10.6 Inverse Proportions

So far our quantities have grown side by side: more parts, more paint. That is direct proportion. Recall the rule of three: if a : b :: c : d then d = \dfrac{bc}{a}, which lets us find a missing fourth quantity.

Worked Example

If 6 workers stack 5400 bricks in a day, how many workers stack 21600 bricks in a day?

More bricks call for more workers, a direct proportion. Write 5400 : 21600 :: 6 : x. Then x = \frac{21600 \times 6}{5400} = 24. So 24 workers are needed.

But sometimes more of one thing means less of another. Suppose a delivery rider covers the route from Indore to Dewas in 4 hours at 24 km/h. If she switches to a scooter at 48 km/h, how long will it take? Here, moving faster should take less time. Look at the table:

Mode Walk Bicycle Auto Scooter
Speed (km/h) 4 12 24 48
Time (hours) 24 8 4 2

A bicycle is 3 times as fast as walking (12 \div 4 = 3), and the time falls to a third (24 \div 8 = 3). The speed climbs by exactly the factor that the time drops. They change by the same factor but in opposite directions, this is inverse proportion.

Math Talk Would it be right to write the travel problem as the direct proportion 24 : 48 :: 4 : x? Try it and you would get x = 8, but that claims the scooter takes longer, which makes no sense. As speed goes up, time must come down. Deciding which way the quantities move is the very first step in any proportion problem.

Notice something elegant: in the table, speed \times time is identical for every row: 4 \times 24 = 12 \times 8 = 24 \times 4 = 48 \times 2 = 96 \text{ km}. That constant 96 is simply the distance from Indore to Dewas, which never changes!

Inverse Proportion Two quantities x and y vary in inverse proportion when their product stays constant: xy = k \quad(\text{a constant}). If x_1, x_2 are two values of x with matching values y_1, y_2 of y, then x_1 y_1 = x_2 y_2 = k, \qquad\text{which rearranges to}\qquad \frac{x_1}{x_2} = \frac{y_2}{y_1}. When one quantity is multiplied by a factor n, the other is multiplied by \dfrac{1}{n}.

So for the scooter: speed \times time = 96, giving 48 \times x = 96, hence x = 2 hours.

Direct or Inverse? A Quick Test Before computing, ask: “If I double the first quantity, does the second double or halve?”

  • Doubles (both rise together) → direct proportion → keep the ratio \dfrac{x_1}{x_2} = \dfrac{y_1}{y_2}.
  • Halves (one rises, one falls) → inverse proportion → keep the product x_1 y_1 = x_2 y_2.

This single question saves you from setting up the wrong equation.

Worked Example

24 workers take 5 days to lay a road. How many days for 12 workers?

Fewer workers means more days, inverse proportion, so x_1 y_1 = x_2 y_2: 24 \times 5 = 12 \times y_2 \quad\Rightarrow\quad y_2 = \frac{24 \times 5}{12} = 10 \text{ days}. Halving the workers doubled the days, exactly as inverse proportion predicts.

Worked Example

3 pumps fill a tank in 16 hours. How long with 6 pumps of the same kind?

More pumps, less time, inverse proportion. With the product constant, 3 \times 16 = 6 \times x \quad\Rightarrow\quad x = \frac{3 \times 16}{6} = 8 \text{ hours}.

Worked Example

A hostel has rations for 90 students for 16 days. If 30 more students arrive, how long will the rations last?

Now there are 90 + 30 = 120 students. More mouths, fewer days, inverse proportion: 90 \times 16 = 120 \times x \quad\Rightarrow\quad x = \frac{90 \times 16}{120} = 12 \text{ days}.

Worked Example

Asha peels a basket of potatoes in 2 hours; Bina peels the same basket in 3 hours. Working together, how long do they take?

Treat the whole job as 1 unit of work. In one hour Asha does \dfrac{1}{2} unit; Bina does \dfrac{1}{3} unit. Together in one hour they do \tfrac{1}{2} + \tfrac{1}{3} = \tfrac{5}{6} \text{ unit}. Work done and time taken are directly proportional, so \tfrac{5}{6} : 1 :: 1 : x, giving \tfrac{5}{6} \times x = 1 \quad\Rightarrow\quad x = \frac{6}{5} \text{ hour}. Together they finish in \tfrac{6}{5} hour, that is 1 hour 12 minutes, faster than either alone.

Worked Example

A printing press needs 60 machines to finish an order in 75 days. How many machines to finish the same order in 50 days?

Fewer days demand more machines, inverse proportion: 60 \times 75 = x \times 50 \quad\Rightarrow\quad x = \frac{60 \times 75}{50} = 90 \text{ machines}.

Use the inverse-proportion explorer below: fix the constant product k, slide one quantity, and watch the other move the opposite way.

Inverse-Proportion Explorer (x times y = k)
Set a constant product k, then drag x. Watch y adjust so that x times y always stays k.

