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Solutions — Full Answer Key

Ecolint Campus des NationsMathematics
Year 10 · Algebra

Solutions · Full Answer Key

Pack A answers · Pack B answers · Problem-solving worked solutions

Pack A — Answers

Bronze
1.6x6x
2.7a+2b7a + 2b
3.3x+123x + 12
4.−2x+10-2x + 10
5.3(2x+3)3(2x + 3)
6.2x(2x+3)2x(2x + 3)
7.16x216 x^2
8.232\sqrt{3}
9.22
10.x=4x = 4
Silver
11.x2+8x+15x^2 + 8x + 15
12.x2−x−12x^2 - x - 12
13.x2+10x+25x^2 + 10x + 25
14.x2−36x^2 - 36
15.(x+3)(x+4)(x + 3)(x + 4)
16.(x+5)(x−2)(x + 5)(x - 2)
17.12
18.525\sqrt{2}
19.x=7x = 7
20.23
Gold
21.(x+7)(x−7)(x + 7)(x - 7)
22.(2x+1)(x+3)(2x + 1)(x + 3)
23.4
24.232\sqrt{3}
25.x=2x = 2, y=5y = 5
26.x=3x = 3, y=3y = 3
27.2(x+3)(x−3)2(x+3)(x-3)
28.x=17x = 17
29.x+1x + 1
30.18 and 12
Platinum
31.x=2x = 2, y=3y = 3
32.(3x−4)(2x+1)(3x - 4)(2x + 1)
33.2(3−5)4=3−52\dfrac{2(3 - \sqrt{5})}{4} = \dfrac{3 - \sqrt{5}}{2}
34.7
35.48 adults, 32 children
36.x=−3x = -3 or x=−4x = -4
37.x−5x - 5
38.x=4,y=16x = 4, y = 16 or x=−1,y=1x = -1, y = 1
39.27
40.x=±52x = \pm 5\sqrt{2}

Pack B — Answers

Bronze
1.5y5y
2.4p+7q4p + 7q
3.5x+105x + 10
4.−4x+12-4x + 12
5.4(2x+3)4(2x + 3)
6.3x(3x+4)3x(3x + 4)
7.49x249 x^2
8.323\sqrt{2}
9.24
10.x=4x = 4
Silver
11.x2+11x+28x^2 + 11x + 28
12.x2−3x−10x^2 - 3x - 10
13.x2+12x+36x^2 + 12x + 36
14.x2−64x^2 - 64
15.(x+4)(x+5)(x + 4)(x + 5)
16.(x+5)(x−3)(x + 5)(x - 3)
17.10
18.535\sqrt{3}
19.x=7x = 7
20.21
Gold
21.(x+10)(x−10)(x + 10)(x - 10)
22.(3x+2)(x+2)(3x + 2)(x + 2)
23.13
24.252\sqrt{5}
25.x=3x = 3, y=11y = 11
26.x=2x = 2, y=5y = 5
27.2(x+5)(x−5)2(x+5)(x-5)
28.x=19x = 19
29.3x+23x + 2
30.30 and 20
Platinum
31.x=3x = 3, y=2y = 2
32.(2x−3)(2x+1)(2x - 3)(2x + 1)
33.4(5−3)22=2(5−3)11\dfrac{4(5 - \sqrt{3})}{22} = \dfrac{2(5 - \sqrt{3})}{11}
34.14
35.30 adults, 30 children
36.x=−3x = -3 or x=−5x = -5
37.x−6x - 6
38.x=6,y=36x = 6, y = 36 or x=−1,y=1x = -1, y = 1
39.18
40.x=±62x = \pm 6\sqrt{2}

Problem-solving — Worked Solutions

1Problem 1
Answer
(a) x2+4x+3=35x^2 + 4x + 3 = 35 → x2+4x−32=0x^2 + 4x - 32 = 0 (b) x=4x = 4 (reject x=−8x = -8) (c) 7 cm × 5 cm
Full working
(a) Area: (x+3)(x+1)=x2+4x+3=35(x+3)(x+1) = x^2 + 4x + 3 = 35. Rearrange: x2+4x−32=0x^2 + 4x - 32 = 0.

(b) Factorise: need two numbers multiplying to −32-32 and summing to +4+4: 88 and −4-4. So (x+8)(x−4)=0(x+8)(x-4) = 0, giving x=−8x = -8 or x=4x = 4. Reject x=−8x = -8 (lengths can't be negative). So x=4x = 4.

