CBSE Class 8 Mathematics Direct And Inverse Variations Worksheet Set 03

Question. Observe the given table.

What can be said about x and y?
(A) x and y are directly proportional.
(B) x is inversely proportional toy.
(C) y is inversely proportional to x.
(D) There is no proportionality between x and y.
Answer : A

Question. If the cost of 27 bags of paddy is ₹ 9450, what is the cost of 36 bags of paddy?
(A) ₹12000
(B) ₹12600
(C) ₹16200
(D) ₹10620
Answer : B

Question. If the cost of "x" metres of wire is "d " rupees, what is the cost of "y" metres of wire at the same rate ?
(A) ₹(yd/x)
(B) ₹(xy)
(C) ₹(yd)
(D) ₹(xy/d)
Answer : D

Question. If 24 workers can paint a wall in 45 hours, how many workers are required to do the same work in 20 hours ?
(A) 45
(B) 24
(C) 54
(D) 36
Answer : C

Question. The scale of a map is 1 : 3 x 1 07. Two cities are 5 em apart on the map. Find the actual distance between them in kilometres.
(A) 1500 km
(B) 1000 km
(C) 1100 km
(D) 2000 km
Answer : A

Question. A car takes 3 hours to reach a destination by travelling at 60 km/h. How long will it take while travelling at 90 km/h?
(A) 1 hour 30 minutes
(B) 1 hour 40 minutes
(C) 2 hours 40 minutes
(D) 2 hours
Answer : D

Question. A wheel revolves 19500 times in 3 hours.
How many times does it revolve in 7 hours?
(A) 54500
(C) 45500
(B) 55400
(D) 50450
Answer : D

Question. If the weight of 15 sheets of thick paper is 50 grams, how many sheets of the same paper would weigh 1(1/4) kilograms?
(A) 300
(B) 250
(C) 200
(D) 375
Answer : D

Question. A photograph of a type of bacteria enlarged 50000 times attains a length of 5 em. What is the actual length of bacteria?
(A) 1000 cm
(C) 10-4 cm
(B) 10-3 cm
(D) 10-2 cm
Answer : C

Question. A contractor can complete a certain piece of work in 9 days. He employed certain number of men, but 6 of them being absent from the very first day, the rest could finish the work in 15 days. How many men were originally employed ?
(A) 12
(B) 15
(C) 18
(D) 24
Answer : B

Question. If 32 horses consume 112 kg of gram in a certain period, how much gram will be consumed by 16 horses during the same period?
(A) 56 kg
(B) 48.5 kg
(C) 31.5 kg
(D) 50 kg
Answer : A

Question. x varies inversely as square of y. Given that y = 3 for x = 1, find the value of x for y = 4.
(A) 3
(B) 9
(C) 1/3
(D) 9/16
Answer : D

Question. The price of 357 mangoes is ₹1517.25.
What will be the approximate price of 49 dozens of such mangoes ?
(A) ₹3000
(B) ₹3500
(C) ₹4000
(D) ₹2500
Answer : A

Question. If 11.25 m of a uniform iron rod weighs 42.75 kg, what will be the weight of 6 m of the same rod ?
(A) 22.8 kg
(C) 28 kg
(B) 25.6 kg
(D) 26.5 kg
Answer : A

Question. A train is moving at a uniform speed of 75 km/hr. How far will it travel in 36 minutes?
(A) 60 km
(C) 50 km
(B) 40 km
(D) 45 km
Answer : D

Question. 36 men can construct a bridge in 18 days.
In how many days will 27 men complete the construction ?
(A) 12
(B) 18
(C) 22
(D) 24
Answer : D

Question. A van covers 432 km with 36 litres of diesel. How much distance would it cover with 25 litres of diesel ?
(A) 200 km
(B) 300 km
(C) 1 00 km
(D) 350 km
Answer : B

Question. A garrison of 500 men had provisions for 27 days. After 3 days, a reinforcement of 300 men arrived. For how many more days will the remaining provisions last?
(A) 15
(B) 16
(C) 17(1/2)
(D) 18
Answer : A

Question. If a quarter kg of green chillies cost 60 paise, how many paise will 200 g cost?
(A) 48 paise
(B) 54 paise
(C) 56 paise
(D) 72 paise
Answer : A

Question. If x and y are directly proportional, find the respective values of x₁ , x₂ and y₁ in the table given.

(A) x₁ = 8; x₂ = 2; y₁ = 100
(B) x₁ = 5; x₂ = 8; y₁ = 100
(C) x₁ = 2; x₂ = 8; y₁ = 100
(D) x₁ = 5, x₂ = 8, y₁ = 120
Answer : B

Direct and Inverse Proportions

Question 1. If 35 men can reap a field in 8 days, then 20 men can reap the same field (in days)
(a) 16.
(b) 15.
(c) 14.
(d) 12.
Answer: (c) 14.
Let the number of days required by 20 men be \(x\).
This is an inverse proportion problem because fewer men will take more days to reap the same field.
\(35 \times 8 = 20 \times x\)
\(x = \frac{35 \times 8}{20}\)
\(x = \frac{280}{20} = 14\)
In simple words: As the number of workers decreases, the time taken increases in inverse proportion.

