From Real Exams Exam Paper

A Level H1 Mathematics Practice Paper 2

Free Exam-Derived Owl Alpha A Level H1 Mathematics Practice Paper 2 practice paper with questions and answers for Singapore students. This page is rendered as a direct URL so the questions and answers can be discovered without pressing in-page buttons.

These static practice materials are generated from the site's syllabus and paper-generation workflow, with source and model context shown so students and parents can evaluate the material before use.

A Level H1 Mathematics From Real Exams Generated by Owl Alpha Updated 2026-06-07

Back to Subject Browse Free Exam Papers

Questions

TuitionGoWhere Practice Paper - Maths H1 A-Level

TuitionGoWhere Secondary School (AI)

Subject: Mathematics
Level: A-Level H1
Paper: Practice Paper (Statistics & Probability)
Version: 2 of 5
Duration: 1 hour 30 minutes
Total Marks: 60

Name: ___________________________
Class: ___________________________
Date: ___________________________

Instructions

Write your answers in the spaces provided.
Show all working clearly. Marks may be awarded for correct working even if the final answer is wrong.
Give non-exact answers correct to 3 significant figures unless otherwise stated.
The use of a graphing calculator is expected.
The total mark for this paper is 60.
You are advised to spend no more than 1 hour 30 minutes on this paper.

Section A: Short Answer Questions [25 marks]

Answer ALL questions in this section.

Question 1 [2 marks]

A random sample of 6 students recorded the number of hours they spent on revision in a week:

$12,\ 15,\ 10,\ 14,\ 11,\ 13$

Calculate the unbiased estimate of the population mean.

$\bar{x} = \frac{\sum x_i}{n}$

Answer: ______________ hours

Question 2 [2 marks]

Using the same data from Question 1, calculate the unbiased estimate of the population variance.

$s^2 = \frac{\sum(x_i - \bar{x})^2}{n-1}$

Answer: ______________ hours²

Question 3 [2 marks]

The random variable $X \sim \mathrm{B}(12, 0.35)$ . Find $\mathrm{P}(X = 5)$ .

Answer: ______________

Question 4 [2 marks]

A fair six-sided die is rolled 8 times. Using a binomial model, find the probability of obtaining exactly 3 sixes.

Answer: ______________

Question 5 [3 marks]

The heights of a certain species of plant are normally distributed with mean 42 cm and standard deviation 5 cm.

(a) Find the probability that a randomly selected plant has a height between 38 cm and 47 cm. [2 marks]

Answer: ______________

(b) A plant is classified as "tall" if its height exceeds 50 cm. In a batch of 200 plants, how many would you expect to be classified as "tall"? [1 mark]

Answer: ______________

Question 6 [3 marks]

A continuous random variable $X$ has probability density function given by

$f(x) = \begin{cases} kx(4-x) & 0 \le x \le 4 \\ 0 & \text{otherwise} \end{cases}$

(a) Show that $k = \frac{3}{32}$ . [2 marks]

(b) Find $\mathrm{E}(X)$ . [1 mark]

Answer: ______________

Question 7 [3 marks]

The following grouped data shows the daily commute times (in minutes) of 50 office workers.

Commute time (min)	Frequency
$0 \le t < 10$	6
$10 \le t < 20$	14
$20 \le t < 30$	16
$30 \le t < 50$	10
$50 \le t < 80$	4

(a) State the modal class. [1 mark]

Answer: ______________

(b) Estimate the mean commute time. [2 marks]

Answer: ______________ minutes

Question 8 [2 marks]

Two events $A$ and $B$ are such that $\mathrm{P}(A) = 0.6$ , $\mathrm{P}(B) = 0.4$ , and $\mathrm{P}(A \cup B) = 0.76$ .

Determine whether $A$ and $B$ are independent. Show your reasoning.

Answer: ______________

Question 9 [3 marks]

A bag contains 5 red balls, 4 blue balls, and 3 green balls. Three balls are drawn at random without replacement.

Find the probability that all three balls are the same colour.

Answer: ______________

Question 10 [3 marks]

The random variable $Y \sim \mathrm{N}(\mu, \sigma^2)$ . It is known that $\mathrm{P}(Y < 30) = 0.1587$ and $\mathrm{P}(Y > 50) = 0.0228$ .

Find $\mu$ and $\sigma$ .

