Formulae and identities – Year 10 GCSE maths worksheet answers to download as PDF

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1. Answer to Question 1

   1. \(v=5+2.5\times6=5+15=\) 20 Substitute the values, then multiply before adding (BIDMAS). So \(v=20\).
   2. \(s=3\times5+\tfrac12\times4\times25=15+50=\) 65 Square \(t\) first: \(t^2=25\). Then \(\tfrac12\times4\times25=50\), so \(s=65\).
   3. \(A=\tfrac12\times18\times6=\) 54 Add the parallel sides first because they are in a bracket: \(7+11=18\). So \(A=54\).
   4. \(v^2=16+2\times3\times10=76\), so \(v=\sqrt{76}=\) 8.72 (3 s.f.) Find \(v^2\) first, then take the positive square root because the question asks for the positive value. So \(v=8.72\) to 3 significant figures.

2. Answer to Question 2

   1. \(y-c=mx\), so \(x=\dfrac{y-c}{m}\) Subtract \(c\) from both sides, then divide both sides by \(m\). The new subject is \(x=\dfrac{y-c}{m}\).
   2. \(P-2w=2l\), so \(l=\dfrac{P-2w}{2}\) Subtract \(2w\) from both sides, then divide by 2. So \(l=\dfrac{P-2w}{2}\).
   3. \(h=\dfrac{V}{lw}\) Here \(h\) is multiplied by \(lw\), so we divide both sides by \(lw\). So \(h=\dfrac{V}{lw}\).
   4. \(r=\dfrac{C}{2\pi}\) Here \(r\) is multiplied by \(2\pi\), so we divide both sides by \(2\pi\). So \(r=\dfrac{C}{2\pi}\).

3. Answer to Question 3

   1. \(2A=(a+b)h\), \(\dfrac{2A}{h}=a+b\), so \(a=\dfrac{2A}{h}-b\) Undo the operations in reverse order: multiply by 2, divide by \(h\), then subtract \(b\).
   2. \(v^2-u^2=2as\), so \(a=\dfrac{v^2-u^2}{2s}\) Subtract \(u^2\) from both sides, then divide by \(2s\).
   3. \(2s=(u+v)t\), \(\dfrac{2s}{t}=u+v\), so \(v=\dfrac{2s}{t}-u\) Multiply by 2, divide by \(t\), then subtract \(u\) from both sides.
   4. Square both sides: \(d^2=2h\), so \(h=\dfrac{d^2}{2}\) Squaring undoes the square root, then dividing by 2 isolates \(h\).

4. Answer to Question 4

   1. An equation: it is true only for \(x=5\).
   2. A formula: it links variables, here the area and radius of a circle.
   3. An identity: it is true for every value of \(x\) (the sign \(\equiv\) shows this).
   4. An expression: it has no equals sign.

5. Answer to Question 5

   1. \(2x+8+3x-3=5x+5\) ✓ Expand both brackets and collect like terms: \(2x+3x=5x\) and \(8-3=5\). The left-hand side equals the right-hand side, so the identity is true.
   2. \((x^2+6x+9)-(x^2-6x+9)=12x\) ✓ The bracket in front of the second square changes every sign: the \(x^2\) terms cancel and so do the 9s, leaving \(6x+6x=12x\). So the identity is true.
   3. \(x^2-2x+5x-10=x^2+3x-10\) ✓ Multiply each term in the first bracket by each term in the second, then collect \(-2x+5x=3x\). So the identity is true.
   4. \((2x-1)(2x-1)=4x^2-2x-2x+1=4x^2-4x+1\) ✓ A squared bracket is the bracket times itself, so there are two middle terms of \(-2x\). So the identity is true for every value of \(x\).

