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Chapter 10: Direct and Inverse Proportions > Moderate Level Worrksheet

Moderate Level Worrksheet

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

Note: Answer each question with steps and explanation. Write down the answers on sheet and submit to the school subject teacher.

In this moderate level, we'll explore more complex proportion problems and understand the mathematical relationships better.

Let's practice solving problems that require deeper analysis and multi-step reasoning.

1. Define constant of proportionality.

The constant of proportionality is between two directly proportional quantities.

Perfect! If x/y = k (constant), then k is the constant of proportionality.

2. What is the condition for two quantities to be in inverse proportion?

Two quantities are in inverse proportion if their .

Excellent! If x₁ × y₁ = x₂ × y₂, they are in inverse proportion.

3. Write two real-life examples of inverse proportion.

Answer:

4. If 20 workers complete a work in 6 days, how many workers are needed for 3 days?

Answer: Answer: workers

Correct! Half the time needs double the workers: 40 workers.

5. The cost of 12 pens is ₹96. Find the cost of 1 pen.

Answer:

Perfect! Cost of 1 pen = 96 ÷ 12 = ₹8.

Drag each relationship to its correct category:

x/y = constant
x × y = constant
More items → More cost
More speed → Less time
Distance ∝ Time (constant speed)
Workers ∝ 1/Time
Direct Proportion
Inverse Proportion

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

1. A car travels 120 km in 3 hours. Find the time taken to travel 200 km at the same speed.

Speed = Distance ÷ Time = 120 ÷ 3 = km/h

Time to travel 200 km = Distance ÷ Speed = 200 ÷ 40 = hours

Excellent! The car will take 5 hours to travel 200 km.

2. A 15 m high pole casts a shadow of 20 m. Find the height of a pole that casts a shadow of 32 m under the same conditions.

This is direct proportion: Height ∝ Shadow length

Using ratio: 15/20 = h/32

h = (15 × 32) ÷ 20 = ÷ 20 = m

Perfect! The height of the pole is 24 m.

3. A factory employs 50 men to produce 1000 units in 6 days. How many days will 75 men take to produce 1500 units (same efficiency)?

Work done by 50 men in 6 days = 1000 units

Work per man per day = 1000 ÷ (50 × 6) = 1000 ÷ = units (approx)

For 75 men to produce 1500 units: Days = 1500 ÷ (75 × 3.33) = 1500 ÷ = days

Excellent! 75 men will take 6 days to produce 1500 units.

4. 8 taps can fill a tank in 27 minutes. How long will 6 taps of the same type take?

This is inverse proportion: More taps → Less time

Using formula: 8 × 27 = 6 × T

T = (8 × 27) ÷ 6 = ÷ 6 = minutes

Great! 6 taps will take 36 minutes to fill the tank.

5. A ship sails 180 km in 4 hours. How much time will it take to sail 450 km at the same speed?

This is direct proportion: More distance → More time

Using ratio: 180/4 = 450/T

T = (450 × 4) ÷ 180 = ÷ 180 = hours

Perfect! The ship will take 10 hours to sail 450 km.

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

1. A certain number of men can finish a piece of work in 8 days. If there were 4 men more, the work could be finished in 2 days less. Find the number of men originally employed.

Let the original number of men = x

Original: x men complete work in days

Total work = x × 8 = man-days

With 4 more men: (x + 4) men complete work in 8 − 2 = days

Total work = (x + 4) × 6 = man-days

Since work is same: 8x = 6x + 24

2x =

x = men

Excellent! Originally 12 men were employed.

2. 5 pumps can fill a tank in 1 hour 20 minutes. How long will it take if only 4 pumps are used?

Time = 1 hour 20 minutes = minutes

This is inverse proportion: Less pumps → More time

Using formula: 5 × 80 = 4 × T

T = (5 × 80) ÷ 4 = ÷ 4 = minutes

Converting: 100 minutes = hour minutes

Perfect! 4 pumps will take 1 hour 40 minutes.

3. 24 men working 8 hours a day can complete a work in 15 days. In how many days can 20 men working 9 hours a day complete it?

Total work = Men × Hours per day × Days

Total work = 24 × 8 × 15 = man-hours

For 20 men working 9 hours/day: 20 × 9 × D = 2880

180 × D = 2880

D = 2880 ÷ 180 = days

Excellent! 20 men working 9 hours/day will take 16 days.

4. A car takes 4 hours to cover a distance at a certain speed. If the speed is increased by 15 km/h, it takes 30 minutes less. Find the distance covered.

Let original speed = s km/h

Distance = Speed × Time = s × 4 = km

New speed = (s + 15) km/h

New time = 4 hours − 30 min = hours

Distance = (s + 15) × 3.5 = km

Since distance is same: 4s = 3.5s + 52.5

0.5s =

s = km/h

Distance = 4s = 4 × 105 = km

Perfect! The distance covered is 420 km.

Part B: Objective Questions - Test Your Knowledge!

Answer these multiple choice questions:

__{.m-red}6. If 4 bags weigh 100 kg, 10 bags will weigh _____

(a) 100 (b) 200 (c) 250 (d) 300

100
kg
200
kg
250
kg
300
kg

Correct! Weight of 1 bag = 100 ÷ 4 = 25 kg. So 10 bags = 10 × 25 = 250 kg.

__{.m-red}7. The constant ratio in direct proportion is called _____

(a) slope (b) constant of proportionality (c) ratio factor (d) multiplier

slope
constant of proportionality
ratio factor
multiplier

Perfect! The constant ratio k in x/y = k is called the constant of proportionality.

__{.m-red}8. The graph of direct proportion is a _____

(a) straight line (b) curve (c) parabola (d) circle

straight line
curve
parabola
circle

Excellent! Direct proportion (y = kx) always gives a straight line passing through origin.

__{.m-red}9. If speed × time = constant, the relation is _____

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

direct
inverse
constant
linear

Correct! When product is constant (speed × time = distance), it's inverse proportion.

10. If 2 workers take 6 hours, 6 workers take ___ hours.

(a) 2 (b) 4 (c) 6 (d) 12

2
hours
4
hours
6
hours
12
hours

Perfect! Triple the workers (2→6) means one-third the time: 6 ÷ 3 = 2 hours.

🎉 Outstanding Work! You've Mastered Moderate Level Proportions!

Here's what you learned:

  • Constant of Proportionality: The fixed ratio (k) in direct proportion where x/y = k

  • Inverse Proportion Condition: Product remains constant (x₁ × y₁ = x₂ × y₂)

  • Complex Problem Types:

    • Shadow and height problems (similar triangles)
    • Work efficiency with varying workers and hours
    • Speed-distance-time relationships
    • Multi-step proportion problems
  • Key Formulas:

    • Direct: x₁/y₁ = x₂/y₂ or y = kx
    • Inverse: x₁ × y₁ = x₂ × y₂ or xy = k
    • Work: Work = Men × Hours per day × Days
  • Problem-Solving Strategy:

    1. Identify the type of proportion
    2. Set up the correct equation
    3. Solve systematically
    4. Verify the answer makes logical sense
  • Graphical Understanding:

    • Direct proportion → Straight line through origin
    • Inverse proportion → Hyperbolic curve

These intermediate skills prepare you for solving complex real-world problems in engineering, economics, and science!