Answer: $\mu =$ ______________ $\sigma =$ ______________

Section B: Structured / Application Questions [35 marks]

Answer ALL questions in this section.

Question 11 [8 marks]

A market researcher collected data on the monthly spending (in dollars) of a random sample of 8 young professionals on dining out.

Person	A	B	C	D	E	F	G	H
Spending ( $x$ )	120	180	95	210	150	165	130	195

(a) Calculate the unbiased estimates of the population mean and population variance of monthly dining spending. [4 marks]

Answer: Mean = $______________ Variance =$ ______________

(b) The researcher claims that the population mean spending exceeds $140. Using your answer to part (a), comment on whether the sample data supports this claim. [2 marks]

(c) Explain why the formula for the unbiased estimate of the population variance uses $n-1$ in the denominator rather than $n$ . [2 marks]

Question 12 [9 marks]

A call centre receives an average of 4.2 calls per minute. The number of calls received in a given minute follows a Poisson distribution.

(a) Find the probability that exactly 5 calls are received in a randomly chosen minute. [2 marks]

Answer: ______________

(b) Find the probability that at least 3 calls are received in a randomly chosen minute. [3 marks]

Answer: ______________

(c) Using a suitable approximation, find the probability that fewer than 35 calls are received in a 10-minute period. State any assumptions you make. [4 marks]

Answer: ______________

Question 13 [9 marks]

A company manufactures light bulbs. The lifetime of a light bulb (in hours) is normally distributed with mean 800 hours and standard deviation 100 hours.

(a) Find the probability that a randomly selected light bulb lasts between 700 and 950 hours. [3 marks]

Answer: ______________

(b) The company offers a warranty that replaces any bulb that fails within the first $k$ hours. If the company wants to replace no more than 5% of bulbs, find the value of $k$ . [3 marks]

Answer: ______________ hours

(c) A quality control inspector takes a random sample of 16 bulbs. Find the probability that the sample mean lifetime exceeds 820 hours. [3 marks]

Answer: ______________

Question 14 [9 marks]

A survey was conducted on 120 university students regarding their preferred mode of transport to campus. The results are summarised below.

	Walk	Bus	MRT	Car	Total
Male	15	20	25	10	70
Female	20	12	15	3	50
Total	35	32	40	13	120

(a) Find the probability that a randomly selected student is female and prefers the MRT. [2 marks]

Answer: ______________

(b) Given that a student prefers to walk, find the probability that the student is male. [2 marks]

Answer: ______________

(c) A student is selected at random. Are the events "the student is male" and "the student prefers the bus" independent? Show your working clearly. [3 marks]

(d) Two students are selected at random without replacement. Find the probability that both prefer the MRT. [2 marks]

Answer: ______________

Answers

TuitionGoWhere Practice Paper - Maths H1 A-Level

Answer Key — Version 2 of 5

Section A

Question 1 [2 marks]

$\bar{x} = \frac{12 + 15 + 10 + 14 + 11 + 13}{6} = \frac{75}{6} = 12.5$

Answer: 12.5 hours

Marking: M1 for correct substitution into the formula. A1 for correct answer.

Teaching note: The sample mean $\bar{x}$ is the best point estimate of the population mean $\mu$ . We sum all observations and divide by the number of observations $n$ . This is the standard unbiased estimator for the population mean.

Question 2 [2 marks]

First compute each deviation from the mean $\bar{x} = 12.5$ :

$x_i$	$x_i - \bar{x}$	$(x_i - \bar{x})^2$
12	$-0.5$	0.25
15	2.5	6.25
10	$-2.5$	6.25
14	1.5	2.25
11	$-1.5$	2.25
13	0.5	0.25

$\sum(x_i - \bar{x})^2 = 0.25 + 6.25 + 6.25 + 2.25 + 2.25 + 0.25 = 17.5$

$s^2 = \frac{17.5}{6-1} = \frac{17.5}{5} = 3.50$

Answer: 3.50 hours²

Marking: M1 for correct deviations/squared deviations. A1 for correct answer with $n-1$ denominator.

Common mistake: Using $n = 6$ in the denominator gives $\frac{17.5}{6} = 2.92$ , which is the biased sample variance. The unbiased estimator uses $n-1 = 5$ because one degree of freedom is "used up" when estimating the sample mean. This correction (Bessel's correction) ensures that on average, $s^2$ equals the true population variance $\sigma^2$ .

