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Pearson Edexcel 4MA1 IGCSE  ·  Topic G5.1

G5.1 Ratio, Fractions and Direct Proportion (y = kx)

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01 — Explanation

How do I turn a ratio into a fraction of the whole?

A ratio A:BA:B compares two parts to each other. A fraction compares one part to the total. To switch between them, find the total number of parts, T=A+BT = A + B, then put each part over that total: the fraction for AA is AT\dfrac{A}{T} and for BB is BT\dfrac{B}{T}.

For a ratio of cement : sand : gravel =3:5:2= 3:5:2, the total is 3+5+2=103+5+2=10, so sand makes up 510=12\dfrac{5}{10} = \dfrac{1}{2} of the mix.

Exam Tip

If red : blue counters are 3:43:4, the fraction red is 37\dfrac{3}{7} (parts over total), not 34\dfrac{3}{4} (part over part). Writing 34\dfrac{3}{4} for "fraction of the whole" is the single most common error on this topic.

02 — Explanation

Why do I add the parts for a fraction but divide them for a rate?

The two uses of A:BA:B look similar but mean different things, and mixing them up is the main source of lost marks:

  • Sharing a total (additive): the denominator is the sum of the parts, A+BA+B.
  • Finding a rate (multiplicative): the ratio becomes the quotient AB\dfrac{A}{B} itself — this is used when AA and BB are in different units (e.g. distance ÷ time = speed, or one currency ÷ another = exchange rate).

Before calculating anything, decide which situation you're in: "Am I splitting a whole into parts?" (add) or "Am I comparing two different quantities to get a rate?" (divide one by the other).

03 — Explanation

How do I turn a numerical ratio into the equation y = kx?

When two quantities are in direct proportion, their relationship is a straight line through the origin, y=kxy = kx, where kk is the constant of proportionality — exactly the unit rate from the paragraph above.

k=yxk = \frac{y}{x}

Here kk is also the gradient of the line: any two values that satisfy the same ratio (e.g. 10:1510:15 and its simplified form 2:32:3) sit on the same line through (0,0)(0,0), since k=1015=23k = \frac{10}{15} = \frac{2}{3} is unchanged.

Exam Tip

Setting up y=kxy = kx correctly depends on correctly deciding which quantity is xx and which is yy. If distance (DD) is proportional to time (TT), then k=D/Tk = D/T is speed — but if you swap the variables by mistake, you'd calculate time-per-unit-distance instead. Always state "let y=…y = \ldots, x=…x = \ldots" before substituting.

04 — Explanation

How do I find and use k when the equation isn't already in that form?

If you're given an equation like 2y=10x2y = 10x, rearrange it first: dividing both sides by 2 gives y=5xy = 5x, so k=5k = 5. If instead you're given one matching pair of values, substitute them directly into k=y/xk = y/x.

Worked Example

A factory produces widgets (WW) in direct proportion to the time (TT) in minutes. In 15 minutes it produces 120 widgets. Find the equation connecting WW and TT, then use it to predict the output in 2 hours.

  1. Find kk: k=WT=12015=8k = \dfrac{W}{T} = \dfrac{120}{15} = 8 widgets per minute.
  2. State the equation: W=8TW = 8T.
  3. Convert 2 hours to minutes: T=120T = 120.
  4. Substitute: W=8×120=960W = 8 \times 120 = 960 widgets.

Exam Tip

Always double-check your units match before substituting — a classic trap is plugging in "2" for 2 hours into an equation defined in minutes.