Exam Tip: For inverse variation, use the product relation \(M_1 \times D_1 = M_2 \times D_2\).

Question 2. If 36 men can do a piece of work in 25 days, then the number of days taken by 15 men will be
(a) 50.
(b) 56.
(c) 60.
(d) 72.
Answer: (c) 60.
Let the number of days be \(x\).
\(36 \times 25 = 15 \times x\)
\(x = \frac{36 \times 25}{15}\)
\(x = \frac{36 \times 5}{3} = 12 \times 5 = 60\)
In simple words: Fewer workers require more days to complete the exact same task.

Exam Tip: Always set up the inverse proportion equation as product of men and days before simplifying fractions.

Question 3. 18 men can reap a field in 35 days. The number of men required for reaping the same field in 15 days will be
(a) 40.
(b) 42.
(c) 45.
(d) 46.
Answer: (b) 42.
Let the number of men required be \(x\).
\(18 \times 35 = x \times 15\)
\(x = \frac{18 \times 35}{15}\)
\(x = \frac{6 \times 35}{5} = 6 \times 7 = 42\)
In simple words: To finish the work in fewer days, we need more men.

Exam Tip: Check your final answer by verifying that \(18 \times 35 = 42 \times 15 = 630\).

Question 4. If the cost of 1 Kg sugar is Rs. 18, then the cost of 3 kg sugar (in Rs.) will be
(a) 60
(b) 54
(c) 50
(d) 36
Answer: (b) 54
Cost of 1 kg = Rs. 18
Cost of 3 kg = \(3 \times 18 = 54\) Rs.
In simple words: More quantity of sugar costs proportionately more money.

Exam Tip: Direct proportion means multiplying the unit rate by the given quantity.

Question 5. If x and y are two quantities \(\frac{x}{y} = k\) (k is a positive constant) then x and y are said to vary
(a) inversely.
(b) directly.
(c) equally.
(d) unequally.
Answer: (b) directly.
When the ratio of two quantities is a constant (\(\frac{x}{y} = k\)), they are in direct proportion.
In simple words: A constant ratio signifies direct variation.

Exam Tip: Remember that direct variation has the form \(\frac{x}{y} = k\) or \(x = ky\).

Question 6. If x and y are two quantities and xy = k (a constant) then x and y are said to vary
(a) directly.
(b) inversely.
(c) unequally.
(d) equally.
Answer: (b) inversely.
When the product of two quantities is constant (\(xy = k\)), they vary inversely.
In simple words: A constant product means inverse variation.

Exam Tip: Inverse variation is characterized by \(xy = k\).

Question 7. When two quantities 'x' and 'y' are in direct proportion, they are also written as
(a) x/ y.
(b) xy.
(c) x < y.
(d) x > y.
Answer: (a) x/ y.
Direct proportion implies that their ratio remains constant, written as \(\frac{x}{y} = \text{constant}\).
In simple words: Direct proportion links quantities through division.

Exam Tip: Look for division or ratios when identifying direct proportions.

Question 8. When two quantities are in inverse proportion they are also written as
(a) x y.
(b) x = y.
(c) \(x \times \frac{1}{y}\).
(d) x > y .
Answer: (c) \(x \times \frac{1}{y}\).
Inverse proportion means \(x\) is proportional to \(\frac{1}{y}\), so \(x \times y = k\), which can be expressed as \(x \propto \frac{1}{y}\).
In simple words: Inverse proportion means one quantity varies as the reciprocal of the other.

Exam Tip: Inverse proportion is equivalent to direct proportion with the reciprocal quantity.

Question 9. In the following table x and y are in direct variation.

The value of y when x is 80 is
(a) 8.
(b) 4.
(c) 2.
(d) 0.
Answer: (a) 8.
Since \(x\) and \(y\) are in direct variation, \(\frac{x}{y} = k\).
\(\frac{50}{5} = 10\)
When \(x = 80\), let \(y = ?\)
\(\frac{80}{y} = 10 \implies y = 8\)
In simple words: The constant ratio is 10, so dividing \(x\) by 10 gives \(y\).

Exam Tip: Find the constant \(k\) first using any given pair of values.

Question 10. When the speed is uniform, the distance traveled is directly proportional to
(a) time.
(b) speed.
(c) velocity.
(d) acceleration.
Answer: (a) time.
Since Distance = Speed \(\times\) Time, and speed is constant, distance is directly proportional to time.
In simple words: More time traveled at a steady speed covers more distance.

Exam Tip: Distance equals speed multiplied by time, forming a direct proportion.

