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Ecolint Campus des NationsMathématiques
Ecolint Campus des NationsMathematics
Year 7 · 7.3 Introduction to Algebra

Solutions · Full Answer Key

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

Pack A — Answers

Bronze
1.b=3b = 3; because 3×4=123 \times 4 = 12 and 12+3=1512 + 3 = 15 ✓
2.3n3n is correct; in algebra the coefficient is written before the variable
3.E.g. 5x+3x5x + 3x and 4x+4x4x + 4x; both simplify to 8x8x ✓
4.E.g. x+4=12x + 4 = 12; solving: x=12−4=8x = 12 − 4 = 8 ✓
5.4n=284n = 28... wait — n=7n = 7 chairs per row; equation from context: "total = rows × per row"
6.3x+123x + 12; when x=2x = 2: bracket gives 3(6)=183(6) = 18; expanded gives 6+12=186 + 12 = 18 ✓
7.x=6x = 6; check: 2(6)+5=172(6) + 5 = 17 ✓
8.4n−34n - 3 and 4×n−34 \times n - 3 are the same; in algebra, 4n4n means 4×n4 \times n
9.Yes, both have yy; 9y−4y=5y9y - 4y = 5y because 9−4=59 - 4 = 5
10.Coefficient of xx is 7 (multiplies xx); constant is 3 (never changes)
11.a=0a = 0. Multiplying anything by 00 gives 00, and 55 on its own is not zero, so aa has to be the 00.
Silver
12.18
13.7x+2y7x + 2y
14.4x+174x + 17
15.x=6x = 6
16.Expression: 5p5p; cost = £40
17.48
18.8a+2b8a + 2b
19.(2x+14)(2x + 14) cm
20.8x−208x - 20
21.x=15x = 15
22.a=4a = 4. The 33 isn't zero, so whatever's in the bracket has to be the zero — meaning 4−a=04 - a = 0, so a=4a = 4.
Gold
23.6x6x
24.8
25.5x+165x + 16
26.x=−2x = -2
27.3p+4q3p + 4q pence (cannot simplify further as pp and qq are different)
28.−18-18
29.x=4x = 4
30.x=4x = 4
31.x−12x - 12
32.53
Platinum
33.n=2n = 2
34.15, 16, 17
35.x=4x = 4
36.The difference is always 5.
37.C=12+3gC = 12 + 3g; g=7g = 7 GB
38.p2−p=p(p−1)p^2 - p = p(p-1) — always even.
39.x=9, y=5x = 9,\ y = 5
40.x=5x = 5; sides: 13 cm, 14 cm, 10 cm
41.c=15c = 15; 4(x−2)+6x=10x−84(x-2) + 6x = 10x - 8
42.Input for output 7: x=5x = 5. Fixed point: x=4x = 4

Pack B — Answers

Bronze
1.b=3b = 3; because 3×6=183 \times 6 = 18 and 18+3=2118 + 3 = 21 ✓
2.n2n^2 is correct; repeated multiplication of the same variable is written using index notation
3.E.g. 7x+4x7x + 4x and 6x+5x6x + 5x; both simplify to 11x11x ✓
4.E.g. 3x=213x = 21; solving: x=21÷3=7x = 21 ÷ 3 = 7 ✓
5.5n=455n = 45... n=9n = 9 chairs per row; equation from context: "total = rows × per row"
6.5x+105x + 10; when x=3x = 3: bracket gives 5(5)=255(5) = 25; expanded gives 15+10=2515 + 10 = 25 ✓
7.x=5x = 5; check: 3(5)+4=193(5) + 4 = 19 ✓
8.6n−56n - 5 and 6×n−56 \times n - 5 are the same; 6n6n is the conventional shorthand
9.Yes, both have yy; 8y−3y=5y8y - 3y = 5y because 8−3=58 - 3 = 5
10.Coefficient of xx is 4 (multiplies xx); constant is 9 (never changes)
11.a=0a = 0. Multiplying anything by 00 gives 00, and 77 on its own is not zero, so aa has to be the 00.
Silver
12.16
13.10x+2y10x + 2y
14.3x+223x + 22
15.x=5x = 5
16.Expression: 7p7p; cost = £42
17.50
18.9a+3b9a + 3b
19.(2x+22)(2x + 22) cm
20.12x−2112x - 21
21.x=20x = 20
22.a=2a = 2. The 55 isn't zero, so the bracket has to be zero — meaning 2−a=02 - a = 0, so a=2a = 2.
Gold
23.7x7x
24.21
25.7x+187x + 18
26.x=−2x = -2
27.5p+2q5p + 2q pence
28.23
29.x=3x = 3
30.x=4x = 4
31.6x−86x - 8
32.32
Platinum
33.n=−2n = -2
34.20, 21, 22
35.x=7x = 7
36.The difference is always 5.
37.C=15+4gC = 15 + 4g; g=9g = 9 GB
38.p2−p=p(p−1)p^2 - p = p(p-1) — always even.
39.x=13, y=7x = 13,\ y = 7
40.x=8x = 8; sides: 26 cm, 13 cm, 7 cm
41.c=24c = 24; 3(x−5)+7x=10x−153(x-5) + 7x = 10x - 15
42.Input for output 6: x=4x = 4. Fixed point: x=103x = \frac{10}{3}

