Rearranging Formulae GCSE Questions

Answer $t = \dfrac{v - u}{a}$

1. $\displaystyle v - u = at$Subtract $u$ from both sides
2. $\displaystyle t = \frac{v - u}{a}$Divide both sides by $a$

• M1 $v - u = at$
• A1 $t = \dfrac{v - u}{a}$

Answer $x = \dfrac{y + 3}{4}$

1. $\displaystyle y + 3 = 4x$Add $3$ to both sides
2. $\displaystyle x = \frac{y + 3}{4}$Divide both sides by $4$

• M1 $y + 3 = 4x$
• A1 $x = \dfrac{y + 3}{4}$

Answer $r = \dfrac{C}{2\pi}$

1. $\displaystyle r = \frac{C}{2\pi}$Divide both sides by $2\pi$

• B1 $r = \dfrac{C}{2\pi}$ oe

Answer $h = \dfrac{2A}{b}$

1. $\displaystyle 2A = bh$Multiply both sides by $2$
2. $\displaystyle h = \frac{2A}{b}$Divide both sides by $b$

• M1 $2A = bh$
• A1 $h = \dfrac{2A}{b}$

Answer $m = \dfrac{P - 2n}{3}$

1. $\displaystyle P - 2n = 3m$Subtract $2n$ from both sides
2. $\displaystyle m = \frac{P - 2n}{3}$

• M1 $P - 2n = 3m$
• A1 $m = \dfrac{P - 2n}{3}$

Answer $x = 2y - 5$

1. $\displaystyle 2y = x + 5$Multiply both sides by $2$ first
2. $\displaystyle x = 2y - 5$

• M1 $2y = x + 5$
• A1 $x = 2y - 5$

Answer $v = \dfrac{2s}{t} - u$

1. $\displaystyle 2s = (u + v)t$Multiply both sides by $2$
2. $\displaystyle \frac{2s}{t} = u + v$Divide both sides by $t$
3. $\displaystyle v = \frac{2s}{t} - u$

• M1 $2s = (u + v)t$
• M1 $\dfrac{2s}{t} = u + v$
• A1 $v = \dfrac{2s}{t} - u$ oe, e.g. $\dfrac{2s - ut}{t}$

Answer $x = \dfrac{7 - y}{2}$

1. $\displaystyle y + 2x = 7$Add $2x$ to both sides, so the $x$ term is positive
2. $\displaystyle 2x = 7 - y$
3. $\displaystyle x = \frac{7 - y}{2}$

• M1 $2x = 7 - y$ or $-2x = y - 7$
• A1 $x = \dfrac{7 - y}{2}$ oe

Answer $c = \dfrac{b - a}{3}$

1. $\displaystyle a + 3c = b$Add $3c$ to both sides
2. $\displaystyle 3c = b - a$
3. $\displaystyle c = \frac{b - a}{3}$

• M1 $3c = b - a$
• A1 $c = \dfrac{b - a}{3}$ oe

Answer $r = \sqrt{\dfrac{A}{\pi}}$

1. $\displaystyle r^2 = \frac{A}{\pi}$Divide both sides by $\pi$
2. $\displaystyle r = \sqrt{\frac{A}{\pi}}$Square root both sides (a radius is positive)

• M1 $r^2 = \dfrac{A}{\pi}$
• A1 $r = \sqrt{\dfrac{A}{\pi}}$

Answer $x = y^2 - 4$

1. $\displaystyle y^2 = x + 4$Square both sides to undo the square root
2. $\displaystyle x = y^2 - 4$

• M1 $y^2 = x + 4$
• A1 $x = y^2 - 4$

Answer $C = \dfrac{5(F - 32)}{9}$

1. $\displaystyle F - 32 = \frac{9C}{5}$Subtract $32$ from both sides
2. $\displaystyle 5(F - 32) = 9C$Multiply both sides by $5$
3. $\displaystyle C = \frac{5(F - 32)}{9}$Divide both sides by $9$

• M1 $F - 32 = \dfrac{9C}{5}$ or $5F = 9C + 160$
• A1 $C = \dfrac{5(F - 32)}{9}$ oe

Answer $25^\circ$C

1. $\displaystyle C = \frac{5(77 - 32)}{9} = \frac{5 \times 45}{9} = 25$

• B1 $25$ — ft their formula

Answer $d = \dfrac{C - 45}{30}$

1. $\displaystyle C - 45 = 30d$
2. $\displaystyle d = \frac{C - 45}{30}$

• M1 $C - 45 = 30d$
• A1 $d = \dfrac{C - 45}{30}$

Answer $5$ days

1. $\displaystyle d = \frac{195 - 45}{30} = \frac{150}{30} = 5$

• B1 $5$ — ft their formula

Answer He divided only $y$ by $3$; the whole of $y - 5$ must be divided by $3$. $x = \dfrac{y - 5}{3}$.