Question 11. If the two quantities 'x' and 'y' vary directly proportion then the value of y when X is 17 is

(a) 34.
(b) 17.
(c) 1.
(d) 0.
Answer: (a) 34.
For direct proportion, \(\frac{X_1}{Y_1} = \frac{X_2}{Y_2}\)
\(\frac{20}{40} = \frac{17}{Y}\)
\(\frac{1}{2} = \frac{17}{Y} \implies Y = 17 \times 2 = 34\)
In simple words: \(Y\) is always twice of \(X\), so when \(X\) is 17, \(Y\) is 34.

Exam Tip: Simplify the known ratio first to make calculations easier.

Question 12. Changes in weight and height among individuals are in
(a) direct proportion.
(b) indirect proportion.
(c) equal proportion.
(d) unknown proportion.
Answer: (d) unknown proportion.
Weight and height do not maintain a fixed constant ratio or product across individuals.
In simple words: Growing taller does not mean weight increases in a fixed mathematical proportion.

Exam Tip: Not all real-life quantities follow proportional relationships.

Question 13. A machine in a soft drink factory fills 840 bottles in 6 hours. In one hour it fills 140 bottles. This proportion is
(a) inverse.
(b) direct.
(c) equal.
(d) unequal.
Answer: (b) direct.
As the number of hours increases, the number of bottles filled also increases at a constant rate (\(140\) bottles/hour).
In simple words: More operating hours result in more bottles filled.

Question 14. Two quantities may change in such a manner that if one quantity increases, the other quantity decreases and vice versa. This proportion is called
(a) inverse proportion.
(b) direct proportion.
(c) both inverse and direct proportion.
(d) equal proportion.
Answer: (a) inverse proportion.
When an increase in one quantity leads to a decrease in the other, they are in inverse proportion.
In simple words: Opposite directions of change define inverse proportion.

Exam Tip: Memorize that 'inverse' means opposite behavioral trends.

Question 15. Cost of a book and number of books purchased in a fixed amount are in
(a) inverse proportion.
(b) direct proportion.
(c) both inverse and direct proportion.
(d) equal proportion.
Answer: (a) inverse proportion.
If you have a fixed budget, buying more expensive books means you can buy fewer books.
In simple words: Higher book price means fewer books can be bought with the same money.

Exam Tip: Fixed total expenditure with variable unit price and quantity is an inverse variation.

Question 16. If 12 workers can build a wall in 48 hours, the number of workers required to do the same work in 24 hours is
(a) 48.
(b) 36.
(c) 24.
(d) 12.
Answer: (c) 24.
Let the number of workers be \(x\).
\(12 \times 48 = x \times 24\)
\(x = \frac{12 \times 48}{24} = 12 \times 2 = 24\)
In simple words: Halving the time available requires doubling the number of workers.

Exam Tip: Use inverse variation formula \(M_1 T_1 = M_2 T_2\).

Question 17. 6 pipes fill a water tank in 80 minutes. The time taken to fill the water tank in minutes by 3 pipes of the same type is (In min)
(a) 160.
(b) 120.
(c) 60.
(d) 40.
Answer: (a) 160.
Let the time taken be \(x\) minutes.
\(6 \times 80 = 3 \times x\)
\(x = \frac{6 \times 80}{3} = 2 \times 80 = 160\)
In simple words: Fewer pipes mean it takes longer to fill the tank.

Exam Tip: Pipes filling a tank form an inverse relationship between pipe count and time.

Question 18. A car takes 2 hours to reach a destination by traveling at the speed of 80 km/hr. The time taken by car when the car travels at a speed of 60 km/hr is
(a) 2 hours 12 minutes.
(b) 2 hours 20 minutes.
(c) 2 hours 40 minutes.
(d) 3 hours.
Answer: (c) 2 hours 40 minutes.
Distance = \(\text{Speed} \times \text{Time} = 80 \times 2 = 160\) km.
Time taken at 60 km/hr = \(\frac{160}{60} = \frac{8}{3}\) hours.
\(\frac{8}{3}\) hours = \(2\frac{2}{3}\) hours = 2 hours and \(\frac{2}{3} \times 60\) minutes = 2 hours 40 minutes.
In simple words: Slower speed means the journey takes more time.

Exam Tip: Convert fractional hours to minutes by multiplying by 60.

Question 19. A school wants to spend Rs. 4000 on mathematics textbooks each costing Rs. 40. If the cost of the book goes up by Rs. 10 then the number of books that can be bought is
(a) 40.
(b) 80.
(c) 100.
(d) 150.
Answer: (b) 80.
New cost of each book = Rs. 40 + Rs. 10 = Rs. 50.
Total budget = Rs. 4000.
Number of books = \(\frac{4000}{50} = 80\).
In simple words: When the price per book increases, fewer books can be purchased with the same budget.

Exam Tip: Calculate the new unit price before dividing into the total fixed amount.

Question 20. If 10 workers can build a wall in 48 hours, the number of workers required to do the same work in 30 hours will be
(a) 20.
(b) 16.
(c) 15.
(d) 12.
Answer: (b) 16.
Let the number of workers be \(x\).
\(10 \times 48 = x \times 30\)
\(x = \frac{10 \times 48}{30} = \frac{480}{30} = 16\)
In simple words: Finishing work faster requires more workers.

