RS Aggarwal Class 8 Mathematics Solutions Chapter 12 Direct and Inverse Proportions

 

Exercise 12A

 

Question 1. Check whether the following are in direct proportion or not.
(i) \( x = 6, 9, 15, 24 \) and \( y = 4, 6, 10, 16 \)
(ii) \( x = 2.5, 4, 7.5, 10 \) and \( y = 10, 16, 30, 40 \)
(iii) \( x = 5, 7, 9, 11 \) and \( y = 15, 21, 27, 33 \)
Answer:
(i) When we examine the ratio \( \frac{x}{y} \), we get \( \frac{6}{4} = \frac{9}{6} = \frac{15}{10} = \frac{24}{16} = \frac{3}{2} \) (constant). This shows that x and y maintain the same proportion throughout.
(ii) When we find the ratio \( \frac{x}{y} \), we get \( \frac{2.5}{10} = \frac{4}{16} = \frac{7.5}{30} = \frac{10}{40} = \frac{1}{4} \), while \( \frac{14}{42} = \frac{1}{3} \). Since the ratio is not the same, x and y are not proportional.
(iii) When we find the ratio \( \frac{x}{y} \), we get \( \frac{5}{15} = \frac{7}{21} = \frac{9}{27} = \frac{11}{33} = \frac{1}{3} \), while \( \frac{15}{60} = \frac{1}{4} \). Since the ratio is not the same, x and y are not proportional.
In simple words: Two quantities are in direct proportion when dividing one by the other always gives the same number. Check this by creating fractions for each pair and seeing if they are all equal.

Exam Tip: Always simplify each ratio completely and compare — if even one ratio differs, the quantities are not in direct proportion. Keep your work organized by listing all ratios side by side.

 

Question 2. If x and y are directly proportional, find the missing values.
\( \frac{3}{72} = \frac{x_1}{120} = \frac{x_2}{192} = \frac{10}{y_1} \)
Answer: Given that x and y are directly proportional, we use the constant ratio.
\( \frac{3}{72} = \frac{x_1}{120} \)
\( \implies x_1 = \frac{120 \times 3}{72} = 5 \)

\( \frac{3}{72} = \frac{x_2}{192} \)
\( \implies x_2 = \frac{192 \times 3}{72} = 8 \)

\( \frac{3}{72} = \frac{10}{y_1} \)
\( \implies y_1 = \frac{72 \times 10}{3} = 240 \)

So, \( x_1 = 5 \), \( x_2 = 8 \), and \( y_1 = 240 \).
In simple words: In direct proportion, the ratio stays the same. Set up the known ratio equal to each unknown and solve by cross-multiplying.

Exam Tip: Cross-multiply carefully and double-check your arithmetic — a single calculation error will give you the wrong answer for all missing values.

 

Question 3. A car travels 300 km on 34 litres of petrol. How far will it travel on 20 litres of petrol?
Answer: Let the required distance be x km. We set up a table of the quantities:

When fuel consumption increases, the distance that can be covered also increases. This is a direct proportion situation.
\( \frac{34}{510} = \frac{20}{x} \)
\( \implies \frac{1}{15} = \frac{20}{x} \)
\( \implies x \times 1 = 20 \times 15 = 300 \)

So, the required distance is 300 km.
In simple words: More petrol means the car travels farther. Divide the distance by the petrol amount to find how far one litre takes you, then multiply by the new amount of petrol.

Exam Tip: Always identify which type of proportion applies (direct or inverse) before setting up your equation — reading the problem carefully tells you whether quantities increase or decrease together.

 

Question 4. A car travels 150 km in a certain time. A taxi travels 124 km in the same time. If the taxi charges Rs 1275, how much should the car charge for travelling 150 km?
Answer: Let the required charge be Rs x. We set up a table:

When distance decreases, the fare should also decrease. This represents a direct proportion.
\( \frac{150}{1275} = \frac{124}{x} \)
\( \implies \frac{2}{17} = \frac{124}{x} \)
\( \implies (2 \times x) = (124 \times 17) \)
\( \implies x = \frac{124 \times 17}{2} \)
\( \implies x = 62 \times 17 = 1054 \)

So, the required charge is Rs 1,054.
In simple words: Shorter distances mean lower fares when the rate per km is fixed. Set up the ratio of distance to fare and cross-multiply to find the missing charge.

Exam Tip: Convert your final answer to a clear monetary value — double-check your simplification of the ratio to avoid calculation mistakes.

