### 2015 A-level H2 Mathematics (9740) Paper 2 Question 12 Suggested Solutions

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

(i)
Let X and Y denote the mass of a apple and pear in grams respectively.
$\mathbb{E}(X_1 + \ldots X_5) = 5(300) = 1500$

$\mathrm{Var}(X_1 + \ldots X_5) = 5(20^2) = 2000$

$X_1 + \ldots X_5 \sim \mathrm{N}(1500, 2000)$

$\mathrm{P}(X_1 + \ldots X_5 > 1600) \approx 0.0127$ (3 SF)

(ii)
$\mathbb{E}[X_1 + \ldots X_5 - (Y_1 + \ldots Y_8)] = 5(300) - 8(200) = -100$

$\mathrm{Var}[X_1 + \ldots X_5 - (Y_1 + \ldots Y_8)] = 5(20^2) + 8(15^2) = 3800$

$X_1 + \ldots X_5 - (Y_1 + \ldots Y_8) \sim \mathrm{N}(-100, 3800)$

$\mathrm{P}(X_1 + \ldots X_5 - (Y_1 + \ldots Y_8) > 0) \approx 0.0524$ (3 SF)

(iii)
Let A and B denote the mass of a peeled apple and peeled pear in grams respectively.
$\mathbb{E}(A)= \mathbb{E}(0.85X) = 0.85(300)=255$

$\mathbb{E}(B)= \mathbb{E}(0.90Y) = 0.90(200)=180$

$\mathrm{Var}(A) = \mathrm{Var}(0.85X) = 0.85^2(20^2) = 289$

$\mathrm{Var}(B) = \mathrm{Var}(0.90Y) = 0.90^2(15^2) = 182.25$

$\mathbb{E}[A_1 + \ldots A_5 + B_1 + \ldots B_8)] = 5(255) + 8(180) = 2715$

$\mathrm{Var}[A_1 + \ldots A_5 + B_1 + \ldots B_8)] = 5(289) + 8(182.25) = 2903$

$A_1 + \ldots A_5 + B_1 + \ldots B_8 \sim \mathrm{N}(2715, 2903)$

$\mathrm{P}(A_1 + \ldots A_5 + B_1 + \ldots B_8 ~ \textless ~ 2750) \approx 0.742$ (3 SF)

### 2015 A-level H2 Mathematics (9740) Paper 2 Question 11 Suggested Solutions

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

(i)
Number of ways $= \frac{8!}{2!2!} = 10080$ ways

(ii)
Number of ways $= 10080 -1 = 10079$ ways

(iii)
Number of ways $= 6! = 720$ ways

(iv)
Case 1: 2 A’s together and B’s separated
$5! \times ^6 C_2 = 1800$
Case 2: 2 B’s together and A’s separated
$5! \times ^6 C_2 = 1800$
Case 3: 2 A’s together and 2 B’s together
720

Number of ways =10080 – 1800 – 1800 – 720 = 5760 ways

### 2015 A-level H2 Mathematics (9740) Paper 2 Question 10 Suggested Solutions

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

(i)
Graph to be inserted
(ii)
(a) $r \approx -0.9807$ (4 DP)
(b) $r \approx -0.9748$ (4 DP)
(c) $r \approx -0.9986$ (4 DP)

(iii)
Using GC, $P = -0.14686 \sqrt{h} + 34.78895$
$\therefore, P = -0.147 \sqrt{h} + 34.8$ (3 SF)

(iv)
$\therefore, P = -0.147 \sqrt{3.28h_{metres}} + 34.8$

$\therefore, P = -0.266 \sqrt{h_{metres}} + 34.8$ (3 SF)

### 2015 A-level H2 Mathematics (9740) Paper 2 Question 9 Suggested Solutions

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

(i)
$\mathrm{P}(B|A) = \mathrm{P}(B)$ since A and B are independent.