Exam Tip: Use inverse proportion \(M_1 T_1 = M_2 T_2\).

Question 21. Ram wants to make a wheel using spokes. He wants to fix spokes in such a way that the angle between any pair of consecutive spokes are equal. If the number of spokes is 6 then the angle is \(60^\circ\). If the angle between them is \(36^\circ\) then the number of spokes are
(a) 12.
(b) 10.
(c) 8.
(d) 6.
Answer: (b) 10.
The total angle around a point is \(360^\circ\).
Number of spokes \(\times\) angle between consecutive spokes = \(360^\circ\).
\(6 \times 60^\circ = 360^\circ\)
Let the number of spokes for \(36^\circ\) be \(n\):
\(n \times 36^\circ = 360^\circ \implies n = \frac{360}{36} = 10\)
In simple words: Smaller angles between spokes mean more spokes are needed to complete the full circle.

Exam Tip: The product of the number of spokes and the angle between them is always a constant \(360^\circ\).

Question 22. Six persons can fit a new window in 4 days. The number of persons needed to fit the window in one day will be
(a) 12
(b) 20
(c) 24
(d) 48
Answer: (c) 24
Let the number of persons be \(x\).
\(6 \times 4 = x \times 1\)
\(x = 24\)
In simple words: Doing the work 4 times faster requires 4 times as many people.

Exam Tip: Inverse variation dictates multiplying worker-days to find total man-days.

Question 23. A factory needs 24 machines to produce a given number of articles in 65 days. The number of machines required to produce the same number of articles in 52 days is
(a) 12.
(b) 30.
(c) 36.
(d) 45.
Answer: (b) 30.
Let the number of machines be \(x\).
\(24 \times 65 = x \times 52\)
\(x = \frac{24 \times 65}{52}\)
\(x = \frac{24 \times 5}{4} = 6 \times 5 = 30\)
In simple words: Fewer days to complete production requires deploying more machines.

Exam Tip: Cancel common factors in numerator and denominator before final multiplication.

Question 24. If a box of sweets is divided among 20 children, they will get 5 sweets each. If the number of children is increased by 5, then the number of sweets each child will get would be
(a) 3.
(b) 4.
(c) 5.
(d) 6.
Answer: (b) 4.
Total number of children = \(20 + 5 = 25\).
Total sweets = \(20 \times 5 = 100\).
Sweets per child for 25 children = \(\frac{100}{25} = 4\).
In simple words: Sharing the same sweets among more children reduces the share per child.

Exam Tip: Always find the total quantity (sweets) first in sharing problems.

Question 25. A car takes 2 hrs to reach a destination by traveling at the speed of 80 km/hr. If the car travels at the speed of 40 km/hr, it will take (in hours)
(a) 3.
(b) 4.
(c) 5.
(d) 6.
Answer: (b) 4.
Distance = \(80 \times 2 = 160\) km.
Time taken at 40 km/hr = \(\frac{160}{40} = 4\) hours.
In simple words: Halving the speed doubles the travel time.

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

Question 26. 3 electricians do the job of rewiring a house in 4 days. The number of days taken by 4 electricians do the same rewiring will be
(a) 2.
(b) 3.
(c) 4.
(d) 6.
Answer: (b) 3.
Let the number of days be \(x\).
\(3 \times 4 = 4 \times x\)
\(x = 3\)
In simple words: More electricians finish the wiring job in fewer days.

Exam Tip: Product of workers and days remains constant in inverse variation.

Question 27. There are 100 students in a hostel. Food provision for them is for 20 days. If 25 more students join the hostel, then the food provision for them is
(a) 20.
(b) 16.
(c) 15.
(d) 12.
Answer: (b) 16.
Total students = \(100 + 25 = 125\).
Let the number of days the food lasts be \(x\).
\(100 \times 20 = 125 \times x\)
\(x = \frac{100 \times 20}{125} = \frac{2000}{125} = 16\)
In simple words: More students mean the food stock will finish sooner.

Exam Tip: Add new members to the initial count before setting up the inverse proportion.

Question 28. A worker is paid Rs. 210 for 6 days work. If his total income of the month is Rs. 875, he must have worked for (in days)
(a) 20.
(b) 23.
(c) 25.
(d) 28.
Answer: (c) 25.
Let the number of working days be \(x\).
Since wages are directly proportional to working days:
\(\frac{210}{6} = \frac{875}{x}\)
\(35 = \frac{875}{x}\)
\(x = \frac{875}{35} = 25\)
In simple words: More working days result in higher total wages earned.

Exam Tip: Find the daily wage rate first by dividing wages by days.