 

Question 5. A car travels 50 km in 60 minutes. How far will it travel in 5 hours?
Answer: Let the required distance be x km. First, convert 5 hours to minutes:
1 h = 60 min
5 h = 5 × 60 = 300 min

As time increases, the distance covered also increases. This is a direct proportion.
\( \frac{16}{25} = \frac{x}{300} \)
\( \implies x = \left(\frac{16 \times 300}{25}\right) \)
\( \implies x = 192 \)

So, the required distance is 192 km.
In simple words: More time means more distance can be covered at the same speed. Divide the known distance by the known time to find speed, then multiply by the new time.

Exam Tip: Always convert time units to match (minutes to minutes, or hours to hours) before setting up your proportion — mismatched units will give a wrong answer.

 

Question 6. At a shop, dolls cost Rs 630 for 18 pieces. How many dolls can be bought for Rs 455?
Answer: Let the required number of dolls be x. We set up a table:

When spending decreases, fewer dolls can be bought. This is a direct proportion situation.
\( \frac{18}{630} = \frac{x}{455} \)
\( \implies \frac{1}{35} = \frac{x}{455} \)
\( \implies x = \frac{455}{35} \)
\( \implies x = 13 \)

So, 13 dolls can be bought for Rs 455.
In simple words: Less money buys fewer items when the price per item stays the same. Divide the cost by the number of items to find the price of one, then divide your budget by that price.

Exam Tip: Always simplify fractions before cross-multiplying to keep your arithmetic manageable and reduce errors.

 

Question 7. The cost of 9 kg of sugar is Rs 166.50. How much sugar can be bought for Rs 259?
Answer: Let the required weight of sugar be x kg. We set up a table:

When you have more money available, you can purchase more sugar. This is a direct proportion.
\( \frac{9}{166.50} = \frac{x}{259} \)
\( \implies x = \frac{9 \times 259}{166.50} \)
\( \implies x = \frac{9 \times 259 \times 100}{16650} \)
\( \implies x = 14 \)

So, 14 kg of sugar can be bought for Rs 259.
In simple words: More money allows you to buy more. Find the cost per kg of sugar, then divide your total budget by that cost per kg to find the amount you can buy.

Exam Tip: When working with decimal costs, multiply both numerator and denominator by 100 to eliminate decimals and make division easier.

 

Question 8. The cost of 15 metres of cloth is Rs 981. How much cloth can be bought for Rs 1308?
Answer: Let the required length of cloth be x m. We set up a table:

When you have more money to spend, you can buy more cloth. This represents a direct proportion situation.
\( \frac{15}{981} = \frac{x}{1308} \)
\( \implies x = \frac{15 \times 1308}{981} \)
\( \implies x = 20 \)

So, 20 m of cloth can be bought for Rs 1,308.
In simple words: Spending more money gets you more fabric when the price per metre is unchanged. Calculate the cost of one metre, then see how many metres your budget will cover.

Exam Tip: Cancel common factors before multiplying to keep the numbers manageable — divide 15 and 981 by 3 to simplify the starting fraction.

 

Question 9. A model of a ship is made on the scale where 1 cm represents 15 m. The length of the mast of the actual ship is 35 m. Find the length of the mast of the model ship.
Answer: Let m be the length of the model ship's mast. We have:
1 m = 100 cm
Therefore, 15 m = 1500 cm and 35 m = 3500 cm

If the actual ship has a larger mast length, the model ship's mast will also be proportionally larger. This is a direct proportion.
\( \frac{1500}{9} = \frac{3500}{x} \)
\( \implies x = \frac{3500 \times 9}{1500} \)
\( \implies x = 21 \text{ cm} \)

So, the length of the model ship's mast is 21 cm.
In simple words: A scale model shrinks all measurements by the same factor. If the real mast is longer, the model mast is also longer in proportion. Divide the model measurement by the real measurement to find the scale, then apply it.

Exam Tip: Always convert all measurements to the same unit (centimetres or metres) before setting up your proportion — mixing units will give an incorrect answer.

 

Question 10. A machine collects dust from air. In 8 days it collects 6.4 × 10^7 kg of dust. How much dust will it collect in 15 days?
Answer: Let x kg be the required amount of dust. We set up a table:

As more days pass, the machine collects more dust. This is a direct proportion situation.
\( \frac{8}{6.4 \times 10^7} = \frac{15}{x} \)
\( \implies x = \frac{15 \times 6.4 \times 10^7}{8} \)
\( \implies x = 12 \times 10^7 \)

So, 12,00,00,000 kg of dust will be collected in 15 days.
In simple words: More days of operation collect more dust at a steady rate. Divide the dust collected by the number of days to find the daily rate, then multiply by the new number of days.