$\therefore, \mathrm{P}(B|A) = \mathrm{P}(B) = 0.4$

(ii)
$\mathrm{P}(A' \cap B' \cap C')$

$= 1 - \mathrm{P}(C \cup B \cup A)$

$= 1 - (0.45 + 0.02 + 0.2 + 0.145)$

$= 0.185$

(iii)

If $\mathrm{P}(A' \cap B' \cap C) = 0$,

$\mathrm{P}(A' \cap B' \cap C')$

$=1 - [0.935 - (0.3 - 0.035 )]$

$= 0.33$ (maximum)

If $\mathrm{P}(A' \cap B \cap C) = 0$,

$\mathrm{P}(A' \cap B' \cap C')$

$= 1 - \mathrm{P}(A \cup B \cup C) - 0.22$

$= 0.165$ (minimum)

$0.165 \le \mathrm{P}(A' \cap B' \cap C') \le 0.33$

### 2015 A-level H2 Mathematics (9740) Paper 2 Question 8 Suggested Solutions

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Let X denote the mass of pineapple tarts and $\mu$ denote the population mean mass of pineapple tarts.

$H_0: \mu = 0.9$

$H_1: \mu ~ \textless ~ 0.9$

Unbiased estimate of population mean = 0.8825

$s^2 = 0.0747854073^2 \approx 0.00559$ (3 SF)

Under $H_0, \bar{X} \sim \mathrm{N} (\mu, \frac{s^2}{n})$

Test Statistic, $T = \frac{\bar{X}- \mu}{\frac{s}{\sqrt{n}}} \sim t(7)$ at 10% level of significance.

Using GC, p-value $= 0.264619 > 0.1$

$\Rightarrow$, do not reject $H_0$

There is insufficient evidence at 10% significance level to reject the stall owner’s claim.

### 2015 A-level H2 Mathematics (9740) Paper 2 Question 7 Suggested Solutions

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(i)
Firstly, the average number of errors per page is constant.
Secondly, the number of errors in a page is independent of the number of errors in another page.

(ii)
Let X denote the number of errors per page.
$X \sim \mathrm{Po}(1.3)$

$X_1 \ldots X_6 \sim \mathrm{Po}(7.8)$

$\mathrm{P}(X_1 \ldots X_6 > 10)$

$= 1 - \mathrm{P}(X_1 \ldots X_6 \le 10)$

$\approx 0.165$ (3 SF)

(iii)
$X_1 \ldots X_n \sim \mathrm{Po}(1.3n)$

$\mathrm{P}(X_1 \ldots X_n ~ \textless ~ 2) ~ \textless ~ 0.05$

$e^{-1.3n} + e^{-1.3n}(1.3n) ~ \textless ~ 0.05$

$e^{-1.3n} (1 + 1.3n) ~ \textless ~ 0.05$

Using GC, $n > 3.6491$

Least $n = 4$

### 2015 A-level H2 Mathematics (9740) Paper 2 Question 6 Suggested Solutions

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(i)
Let X denote the number of red sweets in a small packet of 10 sweets.

$X \sim \mathrm{Bin} (10, 0.25)$

$\mathrm{P}(X \ge 4) = 1 - \mathrm{P}(X \le 3) \approx 0.224$ (3 SF)

(ii)
Let Y denote the number of red sweets in a large packet of 100 sweets.

$Y \sim \mathrm{Bin} (100, 0.25)$

Since n is large, $np = 25 > 5, n(1-p) = 75 > 5$

$Y \sim \mathrm{N}(25, 18.75)$ approximately

$\mathrm{P}( Y \ge 30) = \mathrm{P}(X > 29.5)$ by continuity correction
$\approx 0.149$ (3 SF)

(iii)
Let W denote the number of packets out of 15 packets that contain at least 30 red sweets.

$Y \sim \mathrm{Bin}(15, 0.1493487984)$

$\mathrm{P}( W \le 3) \approx 0.825$ (3 SF)

### 2015 A-level H2 Mathematics (9740) Paper 2 Question 5 Suggested Solutions

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(i)
Manager is unable to obtain an appropriate sampling frame.

(ii)
Manager can consider taking equal number of respondents from adults and aged first. Then, he can stand outside the supermarket, interview whoever walks in, based on his personal preferences until he obtains the number of respondents he require.

(iii)
As he samples based on his own preferences, there might be selection bias.

### 2015 A-level H2 Mathematics (9740) Paper 2 Question 4 Suggested Solutions

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

(a)
Let P(n) be the statement $\sum_{r=1}^n r(r+2)(r+5) = \frac{1}{12}n(n+1)(3n^2+31n+74) \forall n \in \mathbb{Z^+}$

When n=1, LHS $= 1 \times 3 \times 6 = 18$
RHS $= \frac{1}{12}\times 1 \times 2 \times (3+31+74) = 18$

Since LHS = RHS, P(1) is true.

