Year 8 formulae and rearranging worksheet answers to download as PDF

These Year 8 formulae and rearranging worksheet answers give a full worked solution to all 20 questions of the formulae and rearranging worksheet. Try each question yourself first, then check your method step by step.

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Answer to Question 1
1. Expression: no equals sign, it just groups terms.
2. Equation: it can be solved, \(x=5\).
3. Formula: it connects quantities with letters (area, length, width) in a real-life rule.
4. Identity: it is true for every value of \(x\), shown by the sign \(\equiv\).
5. Inequality: it uses \(>\), so the solution is a range of values, \(y>3\).
The key differences: an expression has no equals sign, an equation is true only for particular values, a formula links real quantities, and an identity is true for every value.
Answer to Question 2
1. Substitute \(l=8\), \(w=5\): \(P=2\times8+2\times5=16+10=\)26.
2. Multiplication first: \(v=4+3\times6=4+18=\)22.
3. \(C=5\times12+20=60+20=\)80.
4. \(A=\dfrac{1}{2}\times9\times14=\dfrac{1}{2}\times126=\)63.
In every case write the formula with the numbers in place before calculating, then follow the order of operations.
Writing each substituted formula on a separate line makes your method easy to follow and earns method marks even if the arithmetic slips.
Answer to Question 3
1. \(ut=5\times3=15\) and \(\dfrac{1}{2}at^{2}=\dfrac{1}{2}\times4\times9=18\). So \(s=15+18=\)33.
2. \((-2)^{2}=4\), so \(3\times4=12\); and \(-2\times(-2)=+4\). \(y=12+4=\)16.
3. Square c first: \(10^{2}=100\), then \(E=2\times100=\)200.
4. \(u^{2}=100\) and \(2as=2\times(-3)\times4=-24\). \(V=100-(-24)=\)124.
Brackets round negative values and squaring before multiplying are the two things that matter most here.
Answer to Question 4
1. x has 9 added, so subtract 9 from both sides: \(x=y-9\).
2. x has 4 subtracted, so add 4 to both sides: \(x=y+4\).
3. x is multiplied by 5, so divide both sides by 5: \(x=\dfrac{y}{5}\).
4. x is divided by 3, so multiply both sides by 3: \(x=3y\).
Check (c) with \(x=2\): \(y=10\) and \(\dfrac{10}{5}=2\).
Answer to Question 5
Undo the operations in reverse order: first remove the term that is added or subtracted, then divide by the number multiplying the new subject.
1. \(y-2=3x\), so \(x=\dfrac{y-2}{3}\)
2. \(P-2w=2l\), so \(l=\dfrac{P-2w}{2}\)
3. \(C-20=5n\), so \(n=\dfrac{C-20}{5}\)
4. \(F-32=\dfrac{9C}{5}\). Multiply by 5: \(5(F-32)=9C\). Divide by 9: \(C=\dfrac{5(F-32)}{9}\)
Check (a) with \(x=4\): \(y=14\), and \(\dfrac{14-2}{3}=4\).
Answer to Question 6
When the subject is squared, rearrange to get \(r^{2}\) on its own, then take the square root. When it is under a square root, square both sides.
1. \(r^2=\dfrac{A}{\pi}\), so \(r=\sqrt{\dfrac{A}{\pi}}=\sqrt{50\div\pi}=\) 4.0
2. \(r^2=\dfrac{V}{\pi h}\), so \(r=\sqrt{\dfrac{V}{\pi h}}=\sqrt{300\div(6\pi)}=\) 4.0
3. Square both sides: \(r=s^2\). \(r=7^2=\) 49
4. \(4d=r^2\), so \(r=\sqrt{4d}=\sqrt{36}=\) 6
Answer to Question 7
1. \(F=3.2+1.8\times5=3.2+9=\) £12.20
2. \(F=3.2+1.8\times12.5=3.2+22.5=\) £25.70
3. \(F-3.2=1.8m\), so \(m=\dfrac{F-3.2}{1.8}\)
4. \(m=\dfrac{20-3.2}{1.8}=\dfrac{16.8}{1.8}=\) 9.3 miles (9.3 miles)
