A Level H1 Mathematics Practice Paper 1

TuitionGoWhere Practice Paper - Maths H1 A-Level

Answer Key and Marking Scheme (Version 1)

Section A: Pure Mathematics

Question 1

(a) ( y = x^2 - 6x + 5 = (x - 3)^2 - 4 ). Turning point is ( (3, -4) ). [M1 A1] Since coefficient of ( x^2 ) is positive, it is a minimum point. [A1]

(b) ( x^2 - 6x + 5 = 2x - 3 ) → ( x^2 - 8x + 8 = 0 ) [M1] ( x = \dfrac{8 \pm \sqrt{64 - 32}}{2} = \dfrac{8 \pm \sqrt{32}}{2} = 4 \pm 2\sqrt{2} ) [M1 A1]

Question 2

(a) At ( t = 0 ): ( N = 200e^0 = 200 ) bacteria. [A1]

(b) ( 400 = 200e^{0.4t} ) → ( e^{0.4t} = 2 ) [M1] ( 0.4t = \ln 2 ) → ( t = \dfrac{\ln 2}{0.4} \approx 1.73 ) hours. [M1 A1]

(c) ( \dfrac{dN}{dt} = 200 \times 0.4 \times e^{0.4t} = 80e^{0.4t} ) [M1] At ( t = 5 ): ( \dfrac{dN}{dt} = 80e^2 \approx 591 ) bacteria per hour. [A1]

Question 3

(a) Let ( u = 3x^2 + 1 ), ( v = (x - 2)^4 ). ( u' = 6x ), ( v' = 4(x - 2)^3 ). [M1] ( \dfrac{dy}{dx} = 6x(x - 2)^4 + (3x^2 + 1) \cdot 4(x - 2)^3 ) [M1] ( = 2(x - 2)^3[3x(x - 2) + 2(3x^2 + 1)] = 2(x - 2)^3(3x^2 - 6x + 6x^2 + 2) ) ( = 2(x - 2)^3(9x^2 - 6x + 2) ) [A1]

(b) ( u = e^{2x} ), ( v = x^2 + 1 ). ( u' = 2e^{2x} ), ( v' = 2x ). [M1] ( \dfrac{dy}{dx} = \dfrac{2e^{2x}(x^2 + 1) - e^{2x}(2x)}{(x^2 + 1)^2} ) [M1] ( = \dfrac{2e^{2x}(x^2 + 1 - x)}{(x^2 + 1)^2} = \dfrac{2e^{2x}(x^2 - x + 1)}{(x^2 + 1)^2} ) [A1]

Question 4

(a) ( 4x - x^2 = 3 ) → ( x^2 - 4x + 3 = 0 ) → ( (x - 1)(x - 3) = 0 ). [M1] ( x = 1 ) or ( x = 3 ). [A1]

(b) Area = ( \int_1^3 [(4x - x^2) - 3],dx = \int_1^3 (4x - x^2 - 3),dx ) [M1] ( = \left[ 2x^2 - \dfrac{x^3}{3} - 3x \right]_1^3 ) [M1] ( = \left(18 - 9 - 9\right) - \left(2 - \frac{1}{3} - 3\right) ) [M1] ( = 0 - \left(-\frac{4}{3}\right) = \frac{4}{3} ) units². [A1]

Question 5

(a) ( 2x^2 + 5x - 12 = (2x - 3)(x + 4) \leq 0 ) [M1] Critical values: ( x = \frac{3}{2} ), ( x = -4 ). [A1] Since coefficient of ( x^2 > 0 ), parabola opens upward. Solution: ( -4 \leq x \leq \frac{3}{2} ). [A1]

(b) No real roots when discriminant ( < 0 ): ( k^2 - 4(1)(9) < 0 ) [M1] ( k^2 < 36 ) [M1] ( -6 < k < 6 ). [A1]

Question 6

(a) Volume: ( x^2h = 32,000 ) → ( h = \dfrac{32,000}{x^2} ). [M1] Surface area (open top): ( A = x^2 + 4xh ) [M1] ( = x^2 + 4x\left(\dfrac{32,000}{x^2}\right) = x^2 + \dfrac{128,000}{x} ). [A1]