Question 3 [2 marks]

$X \sim \mathrm{B}(12, 0.35)$

$\mathrm{P}(X = 5) = \binom{12}{5}(0.35)^5(0.65)^7$

$= 792 \times 0.005252 \times 0.049022$

$= 792 \times 0.0002575 = 0.2039$

Answer: 0.204 (3 s.f.)

Marking: M1 for correct binomial probability formula with $\binom{12}{5}$ , $p = 0.35$ , $q = 0.65$ . A1 for correct answer to 3 s.f.

Teaching note: For $X \sim \mathrm{B}(n, p)$ , the probability of exactly $r$ successes is $\mathrm{P}(X = r) = \binom{n}{r}p^r(1-p)^{n-r}$ . The graphing calculator's binomialPdf function can be used directly.

Question 4 [2 marks]

Let $X$ be the number of sixes in 8 rolls. Then $X \sim \mathrm{B}\left(8, \frac{1}{6}\right)$ .

$\mathrm{P}(X = 3) = \binom{8}{3}\left(\frac{1}{6}\right)^3\left(\frac{5}{6}\right)^5$

$= 56 \times \frac{1}{216} \times \frac{3125}{7776}$

$= 56 \times 0.0046296 \times 0.40188$

$= 0.1042$

Answer: 0.104 (3 s.f.)

Marking: M1 for identifying $n = 8$ , $p = \frac{1}{6}$ and using the binomial formula. A1 for correct answer.

Question 5 [3 marks]

Let $X \sim \mathrm{N}(42, 5^2)$ .

(a) [2 marks]

$\mathrm{P}(38 < X < 47) = \mathrm{P}\left(\frac{38-42}{5} < Z < \frac{47-42}{5}\right) = \mathrm{P}(-0.8 < Z < 1.0)$

$= \Phi(1.0) - \Phi(-0.8) = \Phi(1.0) - [1 - \Phi(0.8)]$

$= 0.8413 - (1 - 0.7881) = 0.8413 - 0.2119 = 0.6294$

Answer: 0.629 (3 s.f.)

Marking: M1 for standardising and using normal tables/calculator. A1 for correct answer.

(b) [1 mark]

$\mathrm{P}(X > 50) = \mathrm{P}\left(Z > \frac{50-42}{5}\right) = \mathrm{P}(Z > 1.6) = 1 - \Phi(1.6) = 1 - 0.9452 = 0.0548$

Expected number $= 200 \times 0.0548 = 10.96$

Answer: 11 plants (approximately)

Marking: A1 for correct calculation and rounding.

Question 6 [3 marks]

(a) [2 marks]

Since $f(x)$ is a PDF, $\int_0^4 f(x)\,dx = 1$ :

$\int_0^4 kx(4-x)\,dx = k\int_0^4 (4x - x^2)\,dx = k\left[2x^2 - \frac{x^3}{3}\right]_0^4$

$= k\left[\left(2(16) - \frac{64}{3}\right) - 0\right] = k\left[32 - \frac{64}{3}\right] = k\left[\frac{96-64}{3}\right] = k \cdot \frac{32}{3}$

Setting equal to 1: $k \cdot \frac{32}{3} = 1$ , so $k = \frac{3}{32}$ . $\quad \blacksquare$

(b) [1 mark]

$\mathrm{E}(X) = \int_0^4 x \cdot \frac{3}{32}x(4-x)\,dx = \frac{3}{32}\int_0^4 (4x^2 - x^3)\,dx$

$= \frac{3}{32}\left[\frac{4x^3}{3} - \frac{x^4}{4}\right]_0^4 = \frac{3}{32}\left[\frac{256}{3} - 64\right] = \frac{3}{32}\left[\frac{256 - 192}{3}\right] = \frac{3}{32} \times \frac{64}{3} = 2$

Answer: 2

Marking: M1 for setting up the expectation integral. A1 for correct answer.

Teaching note: For a continuous random variable with PDF $f(x)$ , the expected value is $\mathrm{E}(X) = \int x\,f(x)\,dx$ over the domain. The symmetry of $f(x) = \frac{3}{32}x(4-x)$ about $x = 2$ confirms $\mathrm{E}(X) = 2$ .

Question 7 [3 marks]

(a) [1 mark]

The class with the highest frequency is $20 \le t < 30$ (frequency 16).

