CBSE Class 8 Mathematics Direct And Inverse Variations Worksheet Set 01

CBSE Class 8 Mathematics Worksheet - Direct and Inverse Proportions CBSE,CCE and NCERT students can refer to the attached file. All educational material on the website has been prepared by the best teachers having more than 20 years of teaching experience in various schools. The study material available on our website for all CBSE classes and subjects have been prepared by teachers from all over the country. All study material has been carefully designed keeping into mind the latest CBSE examination trends.

Question. A, B and C are employed to do a piece job for Rs. 529. A and B together are supposed to do 19/23 of the work. C should be paid
(a) Rs. 115
(b) Rs. 92
(c) Rs. 200
(d) Rs. 250

Answer: B

Question. A and B can do a piece of work in 5 days, B and C together can do it in 4 days, If B is twice as good a workman as C, then in how many days A alone can do it?
(a) 30 days
(b) 18 days
(c) 9 days
(d) 10 days

Answer: A

Question. A tank has a capacity of 240 liters. A pipe can empty 1/ 4th of the tank in 5 minutes and another pipe can empty 1/3rd of the tank in 6 minutes. The tank is filled and both the pipes are opened for 3 minutes. How much of the water is now left in the tank?
(a) 160 liters
(b) 164 liters
(c) 196 liters
(d) 200 litres

Answer: B

Question. Rohit can row 18 km in 4 hrs down stream but while returning he takes 12 hours. The speed of the water current is
(a) 1.5 km/hr
(b) 4 km/hr
(c) 2 km/hr
(d) 3 km/hr

Answer: A

Question. Three pipes are connected to a tank. Pipes A and B connected with tubewells can fill the tank in 10 hrs and 12 hrs respectively, while pipe C meant for watering the garden can empty the tank in 6 hrs. At 7 a.m. all the three pipes are opened. One fourth of the tank will be filled at
(a) 9 p.m.
(b) 10 p.m.
(c) 8 p.m.
(d) 11 p.m.

Answer: B

Question. A and B undertook to do a piece of work for Rs. 4500. A alone could do it in 3 days and B in 12 days. With the help of C, they finish the work in 4 days. The share of C in the money is
(a) Rs. 2250
(b) Rs. 1500
(c) Rs. 750
(d) Rs. 375

Answer: C

Question. Shivani can do a piece of work in 16 days, Babu in 24 days and with the help of Sita, they finish it in 8 days. Sita can alone do it in
(a) 40 days
(b) 60 days
(c) 48 days
(d) 24 days

Answer: C

Question. Javed walks at the rate of 3 km/hr for 2 hours and then at the rate of 2 km/hr for 3 hours. His average speed is .... km/hr.
(a) 4
(b) 2.4
(c) 2.5
(d) 5

Answer: B

Question. Sixteen men can complete a work in fifteen days. Twenty four children can do the same work in twenty days. In how many days will eight men and eight children together complete the same job?
(a) 16
(b) 15
(c) 20
(d) None of the above

Answer: C

Question. Kishan cycles at a speed of 8 km/hr. After every 10 km, he rests for 20 minutes. How long will he take to travel a distance 40 km?
(a) 6 hrs 20 min
(b) 8 hrs
(c) 6 hrs
(d) 5 hrs

Answer: C

Question. Two cars start from one point and move along two roads at right angles to each other. Their speeds are 36 km/hr and 48 km/hr respectively. After 15 sec. the distance between them will be
(a) 400 m.
(b) 150 m.
(c) 300 m.
(d) 250 m.

Answer: D

Question. A train running at a speed of 54 km/hour passes a signal post in it 8 seconds. The length of train is
(a) 432 m.
(b) 150 m.
(c) 120 m.
(d) Data inadequate

Answer: C

Question. A and B can do a piece of work in 24 days, B and C can do it in 18 days and A and C in 12 days. Working together they will finish the work in nearly
(a) 9 days
(b) 10 days
(c) 11 days
(d) 8 days

Answer: C

Question. Ravi can finish 12(1)/2 % of a work in day, while Kulwant can finish only 16(2)/3% in a day. With the help of Narender, they finish the work in 3 days.
The percentage of work finished by Narender in a day is nearest to
(a) 5%
(b) 4%
(c) 8%
(d) 10%

Answer: B

Question. Two persons move towards each other from two places 55 km apart. One moves at a speed of 12 km/hr and other at 10 km/hr. They will be first 11 km apart after moving for
(a) 3 hrs
(b) 2hrs
(c) 6hrs
(d) 4hrs

Answer: B

Question. Ram, Mohan and Hari can do a work in 15 days, 6 days and 10 days respectively. All the three can together finish three times of that work in
(a) 21 days
(b) 18 days
(c) 9 days
(d) 15 days

Answer: C

Question. If 6 men or 8 women can finish a work in 12 days, 9 men and 12 women will finish the work in nearly
(a) 4 days
(b) 3 days
(c) 6 days
(d) 5 days

Answer: A

Question. B is twice as fast as A. If A can complete a job in 36 days, how long will it take for both A and B together to complete the same job?
(a) 24 days
(b) 18 days
(c) 12 days
(d) 9 days

Answer: C

Question. Hari singh can cover a circular path of radius 21 m. in 44 sec. He will cover a distance of 3 km in
(a) 16 mt. 20 secs.
(b) 16 mt. 40 secs.
(c) 18 mt. 00 secs.
(d) 18 mt. 30 secs.