(b) ( \dfrac{dA}{dx} = 2x - \dfrac{128,000}{x^2} ) [M1] Set ( \dfrac{dA}{dx} = 0 ): ( 2x = \dfrac{128,000}{x^2} ) → ( 2x^3 = 128,000 ) → ( x^3 = 64,000 ) → ( x = 40 ). [M1 A1] ( \dfrac{d^2A}{dx^2} = 2 + \dfrac{256,000}{x^3} ). At ( x = 40 ): ( \dfrac{d^2A}{dx^2} = 2 + \dfrac{256,000}{64,000} = 6 > 0 ) → minimum. [M1 A1]

(c) ( A = 40^2 + \dfrac{128,000}{40} = 1600 + 3200 = 4800 \text{ cm}^2 ). [A1]

Section B: Probability and Statistics

Question 7

(a) Choose 3 men from 8: ( \binom{8}{3} = 56 ). Choose 2 women from 6: ( \binom{6}{2} = 15 ). [M1] Total = ( 56 \times 15 = 840 ). [A1]

(b) At least 2 women means 2, 3, 4, or 5 women (but max 5 people, max 6 women). Cases: 2W3M, 3W2M, 4W1M, 5W0M. [M1] ( \binom{6}{2}\binom{8}{3} + \binom{6}{3}\binom{8}{2} + \binom{6}{4}\binom{8}{1} + \binom{6}{5}\binom{8}{0} ) [M1] ( = 15 \times 56 + 20 \times 28 + 15 \times 8 + 6 \times 1 = 840 + 560 + 120 + 6 = 1526 ). [A1]

Question 8

(a) Tree diagram with: First draw: R (5/10), B (3/10), G (2/10). Second draw probabilities adjusted for without replacement. [B1 for structure, B1 for first-draw probabilities, B1 for second-draw conditional probabilities]

(b) P(same colour) = P(RR) + P(BB) + P(GG) ( = \frac{5}{10} \times \frac{4}{9} + \frac{3}{10} \times \frac{2}{9} + \frac{2}{10} \times \frac{1}{9} = \frac{20 + 6 + 2}{90} = \frac{28}{90} = \frac{14}{45} ). [M1 A1] P(different colours) = ( 1 - \frac{14}{45} = \frac{31}{45} ). [M1 A1]

Question 9

(a) ( P(A \cap B) = P(A) + P(B) - P(A \cup B) = 0.4 + 0.5 - 0.7 = 0.2 ). [A1]

(b) For independence: ( P(A) \times P(B) = 0.4 \times 0.5 = 0.2 ). [M1] Since ( P(A \cap B) = 0.2 = P(A)P(B) ), events ( A ) and ( B ) are independent. [A1]

(c) ( P(A' \mid B) = \dfrac{P(A' \cap B)}{P(B)} = \dfrac{P(B) - P(A \cap B)}{P(B)} ) [M1] ( = \dfrac{0.5 - 0.2}{0.5} = \dfrac{0.3}{0.5} = 0.6 ). [A1]

Question 10

(a) Assumptions: [2 marks for any two of]

• Each bulb has the same probability (0.08) of being defective (constant probability).
• The bulbs are independent of each other (defect status of one does not affect another).
• Fixed number of trials (n = 20). [B1, B1]

(b) ( X \sim B(20, 0.08) ). ( P(X = 2) = \binom{20}{2}(0.08)^2(0.92)^{18} ) [M1] ( = 190 \times 0.0064 \times 0.92^{18} \approx 0.271 ) (3 s.f.). [A1]

(c) ( P(X \geq 3) = 1 - P(X \leq 2) ) [M1] ( = 1 - [P(X=0) + P(X=1) + P(X=2)] ) ( P(X=0) = (0.92)^{20} \approx 0.1887 ) ( P(X=1) = 20(0.08)(0.92)^{19} \approx 0.3282 ) ( P(X=2) \approx 0.2711 ) ( P(X \leq 2) \approx 0.7880 ) ( P(X \geq 3) \approx 0.212 ) (3 s.f.). [A1]

Question 11

(a) ( X \sim N(500, 12^2) ). ( Z = \dfrac{490 - 500}{12} = -0.8333 ) [M1] ( P(X < 490) = P(Z < -0.8333) = 1 - \Phi(0.8333) \approx 0.202 ) (3 s.f.). [A1]