Answer: $20 \le t < 30$

(b) [2 marks]

Using midpoints: 5, 15, 25, 40, 65.

$\bar{x} = \frac{6(5) + 14(15) + 16(25) + 10(40) + 4(65)}{50} = \frac{30 + 210 + 400 + 400 + 260}{50} = \frac{1300}{50} = 26.0$

Answer: 26.0 minutes

Marking: M1 for using midpoints correctly. A1 for correct answer.

Teaching note: For grouped data, we use the midpoint of each class as the representative value. The last class $50 \le t < 80$ has midpoint $\frac{50+80}{2} = 65$ .

Question 8 [2 marks]

Check independence: $A$ and $B$ are independent if and only if $\mathrm{P}(A \cap B) = \mathrm{P}(A) \times \mathrm{P}(B)$ .

First find $\mathrm{P}(A \cap B)$ using the addition rule:

$\mathrm{P}(A \cup B) = \mathrm{P}(A) + \mathrm{P}(B) - \mathrm{P}(A \cap B)$

$0.76 = 0.6 + 0.4 - \mathrm{P}(A \cap B)$

$\mathrm{P}(A \cap B) = 1.0 - 0.76 = 0.24$

Now check:

$\mathrm{P}(A) \times \mathrm{P}(B) = 0.6 \times 0.4 = 0.24$

Since $\mathrm{P}(A \cap B) = \mathrm{P}(A) \times \mathrm{P}(B) = 0.24$ , the events are independent.

Answer: Yes, A and B are independent because P(A ∩ B) = P(A) × P(B) = 0.24.

Marking: M1 for finding P(A ∩ B) using the addition formula. A1 for correct conclusion with justification.

Question 9 [3 marks]

Total balls = 5 + 4 + 3 = 12. Number of ways to draw 3 balls from 12: $\binom{12}{3} = 220$ .

All red: $\binom{5}{3} = 10$
All blue: $\binom{4}{3} = 4$
All green: $\binom{3}{3} = 1$

$\mathrm{P}(\text{all same colour}) = \frac{10 + 4 + 1}{220} = \frac{15}{220} = \frac{3}{44} = 0.0682$

Answer: $\frac{3}{44}$ or 0.0682 (3 s.f.)

Marking: M1 for using combinations for each colour. M1 for summing and dividing by total. A1 for correct answer.

Teaching note: Since the balls are drawn without replacement, we use combinations (not permutations) because the order of selection does not matter. The events "all red", "all blue", and "all green" are mutually exclusive, so we add their probabilities.

Question 10 [3 marks]

From standard normal tables:

$\mathrm{P}(Y < 30) = 0.1587$ means $\mathrm{P}\left(Z < \frac{30-\mu}{\sigma}\right) = 0.1587$ , so $\frac{30-\mu}{\sigma} = -1.0$ (since $\Phi(-1) = 0.1587$ )
$\mathrm{P}(Y > 50) = 0.0228$ means $\mathrm{P}\left(Z > \frac{50-\mu}{\sigma}\right) = 0.0228$ , so $\frac{50-\mu}{\sigma} = 2.0$ (since $\Phi(2) = 0.9772$ )

Solving the simultaneous equations:

$30 - \mu = -\sigma \quad \Rightarrow \quad \mu = 30 + \sigma \quad \text{...(i)}$

$50 - \mu = 2\sigma \quad \text{...(ii)}$

Substitute (i) into (ii):

$50 - (30 + \sigma) = 2\sigma$

$20 - \sigma = 2\sigma$

$20 = 3\sigma \quad \Rightarrow \quad \sigma = \frac{20}{3} = 6.67$

$\mu = 30 + \frac{20}{3} = \frac{110}{3} = 36.67$

Answer: $\mu = 36\frac{2}{3}$ (or 36.7), $\sigma = 6\frac{2}{3}$ (or 6.67)

Marking: M1 for setting up both equations from the given probabilities. M1 for solving the simultaneous equations. A1 for both correct values.

