Hard Level Worksheet

Part A: Subjective Questions - Very Short Answer (1 Mark Each)

This hard level involves complex problems with multiple variables and mixed proportions requiring advanced problem-solving skills.

Master these challenging concepts to excel in competitive examinations and real-world applications.

1. Define mixed proportion.

Mixed proportion involves situations where one variable is directly proportionalsome quantities vary directly and another is inversely proportionalothers vary inversely simultaneously.

Perfect! Mixed proportion combines both direct and inverse relationships in a single problem.

2. Write one example involving both direct and inverse proportion.

Example: Work completed is directly proportional to workersMore workers and more days complete more work, but inversely proportional to timedirectly proportional to time.

Excellent! Work ∝ (Men × Days), showing both direct relationships.

{.reveal(when="blank-2")}3. If x and y vary inversely, what is the nature of their product?

Their product is .

Correct! In inverse proportion, xy = constant always.

4. When speed is doubled, what happens to the time for a fixed distance?

Time becomes .

Great! Speed and time are inversely proportional for fixed distance.

5. What is the ratio of y₁/y₂ when x₁/x₂ = 3/5 in inverse proportion?

Answer: 5/35:3

Perfect! In inverse proportion, x₁/x₂ = y₂/y₁, so y₁/y₂ = 5/3.

Drag each problem type to its correct category:

Part A: Section B – Short Answer Questions (2 Marks Each)

1. 12 men can dig a canal 3150 m long in 36 days. How many men are required to dig a 3900 m canal in 24 days?

Work done by 1 man in 1 day = 3150 ÷ (12 × 36) = 3150 ÷ = m (approx)

For 3900 m canal in 24 days: Men needed = 3900 ÷ (7.29 × 24)

Men = 3900 ÷ ≈ → men (round up)

Excellent! Approximately 23 men are required.

2. 10 men can lay a road of 75 km in 5 days. In how many days can 15 men lay a 45 km road?

Work done by 1 man in 1 day = 75 ÷ (10 × 5) = 75 ÷ = km

For 15 men to lay 45 km: Days = 45 ÷ (15 × 1.5) = 45 ÷ = days

Perfect! 15 men will take 2 days.

3. If 14 typists working 6 hours a day complete typing in 12 days, how long will 4 typists working 7 hours a day take?

Total work = 14 × 6 × 12 = typist-hours

For 4 typists working 7 hours/day: Days = 1008 ÷ (4 × 7) = 1008 ÷ = days

Excellent! 4 typists will take 36 days.

4. 16 men can build a wall in 24 days working 6 hours daily. How long will 12 men working 8 hours daily take to complete it?

Total work = 16 × 24 × 6 = man-hours

For 12 men working 8 hours/day: Days = 2304 ÷ (12 × 8) = 2304 ÷ = days

Great! 12 men will take 24 days.

5. A pump fills a tank in 8 hours. If 4 such pumps are used, how long will they take?

This is inverse proportion: More pumps → Less time

Using formula: 1 × 8 = 4 × T

T = 8 ÷ 4 = hours

Perfect! 4 pumps will fill the tank in 2 hours.

Part A: Section C – Long Answer Questions (4 Marks Each)

1. A contractor employs 60 men to complete a work in 50 days. After 20 days, he employs 40 more men. Find how many days earlier the work will be completed.

Total work = 60 × 50 = man-days

Work done in first 20 days = 60 × 20 = man-days

Remaining work = 3000 − 1200 = man-days

After 20 days, total men = 60 + 40 = men

Days to complete remaining work = 1800 ÷ 100 = days

Total time taken = 20 + 18 = days

Work completed earlier = 50 − 38 = days

Excellent! The work will be completed 12 days earlier.

2. A train travels 240 km in 4 hours. If its speed increases by 20 km/h, how much time will it save?

Original speed = 240 ÷ 4 = km/h

New speed = 60 + 20 = km/h

Time at new speed = 240 ÷ 80 = hours

Time saved = 4 − 3 = hour

Perfect! The train will save 1 hour.