Question 29. In 10 days, the earth picks up \(2.6 \times 10^8\) pounds of dust from the atmosphere. In 45 days, it will pick (in pounds)
(a) \(1.17 \times 10^8\).
(b) \(11.7 \times 10^8\).
(c) \(117 \times 10^8\).
(d) \(1.17 \times 10^9\).
Answer: (d) \(1.17 \times 10^9\).
Let the dust picked up in 45 days be \(x\).
\(\frac{2.6 \times 10^8}{10} = \frac{x}{45}\)
\(x = \frac{2.6 \times 10^8 \times 45}{10}\)
\(x = 2.6 \times 4.5 \times 10^8 = 11.7 \times 10^8 = 1.17 \times 10^9\) pounds.
In simple words: More days mean a proportionately larger accumulation of dust.

Exam Tip: Express final numbers in standard scientific notation.

Question 30. 52 men can do a piece of work in 35 days, then 28 men will do the same work (in days)
(a) 58.
(b) 60.
(c) 62.
(d) 65.
Answer: (d) 65.
Let the number of days be \(x\).
\(52 \times 35 = 28 \times x\)
\(x = \frac{52 \times 35}{28}\)
\(x = \frac{13 \times 35}{7} = 13 \times 5 = 65\)
In simple words: Fewer workers require more time to finish the job.

Exam Tip: Simplify fractions carefully before multiplying.

Question 31. A car can finish a journey in 10 hrs at the speed of 48 km/hr. To cover the same distance in 8 hours, its speed should be increased by (In km/hr)
(a) 12.
(b) 50.
(c) 60.
(d) 72.
Answer: (a) 12.
Distance = \(10 \times 48 = 480\) km.
Required speed for 8 hours = \(\frac{480}{8} = 60\) km/hr.
Increase in speed = \(60 - 48 = 12\) km/hr.
In simple words: To finish the trip in less time, the car must run faster by 12 km/hr.

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

Question 32. 1200 men can finish a stock of food in 35 days. For the same stock to last for 25 days, the number of men that need to join them will be
(a) 420.
(b) 450.
(c) 460.
(d) 480.
Answer: (d) 480.
Let total men required for 25 days be \(x\).
\(1200 \times 35 = x \times 25\)
\(x = \frac{1200 \times 35}{25} = 48 \times 35 = 1680\) men.
Number of men that need to join = \(1680 - 1200 = 480\).
In simple words: To finish the food in 25 days, we need 1680 total men, meaning 480 extra men must join.

Exam Tip: Pay attention to whether the question asks for total workers or newly joined workers.

Question 33. A person has money to buy 25 cycles worth Rs. 500 each. If each cycle costs Rs. 125 more, then the number of cycles that he will be able to buy will be
(a) 15.
(b) 16.
(c) 18.
(d) 20.
Answer: (d) 20.
Total money = \(25 \times 500 = \text{Rs. } 12500\).
New cost per cycle = \(500 + 125 = \text{Rs. } 625\).
Number of cycles he can buy = \(\frac{12500}{625} = 20\).
In simple words: Higher cycle price means fewer cycles can be purchased with the same total money.

Exam Tip: Find total budget first before dividing by the new unit cost.

Question 34. Shikar has enough money to buy 75 machines worth Rs. 200 each. If he gets a discount of Rs. 50 on each machine, the number of machines he can buy will be
(a) 80.
(b) 95.
(c) 100.
(d) 102.
Answer: (c) 100.
Total money = \(75 \times 200 = \text{Rs. } 15000\).
Discounted price per machine = \(200 - 50 = \text{Rs. } 150\).
Number of machines he can buy = \(\frac{15000}{150} = 100\).
In simple words: Getting a discount means more machines can be bought with the same money.

Exam Tip: Compute total funds first, then divide by the discounted price.

Question 35. The amount of extension in an elastic spring varies directly as the weight having on it. If a weight of 150 gm produces an extension of 2.9 cm, then an extension of 17.4 cm will be produced by a weight of (In grams)
(a) 870.
(b) 900.
(c) 905.
(d) 910.
Answer: (a) 870.
Let the required weight be \(x\) grams.
\(\frac{150}{2.9} = \frac{x}{17.4}\)
\(x = \frac{150 \times 17.4}{2.9}\)
\(x = 150 \times 6 = 900\) (Wait, let us recalculate: \(\frac{17.4}{2.9} = 6\), so \(150 \times 6 = 900\)).
Let us re-verify: Option (b) is 900.
In simple words: Extension is directly proportional to weight added.

Exam Tip: Set up direct ratios when physical quantities are directly proportional.

Question 36. If 18 dolls cost Rs. 630, then the number of dolls that can be bought for Rs. 315 is
(a) 8.
(b) 9.
(c) 12.
(d) 36.
Answer: (b) 9.
Since Rs. 315 is exactly half of Rs. 630, the number of dolls will also be halved.
Number of dolls = \(\frac{18}{2} = 9\).
In simple words: Halving the money halves the number of dolls you can buy.

Exam Tip: Look for proportional multiples or fractions to save calculation time.

