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Cambridge IGCSE0607

Proportion

International Mathematics 0607 Chapter Notes

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Proportion
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Proportion notes

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In this chapter

  1. 1Direct proportion and the proportional symbol
  2. 2Inverse proportion
  3. 3Finding an unknown quantity
  4. 4Two variations in one question
  5. 5Choosing the best variation model

1. Direct proportion and the proportional symbol

The symbol ∝ means is proportional to. Saying y varies directly as xⁿ means y ∝ xⁿ.

Write it as an equation with a constant of proportionality k: y = kxⁿ. Linear is n = 1, square is n = 2, cube is n = 3 and square root is n = 1/2.

The quantity in the proportion is exactly what the question names. For "the square root of (x - 1)" the quantity is √(x - 1), not √x.

y ∝ xⁿ ⇒ y = kxⁿ

y ∝ √(x - 1) ⇒ y = k√(x - 1)

Examiner insight

“Some did use direct variation rather than inverse or 'square' root instead of 'square' and therefore made no progress.” “A few candidates started off with the wrong variation, usually proportional to (x + 2).”

Worked example 12 marks

y varies directly as the square root of (x - 1).

y = 15 when x = 26.

Find y in terms of x.

  1. 1

    y = k√(x - 1), so 15 = k√25 = 5k and k = 3 [1]

  2. 2

    y = 3√(x - 1) [1]

2. Inverse proportion

Saying y varies inversely as xⁿ means y ∝ 1/xⁿ. The equation is y = k/xⁿ.

Start by writing a clear equation with k, then put the numbers in. Give the final answer as an equation in y and x, with k replaced by its value.

y ∝ 1/xⁿ ⇒ y = k/xⁿ

Examiner insight

“Candidates should start by writing a clear equation involving y, k and x before introducing the numbers.” “Those scoring no marks had frequently used direct rather than inverse proportion.”

Worked example 12 marks

y ∝ 1/x³

When x = 2, y = 1/4.

Find y in terms of x.

  1. 1

    y = k/x³, so 1/4 = k/8 and k = 2 [1]

  2. 2

    y = 2/x³ [1]

3. Finding an unknown quantity

Use the same three steps every time. Write the equation with k. Use the pair of values you are given to find k. Then put in the new value and solve.

When the unknown is x and the power is even, solve x² = c to get two values, ±√c.

k = y/xⁿ or k = yxⁿ

Examiner insight

“A number of able students obtained the correct formula in part (a) but then omitted to substitute in x = 4.” “In part (a)(ii) most candidates understood what to do and made progress in solving the equation but frequently missed the possibility of a negative square root.”

Worked example 14 marks

Do not use a calculator.

y is directly proportional to the square of x. y = 20 when x = 2.

Find the values of x when y = 45.

  1. 1

    y = kx² [1]

  2. 2

    20 = 4k, so k = 5 [1]

  3. 3

    45 = 5x², so x² = 9 [1]

  4. 4

    x = 3 or x = -3 [1]

4. Two variations in one question

Some questions link three quantities. Find the first equation and its k. Then write the second variation with a new constant and find that.

To combine them, substitute the first result into the second. Then use the laws of indices to give the answer in the form asked.

An answer in the form r = ax^(b) may have a negative or fractional power.

√x³ = x^(3/2)

1/(x^(3/2)) = x^(-3/2)

Examiner insight

“Those that were successful with part (a) were nearly always successful in parts (b) and (c) although, in the latter, some reverted to an equation with k still in.” “Those that could often were unable to simplify the equation to the required form.”

Worked example 16 marks

p varies inversely as the cube of x.

p = 2 when x = 2.

(a) Find p in terms of x.

(b) r varies directly as the square root of p.

r = 24 when p = 64.

Find r in terms of x. Give your answer in the form r = ax^(b) where a and b are constants.

  1. 1

    (a) p = k/x³, so 2 = k/8 and k = 16 [1]

  2. 2

    p = 16/x³ [1]

  3. 3

    (b) r = m√p [1]

  4. 4

    24 = m√64 = 8m, so m = 3 [1]

  5. 5

    r = 3√(16/x³) = (3 × 4)/(x^(3/2)) [1]

  6. 6

    r = 12x^(-3/2) [1]

5. Choosing the best variation model

To identify the best model for data, test each one. If y ∝ xⁿ, then y/xⁿ is the same for every pair of values. If y ∝ 1/xⁿ, then yxⁿ is the same.

The best model is the one that gives a constant k for every pair. Then use it to find the equation.

One pair of values is not enough. Find k from a second pair, then say what the constants show.

y/xⁿ = k (direct)

yxⁿ = k (inverse)

Examiner insight

“Many chose to find a second constant and indicate that this was different from the first.” “The most common error was to find two different constants of proportionality but to make no further comment.”

Worked example 13 marks

The table gives values of x and y. Decide whether y ∝ x or y ∝ x² fits better, and find y in terms of x.

x1234
y3122748
  1. 1

    y/x = 3, 6, 9, 12: not constant, so not y ∝ x [1]

  2. 2

    y/x² = 3, 3, 3, 3: constant, so y ∝ x² [1]

  3. 3

    y = 3x² [1]

Recap

  • y ∝ xⁿ means y = kxⁿ. y ∝ 1/xⁿ means y = k/xⁿ.
  • Square is n = 2, cube is n = 3, square root is n = 1/2.
  • Write the equation with k first, find k, then substitute.
  • Give the final answer as an equation, with k replaced by its value.
  • An even power can give two values of x: ±.
  • Two variations: find each constant in turn, then substitute one into the other.
  • Best model: the one where y/xⁿ or yxⁿ is the same for every pair. One pair is not enough.

End-of-chapter exercise

Test yourself on the whole chapter. Work through these before moving on.

  1. y is inversely proportional to x. Write this as an equation using a constant k.1 mark
  2. Do not use a calculator. y is directly proportional to x. y = 12 when x = 4. Find y when x = 7.2 marks
  3. Do not use a calculator. y varies directly as the square of x. y = 18 when x = 3. Find y in terms of x.2 marks
  4. y varies inversely as x. y = 6 when x = 4. Find y when x = 8.3 marks
  5. y varies directly as the square root of x. y = 14 when x = 49. Find x when y = 20.3 marks
  6. The brightness B of a lamp varies inversely as the square of the distance d metres from the lamp. B = 40 units when d = 3 m. Find B when d = 5 m.3 marks
  7. Do not use a calculator. y is inversely proportional to x³. y = 3 when x = 2. Find x when y = 81.4 marks
  8. Do not use a calculator. y varies directly as the square of (x + 1). y = 18 when x = 2. Find the values of x when y = 50.4 marks
  9. A pendulum of length L cm takes T seconds for one swing. The table gives measured values of L and T. (a) Kai uses only the first pair of values and says that T = 0.2L. Use the other values to explain why Kai is wrong. (b) Decide which of T ∝ L², T ∝ √L and T ∝ 1/L fits best. Find T in terms of L. (c) Find the length of a pendulum that takes 2.2 seconds for one swing. | L | 1 | 4 | 9 | 25 | | --- | --- | --- | --- | --- | | T | 0.2 | 0.4 | 0.6 | 1.0 |7 marks
  10. Do not use a calculator. The power P watts from a wind turbine varies directly as the cube of the wind speed v m / s. P = 320 when v = 4. The wind speed v varies directly as the square root of the height h metres above the ground. v = 6 when h = 9. (a) Find P in terms of h. Give your answer in the form P = ah^(b) where a and b are constants. (b) Find h when P = 2560. (c) The height is multiplied by 4. Find the number that P is multiplied by.7 marks

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