5. Metered fare for 14 miles: \(3.2+1.8\times14=3.2+25.2=£28.40\). The fixed price of £25 is cheaper by £3.40.
In part (d) the answer is rounded down to 9.3 miles, because 9.4 miles would cost slightly more than £20.
Answer to Question 8
1. Edinburgh: \(1.8\times12+32=\) 53.6°F. Madrid: \(1.8\times35+32=\) 95°F. Reykjavik: \(1.8\times(-5)+32=\) 23°F.
2. \(F-32=1.8C\), so \(C=\dfrac{F-32}{1.8}\)
3. \(C=\dfrac{77-32}{1.8}=\dfrac{45}{1.8}=\) 25°C
4. \(2C=1.8C+32\Rightarrow0.2C=32\Rightarrow C=\) 160°C (and 320°F).
Check part (c) with the original formula: \(1.8\times25+32=45+32=77\).
Part (d) has exactly one answer: below 160°C the Fahrenheit value is more than double the Celsius value, and above 160°C it is less.
Answer to Question 9
1. \(C=45+12\times4=45+48=\)£93.
2. Subtract 45: \(C-45=12h\). Divide by 12: \(h=\dfrac{C-45}{12}\).
3. \(h=\dfrac{150-45}{12}=\dfrac{105}{12}=8.75\). Hiring is by the hour, so she can afford 8 hours (9 hours would cost £153).
4. \(45+12h=20h+15\). Subtract \(12h\) and 15: \(30=8h\), so \(h=3.75\) hours, when both cost £90.
Answer to Question 10
1. \(d=st=80\times2.5=\)200 miles.
2. Divide both sides by t: \(s=\dfrac{d}{t}\).
3. Divide both sides by s: \(t=\dfrac{d}{s}\).
4. \(t=\dfrac{190}{95}=\)2 hours exactly (2 hours 0 minutes).
5. \(s=\dfrac{270}{3}=\)90 mph.
The three forms \(d=st\), \(s=\dfrac{d}{t}\) and \(t=\dfrac{d}{s}\) are the same relationship; choose the one whose subject is the quantity you need.
Answer to Question 11
Substitute the units and the days into the formula. To rearrange, subtract the standing charge \(0.60d\) first, then divide by the price per unit 0.28.
1. March: \(0.28\times150+0.60\times31=42+18.60=\) £60.60. June: \(0.28\times90+0.60\times30=25.20+18=\) £43.20.
2. \(B-0.60d=0.28u\), so \(u=\dfrac{B-0.60d}{0.28}\)
3. \(u=\dfrac{62.40-18}{0.28}=\dfrac{44.40}{0.28}=\) 159 units (about 159)
4. 200 units: \(56+18=£74\). 100 units: \(28+18=£46\), twice that is £92. £74 is not more than £92, because the standing charge is paid once.
Answer to Question 12
1. Subtract \(2l\): \(P-2l=2w\). Divide by 2: \(w=\dfrac{P-2l}{2}\).
2. \(w=\dfrac{360-2\times120}{2}=\dfrac{120}{2}=\)60 m.
3. The fence goes round the perimeter: \(360\times14.50=\)£5 220.
4. Length \(l=2w\): \(P=2(2w)+2w=6w=300\), so \(w=\)50 m and \(l=\)100 m.
Check part (b): \(2\times120+2\times60=240+120=360\) m, the given perimeter.
Answer to Question 13
1. \(P=12+2.5\times16=12+40=\)£52.
2. Subtract 12, then divide by 2.5: \(n=\dfrac{P-12}{2.5}\).
3. \(n=\dfrac{50-12}{2.5}=15.2\). He cannot buy part of a portion, so 15 portions (£49.50).
4. The £8 is added to the fixed part: \(P=20+2.5n\). For 20 portions: \(20+50=\)£70.
Each extra portion adds £2.50, while the £12 (or £20) is paid once, whatever the size of the cake.
Answer to Question 14
1. \(I=\dfrac{800\times3\times4}{100}=\dfrac{9600}{100}=\)£96.
2. Multiply by 100: \(100I=PRT\). Divide by RT: \(P=\dfrac{100I}{RT}\).
3. \(P=\dfrac{100\times45}{2.5\times3}=\dfrac{4500}{7.5}=\)£600.
4. In the same way \(T=\dfrac{100I}{PR}\): \(T=\dfrac{10000}{2000}=\)5 years.
Check (c): \(\dfrac{600\times2.5\times3}{100}=45\), which is Sarah’s interest.
Answer to Question 15
1. Bea is correct.