6. Answer to Question 6

   1. \(y+5=3x^2\), \(x^2=\dfrac{y+5}{3}\), so \(x=\pm\sqrt{\dfrac{y+5}{3}}\) Add 5, divide by 3, then take the square root. A square root can be positive or negative, so we write \(\pm\).
   2. \(y-1=\dfrac{x^2}{4}\), \(x^2=4(y-1)\), so \(x=\pm2\sqrt{y-1}\) Subtract 1, multiply by 4, then take the square root; \(\sqrt{4}=2\) comes out of the root.
   3. \(x^2=\dfrac A\pi\), so \(x=\sqrt{\dfrac A\pi}\) Divide by \(\pi\), then take the positive square root because \(x\) is a length.
   4. \(\sqrt y=\pm(x+2)\), so \(x=-2\pm\sqrt y\) Take the square root of both sides first, then subtract 2 from both sides.

7. Answer to Question 7

   1. London: \(1.8\times25+32=\) 77 °F. Edinburgh: \(1.8\times11+32=\) 51.8 °F. Cardiff: \(1.8\times(-5)+32=\) 23 °F.
   2. \(F-32=1.8C\), so \(C=\dfrac{F-32}{1.8}\). Undo the operations in reverse order: subtract 32 first, then divide by 1.8.
   3. \(C=\dfrac{350-32}{1.8}=176.7\), so 177 °C. So an oven at 350 °F is at about 177 °C.
   4. Set \(C=F\): \(C=1.8C+32\), so \(-0.8C=32\) and \(C=\) −40. At \(-40\) degrees the two scales agree.

8. Answer to Question 8

   1. \(0.28\times160+0.45\times30=44.80+13.50=\) £58.30 Substitute \(u=160\) and \(d=30\). The bill is £58.30.
   2. \(C-0.45d=0.28u\), so \(u=\dfrac{C-0.45d}{0.28}\) Subtract \(0.45d\) from both sides, then divide by 0.28.
   3. \(u=\dfrac{70-0.45\times31}{0.28}=\dfrac{56.05}{0.28}=200.2\), so 200 units. Substitute \(C=70\) and \(d=31\) into the rearranged formula. The customer used about 200 units.
   4. Other firm: \(0.25\times160+0.62\times30=40+18.60=58.60\). The first firm is cheaper by \(58.60-58.30=\) £0.30. So the first firm is cheaper, by 30p.

9. Answer to Question 9

   1. \(T=40\times2.5+20=120\) minutes \(=\) 2 hours. Substitute \(w=2.5\). 120 minutes is 2 hours, so the joint cooks for 2 hours.
   2. \(T-20=40w\), so \(w=\dfrac{T-20}{40}\). Subtract 20 from both sides, then divide by 40.
   3. \(T=40\times3.2+20=148\) minutes \(=2\) h 28 min. Back from 18:30: 16:02 (about 4:02 pm). Count back 2 hours from 18:30 to 16:30, then 28 more minutes to 16:02. The joint should go in at about 4:02 pm.
   4. \(2\tfrac12\) hours \(=150\) min. \(w=\dfrac{150-20}{40}=\) 3.25 kg. Change the time into minutes, then use the rearranged formula. The largest joint is 3.25 kg.

10. Answer to Question 10

    1. \(s=\tfrac12\times9.8\times9=\) 44.1 m. Square the time first: \(t^2=9\). The stone falls 44.1 m in 3 seconds.
    2. \(2s=gt^2\), \(t^2=\dfrac{2s}{g}\), so \(t=\sqrt{\dfrac{2s}{g}}\). Multiply by 2, divide by \(g\), then take the positive square root because time is positive.
    3. \(t=\sqrt{\dfrac{2\times78.4}{9.8}}=\sqrt{16}=\) 4 s. Substitute \(s=78.4\) and \(g=9.8\) into the rearranged formula. The stone takes 4 seconds to reach the water.
    4. \(s=\tfrac12\times9.8\times4=\) 19.6 m. Substitute \(t=2\), so \(t^2=4\). The bridge is 19.6 m high.

