2025 A-level H1 Mathematics (8865) Paper Suggested Solutions

All solutions here are suggested. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions

Numerical Answers (workings/explanations are after the numerical answers.)

Question 1:
Question 2:
Question 3:
Question 4:
Question 5:
Question 6:
Question 7: $r=0.961; y = 0.82x + 24; 82s;$ c decreases by $k$
Question 8: $39916800; 432000$
Question 9: $0.878; 0.849$
Question 10: $0.590; 0.256; 0.236; 0.0734$
Question 11: $\frac{3}{5}; \frac{111}{125}; \frac{56}{125}$
Question 12: $p=0.0233; \{ \sigma \in \mathbb{R}: \sigma_x \ge 0.407\}$
Question 13: $61.44, \frac{16358}{1225}; 0.760; 0.798$

(a)

(b)
Using GC, product moment correlation coefficient $= 0.96145 \approx 0.961$ (3 dp)

(c)
Using GC,
$y = 0.81710x + 24.48456$
$y = 0.82 x + 24$ (2 sf)

(d)
Let $x = 70, y = 81.681 \approx 82$ seconds
The estimate is reliable since $x = 70$ is within the given data range and interpolation produces a reliable estimate.

(e)
$m$ will be unchanged since there are no change in scale of measurement.
$c$ will be reduced by $k$ units.

Note: $r$ is unchanged (since it is independent of scale of measurement)

(a)
Number of ways $= 11! = 39916800$ ways

(b)
Number of ways $= \binom{5}{4} \times 4! \times \binom{6}{2} \times 2! \times 5! = 432000$ ways

(a)
Let $X \sim \text{B}(n, p)$
$\mathbb{E}(X) = np$
$18 = np$
$\text{Var}(X) = np(1 - p)$
$1.8 = np (1 - p)$
$0.1 = 1 - p$
$p = 0.9$
$n = 20$
$\text{P}(X \le 19) = 0.8784233 \approx 0.878$

(b)
$\text{P}(X \ge 17 \vert X \le 19)$
$= \frac{\text{P}(17 \le X \le 19)}{\text{P}(X \le 19)}$
$= \frac{\text{P}(X \le 19) - \text{P}(X \le 16)}{\text{P}(X \le 19)}$
$= 0.848645 \approx 0.849$

Q10

(a)
Let masses of red, green and yellow peppers be $R, G, Y$ respectively.
$R \sim \text{N}(230, 35^2)$
$G \sim \text{N}(225, 30^2)$
$Y \sim \text{N}(220, 32^2)$
$R_1 + R_2 - G - Y \sim \text{N}(15, 4374)$
$\text{P}(R_1 + R_2 > G + Y)$
$= \text{P}(R_1 + R_2 - G - Y > 0)$
$= 0.58971$
$\approx 0.590$

(b)
$R - 1.2Y \sim \text{N}(-34, 2699.56)$
$\text{P}(R \ge 1.2 Y)$
$= \text{P}(R - 1.2Y \ge 0)$
$= 0.2564322$
$\approx 0.256$

(c)
Required probability
$= \frac{0.2 \times \text{P}(Y < 220)}{0.5 \times \text{P}(R < 220) + 0.3 \times \text{P}(G < 220) + 0.2 \times \text{P}(Y < 220)}$
$= 0.2358940$
$\approx 0.236$

(d)
$\text{P}(G > 220) = 0.5661838$
Let $Y$ be the number of green peppers with mass more than 220g, out of 20.
$Y \sim \text{B}(20, 0.5661838)$
$\text{P}(Y \ge 15) = 1 - \text{P}(Y \le 14) = 0.0733942 \approx 0.0734$

Q11

(a)(i)
Since $\text{P}(C) + \text{P}(D) = 1.3 > 1$ , then $\text{P}(C \cap D) \neq 0$ . Thus, both events $C$ and $D$ cannot be mutually exclusive.

