Variation Worked ExamplesIGCSE Maths 0580

Write the proportionality as an equation with a constant $k$. This first line earns a mark on its own.

Substitute the given pair of values to find $k$.

Divide to get $k$, then write the formula.

(b) Substitute $x = 5$. Cube first, then multiply by $2$.

(c) Substitute $y = 128$ and solve for $x^{3}$.

Take the cube root.

"Inversely proportional to $\sqrt{x}$" puts $\sqrt{x}$ in the denominator.

Substitute $x = 4$, $y = 6$. Note $\sqrt{4} = 2$.

Multiply up to find $k$.

(a) Substitute $x = 9$, so $\sqrt{x} = 3$.

(b) Substitute $y = 2$ and rearrange for $\sqrt{x}$.

Square both sides to undo the root.

(c) If $x$ is multiplied by $16$, then $\sqrt{x}$ is multiplied by $\sqrt{16} = 4$. As $\sqrt{x}$ is in the denominator, $y$ is divided by $4$. Check with numbers: $x = 4$ gives $y = 6$, and $x = 64$ gives $y = \frac{12}{8} = 1.5$, which is $6 \div 4$ ✓.

If $y = \dfrac{k}{x^{2}}$ then $k = yx^{2}$, so the product $yx^{2}$ must be the same for every column. Test each one.

First column: $x = 2$, $y = 50$.

Second column: $x = 4$, $y = 12.5$.

Third column: $x = 5$, $y = 8$.

All three give $200$, so the relationship holds with $k = 200$. State the formula — that is what "show that" is asking for.

(b) Substitute $x = 10$.

"Directly with $x$" puts $x$ on the top; "inversely with $y$" puts $y$ on the bottom. One constant $k$ covers both.

Substitute all three known values.

Multiply both sides by $3$, then divide by $4$.

Write the full formula.

Substitute $x = 10$ and $y = 5$.

Start from the relationship. You are not given numbers, so keep $k$ general.

An increase of $20\%$ means multiplying by the factor $1.2$. Write the new value of $x$.

Substitute into the formula. The multiplier gets squared along with $x$.

Compare with the original $y = kx^{2}$ — the $kx^{2}$ cancels.

A multiplier of $1.44$ is an increase of $0.44$, i.e. $44\%$.

Write the relationship.

An increase of $25\%$ is a multiplier of $1.25$.

Substitute. The multiplier is cubed, and it sits in the denominator.

Evaluate $1.25^{3}$.

Express the new $y$ as a multiple of the old $y$.

A multiplier of $0.512$ means $y$ has fallen to $51.2\%$ of its old value.

Write the proportionality with a constant.

Substitute the known pair.

Divide to find $k$ — here it is the cost per item.

(b) Substitute $n = 14$.

(a)(i) $y = kx$ is a straight line through the origin: Graph B.

(a)(ii) $y = kx^{2}$ passes through the origin and gets steeper as $x$ increases (doubling $x$ multiplies $y$ by $4$): Graph D.

(a)(iii) $y = \dfrac{k}{x}$: as $x$ increases, $y$ decreases towards $0$ but never reaches it, and $x = 0$ is not allowed, so the curve never meets either axis: Graph C.

(a)(iv) $y = k\sqrt{x}$ passes through the origin but grows more and more slowly (multiplying $x$ by $4$ only doubles $y$): Graph A.

(b) For inverse proportion $y = \dfrac{k}{x}$. Substitute $(2, 12)$ to find $k$.

Substitute $x = 3$.

(a) Read $P$ from the grid: $v = 20$ and $d = 32$ (the small squares are $1\text{ m/s}$ across and $2\text{ m}$ up). Write the proportionality as an equation.

Substitute the coordinates of $P$ to find $k$.

Write the formula. Check with another point on the curve: $0.08 \times 25^{2} = 50$, which matches the graph ✓.

(b) $35\text{ m/s}$ is off the graph, so the formula is needed.

(c) Substitute $d = 60$ and rearrange for $v$.

Round to $3$ significant figures. The graph confirms it: $d = 60$ is reached a little before $v = 28$.

(d) Replace $v$ with $2v$ in the formula. The factor $2$ is squared as well.

Write the relationship, with $d$ the distance from the lamp. At $O$ the distance is the height of the pole, $3\text{ m}$.

(a) The distance from $L$ to $P$ is not $4\text{ m}$ — it is the hypotenuse of triangle $LOP$.

Substitute $d = 5$.

(b) Work backwards: find the distance from the lamp that gives $B = 10$.

$LQ$ is a hypotenuse, with the pole ($3\text{ m}$) as one side and $OQ$ as the other. Use Pythagoras to find $OQ$.

(c) The lamp is unchanged, so $k = 360$ still. At $O$ the distance is the new height $h$.

Write the relationship with a constant. The whole expression $\sqrt{d + 5}$ goes in the denominator.

Substitute $d = 4$, $h = 20$. Note that $\sqrt{4 + 5} = \sqrt{9} = 3$.

(a) Substitute $d = 10$, so $\sqrt{d + 5} = \sqrt{15}$.

(b) Substitute $h = 12$ and rearrange for the square root.

Square both sides, then subtract $5$.