11. Answer to Question 11

    1. \(B=70\div1.75^2=70\div3.0625=\) 22.9 Square the height first: \(1.75^2=3.0625\). The BMI is 22.9 to 1 decimal place.
    2. \(m=Bh^2\) Multiply both sides by \(h^2\). So \(m=Bh^2\).
    3. \(m=24\times1.6^2=24\times2.56=\) 61.44 kg Substitute into \(m=Bh^2\). The mass is 61.44 kg.
    4. \(h^2=\dfrac mB\), so \(h=\sqrt{\dfrac mB}\). \(h=\sqrt{63\div25}=\sqrt{2.52}=\) 1.59 m. Multiply by \(h^2\), divide by \(B\), then take the positive square root because height is positive. The height is 1.59 m.

12. Answer to Question 12

    1. \(500\times1.04^3=\) £562.43 The multiplier for 4% growth is \(1.04\), and it is applied once for each of the 3 years. The account is worth £562.43.
    2. \(1200\times1.025^5=\) £1357.69 The multiplier for 2.5% is \(1.025\), applied 5 times. The account is worth £1357.69.
    3. \(100I=PRT\), so \(T=\dfrac{100I}{PR}\) Multiply both sides by 100, then divide by \(PR\).
    4. \(T=\dfrac{100\times120}{800\times3}=\dfrac{12\,000}{2400}=\) 5 years. Substitute into the rearranged formula. It takes 5 years to earn £120.

13. Answer to Question 13

    1. \(A=\tfrac12\times3.5\times20=\) 35 m². Add the parallel sides, halve, then multiply by the length. The area is 35 m².
    2. \(2A=(a+b)h\), so \(\dfrac{2A}{h}=a+b\) and \(b=\dfrac{2A}{h}-a\). Multiply by 2, divide by \(h\), then subtract \(a\).
    3. \(b=\dfrac{96}{20}-1.2=4.8-1.2=\) 3.6 m. Substitute \(A=48\), \(h=20\) and \(a=1.2\) into the rearranged formula. The deep end is 3.6 m deep.
    4. Volume \(=35\times8=\) 280 m³. The pool is a prism, so its volume is the cross-section area times the width. The pool holds 280 m³ of water.

14. Answer to Question 14

    1. \(\pi\times49\times10=\) 1539 cm³ Square the radius first: \(7^2=49\). The volume is about 1539 cm³.
    2. \(r^2=\dfrac{V}{\pi h}\), so \(r=\sqrt{\dfrac{V}{\pi h}}\) Divide by \(\pi h\), then take the positive square root because \(r\) is a length.
    3. \(r=\sqrt{\dfrac{750}{12\pi}}=\sqrt{19.89}=\) 4.5 cm Substitute \(V=750\) and \(h=12\). The radius is 4.5 cm.
    4. \(h=\dfrac{V}{\pi r^2}=\dfrac{1000}{36\pi}=\) 8.8 cm Divide both sides by \(\pi r^2\). The tin is 8.8 cm tall.

15. Answer to Question 15

    1. \(ax-cx=d-b\), \(x(a-c)=d-b\), so \(x=\dfrac{d-b}{a-c}\) Collect the \(x\) terms on one side and the other terms on the other, then factorise out \(x\) and divide by \(a-c\).
    2. \(y(x-3)=2x+1\), \(xy-3y=2x+1\), \(xy-2x=3y+1\), \(x(y-2)=3y+1\), so \(x=\dfrac{3y+1}{y-2}\) Multiply both sides by \(x-3\) to clear the fraction, then collect the \(x\) terms on the left. Factorise and divide by \(y-2\).
    3. \(a=S(1-r)\) Multiply both sides by \(1-r\). This is allowed because \(r\) is not 1, so \(1-r\) is not zero.
    4. \(pq+p=3p-6\), \(pq+p-3p=-6\), \(p(q-2)=-6\), so \(p=\dfrac{-6}{q-2}=\dfrac{6}{2-q}\) Expand the bracket, collect the \(p\) terms on one side, factorise and divide by \(q-2\).