(a)(ii)
Since $C$ and $D$ are independent, $\text{P}(C \cap D) = 0.6 \times 0.7 = 0.42$

(b)(i)
$\text{P}(E \cap F') = \frac{7}{25} \times [1 - \text{P}(F)]$
$\text{P}(E \cup F) = \frac{89}{125}$
$\text{P}(F) + \text{P}(E \cap F') = \frac{89}{125}$
$\text{P}(F) + \frac{7}{25} \times [1 - \text{P}(F)] = \frac{89}{125}$
$0.72 \text{P}(F) = 0.432$
$\text{P}(F) = \frac{3}{5}$

(b)(ii)
$\text{P}(E \cap F') = \frac{7}{25} \times (1 - \frac{3}{5}) = \frac{14}{125}$
$\text{P}(E' \cup F) = 1 - \frac{14}{125} = \frac{111}{125}$

(b)(iii)
$\text{P}(F \cap E') = \frac{19}{55} \times \text{P}(E')$
$\text{P}(F) - \text{P}(F \cap E) = \frac{19}{55} \times [ 1 - \text{P}(E)]$
$\frac{3}{5}- \text{P}(F \cap E) = \frac{19}{55} \times [ 1 - \text{P}(F \cap E) - \frac{14}{125}]$
$\frac{36}{55} \text{P}(F \cap E) = \frac{2016}{6875}$
$\text{P}(F \cap E) = \frac{56}{125}$

Q12

(a)
unbiased estimate of population mean, $\bar{x} = \frac{2816}{80} = 35.2$
unbiased estimate of population variance, $s^2 = \frac{1}{79}(99187 - \frac{2816^2}{80}) = 0.80759$
Let $\mu$ be the population mean salinity of water.
Test $H_0: \mu = 35$ against $H_1: \mu > 35$ at 5% significance level.
Under $H_0, \bar{X} \sim \text{N}(35, \frac{0.80759}{80})$ approximately by central limit theorem
Test Statistic, $Z = \frac{\bar{X} - 25}{\sqrt{ \frac{0.80759}{80}}} \sim \text{N}(0, 1)$ approximately
Reject $H_0$ if $p \le 0.05$
Using GC, $p = 0.023263 < 0.05$
Since $p < 0.05$ , we reject $H_0$ at 5\% level of significance and conclude with sufficient evidence that the salinity of seawater has increased.

(b)
Let standard deviation of sample be $\sigma_x$
$H_0: \mu = 35$ against $H_1: \mu \neq 35$ at 5% level of significance.
Under $H_0, \bar{X} \sim \text{N}(35, \frac{s^2}{100})$ approximately by central limit theorem
For $H_0$ is not rejected,
$- 1.9599 < \frac{35.0.8 - 35}{\frac{s}{\sqrt{100}}}< 1.95996$
Since $s > 0$ , $\frac{0.8}{s} < 1.95996 \Rightarrow s^2 > 0.16660$
Since $s^2 = \frac{n}{n-1}\times \text{sample~variance}$ ,
$\frac{100}{99}\times {\sigma_x}^2 > 0.16660$
$\Righarrow \sigma_x > 0.40612$
$\therefore \{ \sigma_x \in \mathbb{R}, \sigma_x \ge 0.407\}$

Q13

(a)
unbiased estimate of population mean $= \frac{72}{50} + 60 = 61.44$
unbiased estimate of population variance $= \frac{1}{50 - 1} ( 758 - \frac{72^2}{50}) = \frac{16358}{1225}$

(b)
Since $n = 36$ is large, by central limit theorem,
$\bar{X} = \frac{X_1 + \ldots + X_{36}}{36} \sim \text{N} ( 61.44, \frac{13.353}{36})$ approximately
Required probability $= \text{P}(61 < \bar{X} < 63 ) \approx 0.760$
Assume that the unbiased estimates of the population mean and variance calculated in (a) are good estimates for the eggs in (b)
Assume that the masses of the eggs in the randomly chosen tray are independently and identically distributed.

(c)
Let $Y$ be the number of trays with mean mass of an egg lying between 61g and 63g, out of 10.
$Y \sim \text{B}(10, 0.75978)$
Required probability $= \text{P}(Y \ge 7) = 1 - \text{P}(Y \le 6) \approx 0.798$

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About

KS Teng

KS has been teaching H2/H1 Mathematics and IB mathematics for almost two decades. Having taught students from all various Junior Colleges, KS adapts to students' abilities and help them better understand the topics. As someone who loves to teach mathematics and sees it as a truly useful tool in life, KS seeks to enable students to appreciate math. Therefore, his tuition mission is to motivate and cultivate students to be independent and confident thinkers.

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2025 A-level H1 Mathematics (8865) Paper Suggested Solutions

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