Chapter focus
Proportional reasoning can describe how one quantity changes when another changes. Rates, equivalent ratios and scale factors help distinguish a genuine proportional relationship from a simple pattern.
These original examples support revision. Work through the complete investigations and exercises in your assigned Ganita Prakash Part II.
1. Recognise direct proportion
Two quantities are directly proportional when their ratio is constant: y/x = k, or equivalently y = kx, for a fixed constant k. If x doubles, y doubles; if x is multiplied by 5, y is multiplied by 5.
A constant difference does not establish direct proportion. The pairs (2, 5), (3, 6), (4, 7) have a constant difference of 3, but the ratios 5/2, 6/3 and 7/4 differ.
Check whether the relationship includes a fixed starting amount. A delivery charge plus a per-kilometre fee is not directly proportional to distance because the total is not zero when distance is zero.
2. Solve direct-proportion problems
Worked example: steady travel
A vehicle travels 150 km in 2.5 hours at a constant speed. Its rate is 150 ÷ 2.5 = 60 km/h. In 4 hours at the same rate it travels 60 × 4 = 240 km.
State the assumption: the speed remains constant. If traffic, stops or changing speed matter, the simple proportional model may not fit.
Worked example: scale a map
On a map, 1 cm represents 5 km. A route measuring 3.6 cm represents 3.6 × 5 = 18 km. Multiply by the scale factor and keep the units clear.
3. Inverse relationships
In an inverse-proportion model, one quantity increases while the other decreases so that their product remains constant: xy = k. For a fixed job, if the number of identical workers doubles, the time may halve—provided all workers work at the same rate and the job can be divided evenly.
For example, 4 identical taps fill a tank in 6 hours. Under the ideal assumption that every tap has the same steady flow, 8 taps would take 4 × 6 ÷ 8 = 3 hours. The product (number of taps) × (time) stays 24 tap-hours.
Real situations may not be perfectly inverse: extra workers can interfere with each other, and more taps may change water pressure. Identify the model’s assumptions.
4. Choose the correct model
Ratio y/x stays constant. Example: cost at a fixed price per item.
Product xy stays constant. Example: time for a fixed task as equal-rate workers increase.
A fixed fee plus a rate, or a changing speed, may not be proportional.
Use a table of values to compare ratios or products. Do not decide from one pair alone.
Common errors to avoid
- For direct proportion, the ratio remains constant; for inverse proportion, the product remains constant.
- Make sure both quantities use consistent units before comparing rates.
- Do not apply inverse proportion unless the total work or product is fixed.
- Describe any assumptions such as steady speed, equal worker rates or unchanged price.
Practice: choose and justify a model
1. A tap fills 24 L in 3 minutes at a steady rate. Find the rate per minute.
Answer: 24 ÷ 3 = 8 L/min.
2. At that rate, how much water flows in 8 minutes?
Answer: 8 × 8 = 64 L, assuming the rate stays steady.
3. Six identical workers finish a task in 10 days. How long would 12 workers take under an inverse-proportion model?
Answer: 6 × 10 = 60 worker-days; 60 ÷ 12 = 5 days.
4. A club charges ₹100 membership plus ₹20 per class. Is total cost directly proportional to the number of classes?
Answer: No. The fixed ₹100 means the cost is not zero at zero classes, and cost/classes is not constant.
5. Two kilograms of apples cost ₹180. What do 5 kg cost at the same price per kilogram?
Answer: Unit price = ₹180 ÷ 2 = ₹90/kg; 5 kg cost ₹450.
Chapter summary
In direct proportion, the ratio of corresponding quantities stays fixed. In inverse proportion, their product stays fixed under a suitable model. Use tables, unit rates and stated assumptions to decide which relationship fits a situation.