3. A job can be completed by 12 men in 10 days, working 6 hours a day. If 9 men work 8 hours a day, in how many days will the work be completed?

Total work = Men × Days × Hours/day = 12 × 10 × 6 = man-hours

For 9 men working 8 hours/day: 9 × D × 8 = 720

72 × D = 720

D = 720 ÷ 72 = days

Excellent! The work will be completed in 10 days.

4. A car travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 48 minutes less. Find the original speed of the car.

Let original speed = s km/h

Original time = 360/s hours

New speed = (s + 5) km/h

New time = 360/(s + 5) hours

Time difference = 48 minutes = 48/60 = hours

Equation: 360/s − 360/(s + 5) = 0.8

360(s + 5 − s) = 0.8 × s(s + 5)

360 × 5 = 0.8(s2 + 5s2s + 5)

= 0.8s2 + 4s

0.8s2 + 4s 1800 = 0

Dividing by 0.8: s2 + 5s − 2250 = 0

Using formula or factoring: (s + 50)(s − 45) = 0

Since speed is positive: s = km/h

Perfect! The original speed of the car is 45 km/h.

Part B: Objective Questions - Test Your Knowledge!

Answer these multiple choice questions:

6. If time increases, speed ___ in inverse proportion.

(a) increases (b) decreases (c) same (d) doubles

Correct! In inverse proportion, if time increases, speed must decrease to cover the same distance.

7. If y ∝ 1/x, then x = 3 gives y = 5, find y when x = 2.

(a) 7.5 (b) 3 (c) 4 (d) 2.5

Perfect! xy = constant, so 3 × 5 = 2 × y, giving y = 15/2 = 7.5.

__{.m-red}8. If the cost of 8 m cloth is ₹400, cost of 12 m cloth = _____

(a) ₹400 (b) ₹600 (c) ₹800 (d) ₹300

Excellent! Cost per meter = 400/8 = ₹50, so 12 m = 12 × 50 = ₹600.

9. If 5 men complete a wall in 6 days, how long will 15 men take?

(a) 2 (b) 4 (c) 6 (d) 3

Perfect! Triple the men (5→15) means one-third the time: 6 ÷ 3 = 2 days.

10. If x/y = 2/3, then x and y are in ___ proportion.

(a) direct (b) inverse (c) equal (d) constant

Correct! When x/y = constant ratio, x and y are in direct proportion.

🎉 Exceptional Achievement! You've Mastered Advanced Proportion Concepts!

Here's what you learned:

• Mixed Proportion: Problems involving both direct and inverse relationships simultaneously

  • Example: Work ∝ (Men × Days) but inversely ∝ 1/Efficiency

• Complex Multi-Variable Problems:

  • Men × Days × Hours per day = Constant Work
  • Length × Workers ÷ Days = Rate of work
  • Speed × Time = Distance (with variable changes)

• Advanced Problem Types:

  • Mid-project changes (adding/removing workers)
  • Time savings with speed adjustments
  • Quadratic equations in speed-time problems
  • Percentage changes in proportional relationships

• Mathematical Relationships:

  • Direct: x₁/x₂ = y₁/y₂ or y = kx
  • Inverse: x₁/x₂ = y₂/y₁ or xy = k
  • Mixed: Combining both relationships

• Problem-Solving Framework:

  1. Identify all variables and their relationships
  2. Determine which are direct and which are inverse
  3. Calculate total work or distance as constant
  4. Set up equations systematically
  5. Solve and verify logical consistency

• Graphical Understanding:

  • Direct proportion → Linear (y = kx)
  • Inverse proportion → Hyperbolic (xy = k)
  • Mixed → Complex curves

• Real-world Applications:

  • Project management (resource optimization)
  • Transportation (time-speed-distance)
  • Production planning (workers, hours, output)
  • Cost analysis and budgeting

These advanced skills are essential for competitive exams, engineering problems, and complex decision-making in business and science!