Problem-solving — Worked Solutions

1Problem 1
Answer
6
Full working
Let the number be nn. Triple it: 3n3n. Add 5: 3n+53n + 5. Subtract twice the original: 3n+5−2n=n+53n + 5 - 2n = n + 5. Set equal to 11: n+5=11n + 5 = 11. Subtract 5: n=6n = 6. Check: triple 6 = 18; add 5 = 23; subtract twice 6 = 23 − 12 = 11 ✓.
2Problem 2
Answer
x=5x = 5; area = 128 cm²
Full working
Perimeter =2(length+width)=2((3x+1)+(x+3))=2(4x+4)=8x+8= 2(\text{length} + \text{width}) = 2((3x+1)+(x+3)) = 2(4x+4) = 8x+8. Set equal to 48: 8x+8=48⇒8x=40⇒x=58x + 8 = 48 \Rightarrow 8x = 40 \Rightarrow x = 5. Length =3(5)+1=16= 3(5)+1 = 16 cm; width =5+3=8= 5+3 = 8 cm. Check perimeter: 2(16+8)=482(16+8) = 48 ✓. Area =16×8=128= 16 \times 8 = 128 cm².
3Problem 3
Answer
Priya: 11, brother: 16, mother: 33
Full working
Priya: xx. Brother: x+5x + 5. Mother: 3x3x. Sum: x+(x+5)+3x=5x+5x + (x+5) + 3x = 5x + 5. Set equal to 60: 5x+5=60⇒5x=55⇒x=115x + 5 = 60 \Rightarrow 5x = 55 \Rightarrow x = 11. Priya is 11, her brother is 11+5=1611+5=16, her mother is 3×11=333 \times 11=33. Check: 11+16+33=6011+16+33=60 ✓.
4Problem 4
Answer
(a) −1 (b) −1 (c) Proof: (n−1)(n+1)−n2=n2−1−n2=−1(n-1)(n+1) - n^2 = n^2 - 1 - n^2 = -1, always.
Full working
(a) Outer two: 5 and 7. 5×7=355 \times 7 = 35. Middle squared: 62=366^2 = 36. 35−36=−135 - 36 = \mathbf{-1} ✓.

(b) Outer two: (−3)(-3) and (−1)(-1). (−3)×(−1)=3(-3) \times (-1) = 3. Middle squared: (−2)2=4(-2)^2 = 4. 3−4=−13 - 4 = \mathbf{-1} ✓. (This also interleaves Y7.2 directed-number multiplication.)

(c) Let the three consecutive integers be n−1n-1, nn, n+1n+1. Multiply the outer two: (n−1)(n+1)=n2−1(n-1)(n+1) = n^2 - 1 (using the difference-of-two-squares pattern). Subtract the middle squared: (n2−1)−n2=−1(n^2 - 1) - n^2 = \mathbf{-1}. This is independent of nn, so the result is always −1.
5Problem 5
Answer
£36
Full working
Let the meal cost £m\pounds m. Total bill including service charge: m+6m + 6. Split three ways: m+63=14\dfrac{m+6}{3} = 14. Multiply both sides by 3: m+6=42m + 6 = 42. Subtract 6: m=36m = 36. The meal cost £36. Check: 36+6=4236 + 6 = 42; 42÷3=1442 \div 3 = 14 ✓.
6Problem 6
Answer
6
Full working
Let the number be nn. "Subtract 8": n−8n - 8. "Multiply by 3": 3(n−8)3(n-8). Set equal to −6-6: 3(n−8)=−63(n-8) = -6. Divide both sides by 3: n−8=−2n - 8 = -2. Add 8: n=6n = 6. Check: 3(6−8)=3(−2)=−63(6-8) = 3(-2) = -6 ✓.
7Problem 7
Answer
(a) 11 (b) 3 (c) x=5x = 5 (d) Inputs above 5 grow without bound; inputs below 5 decrease without bound.
Full working
(a) 3×7−10=21−10=113 \times 7 - 10 = 21 - 10 = \mathbf{11}.

(b) We need 3x−10=−13x - 10 = -1. Add 10: 3x=93x = 9. Divide by 3: x=3x = \mathbf{3}. Check: 3(3)−10=−13(3)-10 = -1 ✓.

(c) Fixed point: output = input, so 3x−10=x3x - 10 = x. Subtract xx: 2x−10=02x - 10 = 0. Add 10: 2x=102x = 10. Divide by 2: x=5x = \mathbf{5}. Check: 3(5)−10=53(5)-10 = 5 ✓.