Figure it Out: Inverse Proportion

Practice

  1. Which of these pairs are in inverse proportion?
    1. Number of taps filling a tank and the time to fill it.
    2. Number of painters hired and the days to paint a fixed wall.
    3. The distance a car can travel and the petrol in the tank.
    4. Speed of a cyclist and the time to cover a fixed route.
    5. Length of cloth bought and the price paid at a fixed rate per metre.
    6. Number of pages in a book and the time to read it at a fixed speed.
  2. If 30 pens cost ₹150, how much will 24 such pens cost?
  3. A tank holds enough water for 18 families for 8 days. If 6 more families move in, how long will the water last? What assumption must you make?
  4. Order these average daily sleep hours from the largest sleeper to the smallest, choosing from: 16,\ 2.5,\ 19,\ 8,\ 4,\ 12,\ 10,\ 14.
  5. A transport survey pie chart has slices: Walk 90°, Bus 120°, Two-wheeler 60°, Car 60°, Cycle 30°. (i) Most common mode? (ii) What fraction travel by car? (iii) If 24 children travel by car, how many took part in the survey? (iv) Which two modes are used by equal numbers?
  6. Four workers paint a fence in 6 days. If two more workers join, how many days? What assumption is needed?
  7. It takes 8 hours to fill 2 tanks of the same size. How long to fill 5 such tanks?
  8. Chairs are arranged in 30 rows of 16 chairs. Rearranged with 24 chairs per row, how many rows?
  9. A school has 9 periods of 40 minutes each. If it switches to 10 periods, how long is each, keeping the same total school time?
  10. A small pump fills a tank in 4 hours, a large pump in 6 hours. Together, how long to fill it?
  11. A workshop needs 60 machines to make a batch in 75 days. How many machines to make the same batch in 50 days?
  12. A car takes 3 hours at 40 km/h. How long at 60 km/h?
  1. Inverse: (i) more taps, less time, yes; (ii) more painters, fewer days, yes; (iv) faster cyclist, less time, yes; (vi) more pages, more time, so direct (not inverse). (iii) more petrol, more distance, direct. (v) more cloth, more price, direct. So (i), (ii) and (iv) are inverse proportions.
  2. Watch out, this is direct, not inverse! More pens cost more. 30 pens cost ₹150, so 1 pen costs ₹5, and 24 pens cost 24 \times 5 = \mathbf{₹120}.
  3. More families, fewer days, inverse. Now 18 + 6 = 24 families: 18 \times 8 = 24 \times x, so x = \dfrac{144}{24} = \mathbf{6} days. Assumption: every family uses the same amount of water each day.
  4. A reasonable matching (largest to smallest sleeper): 19, 16, 14, 12, 10, 8, 4, 2.5, animals that sleep most (e.g. bats/koalas \approx 19) down to those that sleep least (e.g. giraffes/elephants \approx 2.5–4). (Exact pairing depends on the chart shown.)
    1. Bus has the largest slice (120°). (ii) Car = 60° out of 360°, fraction = \dfrac{60}{360} = \mathbf{\tfrac{1}{6}}. (iii) If \tfrac{1}{6} of the children = 24, the total is 24 \times 6 = \mathbf{144} children. (iv) Two-wheeler and Car each have 60°, so equal numbers (\tfrac{1}{6} \times 144 = 24 each).
  6. More workers, fewer days, inverse. Now 4 + 2 = 6 workers: 4 \times 6 = 6 \times x, so x = \mathbf{4} days. Assumption: every worker paints at the same steady rate.
  7. More tanks, more time, direct. 8 hours for 2 tanks \Rightarrow 4 hours per tank, so 5 tanks take 5 \times 4 = \mathbf{20} hours.
  8. Total chairs = 30 \times 16 = 480. With 24 per row: 480 \div 24 = \mathbf{20} rows. (Check: rows \times chairs is constant, 30 \times 16 = 24 \times 20 = 480, inverse proportion.)
  9. Total time is fixed: 9 \times 40 = 360 min. With 10 periods: 360 \div 10 = \mathbf{36} minutes each (more periods, shorter periods, inverse).
  10. In one hour the small pump fills \tfrac{1}{4} of the tank, the large pump \tfrac{1}{6}. Together \tfrac{1}{4} + \tfrac{1}{6} = \tfrac{5}{12} per hour, so the full tank takes \dfrac{1}{5/12} = \dfrac{12}{5} = \mathbf{2.4} hours (2 hours 24 minutes).
  11. Fewer days, more machines, inverse. 60 \times 75 = x \times 50, so x = \dfrac{60 \times 75}{50} = \mathbf{90} machines.
  12. Faster speed, less time, inverse (distance fixed). Distance = 40 \times 3 = 120 km, so 120 = 60 \times x, giving x = \mathbf{2} hours.