Rearranging Formulae
Question 11 mark
Make x the subject of the formula y=x+7
Hint
Do the opposite of adding 7 to both sides.
Worked solution
Make x the subject of the formula y=x+7
- Subtract 7 from both sides: y-7=x.
- So x=y-7.
Question 21 mark
Make a the subject of the formula b=a-4
Hint
The opposite of subtracting 4 is adding 4.
Worked solution
Make a the subject of the formula b=a-4
- Add 4 to both sides: b+4=a.
- So a=b+4.
Question 31 mark
Make h the subject of the formula A=6h
Hint
h is multiplied by 6, so divide both sides by 6.
Worked solution
Make h the subject of the formula A=6h
- Divide both sides by 6: \frac{A}{6}=h.
- So h=\frac{A}{6}.
Question 41 mark
Make m the subject of the formula p=\frac{m}{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}
- Multiply both sides by 3: 3p=m.
- So m=3p.
Question 52 marks
Make n the subject of the formula P=2n+9
Hint
Undo the adding first, then undo the multiplying.
Worked solution
Make n the subject of the formula P=2n+9
- Subtract 9 from both sides: P-9=2n.
- Divide both sides by 2: \frac{P-9}{2}=n.
- So n=\frac{P-9}{2}.
Question 62 marks
Make k the subject of the formula m=5k-8
Hint
Undo the subtracting first, then undo the multiplying.
Worked solution
Make k the subject of the formula m=5k-8
- Add 8 to both sides: m+8=5k.
- Divide both sides by 5: \frac{m+8}{5}=k.
- So k=\frac{m+8}{5}.
Question 72 marks
Make x the subject of the formula y=20-4x
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
- Add 4x to both sides: y+4x=20.
- Subtract y from both sides: 4x=20-y.
- 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.
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.
- Multiply both sides by 2: 2m=a+b.
- Subtract b from both sides: 2m-b=a.
- 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?
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?
- a=Fm
- a=\frac{F}{m}
- a=\frac{m}{F}
- a=F-m
- In F=ma, a is multiplied by m.
- Divide both sides by m: \frac{F}{m}=a.
- So a=\frac{F}{m}.
Question 102 marks
Make x the subject of the formula y=mx+c
Hint
Get the x term on its own first, then divide.
Worked solution
Make x the subject of the formula y=mx+c
- Subtract c from both sides: y-c=mx.
- Divide both sides by m: \frac{y-c}{m}=x.
- So x=\frac{y-c}{m}.
Question 112 marks
Make x the subject of the formula y=3(x+4)
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)
- Divide both sides by 3: \frac{y}{3}=x+4.
- Subtract 4 from both sides: \frac{y}{3}-4=x.
- 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
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
- Subtract 12 from both sides: C-12=\frac{n}{5}.
- Multiply all of the left side by 5: 5(C-12)=n.
- 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.
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.
- Subtract 2w from both sides: P-2w=2l.
- Divide both sides by 2: \frac{P-2w}{2}=l.
- 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
Make h the subject of the formula.
2 marks
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.
- Multiply both sides by 2: 2A=bh.
- Divide both sides by b: \frac{2A}{b}=h.
- So h=\frac{2A}{b}.
Part (b)
A triangle has area 30 cm² and base 12 cm.
Work out the height of the triangle.
- h=\frac{2\times30}{12}=\frac{60}{12}=5.
- 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.
Which of these describes Omar's mistake?
1 mark
Make h the subject of g=6h+18
2 marks
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?
- He did not divide the 18 by 6
- He should have multiplied both sides by 6, not divided
- He should have added 18 to both sides
- He should not have divided g by 6
- Dividing both sides by 6 should give \frac{g}{6}=h+3, because 18\div6=3.
- 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
- Subtract 18 from both sides: g-18=6h.
- Divide both sides by 6: \frac{g-18}{6}=h.
- 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}
Make n the subject of the formula.
2 marks
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.
- Multiply both sides by n: nT=36.
- Divide both sides by T: n=\frac{36}{T}.
Part (b)
The hall must be painted in 4 hours.
How many people are needed?
- n=\frac{36}{4}=9.
- 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.)
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.)
- Divide both sides by 6: \frac{S}{6}=x^2.
- Take the square root of both sides: \sqrt{\frac{S}{6}}=x.
- So x=\sqrt{\frac{S}{6}}.
Question 18Challenge3 marks
Make a the subject of the formula d=\sqrt{5a-4}
Hint
Square both sides first to remove the square root.
Worked solution
Make a the subject of the formula d=\sqrt{5a-4}
- Square both sides: d^2=5a-4.
- Add 4 to both sides: d^2+4=5a.
- Divide both sides by 5: \frac{d^2+4}{5}=a.
- 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.
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.
- Multiply both sides by 2: 2s=(u+v)t.
- Divide both sides by t: \frac{2s}{t}=u+v.
- Subtract u from both sides: \frac{2s}{t}-u=v.
- 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.
Make v the subject of the formula.
3 marks
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.
- Multiply both sides by 2: 2E=mv^2.
- Divide both sides by m: \frac{2E}{m}=v^2.
- Take the square root of both sides: v=\sqrt{\frac{2E}{m}}.
Part (b)
Work out v when E=400 and m=8
- \frac{2\times400}{8}=\frac{800}{8}=100.
- v=\sqrt{100}=10.