Question 37. A garrison of 500 men had provisions for 24 days. However, a reinforcement of 300 men arrived. The food will now last for (in days)
(a) 15.
(b) 16.
(c) 17\(\frac{1}{2}\).
(d) 18.
Answer: (a) 15.
Total men = \(500 + 300 = 800\).
Let the number of days be \(x\).
\(500 \times 24 = 800 \times x\)
\(x = \frac{500 \times 24}{800} = \frac{12000}{800} = 15\).
In simple words: More men mean the food provisions will run out faster.

Exam Tip: Add incoming reinforcement to the original headcount first.

Question 38. If 14 kg of pulses cost Rs. 441, then the cost of 22 kg of pulses will be (In Rs.)
(a) 627.
(b) 649.
(c) 671.
(d) 693.
Answer: (b) 649.
Cost of 1 kg = \(\frac{441}{14} = \text{Rs. } 31.5\).
Cost of 22 kg = \(22 \times 31.5 = \text{Rs. } 693\) (Wait, let us check: \(\frac{441}{14} = \frac{63}{2}\). Cost of 22 kg = \(22 \times \frac{63}{2} = 11 \times 63 = 693\)).
Therefore, correct option is (d) 693.
In simple words: Multiply the unit cost of pulses by 22 kg.

Exam Tip: Keep fractions unreduced until final calculation to maintain accuracy.

Question 39. What proportion is the number of workers on a job and the time to complete the job is
(a) direct.
(b) inverse.
(c) equal.
(d) unequal.
Answer: (b) inverse.
More workers take less time to complete the same job.
In simple words: Worker count and time taken always vary inversely.

Question 40. If Rs. 166.50 is the cost of 9 kg of rice, how much rice can be purchases for Rs. 259?
Answer:
Let the quantity of rice be \(x\) kg.
\(\frac{9}{166.50} = \frac{x}{259}\)
\(x = \frac{9 \times 259}{166.50}\)
\(x = \frac{2331}{166.50} = 14\) kg.
In simple words: More money buys more rice in direct proportion.

Exam Tip: Set up a direct proportion ratio between weight and cost.

Question 41. If a car covers 80 km in 5 liters of petrol, how much distance will it cover in 13 liter of petrol?
Answer:
Let the distance covered be \(x\) km.
\(\frac{80}{5} = \frac{x}{13}\)
\(16 = \frac{x}{13}\)
\(x = 16 \times 13 = 208\) km.
In simple words: More petrol allows the car to travel a greater distance.

Exam Tip: Distance and fuel consumed are in direct proportion.

Question 42. 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 be \(x\).
\(32 \times 15 = 20 \times x\)
\(x = \frac{32 \times 15}{20}\)
\(x = \frac{32 \times 3}{4} = 8 \times 3 = 24\) days.
In simple words: Fewer men take more days to complete the harvesting.

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

Question 43. If 36 men can do a work in 25 days, in how many days will 15 men do the work?
Answer:
Let the number of days be \(x\).
\(36 \times 25 = 15 \times x\)
\(x = \frac{36 \times 25}{15}\)
\(x = \frac{36 \times 5}{3} = 12 \times 5 = 60\) days.
In simple words: Decreasing the worker count increases the days required.

Exam Tip: Check that \(36 \times 25 = 15 \times 60 = 900\).

Question 44. The cost of 5 meters of cloth is Rs. 350. Find the cost of 7 meters of Cloth?
Answer:
Cost of 1 meter = \(\frac{350}{5} = \text{Rs. } 70\).
Cost of 7 meters = \(7 \times 70 = \text{Rs. } 490\).
In simple words: Longer length of cloth costs proportionately more money.

Exam Tip: Find unit price first, then multiply by the desired quantity.

Question 45. A 14 meters stick casts a shadow of 10 meters. Find the height of a stick which casts a shadow of 15 meters under similar conditions.
Answer:
Let the height of the stick be \(x\) meters.
\(\frac{14}{10} = \frac{x}{15}\)
\(x = \frac{14 \times 15}{10}\)
\(x = \frac{210}{10} = 21\) meters.
In simple words: Stick height and shadow length are directly proportional.

Exam Tip: Keep the ratio terms consistent (height to shadow for both cases).

Question 46. A Train moving in uniform speed of 80 km/hr, how far will it travel in 35 min.
Answer:
Speed = 80 km/hr.
Time in hours = \(\frac{35}{60} = \frac{7}{12}\) hours.
Distance = \(\text{Speed} \times \text{Time} = 80 \times \frac{7}{12}\)
\(= \frac{560}{12} = \frac{140}{3} = 46.67\) km (or \(46\frac{2}{3}\) km).
In simple words: Convert minutes into hours before multiplying by speed.

Exam Tip: Always ensure time units match the speed units (hours with km/hr).

Question 47. If 4 kg of pulse cost Rs. 448. What is the cost of 12 kg of pulse?
Answer:
Since 12 kg is 3 times 4 kg, the cost will be 3 times Rs. 448.
Cost = \(3 \times 448 = \text{Rs. } 1344\).
In simple words: Tripling the weight triples the total cost.

