A Level H2 Mathematics Numbers Ratio Proportion Quiz

A-Level Maths H2 Quiz - Numbers Ratio Proportion

Answer Key and Marking Scheme

Total Marks: 50

Section A: Sequences and Series (Questions 1–7)

1. (u_{20} = a + 19d = 5 + 19(3) = 5 + 57 = 62)
[M1] for correct formula (u_n = a + (n-1)d)
[A1] for correct answer 62
[Total: 2 marks]

2. (S_\infty = \frac{a}{1-r} = \frac{48}{1 - \frac{1}{2}} = \frac{48}{\frac{1}{2}} = 96)
[M1] for correct formula and substitution
[A1] for correct answer 96
[Total: 2 marks]

3. (S_n = 3n^2 + 5n)
(S_1 = 3(1)^2 + 5(1) = 8), so (a = u_1 = 8)
(S_2 = 3(4) + 10 = 22), so (u_2 = S_2 - S_1 = 22 - 8 = 14)
(d = u_2 - u_1 = 14 - 8 = 6)
[M1] for finding (S_1) or (u_1)
[M1] for finding (S_2) and (u_2) or using (S_n) formula
[A1] for (a = 8) and (d = 6)
[Total: 3 marks]

4. (S_\infty = \frac{a}{1-r} = 36 \implies a = 36(1-r))
Sum of first two terms: (a + ar = a(1+r) = 27)
Substitute: (36(1-r)(1+r) = 27 \implies 36(1-r^2) = 27 \implies 1-r^2 = \frac{27}{36} = \frac{3}{4})
(r^2 = \frac{1}{4} \implies r = \frac{1}{2}) (since (r > 0))
(a = 36(1 - \frac{1}{2}) = 36 \times \frac{1}{2} = 18)
[M1] for using (S_\infty) formula
[M1] for setting up equation with sum of first two terms
[M1] for solving for (r)
[A1] for (a = 18) and (r = \frac{1}{2})
[Total: 4 marks]

5. (a) (u_5 = \frac{2(5)+1}{5+2} = \frac{11}{7})
[A1] for correct value
[1 mark]

(b) (\lim_{n \to \infty} u_n = \lim_{n \to \infty} \frac{2n+1}{n+2} = \lim_{n \to \infty} \frac{2 + \frac{1}{n}}{1 + \frac{2}{n}} = 2)
The sequence converges to 2.
[M1] for correct limit reasoning
[A1] for stating convergence and limit 2
[Total: 3 marks]

6. AP: (a = 4), (d) unknown. Terms: (4, 4+d, 4+2d)
GP: (a = 4), (r) unknown ((r \neq 1)). Terms: (4, 4r, 4r^2)
Given: (4+d = 4r) and (4+2d = 4r^2)
From first equation: (d = 4r - 4)
Substitute: (4 + 2(4r - 4) = 4r^2 \implies 4 + 8r - 8 = 4r^2 \implies 8r - 4 = 4r^2)
(4r^2 - 8r + 4 = 0 \implies r^2 - 2r + 1 = 0 \implies (r-1)^2 = 0 \implies r = 1)
But (r \neq 1) given. Re-examine: The equations are correct. If (r=1), then (d=0), which gives identical sequences.
Wait—recheck: (4+d = 4r) and (4+2d = 4r^2)
From first: (d = 4(r-1)). Substitute: (4 + 8(r-1) = 4r^2 \implies 4 + 8r - 8 = 4r^2 \implies 8r - 4 = 4r^2 \implies r^2 - 2r + 1 = 0 \implies r = 1).
Since (r \neq 1), there is no solution. However, the problem states they exist. Let me reconsider: perhaps the terms are not corresponding positions.
Actually, the problem says "second term of AP = second term of GP" and "third term of AP = third term of GP". With (a=4) for both, this forces (r=1) or (d=0). The only way to have (r \neq 1) is if the first terms are not equal. But the problem states both have first term 4. This appears to be a contradiction unless (r=1).
Revised answer: The conditions imply (r=1) and (d=0), but since (r \neq 1) is given, no such progressions exist.
Note: This question contains an inherent contradiction. Full marks awarded for identifying the impossibility or for (r=1, d=0) with note.
[M1] for setting up equations
[M1] for substituting and solving
[A1] for identifying (r=1, d=0) or impossibility
[A1] for clear reasoning
[Total: 4 marks]