(c) Length = 4+3=74 + 3 = 7 cm; Width = 4+1=54 + 1 = 5 cm. Check: 7×5=357 \times 5 = 35 ✓.
2Problem 2
Answer
6 fifty-pence and 9 twenty-pence coins
Full working
Let ff = number of 50p coins, tt = number of 20p coins.
f+t=15f + t = 15

0.50f+0.20t=4.800.50 f + 0.20 t = 4.80

Multiply second eq by 10: 5f+2t=485f + 2t = 48. From first: t=15−ft = 15 - f. Substitute: 5f+2(15−f)=485f + 2(15 - f) = 48, so 5f+30−2f=485f + 30 - 2f = 48, 3f=183f = 18, f=6f = 6. Then t=9t = 9.

Check: 6 × £0.50 + 9 × £0.20 = £3.00 + £1.80 = £4.80 ✓.
3Problem 3
Answer
(a) 15\sqrt{15} cm (b) 3 cm² (c) 33+153\sqrt{3} + \sqrt{15} cm
Full working
(a) Hypotenuse: h=(3)2+(12)2=3+12=15h = \sqrt{(\sqrt{3})^2 + (\sqrt{12})^2} = \sqrt{3 + 12} = \sqrt{15} cm.

(b) Area = 12(3)(12)=1236=12(6)=3\frac{1}{2}(\sqrt{3})(\sqrt{12}) = \frac{1}{2}\sqrt{36} = \frac{1}{2}(6) = 3 cm².

(c) Perimeter = 3+12+15\sqrt{3} + \sqrt{12} + \sqrt{15}. Simplify 12=23\sqrt{12} = 2\sqrt{3}. So perimeter =3+23+15=33+15= \sqrt{3} + 2\sqrt{3} + \sqrt{15} = 3\sqrt{3} + \sqrt{15} cm.
4Problem 4
Answer
(a) See working (b) 1200 (c) 12 000
Full working
(a) Expand: (x+y)2=x2+2xy+y2(x+y)^2 = x^2 + 2xy + y^2 and (x−y)2=x2−2xy+y2(x-y)^2 = x^2 - 2xy + y^2. Subtract: (x2+2xy+y2)−(x2−2xy+y2)=4xy(x^2 + 2xy + y^2) - (x^2 - 2xy + y^2) = 4xy ✓.

(b) Treat 103 and 97 as 100+3100 + 3 and 100−3100 - 3. Using (a): (100+3)2−(100−3)2=4(100)(3)=1200(100+3)^2 - (100-3)^2 = 4(100)(3) = 1200.

(c) 2152−1852=(215+185)(215−185)=400×30=12,000215^2 - 185^2 = (215+185)(215-185) = 400 \times 30 = 12{,}000.
5Problem 5
Answer
(a) n+(n+1)+(n+2)=102n + (n+1) + (n+2) = 102 (b) 33, 34, 35 (c) See working
Full working
(a) Let smallest be nn. Then n+(n+1)+(n+2)=102n + (n+1) + (n+2) = 102.

(b) Simplify: 3n+3=1023n + 3 = 102, so 3n=993n = 99, n=33n = 33. Integers: 33, 34, 35. Check: 33+34+35=10233 + 34 + 35 = 102 ✓.

(c) Sum of any three consecutive integers n,n+1,n+2n, n+1, n+2 is 3n+3=3(n+1)3n + 3 = 3(n+1). This is 3×3 \times (an integer), so it is always divisible by 3.
6Problem 6
Answer
(a) (2x+5)(2x−5)(2x+5)(2x-5) (b) x(x+3)(x−3)x(x+3)(x-3) (c) 2(x+1)(x+3)2(x+1)(x+3) (d) (x+2+y)(x+2−y)(x+2+y)(x+2-y)
Full working
(a) Difference of squares: 4x2−25=(2x)2−52=(2x+5)(2x−5)4x^2 - 25 = (2x)^2 - 5^2 = (2x+5)(2x-5).

(b) Common factor xx: x(x2−9)=x(x+3)(x−3)x(x^2 - 9) = x(x+3)(x-3).

(c) HCF 2: 2(x2+4x+3)=2(x+1)(x+3)2(x^2 + 4x + 3) = 2(x+1)(x+3).

(d) Recognise x2+4x+4=(x+2)2x^2 + 4x + 4 = (x+2)^2. So expression is (x+2)2−y2(x+2)^2 - y^2, a difference of squares: ((x+2)+y)((x+2)−y)=(x+2+y)(x+2−y)((x+2)+y)((x+2)-y) = (x+2+y)(x+2-y).
7Problem 7
Answer
(a) b=−9b = -9, c=14c = 14 (b) See working
Full working
(a) If roots are 2 and 7, factorised form is (x−2)(x−7)=x2−9x+14(x-2)(x-7) = x^2 - 9x + 14. So b=−9b = -9, c=14c = 14.