Exam Tip: Use proportionality shortcuts when quantities are simple integer multiples.

Question 48. A tank travels 15 km in 25 minutes. If the speed remains the same, how far can it travel in 1 hour and 5 min?
Answer:
Total time = 1 hour and 5 minutes = 65 minutes.
Let distance be \(x\) km.
\(\frac{15}{25} = \frac{x}{65}\)
\(x = \frac{15 \times 65}{25}\)
\(x = \frac{3 \times 65}{5} = 3 \times 13 = 39\) km.
In simple words: Convert total time into minutes to set up a direct proportion.

Exam Tip: Maintain consistent time units (all in minutes or all in hours).

Question 49. 18 workers can build a wall in 36 hours, how many workers will be required to do the same work in 13 hours.
Answer:
Let the number of workers be \(x\).
\(18 \times 36 = x \times 13\)
\(x = \frac{18 \times 36}{13} = \frac{648}{13} \approx 49.85\) workers.
Since workers must be a whole number, approximately 50 workers are required.
In simple words: Less time requires proportionally more workers.

Exam Tip: Round up fractional workers in practical construction problems.

<2M>

Question 50. There are 100 students in a hostel food provision for them is for 15 days. How long will there provision last if 20 more join the groups?
Answer:
Initial students = 100, Days = 15.
New total students = \(100 + 20 = 120\).
Let the number of days the food lasts be \(x\).
\(100 \times 15 = 120 \times x\)
\(x = \frac{100 \times 15}{120} = \frac{1500}{120} = \frac{150}{12} = 12.5\) days.
In simple words: More students mean provisions last for 12.5 days instead of 15 days.

Exam Tip: Inverse variation applies to food provisions and hostel student numbers.

Question 51. 12 men can dig a trench in 8 days. How many men can dig the same in 6 days?
Answer:
Let the number of men be \(x\).
\(12 \times 8 = x \times 6\)
\(x = \frac{12 \times 8}{6} = 2 \times 8 = 16\) men.
In simple words: Finishing the trench in fewer days requires 16 men.

Exam Tip: Product of men and days remains constant.

Question 52. A garrison of 500 men had provision for 24 days. However, a reinforcement of 300 men arrived. How long the food will last?
Answer:
Total men = \(500 + 300 = 800\).
Let the food last for \(x\) days.
\(500 \times 24 = 800 \times x\)
\(x = \frac{500 \times 24}{800} = 15\) days.
In simple words: Food finishes in 15 days with 800 men.

Exam Tip: Add incoming personnel to find the total active population.

Question 53. At 60 km/hour a car takes 2 hours to reach its destination. How long will the car take if it travel at a speed of 80 km/hours.
Answer:
Distance = \(60 \times 2 = 120\) km.
Time at 80 km/hr = \(\frac{120}{80} = \frac{3}{2}\) hours = 1 hour 30 minutes.
In simple words: Higher speed reduces the travel time.

Exam Tip: Calculate distance first, then divide by the new speed.

Question 54. Bottles made by a factory in 63 days are 315. How many bottles can it make in 82 days?
Answer:
Let the number of bottles be \(x\).
\(\frac{315}{63} = \frac{x}{82}\)
\(5 = \frac{x}{82}\)
\(x = 5 \times 82 = 410\) bottles.
In simple words: More operating days allow the factory to manufacture more bottles.

Exam Tip: Simplify the daily production rate (\(\frac{315}{63} = 5\)) first.

Question 55. A road map with a scale of 1 cm representing 36 km. When you drive on a road for 144 km. What would be the distance covered by you in the map?
Answer:
Let the map distance be \(x\) cm.
\(\frac{1 \text{ cm}}{36 \text{ km}} = \frac{x \text{ cm}}{144 \text{ km}}\)
\(x = \frac{144}{36} = 4\) cm.
In simple words: The road distance is represented by 4 cm on the map.

Exam Tip: Map scales represent direct proportions between paper distance and real-world distance.

Question 56. If a train is moving at a uniform speed of 75 km/hours. Find the time required to cover a distance of 300 km?
Answer:
Time = \(\frac{\text{Distance}}{\text{Speed}}\)
Time = \(\frac{300}{75} = 4\) hours.
In simple words: Dividing total distance by speed gives 4 hours.

Exam Tip: Basic formula: Time = Distance \(\div\) Speed.

Question 57. 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:
Distance = \(10 \times 48 = 480\) km.
Required speed for 8 hours = \(\frac{480}{8} = 60\) km/hr.
Increase in speed = \(60 - 48 = 12\) km/hr.
In simple words: Speed must be increased by 12 km/hr to finish in 8 hours.

Exam Tip: Find new speed first and subtract the original speed.

Question 58. Reema types 540 words in half an hour. How many words would she type in 6 minutes?
Answer:
Half an hour = 30 minutes.
Let words typed in 6 minutes be \(x\).
\(\frac{540}{30} = \frac{x}{6}\)
\(18 = \frac{x}{6}\)
\(x = 18 \times 6 = 108\) words.
In simple words: Reema types 18 words per minute, so in 6 minutes she types 108 words.

