Rearranging Formulae

Question 11 mark

Make x the subject of the formula y=x+7

Type powers with ^ and fractions with /, for example x^2 or (x+1)/3

Hint

Do the opposite of adding 7 to both sides.

Worked solution

Make x the subject of the formula y=x+7

  1. Subtract 7 from both sides: y-7=x.
  2. So x=y-7.

Question 21 mark

Make a the subject of the formula b=a-4

Type powers with ^ and fractions with /, for example x^2 or (x+1)/3

Hint

The opposite of subtracting 4 is adding 4.

Worked solution

Make a the subject of the formula b=a-4

  1. Add 4 to both sides: b+4=a.
  2. So a=b+4.

Question 31 mark

Make h the subject of the formula A=6h

Type powers with ^ and fractions with /, for example x^2 or (x+1)/3

Hint

h is multiplied by 6, so divide both sides by 6.

Worked solution

Make h the subject of the formula A=6h

  1. Divide both sides by 6: \frac{A}{6}=h.
  2. So h=\frac{A}{6}.

Question 41 mark

Make m the subject of the formula p=\frac{m}{3}

Type powers with ^ and fractions with /, for example x^2 or (x+1)/3

Hint

m is divided by 3, so multiply both sides by 3.

Worked solution

Make m the subject of the formula p=\frac{m}{3}

  1. Multiply both sides by 3: 3p=m.
  2. So m=3p.

Question 52 marks

Make n the subject of the formula P=2n+9

Type powers with ^ and fractions with /, for example x^2 or (x+1)/3

Hint

Undo the adding first, then undo the multiplying.

Worked solution

Make n the subject of the formula P=2n+9

  1. Subtract 9 from both sides: P-9=2n.
  2. Divide both sides by 2: \frac{P-9}{2}=n.
  3. So n=\frac{P-9}{2}.

Question 62 marks

Make k the subject of the formula m=5k-8

Type powers with ^ and fractions with /, for example x^2 or (x+1)/3

Hint

Undo the subtracting first, then undo the multiplying.

Worked solution

Make k the subject of the formula m=5k-8

  1. Add 8 to both sides: m+8=5k.
  2. Divide both sides by 5: \frac{m+8}{5}=k.
  3. So k=\frac{m+8}{5}.

Question 72 marks

Make x the subject of the formula y=20-4x

Type powers with ^ and fractions with /, for example x^2 or (x+1)/3

Hint

Add 4x to both sides first, so that the x term is positive.

Worked solution

Make x the subject of the formula y=20-4x

  1. Add 4x to both sides: y+4x=20.
  2. Subtract y from both sides: 4x=20-y.
  3. Divide both sides by 4: x=\frac{20-y}{4}.

Question 82 marks

The mean m of two numbers a and b is given by the formula m=\frac{a+b}{2}

Make a the subject of the formula.

Type powers with ^ and fractions with /, for example x^2 or (x+1)/3

Hint

Multiply both sides by 2 first, to clear the fraction.

Worked solution

The mean m of two numbers a and b is given by the formula m=\frac{a+b}{2}

Make a the subject of the formula.

  1. Multiply both sides by 2: 2m=a+b.
  2. Subtract b from both sides: 2m-b=a.
  3. So a=2m-b.

Question 91 mark

Force, mass and acceleration are linked by the formula F=ma

Which of these shows a as the subject of the formula?

Choose one answer
Hint

a is multiplied by m, so do the opposite to both sides.

Worked solution

Force, mass and acceleration are linked by the formula F=ma

Which of these shows a as the subject of the formula?

  1. a=Fm
  2. a=\frac{F}{m}
  3. a=\frac{m}{F}
  4. a=F-m
  1. In F=ma, a is multiplied by m.
  2. Divide both sides by m: \frac{F}{m}=a.
  3. So a=\frac{F}{m}.

Question 102 marks

Make x the subject of the formula y=mx+c

Type powers with ^ and fractions with /, for example x^2 or (x+1)/3

Hint

Get the x term on its own first, then divide.

