Partial variation and proportion problems · Variation

Many costs have a fixed part and a part that varies. A printer charges a setup fee however many copies you order, plus an amount for each copy. WAEC describes this as “partly constant and partly varies as”:

C = a + bn

where $a$ is the fixed part and $b$ is the rate. There are now two unknown constants, so you need two pairs of values.

Partial: y = a + bxA straight line that crosses the y-axis at the constant a

Direct: y = kxNo constant part, so the line goes through the origin

Try it

Partly constant, partly variesMove a and b until the line passes through both readings

100fixed part a (the intercept)15rate b (the gradient)0 of 2readings on the line

a = 100b = 15

A printer charges ₦450 for 10 copies and ₦950 for 30. The cost is partly fixed (a setup fee) and partly varies with the number of copies: C = a + bn. Change a to move the line up and down, and b to make it steeper, until it passes through both readings. The bar at 30 copies shows the cost as fixed part + varying part: 100 + 15 × 30 = 550.

The constant part $a$ is where the line crosses the $C$ -axis (the cost of 0 copies), and the rate $b$ is how steep it is. Two readings fix the line, just as two equations fix $a$ and $b$ .

The method

Write the equation with two constants: $C = a + bn$ .
Put in both pairs of values to get two equations.
Subtract one from the other: $a$ cancels, leaving $b$ . Then find $a$ .
Write the relationship and use it.

Worked example · WAEC 2019

WAEC 2019 · Paper 2 · Q5

The cost of dinner for a group of tourists is partly constant and partly varies as the number of tourists present. It costs $740.00 when 20 tourists were present and $960.00 when the number of tourists increased by 10. Find the cost of the dinner when only 15 tourists were present.

Write the equation
Let $C$ be the cost in dollars and $n$ the number of tourists: $C = a + bn$ .
Think first. How many constants are there?
Two equations
20 tourists: $a + 20b = 740$ . The number increased by 10, so 30 tourists: $a + 30b = 960$ .
Think first. The number increased by 10. How many tourists is that?
Subtract
$10b = 220$ , so $b = 22$ . Then $a = 740 - 20 \times 22 = 300$ , and $C = 300 + 22n$ .
Think first. Take the first equation from the second.
15 tourists
$C = 300 + 22 \times 15 = 300 + 330 = 630$ dollars, that is $630.00.

The varying part isn’t always “as $x$ ”. Write whatever the question says after “partly constant and partly varies”:

partly constant, partly inversely as $x$ : $y = a + \dfrac{b}{x}$ ;
partly as $x$ and partly as $x^2$ : $y = ax + bx^2$ ;
partly as $M$ and partly inversely as $P$ : $W = aM + \dfrac{b}{P}$ .

More: partial variation

Proportion problems

Some problems use variation without saying so. If more workers finish a job in fewer days, the number of days varies inversely as the number of workers, so workers × days stays the same: it’s the total amount of work.

More: proportion problems

Worked example · WAEC 2019

WAEC 2019 · Paper 2 · Q10 (a)

A twenty-kilogram bag of rice is consumed by $m$ boys in 10 days. When four more boys joined them, the same quantity of rice lasted only 8 days. If the rate of consumption is the same, find the value of $m$ .

What stays the same?
The same bag of rice is eaten either way, so the number of “boy-days” of rice is the same: boys × days is constant.
Think first. The bag is the same. What does boys × days measure?
Equation
$m$ boys for 10 days equals $(m + 4)$ boys for 8 days: $10m = 8(m + 4)$ . So $10m = 8m + 32$ , $2m = 32$ and $m = 16$ .

Your turn

WAEC 2025 · Paper 2 · Q2 ✱✱

Variation

$y$ is partly constant and partly varies inversely as the square of $x$ .

(a)
Write down the relationship between $x$ and $y$ .
Show the answer
$y = a + \dfrac{b}{x^2}$ , where $a$ and $b$ are constants
(b)
When $x = 1$ , $y = 11$ , and when $x = 2$ , $y = 5$ . Find $y$ when $x = 4$ .

Worked solution (try it first)

(a)

Partly constant ( $a$ ) and partly varies inversely as the square of $x$ ( $\frac{b}{x^2}$ ): $y = a + \dfrac{b}{x^2}$ , where $a$ and $b$ are constants.

(b)

$x = 1$ , $y = 11$ : $a + b = 11$ .
$x = 2$ , $y = 5$ : $a + \frac{b}{4} = 5$ .
Subtract: $b - \frac{b}{4} = 6$ , so $\frac{3b}{4} = 6$ and $b = 8$ .
Then $a = 11 - 8 = 3$ , and $y = 3 + \dfrac{8}{x^2}$ .
When $x = 4$ : $y = 3 + \dfrac{8}{16} = 3\frac12$ .

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