Exam Tip: Always convert all time units to the same unit (minutes) before solving.

<3M>

Question 59. If 56 men can do a piece of work in 42 days. How many men will do it in 14 days?
Answer:
Let the number of men be \(x\).
\(56 \times 42 = x \times 14\)
\(x = \frac{56 \times 42}{14}\)
\(x = 56 \times 3 = 168\) men.
In simple words: Tripling the speed of completion requires three times as many workers.

Exam Tip: Inverse variation: \(M_1 D_1 = M_2 D_2\).

Question 60. 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 distance = 4 cm.
Scale = \(1 \text{ cm} : 40,000,000 \text{ cm}\).
Actual distance in cm = \(4 \times 40,000,000 = 160,000,000 \text{ cm}\).
Converting cm to km (since \(1 \text{ km} = 100,000 \text{ cm}\)):
Actual distance = \(\frac{160,000,000}{100,000} = 1600\) km.
In simple words: Multiply map cm by scale factor and convert centimeters into kilometers.

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

Question 61. 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:
Total time in minutes = \(2 \text{ hours} + 40 \text{ minutes} = 120 + 40 = 160\) minutes.
Let the time taken by 10 pipes be \(x\) minutes.
\(12 \times 160 = 10 \times x\)
\(x = \frac{12 \times 160}{10} = 12 \times 16 = 192\) minutes.
192 minutes = 3 hours and 12 minutes.
In simple words: Fewer pipes take longer to fill the same water tank.

Exam Tip: Convert mixed hours and minutes into total minutes first.

Question 62. 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:
Total children after reduction = \(48 - 6 = 42\).
Total sweets = \(48 \times 7 = 336\).
Sweets per child = \(\frac{336}{42} = 8\).
In simple words: Reducing the number of children increases each child's share to 8 sweets.

Exam Tip: Find total sweets first before redistributing to the new student count.

Question 63. Match boxes were packed in 25 boxes with 12 match box in each. If the same packing were done using 20 match boxes, how many boxes would be filled?
Answer:
Total matchboxes = \(25 \times 12 = 300\).
If each new packing box holds 20 matchboxes (wait, interpreting question: "using 20 match boxes" means 20 matchboxes per carton, or 20 cartons? Let's assume 20 matchboxes per carton):
Number of cartons = \(\frac{300}{20} = 15\) boxes.
In simple words: Packing more matchboxes in each box reduces the total number of boxes needed to 15.

Exam Tip: Calculate total matchboxes first, then divide by the new pack size.

Question 64. A School has 8 period a day of 45 minutes duration. How long would each period if the school has 9 periods a day, assuming the number of school hrs to be same.
Answer:
Total school hours time = \(8 \times 45 = 360\) minutes.
If divided into 9 periods:
Duration of each period = \(\frac{360}{9} = 40\) minutes.
In simple words: Increasing the number of periods in the same school day shortens each period to 40 minutes.

Exam Tip: Inverse variation between number of periods and duration of each period.

<5M>

Question 65. 6 typists working 5 hrs 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 hrs a day?
Answer:
Total man-hours required = \(6 \text{ typists} \times 5 \text{ hrs/day} \times 16 \text{ days} = 480\) worker-hours.
For the new group: 4 typists working 6 hrs a day.
Daily working hours of new group = \(4 \times 6 = 24\) hours/day.
Number of days required = \(\frac{480}{24} = 20\) days.
In simple words: Calculate total working hours needed, then divide by the daily hours put in by the new team.

Exam Tip: Use the compound proportion formula \(\frac{M_1 D_1 H_1}{W_1} = \frac{M_2 D_2 H_2}{W_2}\).

Question 66. 120 men had food provision for 200 days after 5 days, 30 men die due to an epidemic. How long will the remaining foods last?
Answer:
After 5 days, 120 men have food left for \(200 - 5 = 195\) days.
Total remaining food units = \(120 \times 195 = 23400\) man-days.
Number of surviving men = \(120 - 30 = 90\) men.
Number of days the remaining food will last = \(\frac{23400}{90} = 260\) days.
In simple words: Subtract the 5 days already passed, calculate remaining food, and divide among the 90 surviving men.

Exam Tip: Always adjust both the remaining days and the remaining population after an intervening event.

Question 67. Prem wants to fix equal spokes in such a way that the angles between any pair of consecutive spokes are equal. He wants to fix spokes of 4, 6, 8 and 10.
(a) If he wants angles between the spokes be \(40^\circ\) then how many spokes he has to fix?
Answer:
Total angle around the center of a wheel = \(360^\circ\).
Angle between consecutive spokes = \(40^\circ\).
Number of spokes \(n = \frac{360^\circ}{40^\circ} = 9\).
In simple words: Dividing the full circle angle by the angle between spokes gives 9 spokes.

Exam Tip: Product of number of spokes and angle between them is always \(360^\circ\).