Answer: B

Question. Ram can complete a work in 12 days, while Rajan can finish it in 18 days. Ram works on it for 8 days and then Rajan starts working on it. The work will be finished in
(a) 6 days
(b) 7 days
(c) 8 days
(d) 9 days

Answer: A

Question. A contractor undertakes to make a road in 72 days and employs 24 men. After 48 days, he finds that only half of the road is made. How many extra men should he now employ to complete the road in time?
(a) 36 men
(b) 24 men
(c) 21 men
(d) None of these

Answer: B

Question. Hira, Radhu and Kalu working alone can do a piece of work in 9, 8 and 6 days respectively. They jointly finish the work and earn Rs. 522. The earnings should be divided in the ratio of
(a) 6 : 9 : 8
(b) 9 : 8 : 6
(c) 8 : 9 : 12
(d) 6 : 8 : 9

Answer: C

Question. Two pipes A and B can fill an empty tank in 8 hours and 12 hours respectively, If the pipes are opened alternately every hour, the tank will be full earlier if
(a) A is opened first
(b) B is opened first
(c) It will take same time no mater which pipe is opened first
(d) None of these

Answer: A

Question. A car is 25 km ahead of a scooter. The car is travelling at 40 km/hr and the scooter at 50 km/hr. The scooter will overtake the car after
(a) 1(1)/2 hrs
(b) 2(1)/2 hrs
(c) 3(1)/2 hrs
(d) 3 hrs

Answer: B

Question. Two pipes A and B can fill a cistern in 24, minutes and 32 minutes respectively, If both the pipes are opened together, then after how much time the pipe B be closed so that the cistern is full in 18 minutes?
(a) 6 minutes
(b) 8 minutes
(c) 10 minutes
(d) 14 minutes

Answer: B

Question 1. If 9 kg of rice costs Rs 166.50, how much rice can be purchased for Rs 259?
Answer: Let the quantity of rice that can be purchased for Rs 259 be \(x\) kg.
Since more rice can be purchased for more money, this is a case of direct proportion.
\(\frac{9}{166.50} = \frac{x}{259}\)
\(x = \frac{9 \times 259}{166.50}\)
\(x = \frac{2331}{166.50} = 14 \text{ kg}\)
In simple words: As the amount of money increases, the quantity of rice you can buy also increases proportionately.
Exam Tip: For direct proportion problems, set up the ratio as \(\frac{x_1}{y_1} = \frac{x_2}{y_2}\) and cross-multiply to solve for the unknown.

Question 2. If a car covers 80 km in 5 litres of petrol, how much distance will it cover in 13 litres of petrol?
Answer: Let the distance covered in 13 litres of petrol be \(x\) km.
More petrol will cover more distance, so this is direct proportion.
\(\frac{80}{5} = \frac{x}{13}\)
\(16 = \frac{x}{13}\)
\(x = 16 \times 13 = 208 \text{ km}\)
In simple words: More fuel allows the car to travel a greater distance at a constant mileage.

Exam Tip: Always verify if the relationship is direct (both quantities increase together) before applying ratio formulas.

Question 3. If 32 men can reap a field in 15 days, in how many days can 20 men reap the same field?
Answer: Let the number of days required by 20 men be \(x\).
Fewer men will take more days to complete the same work, so this is a case of inverse proportion.
\(32 \times 15 = 20 \times x\)
\(480 = 20x\)
\(x = \frac{480}{20} = 24 \text{ days}\)
In simple words: Decreasing the number of workers increases the time needed to finish the job.

Exam Tip: In inverse proportion, the product of the two quantities remains constant (\(x_1 y_1 = x_2 y_2\)), unlike direct proportion where the ratio is constant.

Question 4. If 36 men can do a work in 25 days, in how many days will 15 men do the same work?
Answer: Let the required number of days be \(x\).
This is an inverse proportion problem because fewer men require more days.
\(36 \times 25 = 15 \times x\)
\(900 = 15x\)
\(x = \frac{900}{15} = 60 \text{ days}\)
In simple words: Less manpower means the work takes longer to finish.

Exam Tip: Double-check your multiplication product before dividing to isolate the variable.

Question 5. If 4 kg of potatoes cost Rs 60, what is the cost of 12 kg of potatoes?
Answer: Let the cost of 12 kg of potatoes be Rs \(x\).
This is direct proportion since buying more potatoes costs more money.
\(\frac{4}{60} = \frac{12}{x}\)
\(\frac{1}{15} = \frac{12}{x}\)
\(x = 12 \times 15 = \text{Rs } 900\)
In simple words: Triple the weight means triple the total cost.