Exam Tip: When multiplying with scientific notation, handle the coefficients and powers of 10 separately to avoid mistakes with large numbers.

 

Question 11. A car travels 50 km in 60 minutes. How far will it travel in 1 hour 12 minutes?
Answer: Let x km be the required distance. First, convert 1 hour 12 minutes to minutes:
1 h = 60 min
1 h 12 min = (60 + 12) min = 72 min

When more time passes, the car travels a greater distance. This is a direct proportion.
\( \frac{50}{60} = \frac{x}{72} \)
\( \implies x = \frac{50 \times 72}{60} \)
\( \implies x = 60 \)

So, the distance travelled by the car in 1 h 12 min is 60 km.
In simple words: At the same speed, travelling for longer takes you farther. Find how far the car goes in one minute, then multiply by the total number of minutes.

Exam Tip: Convert all times to the same unit before setting up the ratio - mixing hours and minutes will lead to calculation errors.

 

Question 12. Ravi can cover a distance of 5 km in 60 minutes. How far will he travel in 2 hours 24 minutes?
Answer: Let x km be the required distance covered by Ravi in 2 h 24 min. We have:
1 h = 60 min
2 h 24 min = (120 + 24) min = 144 min

As time increases, the distance Ravi covers also increases. This represents a direct proportion.
\( \frac{5}{60} = \frac{x}{144} \)
\( \implies x = \frac{5 \times 144}{60} \)
\( \implies x = 12 \)

So, the distance covered by Ravi in 2 h 24 min is 12 km.
In simple words: Travelling for a longer period at the same speed covers more ground. Calculate the speed per minute, then multiply by the total minutes in the new time period.

Exam Tip: Always express mixed time units (hours and minutes) in a single unit (total minutes) before using them in a proportion.

 

Question 13. A pile of 12 cardboards, each having thickness 65 mm, will reach what height? If the thickness of each cardboard in a pile of 312 cardboards is x mm, find x.
Answer: Let x mm be the required thickness of each cardboard. We set up a table:

If you have more cardboards stacked, each individual cardboard must be thinner to keep the total pile height the same. This is an inverse proportion situation.
\( \frac{65}{12} = \frac{x}{312} \)
\( \implies x = \frac{65 \times 312}{12} \)
\( \implies x = 1690 \)

So, the thickness of the pile of 312 cardboards is 1690 mm.
In simple words: When you stack more items to reach the same height, each item must be thinner. Multiply the original thickness by the original count, then divide by the new count to find the new thickness.

Exam Tip: Recognise inverse proportion when increasing one quantity requires decreasing the other to achieve a fixed total - here the total height stays constant.

 

Question 14. 11 men dig a trench of length \( 6\frac{3}{4} \) m in a day. How many men should be employed to dig a trench of length 27 m in a day?
Answer: Let x be the required number of men. We have:
\( 6\frac{3}{4} \text{ m} = \frac{27}{4} \text{ m} \)

Then, we set up a table:

When you need to dig a longer trench, more workers are required. This is a direct proportion.
\( \frac{11}{\frac{27}{4}} = \frac{x}{27} \)
\( \implies \frac{11 \times 4}{27} = \frac{x}{27} \)
\( \implies \frac{44}{27} = \frac{x}{27} \)
\( \implies x = 44 \)

So, 44 men should be employed to dig a trench of length 27 m.
In simple words: Longer trenches need more workers to finish in the same time. Divide the new trench length by the original trench length, then multiply by the original number of workers.

Exam Tip: Convert mixed numbers to improper fractions before setting up the proportion to avoid careless errors.

 

Question 15. Reenu types 540 words in 30 minutes. How many words will she type in 8 minutes?
Answer: Let x be the number of words Reenu types in 8 minutes. We set up a table:

When you type for less time, you complete fewer words at the same typing speed. This is a direct proportion.
\( \frac{540}{30} = \frac{x}{8} \)
\( \implies x = \frac{540 \times 8}{30} \)
\( \implies x = 144 \)

So, Reenu will type 144 words in 8 minutes.
In simple words: Less time means fewer words typed when your speed stays the same. Find how many words you type per minute, then multiply by the new time in minutes.

Exam Tip: Simplify the ratio before cross-multiplying - here 540 divided by 30 equals 18, making the multiplication much simpler.