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Lesson 7 of 8

Proportional Reasoning-1 · Lesson 7 of 8

Applying Proportional Reasoning

“Solve all twelve final problem types by combining ratio simplification, fair comparisons, sharing, unit conversion and interpretation of practical answers.”

Learning Objectives

• Select a method from the meaning of the quantities. • Interpret fractional answers in practical situations. • Combine conversions with sharing and proportion. • Solve changing-age and different-rate problems. • Explain and check each stage of a mixed solution.

In a mixed problem, the most important decision often comes before the calculation. Do you need to simplify a comparison, share a total, convert a unit or find how much corresponds to one unit? Different questions may contain similar-looking numbers but require different starting points. This lesson brings the earlier ideas together and gives you time to explain the choice of method, including what the final answer means in the situation.

The figures about populations, materials, farming and prices below are the supplied exercise data. Use them as stated mathematical models; they are not claims about current prices or general instructions for real-world use.

Figure it Out

Start by writing what is known and what is requested. Add units to every measured amount. Then look for the relationship that connects the quantities. A final check should cover both the arithmetic and the interpretation: a fraction of a bus may be a useful intermediate result, but it is not a bus you can hire.

Fruit-drink ratio

Problem
Anagh combines 600 mL orange juice with 900 mL apple juice. Find orange : apple in simplest form.

How many buses?

Problem
Three identical full buses carry 162 people. How many such buses are needed for 204 people, and how many seats remain empty?

Capacity: 4 × 54 = 216 seats54545454Passengers: 204Unused seats: 216 − 204 = 12
Four 54-seat buses accommodate 204 people

To compare crowding, population alone is insufficient. A larger population spread over much more land may have fewer people per square kilometre. Compare each city at the same one-square-kilometre amount of area.

The same principle appears when comparing prices per kilogram or material quantities per acre: a common comparison amount makes the relationship visible.

Which city is more crowded in the given data?

Problem
Use Delhi: area 1484 km² and population 30 million; Mumbai: area 550 km² and population 20 million. Compare people per km².

A hypothetical neck-to-body proportion

Problem
A crane’s neck : rest of body is 4 : 6. If a person of height h had this same proportion, what would the neck length be?

The Lilavati saffron problem

Problem
If 2½ palas of saffron cost 3/7 niska, how much can be bought for 9 niskas at the same price per pala?

For an age problem, keep the age difference fixed. You may reason directly from that difference or let t represent elapsed years. Do not multiply the two current ages by the future ratio, because both people age by the same number of years.

When does an age ratio become 1 : 2?

Problem
Harmain is 1 year old and her brother is 5. How old is Harmain when her age : his age is 1 : 2?

Using a given equal-volume mass ratio

Problem
For this exercise, equal volumes of gold and water have mass ratio 37 : 2, and one litre of water has mass 1 kg. Find the stated model’s mass for one litre of gold.

Manure for a rectangular plot

Problem
Using the exercise rate of 10 tonnes per acre, find the amount for a 200 ft by 500 ft plot.

100,000 ft²÷ 43,560≈ 2.29568 acres2.29568 acres× 10 tonnes/acre≈ 22.9568 tonnes
Area and rate calculation for the plot
A bucket filled at constant flow

Problem
A tap fills a 500 mL mug in 15 seconds. How long will it take to fill a 10 L bucket at the same flow?

Cost of a smaller area

Problem
Land costs ₹15,00,000 per acre. At the same price per area, what is the cost of 2400 ft²?

Working faster changes the time required for the same job. If one machine covers four times the area in the same time, it needs one quarter of the time for a fixed area. Explain what “four times faster” means in the exercise before choosing multiplication or division.

Oxen and tractor

Problem
Oxen take six hours per acre. A tractor works at four times their area-per-hour rate. Find the time each needs for 20 acres.

Cost of the metals in the stated coin model

Problem
A coin model has total mass 7.74 g and copper : nickel = 3 : 1. Copper costs ₹906/kg and nickel ₹1341/kg. Find the material cost.

Total = 7.74 gCopper: 3 partsNickel: 1 parts4 equal parts altogether
Split the stated 7.74 g coin mass before pricing the metals
Choose the method from the question

Use HCF for simplifying a ratio; common-amount comparison for prices or crowding; total ratio parts for sharing; fixed differences for changing ages; and unit conversion before comparing mismatched measures. Practical answers may need rounding up or an explicit approximation.

Quiz

Quick check

Three full buses hold 162 people. How many identical buses carry 204 people?

Quick check

Which comparison measures average crowding?

Quick check

Harmain is 1 and her brother is 5. When their ratio is 1 : 2, how old is Harmain?

Quick check

At constant flow, how long does a tap take for 10 L if 500 mL takes 15 seconds?

Quick check

What should you do before multiplying a metal’s mass in grams by a price per kilogram?

Practice Problems

Practice Problems
  1. A drink contains 600 mL orange juice and 900 mL apple juice. Find the ratio and the orange fraction of the total.
  2. Three full 54-seat buses carried a group. This time 204 people travel. Find buses needed and unused seats.
  3. At the exercise rate of ten tonnes per acre, calculate the amount for a 200 ft by 500 ft plot.
  4. If 2½ palas cost 3/7 niska, find how many palas cost 9 niskas.
  5. Find the copper and nickel masses and total metal cost for the 7.74 g, 3 : 1 coin model with prices ₹906/kg and ₹1341/kg.
Practice 1: worked solution

1. The ratio is 600 : 900 = 2 : 3. 2. The whole contains five ratio parts, so orange is 2/5 of the total.

Practice 2: worked solution

1. 204/54 lies between three and four, so four buses are needed. 2. Their total capacity is 216; unused seats = 216 − 204 = 12.

Practice 3: worked solution

1. Area = 100,000 ft². In acres it is 100,000/43,560. 2. Multiply by ten to obtain approximately 22.9568 tonnes, or 22.96 tonnes to two decimal places.

Practice 4: worked solution

1. Cost factor = 9 ÷ (3/7) = 21. 2. Weight = 2½ × 21 = 52½ palas. 3. Both corresponding quantities have been multiplied by the same factor.

Practice 5: worked solution

1. Masses: 5.805 g copper and 1.935 g nickel. 2. In kilograms these are 0.005805 and 0.001935. 3. Costs are ₹5.25933 and ₹2.594835, totalling ₹7.854165 ≈ ₹7.85. 4. Check the masses add to 7.74 g and their ratio is 3 : 1 before pricing.

Key Takeaways

Key Takeaways

• Choose a method from the quantities and relationships, not just the appearance of the numbers. • A fair comparison often uses the same unit amount, such as one kilogram or one square kilometre. • Round capacity requirements up when all people or objects must fit. • Age differences remain constant even when age ratios change. • Convert units before applying a rate or comparing ratios. • Label approximate results and explain the assumptions behind practical models.