Section B

Question 11 [8 marks]

(a) [4 marks]

$\bar{x} = \frac{120 + 180 + 95 + 210 + 150 + 165 + 130 + 195}{8} = \frac{1245}{8} = 155.625$

For the unbiased variance, compute $\sum(x_i - \bar{x})^2$ :

$x_i$	$x_i - \bar{x}$	$(x_i - \bar{x})^2$
120	$-35.625$	1269.14
180	24.375	594.14
95	$-60.625$	3675.39
210	54.375	2956.64
150	$-5.625$	31.64
165	9.375	87.89
130	$-25.625$	656.64
195	39.375	1550.39

$\sum(x_i - \bar{x})^2 = 10821.875$

$s^2 = \frac{10821.875}{8-1} = \frac{10821.875}{7} = 1545.98$

Answer: Mean = $155.6 (or $155.63), Variance = 1550 (3 s.f.)

Marking: M1 for correct mean. M1 for correct squared deviations (at least 3 correct). M1 for using $n-1 = 7$ in the denominator. A1 for correct variance to 3 s.f.

(b) [2 marks]

The sample mean is $\bar{x} = \$ 155.63 $, which is greater than \$ 140. This provides some support for the researcher's claim that the population mean exceeds $140. However, with only a sample of 8, there is considerable sampling variability, and a formal hypothesis test would be needed to draw a statistically significant conclusion.

Marking: B1 for noting the sample mean exceeds $140. B1 for acknowledging the limitation of the small sample size.

(c) [2 marks]

Using $n$ in the denominator gives the average of the squared deviations from the sample mean. However, the sample mean $\bar{x}$ is itself estimated from the same data, so the deviations $(x_i - \bar{x})$ tend to be slightly smaller than the true deviations $(x_i - \mu)$ from the population mean. Dividing by $n-1$ instead of $n$ corrects this downward bias, making $s^2$ an unbiased estimator of $\sigma^2$ . This correction is known as Bessel's correction.

Marking: B1 for explaining that the sample mean causes deviations to be smaller. B1 for stating that $n-1$ corrects the bias (Bessel's correction).

Question 12 [9 marks]

Let $X$ be the number of calls per minute. $X \sim \mathrm{Po}(4.2)$ .

(a) [2 marks]

$\mathrm{P}(X = 5) = \frac{e^{-4.2}(4.2)^5}{5!} = \frac{e^{-4.2} \times 1306.91}{120} = \frac{1306.91}{120} \times 0.0150 = 0.1634$

Answer: 0.163 (3 s.f.)

Marking: M1 for correct Poisson formula with $\lambda = 4.2$ . A1 for correct answer.

(b) [3 marks]

$\mathrm{P}(X \ge 3) = 1 - \mathrm{P}(X \le 2) = 1 - [\mathrm{P}(X=0) + \mathrm{P}(X=1) + \mathrm{P}(X=2)]$

$\mathrm{P}(X = 0) = e^{-4.2} = 0.0150$

$\mathrm{P}(X = 1) = 4.2 \times e^{-4.2} = 4.2 \times 0.0150 = 0.0630$

$\mathrm{P}(X = 2) = \frac{4.2^2}{2} \times e^{-4.2} = 8.82 \times 0.0150 = 0.1323$

$\mathrm{P}(X \ge 3) = 1 - (0.0150 + 0.0630 + 0.1323) = 1 - 0.2103 = 0.7897$

Answer: 0.790 (3 s.f.)

Marking: M1 for using the complement method. M1 for calculating individual probabilities correctly. A1 for correct final answer.

(c) [4 marks]

Over 10 minutes, $\lambda = 4.2 \times 10 = 42$ . Let $Y \sim \mathrm{Po}(42)$ .

Since $\lambda = 42$ is large, use the normal approximation: $Y \approx \mathrm{N}(42, 42)$ .

Assumption: The Poisson rate is constant over time, and calls in different minutes are independent.

With continuity correction:

$\mathrm{P}(Y < 35) = \mathrm{P}(Y \le 34) \approx \mathrm{P}\left(Z < \frac{34.5 - 42}{\sqrt{42}}\right) = \mathrm{P}\left(Z < \frac{-7.5}{6.481}\right) = \mathrm{P}(Z < -1.157)$

$= 1 - \Phi(1.157) = 1 - 0.8764 = 0.1236$

Answer: 0.124 (3 s.f.)

Marking: M1 for $\lambda = 42$ and stating the normal approximation. M1 for stating assumptions (constant rate, independence). M1 for continuity correction. A1 for correct answer.

Question 13 [9 marks]

Let $X \sim \mathrm{N}(800, 100^2)$ .