(b) P(underweight) = ( P(X < 485) ). ( Z = \dfrac{485 - 500}{12} = -1.25 ). [M1] ( p = P(Z < -1.25) = 0.1056 ). [A1] Let ( Y \sim B(30, 0.1056) ) be number underweight. [M1] ( P(Y \leq 2) = P(Y=0) + P(Y=1) + P(Y=2) ) ( = (0.8944)^{30} + 30(0.1056)(0.8944)^{29} + \binom{30}{2}(0.1056)^2(0.8944)^{28} ) ( \approx 0.0352 + 0.1247 + 0.2135 = 0.373 ) (3 s.f.). [A1]

Question 12

(a) Let ( P \sim N(150, 18^2) ), ( Q \sim N(140, 15^2) ). ( D = P - Q ). ( E(D) = 150 - 140 = 10 ). [M1] ( \text{Var}(D) = 18^2 + 15^2 = 324 + 225 = 549 ). [M1] ( D \sim N(10, 549) ). ( P(D > 0) ): ( Z = \dfrac{0 - 10}{\sqrt{549}} = -0.4267 ). [M1] ( P(Z > -0.4267) = \Phi(0.4267) \approx 0.665 ) (3 s.f.). [A1]

(b) ( T = P + Q ). ( E(T) = 150 + 140 = 290 ). [M1] ( \text{Var}(T) = 18^2 + 15^2 = 549 ). [M1] ( P(T > 310) ): ( Z = \dfrac{310 - 290}{\sqrt{549}} = 0.8535 ). ( P(Z > 0.8535) = 1 - \Phi(0.8535) \approx 0.197 ) (3 s.f.). [A1]

Question 13

(a) ( \bar{x} = \dfrac{1275}{15} = 85 ) minutes. [A1] Unbiased variance: ( s^2 = \dfrac{1}{14}\left[114,375 - \dfrac{1275^2}{15}\right] ) [M1] ( = \dfrac{1}{14}[114,375 - 108,375] = \dfrac{6000}{14} \approx 429 ) minutes² (3 s.f.). [A1]

(b) "Unbiased" means the expected value of the estimator equals the true population parameter. For variance, dividing by ( n-1 ) gives an unbiased estimator of ( \sigma^2 ). [B1]

Question 14

(a) ( H_0: \mu = 50.0 ); ( H_1: \mu < 50.0 ) (one-tail). [B1] Test statistic: ( Z = \dfrac{49.85 - 50.0}{0.4/\sqrt{25}} = \dfrac{-0.15}{0.08} = -1.875 ). [M1 A1] Critical value at 5% (one-tail): ( z = -1.645 ). [B1] Since ( -1.875 < -1.645 ), reject ( H_0 ). There is sufficient evidence at the 5% level that the mean length is less than 50.0 cm. [A1]

(b) At 1% significance level, critical value is ( -2.326 ). [M1] Since ( -1.875 > -2.326 ), we do not reject ( H_0 ). The conclusion would be different. [A1]

Question 15

(a) ( r = \dfrac{n\sum xy - \sum x \sum y}{\sqrt{[n\sum x^2 - (\sum x)^2][n\sum y^2 - (\sum y)^2]}} ) [M1] ( = \dfrac{10(20,500) - 350(520)}{\sqrt{[10(14,500) - 350^2][10(30,400) - 520^2]}} ) ( = \dfrac{205,000 - 182,000}{\sqrt{[145,000 - 122,500][304,000 - 270,400]}} ) ( = \dfrac{23,000}{\sqrt{22,500 \times 33,600}} = \dfrac{23,000}{\sqrt{756,000,000}} ) [M1] ( = \dfrac{23,000}{27,495.45} \approx 0.837 ) (3 s.f.). There is a strong positive linear correlation between advertising expenditure and sales revenue. [A1]

(b) ( b = \dfrac{n\sum xy - \sum x \sum y}{n\sum x^2 - (\sum x)^2} = \dfrac{23,000}{22,500} \approx 1.0222 ) [M1] ( a = \bar{y} - b\bar{x} = 52 - 1.0222(35) = 52 - 35.777 = 16.223 ) [M1] Equation: ( y = 16.2 + 1.02x ) (3 s.f.). [A1]