(b) Check x=2x = 2: 22+(−9)(2)+14=4−18+14=02^2 + (-9)(2) + 14 = 4 - 18 + 14 = 0 ✓.
8Problem 8
Answer
(a) 626\sqrt{2} (b) 4(2−1)=42−44(\sqrt{2} - 1) = 4\sqrt{2} - 4 (c) 4 ✓
Full working
(a) 50=52\sqrt{50} = 5\sqrt{2}; 18=32\sqrt{18} = 3\sqrt{2}; 8=22\sqrt{8} = 2\sqrt{2}. Sum: 52+32−22=625\sqrt{2} + 3\sqrt{2} - 2\sqrt{2} = 6\sqrt{2}.

(b) Multiply top and bottom by conjugate: 42+1⋅2−12−1=4(2−1)2−1=4(2−1)\frac{4}{\sqrt{2}+1} \cdot \frac{\sqrt{2}-1}{\sqrt{2}-1} = \frac{4(\sqrt{2}-1)}{2 - 1} = 4(\sqrt{2} - 1).

(c) Product: (2+1)⋅4(2−1)=4((2)2−12)=4(2−1)=4(\sqrt{2}+1) \cdot 4(\sqrt{2}-1) = 4((\sqrt{2})^2 - 1^2) = 4(2 - 1) = 4. This confirms the rationalising step: the original was indeed 42+1\frac{4}{\sqrt{2}+1} rewritten without surd in denominator.
9Problem 9
Answer
(a) 4x+5=204x + 5 = 20 (b) x=3.75x = 3.75; sides 6.75,6.75,6.56.75, 6.75, 6.5 cm (c) No — sides would be negative or violate triangle inequality
Full working
(a) Perimeter: (x+3)+(x+3)+(2x−1)=4x+5=20(x+3) + (x+3) + (2x-1) = 4x + 5 = 20.

(b) 4x=154x = 15, x=3.75x = 3.75. Sides: x+3=6.75x + 3 = 6.75 cm (twice), 2x−1=6.52x - 1 = 6.5 cm. Total =6.75+6.75+6.5=20= 6.75+6.75+6.5 = 20 ✓.

(c) If perimeter =5= 5: 4x+5=54x + 5 = 5, so x=0x = 0. Sides: 3, 3, −1-1 cm. A negative side length is impossible. Also even with xx slightly positive (x=0.5x = 0.5), sides 3.5,3.5,03.5, 3.5, 0 — degenerate. The triangle does **not** exist.
10Problem 10
Answer
x=6x = 6 or x=−1x = -1
Full working
Rearrange to standard form: x2−5x−6=0x^2 - 5x - 6 = 0. Factorise: two numbers ×(−6)\times (-6) and sum −5-5: −6-6 and 11. So (x−6)(x+1)=0(x-6)(x+1) = 0. Therefore x=6x = 6 or x=−1x = -1. Check: 36−30=636 - 30 = 6 ✓; 1+5=61 + 5 = 6 ✓.
11Problem 11
Answer
(a) f+3r=45f + 3r = 45, f+5r=67f + 5r = 67 (b) f=12f = 12, r=11r = 11 (c) £100
Full working
(a) Cost = fixed + (rate × hours). So f+3r=45f + 3r = 45 and f+5r=67f + 5r = 67.

(b) Subtract: 2r=222r = 22, so r=11r = 11. Then f+33=45f + 33 = 45, so f=12f = 12.

(c) 8 hours: 12+8×11=12+88=£10012 + 8 \times 11 = 12 + 88 = \mathbf{\pounds 100}.
12Problem 12
Answer
(a) 424\sqrt{2} cm (b) 32 cm² (c) 16216\sqrt{2} cm
Full working
(a) Let side be ss. Diagonal of square: d=s2d = s\sqrt{2}. So s2=8s\sqrt{2} = 8, giving s=82=822=42s = \frac{8}{\sqrt{2}} = \frac{8\sqrt{2}}{2} = 4\sqrt{2} cm.

(b) Area = s2=(42)2=16×2=32s^2 = (4\sqrt{2})^2 = 16 \times 2 = 32 cm².

(c) Perimeter = 4s=4×42=1624s = 4 \times 4\sqrt{2} = 16\sqrt{2} cm.