7. (a) (u_2 = 3(2) - 4 = 2)
(u_3 = 3(2) - 4 = 2)
[A1] for both correct
[2 marks]

(b) Let (P(n)) be the statement (u_n = 3^{n-1} + 1).
Base case (n=1): (u_1 = 3^0 + 1 = 1 + 1 = 2). True.
Inductive step: Assume (P(k)) true, i.e., (u_k = 3^{k-1} + 1).
Then (u_{k+1} = 3u_k - 4 = 3(3^{k-1} + 1) - 4 = 3^k + 3 - 4 = 3^k + 1 = 3^{(k+1)-1} + 1).
Thus (P(k+1)) is true.
By mathematical induction, (P(n)) is true for all (n \in \mathbb{Z}^+).
[M1] for base case
[M1] for inductive hypothesis
[M1] for inductive step algebra
[A1] for correct conclusion
[Total: 6 marks]

Section B: Ratio, Proportion, and Applications (Questions 8–13)

8. Ratio boys : girls = (5 : 3). Boys = 480.
(5 \text{ parts} = 480 \implies 1 \text{ part} = 96)
Total parts = (5 + 3 = 8). Total students = (8 \times 96 = 768)
[M1] for finding value of one part
[A1] for 768
[Total: 2 marks]

9. Ratio = (3 : 5 : 7). Largest share (7 parts) = 420. \(1 \text{ part} = 420 \div 7 = 60\) Total parts = \(3 + 5 + 7 = 15\). Total = \(15 \times 60 = 900\) **[M1]** for finding value of one part **[A1]** for 900
[Total: 2 marks]

10. (a) (y = kx^2). When (x=4, y=96): (96 = k(16) \implies k = 6)
Equation: (y = 6x^2)
[M1] for setting up proportionality
[A1] for correct equation
[2 marks]

(b) When (x = 7): (y = 6(49) = 294)
[A1] for 294
[Total: 3 marks]

11. (a) (V = kr^3). When (r=3, V=36\pi): (36\pi = k(27) \implies k = \frac{36\pi}{27} = \frac{4\pi}{3})
(V = \frac{4\pi}{3}r^3)
[M1] for setting up proportionality
[A1] for correct expression
[2 marks]

(b) (288\pi = \frac{4\pi}{3}r^3 \implies r^3 = 288\pi \times \frac{3}{4\pi} = 216 \implies r = 6) cm
[M1] for substitution and solving
[A1] for (r = 6)
[Total: 4 marks]

12. (a) (T = \frac{k}{n}). When (n=6, T=10): (10 = \frac{k}{6} \implies k = 60)
(T = \frac{60}{n})
[M1] for setting up inverse proportion
[A1] for correct equation
[2 marks]

(b) (4 = \frac{60}{n} \implies n = 15) workers
[M1] for substitution
[A1] for 15
[Total: 4 marks]

13. Let cost per book (C = a + \frac{b}{n}), where (a) is constant cost and (\frac{b}{n}) is the inversely proportional part.
When (n = 500, C = 8): (8 = a + \frac{b}{500}) ... (1)
When (n = 2000, C = 5): (5 = a + \frac{b}{2000}) ... (2)
(1) - (2): (3 = b(\frac{1}{500} - \frac{1}{2000}) = b(\frac{4-1}{2000}) = \frac{3b}{2000} \implies b = 2000)
From (1): (8 = a + \frac{2000}{500} = a + 4 \implies a = 4)
So (C = 4 + \frac{2000}{n})
When (n = 1000): (C = 4 + \frac{2000}{1000} = 4 + 2 = 6)
Cost per book = 6. **[M1]** for setting up model \(C = a + \frac{b}{n}\) **[M1]** for solving for \(a\) and \(b\) **[A1]** for 6
[Total: 3 marks]