Assume P(k) is true for some k, $k \in \mathbb{Z^+}$

$\sum_{r=1}^k r(r+2)(r+5) = \frac{1}{12}k(k+1)(3k^2+31k+74)$

Want to show P(k+1) is true, $\sum_{r=1}^{k+1} r(r+2)(r+5) = \frac{1}{12}(k+1)(k+2)(3k^3+43k^2+182k+216)$

LHS
$= \sum_{r=1}^{k+1} r(r+2)(r+5)$

$= \frac{1}{12}k(k+1)(3k^2+31k+74) + (k+1)(k+3)(k=6)$

$= \frac{1}{12}(k+1)[3k^3 + 321k^2 + 74k + 12(k^2 + 9k +18)]$

$= \frac{1}{12}(k+1)(3k^3+43k^2+182k+216)$

= RHS

Since P(1) is true, P(k) is true $\Rightarrow$ P(k+1) is true, hence, by Mathematical Induction, P(n) is true for all $n \in \mathbb{Z^+}$

(b)
(i)
$A = 1, B = -1$

(ii)
$\sum_{r=1}^n \frac{1}{2r+1} - \frac{1}{2r+3}$

$= \frac{1}{3} - \frac{1}{5}$

$+ \frac{1}{5} - \frac{1}{7}$

$\ldots$

$+ \frac{1}{2n-1} - \frac{1}{2n+1}$

$+ \frac{1}{2n+1} - \frac{1}{2n+3}$

$= \frac{1}{3} - \frac{1}{2n+3}$

(iii)

$S_{\infty} = \frac{1}{3}$

$S_n - S_{\infty} = \frac{1}{2n+3} ~ \textless ~10^{-3}$

Least n = 499

### 2015 A-level H2 Mathematics (9740) Paper 2 Question 2 Suggested Solutions

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

(i)
$\theta = \mathrm{cos^{-1}} |\frac{\begin{pmatrix}2\\3\\{-6}\end{pmatrix}\cdot \begin{pmatrix}1\\0\\{0}\end{pmatrix}}{\sqrt{49} \cdot 1}| = 73.4^{\circ}$

(ii)
Let $\vec{ON}$ be point on L that makes $\sqrt{33}$ from P.

$\vec{ON} = \begin{pmatrix}1\\{-2}\\{-4}\end{pmatrix} + \lambda \begin{pmatrix}2\\3\\{-6}\end{pmatrix}$ for some $\lambda$

$\vec{PN} = \begin{pmatrix}{-1}\\{-7}\\{2}\end{pmatrix} + \lambda \begin{pmatrix}2\\3\\{-6}\end{pmatrix}$

$|\vec{PN}| = \sqrt{(-1+2\lambda)^2 + (-7+3\lambda)^2 + (2-6\lambda)^2}$

$33 = 49 \lambda^2 - 70 \lambda + 54$

$\lambda = 1 \mathrm{~or~} \frac{3}{7}$

$\vec{ON} = \begin{pmatrix}3\\{1}\\{-10}\end{pmatrix} \mathrm{~or~} \frac{1}{7}\begin{pmatrix}13\\{-5}\\{-46}\end{pmatrix}$

$L = |\vec{PN}|^2 = (-1+2\lambda)^2 + (-7+3\lambda)^2 + (2-6\lambda)^2 = 49 \lambda^2 - 70 \lambda +54$

$\frac{dL}{d\lambda} = 98 \lambda - 70$

$\frac{d^2L}{d\lambda ^2} = 98 > 0$

So when $\lambda = \frac{70}{98} = \frac{5}{7}$, L is minimum.

$\vec{ON} = \frac{1}{7} \begin{pmatrix}{17}\\{1}\\{-58}\end{pmatrix}$

(iii)
$\begin{pmatrix}{-1}\\{-7}\\{2}\end{pmatrix} \times \begin{pmatrix}{2}\\{3}\\{-6}\end{pmatrix} = \begin{pmatrix}{36}\\{-2}\\{11}\end{pmatrix}$

$\begin{pmatrix}{1}\\{-2}\\{-4}\end{pmatrix} \bullet \begin{pmatrix}{36}\\{-2}\\{11}\end{pmatrix} = -4$

$\therefore, \pi : 36x - 2y +11z=-4$