1. $\displaystyle y - 5 = 3x$Subtract $5$ first
2. $\displaystyle x = \frac{y - 5}{3}$Then divide the whole of $y - 5$ by $3$

• B1 Explains that $5$ must be subtracted before dividing, or that all of $y - 5$ is divided by $3$
• B1 $x = \dfrac{y - 5}{3}$

Answer $m = \dfrac{2E}{v^2}$

1. $\displaystyle 2E = mv^2$Multiply both sides by $2$
2. $\displaystyle m = \frac{2E}{v^2}$Divide both sides by $v^2$

• M1 $2E = mv^2$
• A1 $m = \dfrac{2E}{v^2}$

Answer $u = \sqrt{v^2 - 2as}$

1. $\displaystyle u^2 = v^2 - 2as$Subtract $2as$ from both sides
2. $\displaystyle u = \sqrt{v^2 - 2as}$Square root the whole side: it is not $v - \sqrt{2as}$

• M1 $u^2 = v^2 - 2as$
• M1 Square roots the whole of the right-hand side
• A1 $u = \sqrt{v^2 - 2as}$

Answer $x = \dfrac{d - b}{a - c}$

1. $\displaystyle ax - cx = d - b$Get both $x$ terms on the same side
2. $\displaystyle x(a - c) = d - b$Factorise out $x$
3. $\displaystyle x = \frac{d - b}{a - c}$Divide by the bracket

• M1 Collects the $x$ terms: $ax - cx = d - b$
• M1 $x(a - c) = d - b$
• A1 $x = \dfrac{d - b}{a - c}$ oe

Answer $W = \dfrac{T - 20}{40}$

1. $\displaystyle T - 20 = 40W$
2. $\displaystyle W = \frac{T - 20}{40}$

• M1 $T - 20 = 40W$
• A1 $W = \dfrac{T - 20}{40}$

Answer $2.5$ kg

1. $\displaystyle T = 2 \times 60 = 120$The formula uses minutes, so convert the hours first
2. $\displaystyle W = \frac{120 - 20}{40} = \frac{100}{40} = 2.5$

• M1 $T = 120$ substituted — ft their formula
• A1 $2.5$ kg

Answer $m = \dfrac{y - c}{x}$

1. $\displaystyle y - c = mx$
2. $\displaystyle m = \frac{y - c}{x}$

• M1 $y - c = mx$
• A1 $m = \dfrac{y - c}{x}$

Answer $m = 3$

1. $\displaystyle m = \frac{17 - 5}{4} = \frac{12}{4} = 3$

• B1 $3$ — ft their formula

Answer $h = \dfrac{2A}{a + b}$

1. $\displaystyle 2A = (a + b)h$Multiply both sides by $2$
2. $\displaystyle h = \frac{2A}{a + b}$Divide by the whole bracket $(a + b)$

• M1 $2A = (a + b)h$
• A1 $h = \dfrac{2A}{a + b}$

Answer $x = \dfrac{3y + 2}{y - 1}$

1. $\displaystyle y(x - 3) = x + 2$Multiply both sides by $(x - 3)$
2. $\displaystyle xy - 3y = x + 2$
3. $\displaystyle xy - x = 3y + 2$Get both $x$ terms on one side
4. $\displaystyle x(y - 1) = 3y + 2$Factorise out $x$
5. $\displaystyle x = \frac{3y + 2}{y - 1}$

• M1 $y(x - 3) = x + 2$
• M1 $xy - x = 3y + 2$
• M1 $x(y - 1) = 3y + 2$
• A1 $x = \dfrac{3y + 2}{y - 1}$ oe

Answer $l = \dfrac{gT^2}{4\pi^2}$

1. $\displaystyle \frac{T}{2\pi} = \sqrt{\frac{l}{g}}$Divide both sides by $2\pi$
2. $\displaystyle \frac{T^2}{4\pi^2} = \frac{l}{g}$Square both sides: $(2\pi)^2 = 4\pi^2$
3. $\displaystyle l = \frac{gT^2}{4\pi^2}$Multiply both sides by $g$

• M1 $\dfrac{T}{2\pi} = \sqrt{\dfrac{l}{g}}$
• M1 $\dfrac{T^2}{4\pi^2} = \dfrac{l}{g}$
• A1 $l = \dfrac{gT^2}{4\pi^2}$

Answer $x = \dfrac{ab}{a + b}$

1. $\displaystyle bx + ax = ab$Multiply every term by $ab$
2. $\displaystyle x(a + b) = ab$Factorise out $x$
3. $\displaystyle x = \frac{ab}{a + b}$

• M1 Multiplies by $ab$: $bx + ax = ab$
• M1 $x(a + b) = ab$
• A1 $x = \dfrac{ab}{a + b}$

Answer $x = \sqrt{b^2 - a}$

1. $\displaystyle x^2 + a = b^2$Square both sides to remove the root
2. $\displaystyle x^2 = b^2 - a$
3. $\displaystyle x = \sqrt{b^2 - a}$Square root the whole of the right-hand side

• M1 $x^2 + a = b^2$
• M1 $x^2 = b^2 - a$
• A1 $x = \sqrt{b^2 - a}$