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Variation · Form 5

Variation: Worked Examples (KBAT)

This set hides the variation inside real situations - fuel, workers, and gas pressure - where students must first decide whether it is direct, inverse, or joint before finding k and answering. It helps students who can compute but struggle to translate a word problem into the right variation.

Written by the spmmath.com.my editorial team.

Worked example 1

A car's fuel used varies directly as the distance travelled. A trip from Ipoh to Kuala Lumpur of 205 km uses 15 litres.

On a road trip the tank holds 44 litres when full. How far, to the nearest km, can the car travel on a full tank at the same rate?

  1. Fuel used varies directly as distance, so let f = kd, where f is litres and d is km.
  2. Substitute f = 15, d = 205: 15 = k × 205.
  3. Solve for k: k = 15 ÷ 205 = 3/41 litre per km.
  4. We need distance for f = 44, so use d = f/k = 44 ÷ (3/41).
  5. Divide by the fraction: d = 44 × 41 / 3 = 1804 ÷ 3 = 601.33... km.
  6. Round to the nearest km: d ≈ 601 km.

Worked example 2

The time to finish painting a hall varies inversely as the number of painters working. With 6 painters the job takes 10 days.

The contractor is offered the job on condition it is done in at most 4 days. What is the least number of painters needed?

  1. Time varies inversely as number of painters, so let t = k/n, where t is days and n is painters.
  2. Substitute t = 10, n = 6: 10 = k/6.
  3. Solve for k: k = 10 × 6 = 60.
  4. Equation is t = 60/n.
  5. The job must satisfy t ≤ 4, so 60/n ≤ 4.
  6. Rearrange: n ≥ 60 ÷ 4 = 15.
  7. n must be a whole number of painters, and 15 already meets n ≥ 15, so the least is 15.

Worked example 3

For a fixed mass of gas, the pressure P varies directly as the temperature T (in kelvin) and inversely as the volume V. A sealed cylinder holds gas at P = 120 kPa, T = 300 K and V = 5 litres.

The gas is warmed to T = 360 K and the piston is pushed so V = 4 litres. Find the new pressure.

  1. P varies directly as T and inversely as V, so write P = kT/V.
  2. Substitute P = 120, T = 300, V = 5: 120 = k × 300 / 5.
  3. Simplify the right side: 120 = 60k.
  4. Solve for k: k = 120 ÷ 60 = 2.
  5. Equation is P = 2T/V.
  6. Substitute T = 360, V = 4: P = 2 × 360 / 4 = 720 ÷ 4 = 180.

Worked example 4

The cost of printing one booklet varies directly as the number of pages in it. A 120-page booklet costs RM4.80 to print.

A school wants to print 500 copies of a booklet and has a budget of RM3000. Find the maximum number of pages each booklet can have within the budget.

  1. Cost per booklet C = k × p (p = number of pages). Find k: 4.80 = k × 120, so k = 0.04.
  2. Cost of one booklet with p pages = 0.04p.
  3. Cost of 500 copies = 500 × 0.04p = 20p.
  4. Within budget: 20p ≤ 3000, so p ≤ 150.

Worked example 5

The braking distance of a car varies directly as the square of its speed. At 40 km/h the braking distance is 8 m.

Find the braking distance at 100 km/h and state by what factor the braking distance increases when the speed rises from 40 km/h to 100 km/h.

  1. Variation: d = k v².
  2. Find k: 8 = k × 40² = 1600k, so k = 0.005.
  3. Equation: d = 0.005 v².
  4. At v = 100: d = 0.005 × 100² = 0.005 × 10000 = 50 m.
  5. Factor of increase = 50 ÷ 8 = 6.25.

Worked example 6

The safe load a wooden beam can support varies directly as its width and the square of its depth, and inversely as its length. A beam 6 cm wide, 10 cm deep and 4 m long supports 600 kg.

A second beam of the same wood is 8 cm wide, 12 cm deep and 6 m long. Find the safe load the second beam can support.

  1. Combined variation: L = k × (w × d²) / ℓ.
  2. Find k using the first beam: 600 = k × (6 × 10²) / 4 = k × 600 / 4 = 150k, so k = 4.
  3. Equation: L = 4 × (w × d²) / ℓ.
  4. Second beam: L = 4 × (8 × 12²) / 6 = 4 × (8 × 144) / 6 = 4 × 1152 / 6 = 4 × 192 = 768.

Worked example 7

The light intensity I received from a lamp varies inversely as the square of the distance d from the lamp. At a distance of 2 m, the intensity is 80 lux.

A student needs at least 20 lux of light to read comfortably. Find the greatest distance from the lamp at which the light is still strong enough to read by.

  1. Write the equation: I = k/d².
  2. Substitute d = 2, I = 80: 80 = k/(2)² = k/4.
  3. Solve for k: k = 80 × 4 = 320.
  4. Equation is I = 320/d². For the greatest distance, set I = 20 (the minimum usable intensity).
  5. 20 = 320/d², so d² = 320 ÷ 20 = 16, giving d = 4.
  6. The greatest usable distance is 4 m.

Worked example 8

The electrical resistance R (in ohms) of a wire varies directly as its length L (in metres) and inversely as the square of its diameter d (in mm). A wire of length 4 m and diameter 1 mm has a resistance of 8 Ω.

An electrician needs a wire of length 6 m with resistance not exceeding 3 Ω. Find the minimum diameter of wire needed.

  1. Write the equation: R = kL/d².
  2. Substitute L = 4, d = 1, R = 8: 8 = k(4)/(1)² = 4k.
  3. Solve for k: k = 8 ÷ 4 = 2.
  4. Equation is R = 2L/d². For L = 6, set R = 3 (the maximum allowed resistance): 3 = 2(6)/d² = 12/d².
  5. d² = 12 ÷ 3 = 4, giving d = 2.
  6. The minimum diameter needed is 2 mm.

Source:DSKP KSSM Mathematics Form 4 and 5 (Versi English)

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Frequently asked questions

What do KBAT questions on variation usually involve?

KBAT questions embed variation inside a real-world context, like cost, speed, or resource usage, described entirely in words, without stating "y" or "x" directly. You must first identify which quantity is which variable and how they relate before you can even write the variation equation.

What does the examiner reward in a variation KBAT answer?

Marks go to correctly translating the word problem into a variation equation with the right variables and relationship (direct, inverse, or joint), shown as a clear first step. An answer that reaches the right number without this translation step visible often loses marks even if correct.

What's the most common error in variation KBAT questions?

Students often misidentify which quantity varies directly and which varies inversely, especially when the context uses unfamiliar wording instead of "varies as". Underline the two changing quantities in the question first, decide their relationship carefully, then set up the equation, don't guess the formula from memory alone.

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