Question 6. A tank travels 15 km in 25 minutes. If the speed remains the same, how far can it travel in 1 hour and 5 minutes?
Answer: Convert 1 hour and 5 minutes completely into minutes:
\(1 \text{ hour} + 5 \text{ minutes} = 60 + 5 = 65 \text{ minutes}\)
Let the distance travelled in 65 minutes be \(x\) km.
Since speed is uniform, distance and time are in direct proportion.
\(\frac{15}{25} = \frac{x}{65}\)
\(\frac{3}{5} = \frac{x}{65}\)
\(x = \frac{3 \times 65}{5} = 3 \times 13 = 39 \text{ km}\)
In simple words: More travel time at the same speed results in a greater distance covered.

Exam Tip: Always convert mixed time units into a single uniform unit (minutes or hours) before setting up proportions.

Question 7. The cost of 5 metres of cloth is Rs 350. Find the cost of 7 metres of cloth.
Answer: Let the cost of 7 metres of cloth be Rs \(x\).
More cloth means more cost, representing a direct proportion.
\(\frac{5}{350} = \frac{7}{x}\)
\(\frac{1}{70} = \frac{7}{x}\)
\(x = 7 \times 70 = \text{Rs } 490\)
In simple words: Calculate the price per metre first, then multiply by the required length.

Exam Tip: Unitary method is often the quickest path for simple direct proportion word problems.

Question 8. A 14 metres stick casts a shadow of 10 metres. Find the height of a stick which casts a shadow of 15 metres under similar conditions.
Answer: Let the height of the stick be \(x\) metres.
Object height and shadow length are directly proportional.
\(\frac{14}{10} = \frac{x}{15}\)
\(\frac{7}{5} = \frac{x}{15}\)
\(x = \frac{7 \times 15}{5} = 7 \times 3 = 21 \text{ m}\)
In simple words: Taller objects cast longer shadows under identical lighting conditions.

Exam Tip: Keep the corresponding quantities in the same positions in the numerator and denominator.

Question 9. A train is moving with uniform speed of 80 km/hr. How for will it travel in 35 minutes?
Answer: Speed = \(80 \text{ km/hr}\)
Time = \(35 \text{ minutes} = \frac{35}{60} \text{ hours} = \frac{7}{12} \text{ hours}\)
Distance = \(\text{Speed} \times \text{Time}\)
Distance = \(80 \times \frac{7}{12} = \frac{560}{12} = \frac{140}{3} = 46.67 \text{ km}\)
In simple words: Convert minutes into hours, then multiply by speed to find distance.

Exam Tip: When speed is in km/hr, time must always be expressed in hours.

Question 10. 18 workers can build a wall in 26 hours, how many workers will be required to do the same work in 13 hours?
Answer: Let the required number of workers be \(x\).
Fewer hours to finish the work means more workers are needed, indicating inverse proportion.
\(18 \times 26 = x \times 13\)
\(x = \frac{18 \times 26}{13}\)
\(x = 18 \times 2 = 36 \text{ workers}\)
In simple words: Cutting the required time in half means you need twice as many workers.

Exam Tip: Inverse variation means worker count multiplied by time taken remains constant.

Question 11. There are 100 students in a hostel. The food provision for them is for 15 days. How long will there provision last if 20 more join the group?
Answer: Initial number of students = 100
New total number of students = \(100 + 20 = 120\)
Let the food last for \(x\) days.
More students mean the food will run out faster, which is an inverse proportion.
\(100 \times 15 = 120 \times x\)
\(1500 = 120x\)
\(x = \frac{1500}{120} = \frac{150}{12} = 12.5 \text{ days}\)
In simple words: An increase in consumers reduces the number of days food supplies will last.

Exam Tip: Remember to calculate the total new population (old students plus newcomers) before setting up the equation.

Question 12. 12 men can dig a trench in 8 days. How many men can dig the same in 6 days?
Answer: Let the required number of men be \(x\).
Fewer days mean more men are required (inverse proportion).
\(12 \times 8 = 6 \times x\)
\(96 = 6x\)
\(x = \frac{96}{6} = 16 \text{ men}\)
In simple words: To finish a task quicker, you must deploy more workers.

Question 13. A garrison of 500 men had provision for 24 days. However, a reinforcement of 300 men arrived. How long the food will last?
Answer: Initial men = 500
Reinforcement = 300 men
Total men now = \(500 + 300 = 800\)
Let the food last for \(x\) days.
This is an inverse proportion.
\(500 \times 24 = 800 \times x\)
\(12000 = 800x\)
\(x = \frac{12000}{800} = \frac{120}{8} = 15 \text{ days}\)
In simple words: Additional arrivals consume the food stock faster, shortening the number of days it lasts.

Exam Tip: Always add reinforcements to the original headcount to find the final total.

Question 14. At 60 km/hour a car takes 2 hours to reach its destination. How long will the car take if it travels at a speed of 80 km/hr?
Answer: Let the time taken at 80 km/hr be \(x\) hours.
Higher speed means less time taken (inverse proportion).
\(60 \times 2 = 80 \times x\)
\(120 = 80x\)
\(x = \frac{120}{80} = \frac{3}{2} = 1.5 \text{ hours}\)
In simple words: Driving faster decreases the travel time for the same distance.

