1 kg$ Let the mass of 1 litre of gold $= x kg$ . ∴ With equal volumes, the ratio of masses of gold and water is $x : 1$ . These ratios are in proportion. $37 : 2 ::$ x : 1 $\Rightarrow \frac{37}{2} = \frac{x}{1}$ $\Rightarrow x = \frac{37}{2}$ Thus, the mass of 1 litre of gold is $\frac{37}{2} kg$ . 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).

Sol. We have 1 ton $= 1, 000 kg$ . $∴ 10$ tonnes $= 10 \times 1, 000 = 10, 000 kg$ Also, 1 acre $= 43, 560$ sq. ft. ∴ Ratio of cow manure to area of land in kg and sq. ft. $= 10, 000 : 43, 560$ Size of plot $= 200 ft$ . by 500 ft . ∴ Area of plot $= 200 \times 500 = 1, 00, 000 sq . ft$ . Let cow manure be x kg . ∴ Ratio of cow manure to area of plot $= x : 1, 00, 000$ These ratios are in proportion. $10, 000 : 43, 560 :: x : 1, 00, 000$ $\Rightarrow \frac{10000}{43560} = \frac{x}{100000}$ $\Rightarrow 43, 560 x = 10, 000 \times 1, 00, 000 = 1, 00, 00, 00, 000$ $\Rightarrow x = \frac{1000000000}{43560}$ $\Rightarrow x = 22956.84$ $∴$ Required cow manure $= 22956.84 kg = 22.95684$ tonnes. 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?

Sol. Time taken by the tap for 500 mL of water $= 15$ seconds ∴ Ratio of volume to time $= 500 : 15$ We know 1 litre $= 1, 000 mL$ 10 litres $= 10 \times 1, 000 = 10, 000 mL$ Let the time taken to fill a bucket of $10, 000 mL$ be x seconds. ∴ Ratio of volume to time $= 10, 000 : x$ These ratios are proportional.

$500 : 15 :: 10, 000 : x$

$\Rightarrow \frac{500}{15} = \frac{10000}{x}$ $\Rightarrow 500 x = 150, 000$ $\Rightarrow x = 300$ ∴ Time to fill bucket $= 300$ seconds

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

Sol. We know that 1 acre $= 43, 560$ square feet. ∴ Cost of 43,560 sq. ft. land $= ₹15, 00, 000$ ∴ Ratio of area of land to cost $= 43, 560 : 15, 00, 000$ Let the cost of $2, 400 sq . ft$ . of land be ₹x. ∴ Ratio of area of land to cost $= 2, 400 : x$ These ratios are proportional. $∴ 43, 560 : 15, 00, 000 :: 2, 400 : x$ $\Rightarrow \frac{43560}{1500000} = \frac{2400}{x}$ $\Rightarrow 43, 560 x = 2, 400 \times 15, 00, 000$ $\Rightarrow x = \frac{2400 \times 1500000}{43560}$ $\Rightarrow x = 82, 664.63$ ∴ Cost of land $= ₹82, 664.63$ . 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?

Sol. Ratio of efficiency of a tractor to a pair of oxen $= 4 : 1$ Time taken by a pair of oxen to plough 1 acre of field $= 6$ hours ∴ Time taken by a tractor to plough 1 acre field $= \frac{6}{4} = 1.5$ hours ∴ Time taken by a pair of oxen to plough 20 20-acre field $= 20 \times 6 = 120$ hours ∴ Time taken by a tractor to plough a 20 acre field $= 20 \times 1.5 = 30$ hours 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 ₹10coin?

Sol. Ratio of copper and nickel in ₹10 coin $= 3 : 1$ Mass of one ₹10 coin = 7.74 grams ∴ Mass of copper in one $₹10 coin = \frac{3}{3 + 1} \times 7.74 = \frac{3}{4} \times 7.74 = 5.805$ grams Mass of nickel in one $₹10$ coin $= \frac{1}{3 + 1} \times 7.74 = \frac{1}{4} \times 7.74 = 1.935$ grams Cost of 1 kg copper $= ₹906$ ∴ Cost of 1000 grams of copper $= ₹906$ ∴ Cost of 5.805 grams copper $\frac{906}{1000} \times 5.805 = ₹5.26$ Cost of $1 kg = ₹1341$ ∴ Cost of 1000 grams of nickel $= ₹1341$ ∴ Cost of $1.935 \frac{1341}{1000} \times 1.935 = ₹2.59$ ∴ In one ₹10 coin, the cost of copper and the cost of nickel are respectively ₹5.26 and ₹2.59.

3.0Key Concepts of Chapter 7 Proportional Reasoning 1 Exercise 7.4

Time and Speed Relationship: For a constant distance, speed and time are inversely proportional.
Pipe and Cistern Logic: The number of pipes used to fill a tank is inversely proportional to the time taken to fill it. If more pipes are open, the tank fills faster.
Rate of Work: Understanding that "Rate" is the amount of work done per unit of time.

4.0NCERT Solutions for Class 8 Maths Chapter 7 Proportional Reasoning 1 : All Exercises

NCERT Solutions for Class 8 Maths Chapter 7 Proportional Reasoning 1 - Exercise 7.1

NCERT Solutions for Class 8 Maths Chapter 7 Proportional Reasoning 1 - Exercise 7.2

NCERT Solutions for Class 8 Maths Chapter 7 Proportional Reasoning 1 - Exercise 7.3

NCERT Solutions for Class 8 Maths Chapter 7 Proportional Reasoning 1 - Exercise 7.4

5.0Benefits of NCERT Solutions for Class 8 Maths Chapter 7 Exercise 7.4

Conceptual Clarity: Helps students visualize "Flow Rates" and "Travel Efficiency."
Logical Structuring: Emphasizes converting units (like hours to minutes) before applying formulas.
Exam Readiness: Prepares students for the logic used in physics problems regarding velocity and time.

NCERT Class 8 Maths Ch. 7 Proportional Reasoning Other Exercises:-

Exercise 7.1

Exercise 7.2

Exercise 7.3

Exercise 7.4

NCERT Solutions Class 8 Maths All Chapters:-

Chapter 1 - A Square and a Cube

Chapter 2 - Power play

Chapter 3 - A story of Numbers

Chapter 4 - Quadrilaterals

Chapter 5 - Number Play

Chapter 6 - We distribute, yet things multiply

Chapter 7 - Proportional Reasoning