Section C: Financial Mathematics and Modelling (Questions 14–20)

14. (A = P(1 + r)^t = 5000(1.04)^6)
(= 5000 \times 1.265319... = 6326.60) (to 2 d.p.)
Value = $6326.60
[M1] for compound interest formula
[A1] for correct value
[Total: 2 marks]

15. (V = 30000(1 - 0.15)^4 = 30000(0.85)^4)
(= 30000 \times 0.52200625 = 15660.19)
Value = $15,700 (3 s.f.) or$15,660.19
[M1] for depreciation formula
[A1] for correct value
[Total: 2 marks]

16. This is an annuity due (payments at start of each year).
Amount = (200(1.03)\frac{(1.03^{10} - 1)}{0.03})
(= 200 \times 1.03 \times \frac{1.343916 - 1}{0.03})
(= 206 \times \frac{0.343916}{0.03} = 206 \times 11.46387 = 2361.56)
Alternatively: (200 \times \frac{1.03(1.03^{10} - 1)}{0.03} = 200 \times 11.8078 = 2361.56)
Amount = $2,360 (3 s.f.)
[M1] for identifying annuity due
[M1] for correct formula and substitution
[A1] for correct amount
[Total: 3 marks]

17. (S_n = \frac{100(1.05^n - 1)}{1.05 - 1} = \frac{100(1.05^n - 1)}{0.05} = 2000(1.05^n - 1))
Need (2000(1.05^n - 1) > 1000 \implies 1.05^n - 1 > 0.5 \implies 1.05^n > 1.5)
(n \ln 1.05 > \ln 1.5 \implies n > \frac{\ln 1.5}{\ln 1.05} \approx \frac{0.405465}{0.048790} \approx 8.31)
Least integer (n = 9)
[M1] for correct (S_n) formula
[M1] for setting up inequality and using logarithms
[A1] for (n = 9)
[Total: 3 marks]

18. (a) Population = (50000(1.02)^{10} = 50000 \times 1.218994 = 60949.7)
Population ≈ 60,900 (3 s.f.)
[M1] for geometric growth formula
[A1] for correct population
[2 marks]

(b) (50000(1.02)^t > 80000 \implies 1.02^t > 1.6)
(t \ln 1.02 > \ln 1.6 \implies t > \frac{\ln 1.6}{\ln 1.02} \approx \frac{0.470004}{0.019803} \approx 23.73)
So (t = 24) years after 2020, i.e., year 2044.
[M1] for setting up inequality
[A1] for year 2044
[Total: 4 marks]

19. Total distance = initial drop + sum of all upward and downward bounces after first drop.
First drop: 10 m
First bounce up: (10 \times 0.6 = 6) m, down: 6 m
Second bounce up: (6 \times 0.6 = 3.6) m, down: 3.6 m
And so on.
Total distance = (10 + 2(6 + 3.6 + 2.16 + ...))
The series (6 + 3.6 + 2.16 + ...) is a GP with (a = 6, r = 0.6).
Sum to infinity = (\frac{6}{1 - 0.6} = \frac{6}{0.4} = 15)
Total distance = (10 + 2(15) = 10 + 30 = 40) metres.
[M1] for identifying the GP of bounce heights
[M1] for using sum to infinity
[A1] for 40 metres
[Total: 3 marks]

20. GP: (T_1 = 100,000), (T_3 = 144,000)
(T_3 = ar^2 \implies 100,000r^2 = 144,000 \implies r^2 = 1.44 \implies r = 1.2) (since growth is positive)
(S_8 = \frac{100,000(1.2^8 - 1)}{1.2 - 1} = \frac{100,000(4.299817 - 1)}{0.2} = \frac{100,000 \times 3.299817}{0.2} = 500,000 \times 3.299817 = 1,649,908.50)
Total profit ≈ $1,650,000 (3 s.f.)
[M1] for finding (r)
[M1] for sum formula and substitution
[A1] for correct total
[Total: 3 marks]

END OF ANSWER KEY