Exam Tip: Speed and time taken for a fixed distance are always inversely proportional.

Question 15. The number of bottles made by a factory in 63 days is 315. How many bottles can it make in 82 days?
Answer: Let the number of bottles made in 82 days be \(x\).
More days mean more production, which is direct proportion.
\(\frac{315}{63} = \frac{x}{82}\)
\(5 = \frac{x}{82}\)
\(x = 5 \times 82 = 410 \text{ bottles}\)
In simple words: Production volume increases uniformly with operating time.

Exam Tip: Simplifying \(\frac{315}{63}\) to 5 makes the calculation much easier.

Question 16. A road map with a scale of 1 cm represents 36 km. When Ravi drives on a road for 144 km, what would be the distance covered by him on the map?
Answer: Let the map distance be \(x\) cm.
Map scale and actual distance are directly proportional.
\(\frac{1}{36} = \frac{x}{144}\)
\(x = \frac{144}{36} = 4 \text{ cm}\)
In simple words: Actual distance scales down uniformly according to the map ratio.

Exam Tip: Keep units consistent (cm for map, km for actual) when setting up proportions.

Question 17. If a train is moving at a uniform speed of 75 km/hour, find the time required to cover the distance of 300 km.
Answer: Distance = \(300 \text{ km}\)
Speed = \(75 \text{ km/hr}\)
Time = \(\frac{\text{Distance}}{\text{Speed}}\)
Time = \(\frac{300}{75} = 4 \text{ hours}\)
In simple words: Divide the total distance by the speed to find the time taken.

Exam Tip: Memorize the speed-distance-time triangle: Distance = Speed \(\times\) Time.

Question 18. A car can finish a journey in 10 hours at the speed of 48 km/hr. By how much should its speed be increased so that it may take only 8 hours to cover the same distance?
Answer: Let the new speed be \(S\) km/hr.
Since distance is constant, speed and time are inversely proportional.
\(48 \times 10 = S \times 8\)
\(480 = 8S\)
\(S = \frac{480}{8} = 60 \text{ km/hr}\)
Increase in speed = New speed - Original speed
Increase = \(60 - 48 = 12 \text{ km/hr}\)
In simple words: Find the required new speed first, then subtract the original speed to find the increase.

Exam Tip: Read carefully whether the question asks for the new speed or the increase in speed.

Question 19. Reema types 540 words in half-hour. How many words would she type in 6 minutes?
Answer: Half-hour = 30 minutes.
Let the number of words typed in 6 minutes be \(x\).
Typing speed is constant, so time and word count are in direct proportion.
\(\frac{540}{30} = \frac{x}{6}\)
\(18 = \frac{x}{6}\)
\(x = 18 \times 6 = 108 \text{ words}\)
In simple words: Less time means fewer words typed.

Exam Tip: Ensure both time values are in the same unit (minutes) before creating the fraction.

Question 20. If 56 men can do a piece of work in 42 days, how many men are required to complete it in 14 days?
Answer: Let the required number of men be \(x\).
Fewer days demand more workers (inverse proportion).
\(56 \times 42 = x \times 14\)
\(x = \frac{56 \times 42}{14}\)
\(x = 56 \times 3 = 168 \text{ men}\)
In simple words: Finishing work three times faster requires three times as many workers.

Exam Tip: Cancel common factors before multiplying to simplify your calculation.

Question 21. The scale of a map is given as 1 : 40000000. Two cities are 4 cm apart on the map. Find the actual distance between them.
Answer: Map scale = \(1 : 40000000\)
Map distance = 4 cm
Actual distance = \(4 \times 40000000 = 160000000 \text{ cm}\)
Convert cm to km (since \(1 \text{ km} = 100,000 \text{ cm}\))
Actual distance = \(\frac{160000000}{100000} = 1600 \text{ km}\)
In simple words: Multiply map distance by the scale factor, then convert centimetres to kilometres.

Question 22. 12 pipes are required to fill a tank in 2 hours 40 minutes. How long will it take if only 10 pipes of the same type are used?
Answer: Convert total time into minutes:
\(2 \text{ hours} + 40 \text{ minutes} = 120 + 40 = 160 \text{ minutes}\)
Let the time taken by 10 pipes be \(x\) minutes.
Fewer pipes mean more time to fill the tank (inverse proportion).
\(12 \times 160 = 10 \times x\)
\(1920 = 10x\)
\(x = \frac{1920}{10} = 192 \text{ minutes}\)
Convert 192 minutes back into hours and minutes:
\(192 \text{ minutes} = 3 \text{ hours and } 12 \text{ minutes}\)
In simple words: Removing pipes slows down the filling process, increasing the total time required.

Exam Tip: Convert time into minutes at the start and back into hours/minutes at the end.