2. Ali subtracted 6 instead of adding. Cal added 6 but forgot to divide the 6 by 4. Dev multiplied by 4 instead of dividing.
3. \(y=4\times3-6=6\), then \(x=\dfrac{6+6}{4}=3\). It works.
4. \(x=\dfrac{22+6}{4}=\dfrac{28}{4}=\) 7
The correct order is: add 6 to both sides first, then divide everything by 4.
Answer to Question 16
1. \(A=\frac12(6+10)\times5=\frac12\times16\times5=\) 40 cm²
2. \(2A=(a+b)h\), so \(h=\dfrac{2A}{a+b}\)
3. \(h=\dfrac{108}{18}=\) 6 cm
4. \(a+b=\dfrac{2A}{h}\), so \(a=\dfrac{2A}{h}-b\)
5. \(a=\dfrac{80}{5}-9=16-9=\) 7
To make h the subject, first multiply both sides by 2 to remove the half, then divide by the bracket \((a+b)\). To make a the subject, divide by h and then subtract b.
Check (e): \(\dfrac{1}{2}(7+9)\times5=\dfrac{1}{2}\times16\times5=40\), as required.
Answer to Question 17
1. Ali: \(C=40+30h\). Beth: \(C=70+20h\).
2. Ali: \(40+60=\) £100. Beth: \(70+40=\) £110.
3. \(40+30h=70+20h\Rightarrow10h=30\Rightarrow h=\) 3 hours (both £130).
4. For more than 3 hours Beth is cheaper, because her hourly rate is lower.
5. \(h=\dfrac{C-70}{20}=\dfrac{150-70}{20}=\) 4 hours
The cheaper choice depends on the length of the job: Ali is cheaper below 3 hours because his callout fee is lower, Beth is cheaper above 3 hours because her hourly rate is lower.
Answer to Question 18
1. Distance on a full tank = 10 × miles per gallon: Fiesta 500 miles, Estate 350 miles, Van 280 miles.
2. Divide both sides of \(d=Gm\) by m: \(G=\dfrac{d}{m}\).
3. \(G=\dfrac{245}{35}=\)7 gallons, leaving 3 gallons in the tank.
4. \(d=Gm=6.5\times28=\)182 miles.
Answer to Question 19
1. Plan A: \(C=25m\). Plan B: \(C=60+15m\).
2. A: \(25\times6=\) £150. B: \(60+90=\) £150.
3. \(25m=60+15m\Rightarrow10m=60\Rightarrow m=\) 6 months
4. A: \(25\times12=£300\). B: \(60+180=£240\). Plan B is cheaper by £60.
Before 6 months Plan A is cheaper; after 6 months Plan B is cheaper, because its monthly cost is £10 lower.
Answer to Question 20
1. \(S=2\times9+4\times3\times5=18+60=\) 78
2. \(S-2x^2=4xh\), so \(h=\dfrac{S-2x^2}{4x}\)
3. \(h=\dfrac{150-50}{20}=\dfrac{100}{20}=\) 5
4. \(h=\dfrac{240-72}{24}=\dfrac{168}{24}=\) 7 cm
5. The letter \(x\) appears twice, once squared and once as a plain term, so it cannot be isolated with the operations met in Year 8 (it needs a quadratic equation).
Check (d): \(2\times36+4\times6\times7=72+168=240\) cm², the given surface area.

How to use these Year 8 formulae and rearranging worksheet answers

First, try every question on the worksheet without looking. Then compare your working with the solution line by line, not only the final answer. When your answer is different, find the first step that changes, because that is where the mistake is. After that, read the matching part of the formulae and rearranging lesson again. Finally, check that the method is secure with the quiz or the test.

Also, the Year 8 formulae and rearranging worksheet answers are available as a PDF, so you can mark printed work quickly. Moreover, all the pages of this chapter work together, so you can move between them at any time. For example, many pupils read the lesson on Monday, practise on Wednesday and test themselves on Friday. In short, these Year 8 formulae and rearranging worksheet answers show every step, so nothing is left unexplained.