(d) Try x=6x = 6 (above 5): 3(6)−10=83(6)-10=8; 3(8)−10=143(8)-10=14; the outputs grow larger. Try x=4x = 4 (below 5): 3(4)−10=23(4)-10=2; 3(2)−10=−43(2)-10=-4; 3(−4)−10=−223(-4)-10=-22; the outputs decrease (and become very negative). The fixed point x=5x=5 is an **unstable** equilibrium — inputs close to it move away from it under repeated application.
8Problem 8
Answer
t=−3t = -3 °C
Full working
At 6 am: t+7t + 7. At noon (doubled from 6 am): 2(t+7)2(t + 7). Set equal to 8: 2(t+7)=82(t+7) = 8. Divide by 2: t+7=4t + 7 = 4. Subtract 7: t=−3t = -3. The midnight temperature was −3-3 °C. Check: −3+7=4-3 + 7 = 4; 4×2=84 \times 2 = 8 ✓.
9Problem 9
Answer
(a) 4n+14n + 1; (b) pattern 10; (c) No
Full working
(a) The sequence 5, 9, 13, … increases by 4 each time. This is linear: tiles=4n+1\text{tiles} = 4n + 1 (check: n=1⇒5n=1 \Rightarrow 5 ✓, n=2⇒9n=2 \Rightarrow 9 ✓). (b) Set 4n+1=414n + 1 = 41: 4n=404n = 40, n=10n = 10 ✓. (c) Set 4n+1=1004n + 1 = 100: 4n=994n = 99, n=24.75n = 24.75. Since nn must be a whole number, no pattern has exactly 100 tiles.
10Problem 10
Answer
(a) Three ways for 15: see working (b) 16 is not a staircase number (c) See algebraic derivation (d) No power of 2 is a staircase number.
Full working
(a) 15=7+815 = 7 + 8 (two consecutive). 15=4+5+615 = 4 + 5 + 6 (three consecutive). 15=1+2+3+4+515 = 1 + 2 + 3 + 4 + 5 (five consecutive). ✓

(b) Two consecutive: n+(n+1)=2n+1n + (n+1) = 2n+1 — always odd. 16 is even, so no. Three consecutive: n+(n+1)+(n+2)=3n+3=16⇒3n=13n + (n+1) + (n+2) = 3n + 3 = 16 \Rightarrow 3n = 13 — not an integer. Four consecutive: 4n+6=16⇒4n=104n + 6 = 16 \Rightarrow 4n = 10 — not an integer. Five consecutive: 5n+10=16⇒5n=65n + 10 = 16 \Rightarrow 5n = 6 — not an integer. **16 is not a staircase number.**

(c) Consecutive integers from nn to n+(k−1)n+(k-1): sum = n+(n+1)+⋯+(n+k−1)n + (n+1) + \cdots + (n+k-1). There are kk terms each contributing nn, giving knkn, plus the extras 0+1+2+⋯+(k−1)=(k−1)k20 + 1 + 2 + \cdots + (k-1) = \dfrac{(k-1)k}{2}. Total: kn+k(k−1)2kn + \dfrac{k(k-1)}{2}.

(d) Powers of 2: 1, 2, 4, 8, 16, 32, … Testing each (as in part b), none can be written as a staircase. The sum kn+k(k−1)2=k(2n+k−1)2kn + \frac{k(k-1)}{2} = \frac{k(2n+k-1)}{2}. For a power of 2, both factors kk and (2n+k−1)(2n+k-1) must be powers of 2 — but they have opposite parity (one odd, one even), so the product can only produce a power of 2 if one of them equals 1, giving trivial cases. **Conjecture: no power of 2 is a staircase number.** (This is a famous result in number theory.)
11Problem 11
Answer
Each box: 1.25 kg; total each side: 8.75 kg
Full working
Let the mass of one box be mm kg. Balance equation: 3m+5=7m3m + 5 = 7m. Subtract 3m3m: 5=4m5 = 4m. Divide by 4: m=1.25m = 1.25. Total on each side: 7×1.25=8.757 \times 1.25 = 8.75 kg. Check left: 3×1.25+5=3.75+5=8.753 \times 1.25 + 5 = 3.75 + 5 = 8.75 ✓.
12Problem 12
Answer
(a) 12 (even); (b) proof below; (c) n=6n = 6 and n=−5n = -5
Full working
(a) (−3)2−(−3)=9+3=12(-3)^2 - (-3) = 9 + 3 = 12 ✓ (even — interleaves Y7.2 directed numbers). (b) Factorise: n2−n=n(n−1)n^2 - n = n(n-1). nn and n−1n-1 are consecutive integers, so one is always even. The product of an even and any integer is even. Therefore n2−nn^2 - n is always even ✓. (c) n(n−1)=30n(n-1) = 30. Look for factor pairs of 30 where one factor is one more than the other: 6×5=306 \times 5 = 30 ✓ (so n=6n = 6); (−5)×(−6)=30(-5) \times (-6) = 30 ✓ (so n=−5n = -5). Check: 62−6=306^2 - 6 = 30 ✓; (−5)2−(−5)=25+5=30(-5)^2 - (-5) = 25 + 5 = 30 ✓.