Question 23. If a box of sweets is divided among 48 children, they will get 7 sweets each. How many would each get, if the number of the children is reduced by 6?
Answer: Initial children = 48
New number of children = \(48 - 6 = 42\)
Let each child get \(x\) sweets.
Fewer children mean each child gets more sweets (inverse proportion).
\(48 \times 7 = 42 \times x\)
\(336 = 42x\)
\(x = \frac{336}{42} = 8 \text{ sweets}\)
In simple words: When the same number of sweets is shared among fewer people, each person's share increases.

Question 24. A batch of bottles were packed in 25 boxes with 12 bottles in each box. If the same batch is packed using 20 bottles in each box, how many boxes would be filled?
Answer: Let the number of boxes needed be \(x\).
Putting more bottles per box means fewer boxes will be needed (inverse proportion).
\(25 \times 12 = 20 \times x\)
\(300 = 20x\)
\(x = \frac{300}{20} = 15 \text{ boxes}\)
In simple words: Packing denser reduces the total count of boxes required.

Exam Tip: Total number of bottles is constant (\(25 \times 12 = 300\)).

Question 25. A school has 8 periods a day each of 45 minutes duration. How long would each period be, if the school has 9 periods a day, assuming the number of school hours to be same?
Answer: Let the duration of each period be \(x\) minutes.
More periods in the same school day mean each period must be shorter (inverse proportion).
\(8 \times 45 = 9 \times x\)
\(360 = 9x\)
\(x = \frac{360}{9} = 40 \text{ minutes}\)
In simple words: Increasing the daily number of periods cuts down the duration of each individual period.

Exam Tip: Total school hours remain constant, equating the product of periods and duration.

Question 26. 120 men had food provision for 195 days. How long would the food last if there are 90 men?
Answer: Let the food last for \(x\) days.
Fewer men mean the food will last longer (inverse proportion).
\(120 \times 195 = 90 \times x\)
\(23400 = 90x\)
\(x = \frac{23400}{90} = 260 \text{ days}\)
In simple words: Fewer mouths to feed extends the lifespan of existing provisions.

Exam Tip: Cancel zeroes on both sides to simplify \(\frac{2340}{9}\).

Question 27. 18 men can reap a field in 35 days. Find the number of men required to reap the same field in 15 days.
Answer: Let the required number of men be \(x\).
Fewer days mean more men are needed (inverse proportion).
\(18 \times 35 = 15 \times x\)
\(630 = 15x\)
\(x = \frac{630}{15} = 42 \text{ men}\)
In simple words: Speeding up the harvest requires deploying more workers.

Exam Tip: Maintain the inverse relation: \(M_1 \times D_1 = M_2 \times D_2\).

Question 28. Working for 8 hours daily, 40 people can dig the foundation of a building in 21 days. Working for 10 hours daily, if the work is to be finished in 14 days, how many people are needed to do the same work?
Answer: Let the required number of people be \(x\).
Using compound proportion (Men, Daily Hours, Days):
More hours daily means fewer people needed (inverse).
Fewer days mean more people needed (inverse).
\(40 \times 8 \times 21 = x \times 10 \times 14\)
\(6720 = 140x\)
\(x = \frac{6720}{140} = 48 \text{ people}\)
In simple words: Combine all variables into a single product equation where total man-hours remain constant.

Exam Tip: For compound proportions, equate \(M_1 \times H_1 \times D_1 = M_2 \times H_2 \times D_2\).

Question 29. 6 typists working 5 hours a day can type the manuscript of a book in 16 days. How many days will 4 typists take to do the same job each working 6 hours a day?
Answer: Let the required number of days be \(x\).
Using compound proportion:
\(6 \text{ typists} \times 5 \text{ hours} \times 16 \text{ days} = 4 \text{ typists} \times 6 \text{ hours} \times x\)
\(480 = 24x\)
\(x = \frac{480}{24} = 20 \text{ days}\)
In simple words: Equate total man-hours (\(\text{Typists} \times \text{Hours} \times \text{Days}\)) on both sides.

Exam Tip: Always multiply all parameters of the first scenario together and equate to the product of the second scenario.

Question 30. A hostel spends Rs. 3200 as cost of rice for 20 students for 30 days. If the hostel spends Rs. 2400 for 30 students, for how many days will the rice last?
Answer: Let the number of days be \(x\).
Using compound proportion where cost is directly proportional and students are inversely proportional to days:
\(\frac{\text{Cost}_1}{\text{Students}_1 \times \text{Days}_1} = \frac{\text{Cost}_2}{\text{Students}_2 \times \text{Days}_2}\)
\(\frac{3200}{20 \times 30} = \frac{2400}{30 \times x}\)
\(\frac{3200}{600} = \frac{2400}{30x}\)
Cross-multiply to solve:
\(3200 \times 30x = 2400 \times 600\)
\(96000x = 1440000\)
\(x = \frac{1440000}{96000} = 15 \text{ days}\)
In simple words: Total consumption depends on budget, number of students, and duration.

Exam Tip: Put direct variables in the numerator/denominator ratio and inverse variables crosswise.