(a) [3 marks]

$\mathrm{P}(700 < X < 950) = \mathrm{P}\left(\frac{700-800}{100} < Z < \frac{950-800}{100}\right) = \mathrm{P}(-1.0 < Z < 1.5)$

$= \Phi(1.5) - \Phi(-1.0) = \Phi(1.5) - [1 - \Phi(1.0)]$

$= 0.9332 - (1 - 0.8413) = 0.9332 - 0.1587 = 0.7745$

Answer: 0.775 (3 s.f.)

Marking: M1 for standardising both bounds. M1 for correct use of $\Phi$ values. A1 for correct answer.

(b) [3 marks]

We need $\mathrm{P}(X < k) = 0.05$ .

From tables, $\Phi(-1.645) = 0.05$ , so:

$\frac{k - 800}{100} = -1.645$

$k = 800 - 164.5 = 635.5$

Answer: 636 hours (or 635 hours)

Marking: M1 for setting up $\mathrm{P}(X < k) = 0.05$ . M1 for using the correct $z$ -value ( $-1.645$ ). A1 for correct answer.

(c) [3 marks]

By the Central Limit Theorem, the sample mean $\bar{X} \sim \mathrm{N}\left(\mu, \frac{\sigma^2}{n}\right) = \mathrm{N}\left(800, \frac{100^2}{16}\right) = \mathrm{N}(800, 625)$ , so $\sigma_{\bar{X}} = 25$ .

$\mathrm{P}(\bar{X} > 820) = \mathrm{P}\left(Z > \frac{820 - 800}{25}\right) = \mathrm{P}(Z > 0.8) = 1 - \Phi(0.8) = 1 - 0.7881 = 0.2119$

Answer: 0.212 (3 s.f.)

Marking: M1 for correct distribution of sample mean with $\sigma/\sqrt{n} = 25$ . M1 for standardising. A1 for correct answer.

Teaching note: The distribution of the sample mean has the same mean $\mu$ but a reduced standard deviation $\sigma/\sqrt{n}$ (the standard error). This reflects the fact that sample means vary less than individual observations.

Question 14 [9 marks]

(a) [2 marks]

$\mathrm{P}(\text{Female and MRT}) = \frac{15}{120} = \frac{1}{8}$

Answer: $\frac{1}{8}$ or 0.125

Marking: M1 for identifying the correct cell (Female, MRT = 15). A1 for correct probability.

(b) [2 marks]

$\mathrm{P}(\text{Male} \mid \text{Walk}) = \frac{\mathrm{P}(\text{Male and Walk})}{\mathrm{P}(\text{Walk})} = \frac{15/120}{35/120} = \frac{15}{35} = \frac{3}{7}$

Answer: $\frac{3}{7}$ or 0.429 (3 s.f.)

Marking: M1 for using conditional probability formula. A1 for correct answer.

(c) [3 marks]

Check if $\mathrm{P}(\text{Male} \cap \text{Bus}) = \mathrm{P}(\text{Male}) \times \mathrm{P}(\text{Bus})$ .

$\mathrm{P}(\text{Male} \cap \text{Bus}) = \frac{20}{120} = \frac{1}{6} = 0.1667$

$\mathrm{P}(\text{Male}) \times \mathrm{P}(\text{Bus}) = \frac{70}{120} \times \frac{32}{120} = \frac{70 \times 32}{14400} = \frac{2240}{14400} = 0.1556$

Since $0.1667 \ne 0.1556$ , the events are not independent.

Answer: The events "male" and "prefers bus" are NOT independent because P(Male ∩ Bus) ≠ P(Male) × P(Bus).

Marking: M1 for calculating both sides. M1 for comparing. A1 for correct conclusion.

(d) [2 marks]

$\mathrm{P}(\text{both MRT}) = \frac{40}{120} \times \frac{39}{119} = \frac{1}{3} \times \frac{39}{119} = \frac{39}{357} = \frac{13}{119} = 0.1092$

Answer: $\frac{13}{119}$ or 0.109 (3 s.f.)

Marking: M1 for multiplying probabilities without replacement. A1 for correct answer.

Mark Summary

Question	Marks
1	2
2	2
3	2
4	2
5	3
6	3
7	3
8	2
9	3
10	3
Section A Total	25
11	8
12	9
13	9
14	9
Section B Total	35
Grand Total	60