Worked solution

Make x the subject of the formula y=mx+c

  1. Subtract c from both sides: y-c=mx.
  2. Divide both sides by m: \frac{y-c}{m}=x.
  3. So x=\frac{y-c}{m}.

Question 112 marks

Make x the subject of the formula y=3(x+4)

Type powers with ^ and fractions with /, for example x^2 or (x+1)/3

Hint

You can divide both sides by 3 first, or multiply out the bracket first.

Worked solution

Make x the subject of the formula y=3(x+4)

  1. Divide both sides by 3: \frac{y}{3}=x+4.
  2. Subtract 4 from both sides: \frac{y}{3}-4=x.
  3. So x=\frac{y}{3}-4, which can also be written x=\frac{y-12}{3}.

Question 122 marks

Make n the subject of the formula C=\frac{n}{5}+12

Type powers with ^ and fractions with /, for example x^2 or (x+1)/3

Hint

Take away 12 first, then undo the dividing by 5.

Worked solution

Make n the subject of the formula C=\frac{n}{5}+12

  1. Subtract 12 from both sides: C-12=\frac{n}{5}.
  2. Multiply all of the left side by 5: 5(C-12)=n.
  3. So n=5(C-12), which is the same as n=5C-60.

Question 132 marks

The perimeter P of a rectangle with length l and width w is given by the formula P=2l+2w

Make l the subject of the formula.

Type powers with ^ and fractions with /, for example x^2 or (x+1)/3

Hint

Move the 2w term to the other side first.

Worked solution

The perimeter P of a rectangle with length l and width w is given by the formula P=2l+2w

Make l the subject of the formula.

  1. Subtract 2w from both sides: P-2w=2l.
  2. Divide both sides by 2: \frac{P-2w}{2}=l.
  3. So l=\frac{P-2w}{2}, which is the same as l=\frac{P}{2}-w.

Question 143 marks

The area A of a triangle with base b and height h is given by the formula A=\frac{1}{2}bh

(a)

Make h the subject of the formula.

2 marks

Type powers with ^ and fractions with /, for example x^2 or (x+1)/3

(b)

A triangle has area 30 cm² and base 12 cm.

Work out the height of the triangle.

1 mark

Hint

Clear the half first, then divide by the other letter.

Worked solution

The area A of a triangle with base b and height h is given by the formula A=\frac{1}{2}bh

Part (a)

Make h the subject of the formula.

  1. Multiply both sides by 2: 2A=bh.
  2. Divide both sides by b: \frac{2A}{b}=h.
  3. So h=\frac{2A}{b}.

Part (b)

A triangle has area 30 cm² and base 12 cm.

Work out the height of the triangle.

  1. h=\frac{2\times30}{12}=\frac{60}{12}=5.
  2. The height is 5 cm.

Question 153 marks

Omar wants h on its own in the formula g=6h+18

These are the two lines he writes.

\frac{g}{6}=h+18

h=\frac{g}{6}-18

Omar's answer is not correct.

(a)

Which of these describes Omar's mistake?

1 mark

Choose one answer
(b)

Make h the subject of g=6h+18

2 marks

Type powers with ^ and fractions with /, for example x^2 or (x+1)/3

Hint

Check each line of the working: was every term dealt with?

Worked solution

Omar wants h on its own in the formula g=6h+18

These are the two lines he writes.

\frac{g}{6}=h+18

h=\frac{g}{6}-18

Omar's answer is not correct.

Part (a)

Which of these describes Omar's mistake?

  1. He did not divide the 18 by 6
  2. He should have multiplied both sides by 6, not divided
  3. He should have added 18 to both sides
  4. He should not have divided g by 6
  1. Dividing both sides by 6 should give \frac{g}{6}=h+3, because 18\div6=3.
  2. Omar divided g and 6h by 6 but did not divide the 18 by 6.