Question 31. A school hostel with 40 children has enough provisions for food to last for 15 days. If 10 children left the hostel, how many days will the provisions last now?
Answer: Remaining children = \(40 - 10 = 30\)
Let the provisions last for \(x\) days.
Fewer children mean provisions last longer (inverse proportion).
\(40 \times 15 = 30 \times x\)
\(600 = 30x\)
\(x = \frac{600}{30} = 20 \text{ days}\)
In simple words: When children leave, food consumption drops, stretching the supplies over more days.

Exam Tip: Always subtract departing members to get the correct active population count.

Question 32. A train can finish a journey in 10 hours, travelling at a speed of 56 km/h. If another faster train is to cover the same journey in 8 hours, what would be the average speed of the new train?
Answer: Let the new speed be \(S\) km/h.
Distance is constant, so speed and time are inversely proportional.
\(56 \times 10 = S \times 8\)
\(560 = 8S\)
\(S = \frac{560}{8} = 70 \text{ km/h}\)
In simple words: Finishing a trip in less time requires travelling at a higher average speed.

Exam Tip: Use the inverse relation \(S_1 \times T_1 = S_2 \times T_2\) whenever distance is unchanged.

Question 33. 10 men working for 6 days mow an area of 5 acres. If there are 8 men working to mow 4 acres of land, how many days will it take?
Answer: Let the required number of days be \(x\).
Using compound proportion:
Men and days are inverse; Area (acres) and days are direct.
\(\frac{\text{Men}_1 \times \text{Days}_1}{\text{Area}_1} = \frac{\text{Men}_2 \times \text{Days}_2}{\text{Area}_2}\)
\(\frac{10 \times 6}{5} = \frac{8 \times x}{4}\)
\(\frac{60}{5} = \frac{8x}{4}\)
\(12 = 2x\)
\(x = \frac{12}{2} = 6 \text{ days}\)
In simple words: Set up the ratio with work (area) in the denominator and workers in the numerator.

Exam Tip: Remember: Work done is always placed in the denominator for compound proportion formulas.

Question 34. A family of 8 people has enough stock of rice to last for 30 days. Due to the arrival of some relatives, this food was consumed in 20 days. How many guests joined the family?
Answer: Let the total number of people after relatives arrived be \(P\).
\(8 \times 30 = P \times 20\)
\(240 = 20P\)
\(P = \frac{240}{20} = 12 \text{ people}\)
Number of guests joined = Total people - Original family members
Guests = \(12 - 8 = 4\)
In simple words: First find the total final population, then subtract the original family size to find the number of guests.

Exam Tip: Read carefully whether the question asks for the total final population or the number of guests.

Question 35. In an army camp, there are 800 soldiers. There is enough food for them for 60 days. If 400 more soldiers arrive at the camp, how many days will the food last?
Answer: Total soldiers after arrival = \(800 + 400 = 1200\)
Let the food last for \(x\) days.
More soldiers mean food lasts for fewer days (inverse proportion).
\(800 \times 60 = 1200 \times x\)
\(48000 = 1200x\)
\(x = \frac{48000}{1200} = \frac{480}{12} = 40 \text{ days}\)
In simple words: A 50 percent increase in soldiers reduces the duration of food provisions proportionally.

Exam Tip: Always add arriving personnel to the base count before solving inverse food supply problems.

Most Important Questions

Question 1. The variables x and y are in direct proportion, based on this relation fill the table

Answer: Since \(x\) and \(y\) are in direct proportion, \(\frac{x}{y} = k\) (constant ratio).
Using the first column (\(x = 12, y = 60\)):
\(k = \frac{12}{60} = \frac{1}{5} = 0.2\)

For \(x_2\) when \(y = 5\):
\(\frac{x_2}{5} = \frac{1}{5} \implies x_2 = 1\)

For \(y_3\) when \(x = 8\):
\(\frac{8}{y_3} = \frac{1}{5} \implies y_3 = 8 \times 5 = 40\)

For \(y_4\) when \(x = 15\):
\(\frac{15}{y_4} = \frac{1}{5} \implies y_4 = 15 \times 5 = 75\)
In simple words: Find the constant ratio \(\frac{x}{y}\) first, then use it to fill in the missing values.
Exam Tip: In direct proportion tables, all column ratios \(\frac{x}{y}\) are equal.

Question 2. If the cost of 5 pens is Rs. 20, what will the cost of 9 such pens?
Answer: Let the cost of 9 pens be Rs \(x\).
This is a direct proportion.
\(\frac{5}{20} = \frac{9}{x}\)
\(\frac{1}{4} = \frac{9}{x}\)
\(x = 9 \times 4 = \text{Rs } 36\)
In simple words: Buying more pens increases the total price proportionally.

Exam Tip: Unit price is Rs 4 per pen, so 9 pens cost \(9 \times 4 = 36\).

Question 3. Raja covers a distance of 18 km in 1 hour, how much time will he need to travel a distance of 8 Km.
Answer: Let the time required be \(x\) hours.
Distance and time are in direct proportion.
\(\frac{18}{1} = \frac{8}{x}\)
\(18x = 8\)
\(x = \frac{8}{18} = \frac{4}{9} \text{ hours}\)
In minutes: \(\frac{4}{9} \times 60 = \frac{80}{3} = 26.67 \text{ minutes}\)
In simple words: Covering a shorter distance takes less time at a constant speed.