16. Answer to Question 16

    1. \((n^2+2n+1)-(n^2-2n+1)=4n\). ✓
    2. \(n^2+(n+1)^2=2n^2+2n+1=2(n^2+n)+1\), which is 2 times a whole number plus 1: odd.
    3. \((2n+1)^2-1=4n^2+4n+1-1=4n^2+4n=4n(n+1)\), a multiple of 4.
    4. \(4n(n+1)\): one of \(n\) and \(n+1\) is even, so \(n(n+1)\) is even \(=2k\), and \(4\times2k=8k\). So it is a multiple of 8.

17. Answer to Question 17

    1. \(d=\dfrac{900}{60}+9=15+9=\) 24 m. Substitute \(v=30\): \(30^2=900\) and \(0.3\times30=9\). The stopping distance is 24 m.
    2. \(d=\dfrac{4900}{60}+21=81.67+21=102.67\), so 103 m. Substitute \(v=70\): \(70^2=4900\). The stopping distance is about 103 m.
    3. \(\dfrac{v^2}{60}+0.3v=60\). Multiply by 60: \(v^2+18v=3600\), so \(v^2+18v-3600=0\). Multiplying every term by 60 clears the fraction, then subtract 3600 from both sides.
    4. \(v=\dfrac{-18+\sqrt{18^2+4\times3600}}{2}=\dfrac{-18+\sqrt{14724}}{2}=\dfrac{-18+121.35}{2}=\) 51.7 mph. Use the quadratic formula with \(a=1\), \(b=18\), \(c=-3600\). We reject the negative root because a speed cannot be negative.

18. Answer to Question 18

    1. Expand: \(x^2+2ax+a^2+b\). Compare \(x\) terms: \(2a=6\), so \(a=3\). Constant: \(9+b=13\), so \(b=\) 4. So \(a=\) 3, \(b=\) 4.
    2. \(px+3p+q\). Compare \(x\): \(p=\) 5. Constant: \(15+q=19\), so \(q=\) 4.
    3. \(x^2+(k-4)x-4k\). Compare \(x\): \(k-4=2\), so \(k=6\); check the constant: \(-4\times6=-24\) ✓. So \(k=\) 6.
    4. The left side has \(2x\) and the right side has \(3x+b\). The \(x\) terms \(2x\) and \(3x\) cannot match, so no values of \(a\) and \(b\) work.

19. Answer to Question 19

    1. \(\theta=\dfrac{360L}{2\pi r}=\) \(\dfrac{180L}{\pi r}\) Simplify \(\tfrac{360}{2}=180\), then multiply both sides by 360 and divide by \(2\pi r\).
    2. \(r=\dfrac{360L}{2\pi\theta}=\dfrac{180L}{\pi\theta}\) In the same way, multiply by 360 and divide by \(2\pi\theta\).
    3. \(\theta=\dfrac{180\times15.7}{\pi\times10}=\dfrac{2826}{31.416}=89.96\), so 90°. Substitute \(L=15.7\) and \(r=10\). The angle is \(90^\circ\) to the nearest degree.
    4. \(r=\dfrac{180\times22}{\pi\times90}=\dfrac{3960}{282.74}=\) 14.0 cm. Substitute \(L=22\) and \(\theta=90\). The radius is 14.0 cm to 1 decimal place.

20. Answer to Question 20

    1. \(T_6=\dfrac{6\times7}{2}=\) 21 and \(T_{10}=\dfrac{10\times11}{2}=\) 55.
    2. \(\dfrac{n(n+1)}{2}+\dfrac{(n+1)(n+2)}{2}=\dfrac{(n+1)(n+n+2)}{2}=\dfrac{(n+1)(2n+2)}{2}=(n+1)^2\). Write both fractions over 2, then take out the common factor \((n+1)\).
    3. The result \((n+1)^2\) is the square of a whole number, so it is a square number. For example \(6+10=16=4^2\) and \(10+15=25=5^2\).
    4. \(n(n+1)=240\), so \(n^2+n-240=0\), \((n-15)(n+16)=0\) and \(n=\) 15.