Part (b)

Make h the subject of g=6h+18

  1. Subtract 18 from both sides: g-18=6h.
  2. Divide both sides by 6: \frac{g-18}{6}=h.
  3. So h=\frac{g-18}{6}, which is the same as h=\frac{g}{6}-3.

Question 16Challenge3 marks

It takes n people T hours to paint a hall, where T=\frac{36}{n}

(a)

Make n the subject of the formula.

2 marks

Type powers with ^ and fractions with /, for example x^2 or (x+1)/3

(b)

The hall must be painted in 4 hours.

How many people are needed?

1 mark

Hint

When the letter you want is on the bottom of a fraction, multiply both sides by it first.

Worked solution

It takes n people T hours to paint a hall, where T=\frac{36}{n}

Part (a)

Make n the subject of the formula.

  1. Multiply both sides by n: nT=36.
  2. Divide both sides by T: n=\frac{36}{T}.

Part (b)

The hall must be painted in 4 hours.

How many people are needed?

  1. n=\frac{36}{4}=9.
  2. Check: \frac{36}{9}=4 hours. 9 people are needed.

Question 17Challenge3 marks

The total surface area S of a cube with edges of length x is given by the formula S=6x^2

Make x the subject of the formula. (x is positive.)

Type powers with ^ and fractions with /, for example x^2 or (x+1)/3

Hint

Get x squared on its own first. The opposite of squaring is taking the square root.

Worked solution

The total surface area S of a cube with edges of length x is given by the formula S=6x^2

Make x the subject of the formula. (x is positive.)

  1. Divide both sides by 6: \frac{S}{6}=x^2.
  2. Take the square root of both sides: \sqrt{\frac{S}{6}}=x.
  3. So x=\sqrt{\frac{S}{6}}.

Question 18Challenge3 marks

Make a the subject of the formula d=\sqrt{5a-4}

Type powers with ^ and fractions with /, for example x^2 or (x+1)/3

Hint

Square both sides first to remove the square root.

Worked solution

Make a the subject of the formula d=\sqrt{5a-4}

  1. Square both sides: d^2=5a-4.
  2. Add 4 to both sides: d^2+4=5a.
  3. Divide both sides by 5: \frac{d^2+4}{5}=a.
  4. So a=\frac{d^2+4}{5}.

Question 19Challenge3 marks

The distance s travelled by a car is given by the formula s=\frac{(u+v)t}{2}

Make v the subject of the formula.

Type powers with ^ and fractions with /, for example x^2 or (x+1)/3

Hint

Clear the fraction first, then deal with the t, then the u.

Worked solution

The distance s travelled by a car is given by the formula s=\frac{(u+v)t}{2}

Make v the subject of the formula.

  1. Multiply both sides by 2: 2s=(u+v)t.
  2. Divide both sides by t: \frac{2s}{t}=u+v.
  3. Subtract u from both sides: \frac{2s}{t}-u=v.
  4. So v=\frac{2s}{t}-u, which is the same as v=\frac{2s-ut}{t}.

Question 20Challenge4 marks

The energy E of a moving object with mass m and speed v is given by the formula E=\frac{1}{2}mv^2

v is positive.

(a)

Make v the subject of the formula.

3 marks

Type powers with ^ and fractions with /, for example x^2 or (x+1)/3

(b)

Work out v when E=400 and m=8

1 mark

Hint

Undo the steps in the right order: clear the half and the m, then undo the squaring.

Worked solution

The energy E of a moving object with mass m and speed v is given by the formula E=\frac{1}{2}mv^2

v is positive.

Part (a)

Make v the subject of the formula.

  1. Multiply both sides by 2: 2E=mv^2.
  2. Divide both sides by m: \frac{2E}{m}=v^2.
  3. Take the square root of both sides: v=\sqrt{\frac{2E}{m}}.

Part (b)

Work out v when E=400 and m=8

  1. \frac{2\times400}{8}=\frac{800}{8}=100.
  2. v=\sqrt{100}=10.