Exam Tip: Fractions can be left as hours or converted to minutes depending on standard preference.

Question 4. A milk factory can fill 480 bottles in 6 hours. How many bottles will it fill in 3 hours?
Answer: Let the number of bottles filled in 3 hours be \(x\).
Direct proportion applies here.
\(\frac{480}{6} = \frac{x}{3}\)
\(80 = \frac{x}{3}\)
\(x = 80 \times 3 = 240 \text{ bottles}\)
In simple words: Halving the operating time cuts the total bottle production in half.

Exam Tip: Notice that 3 hours is half of 6 hours, so the answer is simply half of 480.

Question 5. The scale of a map is given as 1:30000000. Two cities are 4 cm apart on the map. What is the actual distance between them?
Answer: Map distance = 4 cm
Scale = \(1 : 30,000,000\)
Actual distance = \(4 \times 30,000,000 = 120,000,000 \text{ cm}\)
Convert to km: \(\frac{120,000,000}{100,000} = 1200 \text{ km}\)
In simple words: Multiply map measurement by the scale factor and convert units to kilometres.

Exam Tip: Remember that \(1 \text{ km} = 100,000 \text{ cm}\).

Question 6. A car is moving at a uniform speed of 75 km/hr.
How far will it travel in 20 minutes?
Find the time required to cover a distance of 250 km.
Answer: Part 1: Distance in 20 minutes
Time = \(\frac{20}{60} \text{ hours} = \frac{1}{3} \text{ hours}\)
Distance = \(\text{Speed} \times \text{Time} = 75 \times \frac{1}{3} = 25 \text{ km}\)

Part 2: Time required for 250 km
Time = \(\frac{\text{Distance}}{\text{Speed}} = \frac{250}{75} = \frac{10}{3} \text{ hours} = 3 \text{ hours and } 20 \text{ minutes}\)
In simple words: Use distance = speed \(\times\) time for part one, and time = distance \(\div\) speed for part two.

Exam Tip: Always keep units consistent between hours and minutes during speed calculations.

Question 7. In the model of a fort, the flag is 9 m high, however the flag of the actual ship is 12 m high. If the height of the actual fort is 28 m what is the height of the model fort.
Answer: Let the height of the model fort be \(x\) metres.
Model dimensions and actual dimensions are in direct proportion.
\(\frac{\text{Model Flag}}{\text{Actual Flag}} = \frac{\text{Model Fort}}{\text{Actual Fort}}\)
Note: The problem mentions "actual ship", treating it as the scaling reference for the model.
\(\frac{9}{12} = \frac{x}{28}\)
\(\frac{3}{4} = \frac{x}{28}\)
\(x = \frac{3 \times 28}{4} = 3 \times 7 = 21 \text{ m}\)
In simple words: Scale models maintain a constant ratio with actual object dimensions.

Exam Tip: Set up corresponding ratios carefully: (Model height) / (Actual height) = constant.

Question 8. If the weight of 7 bags is 560 kg, what is the weight of 14 such bags?
Answer: Let the weight of 14 bags be \(x\) kg.
Direct proportion applies since more bags weigh more.
\(\frac{7}{560} = \frac{14}{x}\)
\(\frac{1}{80} = \frac{14}{x}\)
\(x = 14 \times 80 = 1120 \text{ kg}\)
In simple words: Doubling the number of bags doubles their total weight.

Exam Tip: Since 14 is exactly twice of 7, you can directly multiply 560 by 2.

Question 9. If 18 earrings cost Rs 1,530 what is the cost of 13 such earrings?
Answer: Let the cost of 13 earrings be Rs \(x\).
Direct proportion applies.
\(\frac{18}{1530} = \frac{13}{x}\)
First find unit cost: \(\frac{1530}{18} = 85 \text{ Rs per earring}\)
\(x = 13 \times 85 = \text{Rs } 1105\)
In simple words: Find the price of a single earring first, then multiply by 13.

Question 10. The variables x and y are in indirect proportion, based on this relation fill the table

Answer: Since \(x\) and \(y\) are in indirect proportion, their product is constant: \(x \times y = k\).
Using the first column (\(x = 12, y = 60\)):
\(k = 12 \times 60 = 720\)

For \(x_2\) when \(y = 5\):
\(x_2 \times 5 = 720 \implies x_2 = \frac{720}{5} = 144\)

For \(y_3\) when \(x = 8\):
\(8 \times y_3 = 720 \implies y_3 = \frac{720}{8} = 90\)

For \(y_4\) when \(x = 15\):
\(15 \times y_4 = 720 \implies y_4 = \frac{720}{15} = 48\)
In simple words: Multiply the given pair (\(12 \times 60 = 720\)) to find the constant product, then divide by each given value to find the missing ones.
Exam Tip: In indirect proportion tables, the product \(x \times y\) remains constant across every column.

