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Variation · Form 5

Variation: Worked Examples (Easier)

This set covers the basics of variation: writing the equation, finding the constant k, and solving for one unknown in direct and inverse variation. It helps students who are just starting the chapter and want the single-step method solid.

Written by the spmmath.com.my editorial team.

Worked example 1

y varies directly as x. When x = 6, y = 24.

Find the value of y when x = 10.

  1. Direct variation, so write y = kx.
  2. Substitute x = 6, y = 24: 24 = k × 6.
  3. Solve for k: k = 24 ÷ 6 = 4.
  4. Equation is y = 4x.
  5. Substitute x = 10: y = 4 × 10 = 40.

Worked example 2

y varies inversely as x. When x = 4, y = 15.

Find the value of y when x = 12.

  1. Inverse variation, so write y = k/x.
  2. Substitute x = 4, y = 15: 15 = k/4.
  3. Solve for k: k = 15 × 4 = 60.
  4. Equation is y = 60/x.
  5. Substitute x = 12: y = 60 ÷ 12 = 5.

Worked example 3

The table shows that p varies directly as q. Find the value of a.

q: 5, 8; p: 35, a.

  1. Direct variation, so write p = kq.
  2. Use the first pair p = 35, q = 5: 35 = k × 5.
  3. Solve for k: k = 35 ÷ 5 = 7.
  4. Equation is p = 7q.
  5. Substitute q = 8: a = 7 × 8 = 56.

Worked example 4

y varies directly as x. When x = 3, y = 21.

Find y when x = 8.

  1. Direct variation: y = kx.
  2. Find k: 21 = k × 3, so k = 7.
  3. Equation: y = 7x.
  4. When x = 8: y = 7 × 8 = 56.

Worked example 5

y varies inversely as x. When x = 5, y = 8.

Find y when x = 2.

  1. Inverse variation: y = k/x.
  2. Find k: 8 = k/5, so k = 40.
  3. Equation: y = 40/x.
  4. When x = 2: y = 40 ÷ 2 = 20.

Worked example 6

d varies directly as w. When w = 9, d = 54.

Find d when w = 15.

  1. Direct variation: d = kw.
  2. Find k: 54 = k × 9, so k = 6.
  3. Equation: d = 6w.
  4. When w = 15: d = 6 × 15 = 90.

Worked example 7

s varies directly as t. When t = 12, s = 30.

Find the value of s when t = 20.

  1. Direct variation, so write s = kt.
  2. Substitute t = 12, s = 30: 30 = k × 12.
  3. Solve for k: k = 30 ÷ 12 = 2.5.
  4. Equation is s = 2.5t.
  5. Substitute t = 20: s = 2.5 × 20 = 50.

Worked example 8

h varies inversely as m. When m = 6, h = 9.

Find the value of h when m = 18.

  1. Inverse variation, so write h = k/m.
  2. Substitute m = 6, h = 9: 9 = k/6.
  3. Solve for k: k = 9 × 6 = 54.
  4. Equation is h = 54/m.
  5. Substitute m = 18: h = 54 ÷ 18 = 3.

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

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Frequently asked questions

What's covered at the easy level of variation?

Easy questions test direct variation (y = kx) and inverse variation (y = k/x) separately. You'll be given a pair of values to find the constant k, then use that k to find another unknown value.

Reading whether a statement means direct or inverse variation correctly is the first key skill.

How do I tell direct variation apart from inverse variation in a word problem?

Look at the relationship described: if one quantity increases as the other increases, it's direct variation (y = kx). If one increases as the other decreases, it's inverse variation (y = k/x).

Phrases like "varies directly as" or "varies inversely as" state this outright, read them carefully before writing an equation.

What mistakes do students make at the easy level of variation?

The most common mistake is using the wrong formula, writing y = kx for an inverse relationship or y = k/x for a direct one. Another is forgetting to find k first before substituting to find the unknown, which leads to an equation with two unknowns and no way to solve it.

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