(c) When ( x = 45 ): ( y = 16.2 + 1.02(45) = 62.1 ) thousand dollars. [M1 A1] This is extrapolation since ( x = 45 ) is outside the range of observed ( x ) values (mean 35, max likely around 50–60). The estimate may be unreliable. [B1]

Question 16

(a) ( P(C \cup B) = P(C) + P(B) - P(C \cap B) = 0.60 + 0.35 - 0.20 = 0.75 ). [A1]

(b) ( P(B \mid C) = \dfrac{P(B \cap C)}{P(C)} = \dfrac{0.20}{0.60} = \dfrac{1}{3} \approx 0.333 ). [M1 A1]

(c) ( P(C) \times P(B) = 0.60 \times 0.35 = 0.21 ). [M1] Since ( P(C \cap B) = 0.20 \neq 0.21 ), the events are not independent. [A1]

Question 17

(a) ( P(X < 2.5) = 0.10 ). ( Z = \dfrac{2.5 - \mu}{1.2} ). [M1] From normal tables, ( P(Z < -1.2816) = 0.10 ). [M1] ( \dfrac{2.5 - \mu}{1.2} = -1.2816 ) → ( 2.5 - \mu = -1.5379 ) → ( \mu = 4.0379 \approx 4.04 ) minutes (3 s.f.). [A1]

(b) ( P(X > 5) ): ( Z = \dfrac{5 - 4.0379}{1.2} = 0.8018 ). [M1] ( P(Z > 0.8018) = 1 - \Phi(0.8018) \approx 0.211 ) (3 s.f.). [A1]

Question 18

(a) ( \bar{X} \sim N\left(62, \dfrac{8^2}{16}\right) = N(62, 4) ). [M1] ( P(60 < \bar{X} < 64) = P\left(\dfrac{60-62}{2} < Z < \dfrac{64-62}{2}\right) = P(-1 < Z < 1) ) [M1] ( = \Phi(1) - \Phi(-1) = 0.8413 - 0.1587 = 0.6826 \approx 0.683 ) (3 s.f.). [A1]

(b) The Central Limit Theorem states that for a large sample size (typically ( n > 30 )), the sample mean is approximately normally distributed regardless of the population distribution. [B1] It is not needed in part (a) because the population is already normally distributed, so the sample mean is exactly normally distributed for any sample size. [B1]

Question 19

( H_0: \mu = 85 ); ( H_1: \mu > 85 ) (one-tail). [B1] Test statistic: ( Z = \dfrac{88.40 - 85}{12.50/\sqrt{50}} = \dfrac{3.40}{1.7678} = 1.923 ). [M1 A1] Critical value at 5% (one-tail): ( z = 1.645 ). [B1] Since ( 1.923 > 1.645 ), reject ( H_0 ). [M1] There is sufficient evidence at the 5% significance level to support the manager's belief that the mean amount spent is more than $85. [A1]

Question 20

(a) Let ( R \sim N(120, 15^2) ), ( P \sim N(45, 10^2) ). ( T = R + P ). ( E(T) = 120 + 45 = 165 ). [M1] ( \text{Var}(T) = 15^2 + 10^2 = 225 + 100 = 325 ). [M1] ( P(T > 180) ): ( Z = \dfrac{180 - 165}{\sqrt{325}} = 0.8321 ). [M1] ( P(Z > 0.8321) = 1 - \Phi(0.8321) \approx 0.203 ) (3 s.f.). [A1]

(b) Profit = ( 2.50R + 4.00P ). [M1] ( E(\text{Profit}) = 2.50(120) + 4.00(45) = 300 + 180 = $480 ). [A1] ( \text{Var}(\text{Profit}) = (2.50)^2(225) + (4.00)^2(100) = 6.25(225) + 16(100) = 1406.25 + 1600 = 3006.25 ). [M1] ( \text{SD}(\text{Profit}) = \sqrt{3006.25} \approx $54.83 ) (2 d.p.). [A1]

(c) Total demand ( T \sim N(165, 325) ). Need ( c ) such that ( P(T \leq c) = 0.95 ). [M1] ( Z = 1.645 ) for 95th percentile. [M1] ( c = 165 + 1.645\sqrt{325} = 165 + 1.645(18.0278) = 165 + 29.66 = 194.66 ). [M1] The café should prepare approximately 195 cups of coffee. [A1]

END OF ANSWER KEY