Question 11. A fort had provisions for food for 300 men over a period of 90 days. After 20 days, 50 men left the fort. How long would the food last at the same rate?
Answer: Initial men = 300
Food lasted for 90 days initially.
After 20 days, food consumed is for 20 days by 300 men.
Remaining days of food for 300 men = \(90 - 20 = 70 \text{ days}\)
Now, 50 men left, so remaining men = \(300 - 50 = 250\)
Let the remaining food last for \(x\) days.
Using inverse proportion for the remaining stock:
\(300 \times 70 = 250 \times x\)
\(21000 = 250x\)
\(x = \frac{21000}{250} = \frac{2100}{25} = 84 \text{ days}\)
In simple words: Calculate the remaining food days for the original crew first, then adjust for the smaller crew size.

Exam Tip: Always subtract the days that have already passed before calculating remaining provisions.

Question 12. A hostel had provisions for 75 students for 30 days. After 6 days, 15 more students come to the hostel. How long will the remaining provisions last at the same rate?
Answer: Initial students = 75
After 6 days, remaining days of food for 75 students = \(30 - 6 = 24 \text{ days}\)
15 more students arrive, so total students now = \(75 + 15 = 90\)
Let the remaining food last for \(x\) days.
Using inverse proportion:
\(75 \times 24 = 90 \times x\)
\(1800 = 90x\)
\(x = \frac{1800}{90} = 20 \text{ days}\)
In simple words: Subtract elapsed days, account for new arrivals, and apply the inverse proportion formula.

Exam Tip: Total food units remaining equal \(75 \times 24\), which is then shared among 90 students.

Question 13. 6 pipes are required to fill a tank in 1 hour 20 min. How long will it take if only 5 pipes of the same type are used?
Answer: Convert time into minutes: \(1 \text{ hour} + 20 \text{ minutes} = 60 + 20 = 80 \text{ minutes}\)
Let the time taken by 5 pipes be \(x\) minutes.
Fewer pipes mean more time required (inverse proportion).
\(6 \times 80 = 5 \times x\)
\(480 = 5x\)
\(x = \frac{480}{5} = 96 \text{ minutes}\)
Convert 96 minutes back: \(96 \text{ minutes} = 1 \text{ hour and } 36 \text{ minutes}\)
In simple words: Fewer filling pipes result in a longer filling duration.

Exam Tip: Convert time to minutes before calculation to avoid fractional hours.

Question 14. If 15 workers can build a wall in 48 hours, how many workers will be required to do the same work in 30 hours?
Answer: Let the required number of workers be \(x\).
Fewer hours mean more workers needed (inverse proportion).
\(15 \times 48 = x \times 30\)
\(720 = 30x\)
\(x = \frac{720}{30} = 24 \text{ workers}\)
In simple words: To finish work quicker, increase the number of workers.

Exam Tip: Product of workers and hours remains constant (\(15 \times 48 = 720\)).

Question 15. If a box of toffees is divided among 24 children, they will get 5 toffees each. How many toffees will each get if the number of children is reduced by 4?
Answer: New number of children = \(24 - 4 = 20\)
Let each child get \(x\) toffees.
Fewer children mean each child gets more toffees (inverse proportion).
\(24 \times 5 = 20 \times x\)
\(120 = 20x\)
\(x = \frac{120}{20} = 6 \text{ toffees}\)
In simple words: Sharing a fixed quantity among fewer people increases individual shares.

Exam Tip: Total toffees = \(24 \times 5 = 120\), divided by 20 gives 6.

Question 16. A factory of socks requires 42 machines to produce a given number of socks in 63 days, how many machines are required to produce the same number of socks in 54 days?
Answer: Let the required number of machines be \(x\).
Fewer days mean more machines needed (inverse proportion).
\(42 \times 63 = x \times 54\)
\(2646 = 54x\)
\(x = \frac{2646}{54} = 49 \text{ machines}\)
In simple words: Speeding up production deadlines requires operating more machines simultaneously.

Exam Tip: Use the inverse variation formula \(M_1 \times D_1 = M_2 \times D_2\).

Question 17. A farmer has enough grains to feed 20 hens for a period of 6 days. How long will the food last if there were 10 more hens with him?ర్థ?
Answer: New total hens = \(20 + 10 = 30\)
Let the food last for \(x\) days.
More hens mean food lasts for fewer days (inverse proportion).
\(20 \times 6 = 30 \times x\)
\(120 = 30x\)
\(x = \frac{120}{30} = 4 \text{ days}\)
In simple words: Adding more hens increases consumption rate, shortening grain supply duration.

Exam Tip: Always calculate the total new animal count before setting up the inverse proportion.

Question 18. A car takes 2 hours to reach a destination by traveling 60 km/hr. How long will it take when the car travels at the speed of 80 km/hr.
Answer: Let the time taken be \(x\) hours.
Higher speed means less time taken (inverse proportion).
\(60 \times 2 = 80 \times x\)
\(120 = 80x\)
\(x = \frac{120}{80} = \frac{3}{2} = 1.5 \text{ hours}\)
In simple words: Increasing your speed shortens the travel time for any fixed journey.

Exam Tip: Speed and time are inversely proportional when the destination distance is constant.