# Bohdan DSE 2022

# 2021-07-15
1. [3M] Simplify $\frac{(x^{-2}y^3)^2}{xy^5}$ and express your answer with positive indices.

$$ \begin{align}
\frac{(x^{-2}y^3)^2}{xy^5} &= \frac{x^{-4}y^6}{xy^5}\\
&= \frac{y}{x^5}
\end{align} $$

2. [4M]
a) Make $b$ the subject of the formula $\frac{3}{2} = \frac{2b-1}{a+b}$.
b) If the value of $a$ is increased by $5$, find the corresponding increase in the value of $b$.
a)
$$\begin{align}
\frac{3}{2} &= \frac{2b-1}{a+b}\\
3a+3b &= 4b-2\\
b &= 3a+2
\end{align}$$
b)
$$\begin{align}
b' &= 3(a+5)+2\\
&= 3a+17\\
Increase &= (3a+17)-(3a+2)\\
&= 15
\end{align}$$

3. [4M] The cost of a watch is \$180. The watch is now sold and the percentage profit is 25%.
a) Find the selling price of the watch.
b) If the watch is sold at a discount of 40% on its marked price, find the marked price of the watch
a)
$$\begin{align}
Selling\ price &= $180 \times (1+25\%)\\
&= $225
\end{align}$$
b)
$$\begin{align}
Marked\ price &= $225 \div (1-40\%)\\
&= $375
\end{align}$$

4. [3M]
a) Round up 519.39825 to the nearest integer.
b) Round down 519.39825 to 1 decimal place.
c) Round off 519.39825 to 5 significant figures.
a) $519$
b) $519.4$
c) $519.40$

5. [4M] Factorize
a) $16a^2-49$
b) $4a^2b+5ab-21b$
c) $16a^2-49-4a^2b-5ab-21b$
a)
$$\begin{align}
16a^2-49 &= (4a+7)(4a-7)
\end{align}$$
b)
$$\begin{align}
4a^2b+5ab-21b &= b(4a^2+5a-21)\\
&= b(4a-7)(a+3)
\end{align}$$
c)
$$\begin{align}
16a^2-49-4a^2b-5ab-21b &= (4a+7)(4a-7) - b(4a-7)(a+3)\\
&= (4a-7)(4a+7-ab-3b)
\end{align}$$

6. [3M] The following table shows the distribution of the heights of students.

| Height(cm) | 140-149 | 150-159 | 160-169 | 170-179 | 180-189 |
| - | - | - | - | - | - |
| Frequency | 6 | 18 | 34 | 28 | 14 |

Find the modal class, the mean and the standard deviation of the above distribution.

$Modal\ class\ is\ 160\ -\ 169\ cm$
$$\begin{align}
Mean &= \frac{144.5\times6 + 154.5\times18 + 164.5\times34 + 174.5\times28 + 184.5\times14}{6+18+34+28+14} \\
&= 167.1\\
\sigma &= 10.9
\end{align}$$

7. [4M]
a) Find the range of values of $x$ which satisfy $3x-2\le\frac{10-x}{3}$ or $-x+3<7$
b) Write down the least integer which satisfies the compound inequality in (a)
a)
$$\begin{align}
3x-2&\ge\frac{10-x}{3}\ &or&&\ -x+3&<7\\
10x&\ge16\ &or&&\ -4&<x\\
x&\ge1.6\ &or&&\ x&>-4\\
&\therefore x>-4
\end{align}$$
b) $-3$

8. [5M] Mr Wong takes 135 minutes to drive from city A to city C via city B. He drives from city A to city B at a speed of 54 km/h and drives from city B to city C at a speed of 72 km/h. If the total distance of the journey is 135 km, find the time taken for him to drive from city B to city C.

$$\begin{align}
Let&\ time\ taken\ be\ x\ hours.\\
72x &= 135 - 54 \times (\frac{135}{60} - x)\\
18x &= 135 - \frac{9\times27}{2}\\
x &= \frac{3}{4}
\end{align}$$

10. a)[1M] Prove that $x+3$ is a factor of $P(x) = x^3 + 4x^2 -27x -90$
b)[3M] It is given that $x^2 -2x -15$ is a factor of $Q(x)=x^3+ax^2+bx+30$, find the values of $a$ and $b$
c)[3M] Hence, solve $P(x)+Q(x)=0$
a)
$$\begin{align}
P(-3) &= (-3)^3+4(-3)^2-27(-3)-90 = 0\\
By&\ Factor\ Thm,\ x+3\ is\ a\ factor\ of\ P(x)
\end{align}$$
b)
$$\begin{align}
x^2-2x-15 &= (x-5)(x+3)\\
By&\ Factor\ Thm,\\
Q(5)&=5^3+a(5)^2+b(5)+30\\
0&=5a+b+31\ \ \ -(i)\\
Q(-3)&=(-3)^3+a(-3)^2+b(-3)+30\\
0&=3a-b+1\ \ \ -(ii)\\
(i)&+(ii)\:\\
0&=8a+32\\
a&=-4\\
b&=-11
\end{align}$$
c)
$$\begin{align}
0&=P(x)+Q(x)\\
&=(x+3)(x^2+x-30)+(x+3)(x-5)(x-2)\\
&=(x+3)(x-5)(2x+4)\\
\therefore\ x&= -3,\ -2\ or\ 5
\end{align}$$

11. An inverted right circular conical vessel is held vertically and fully filled with soft drink. The depth of soft drink in the vessel is $18\ cm$. Patrick drinks $364\pi\ cm^3$ of the soft drink and now he finds that the depth of the soft drink in the vessel is $15\ cm$.
a)[3M] Express the volume of remaining soft drink in terms of $\pi$
b)[3M] Patrick then pours the remaining soft drink into a right cylinderical vessel without overflowing. The base radius of the cylindrical vessesl is $5 cm$. Patrick claims that the total area of the wet surface of the cylindrical vessel is greater than $700 cm^2$. Do you agree? Explain your answer.
a)
$$\begin{align}
Ratio\ of\ heights&=15:18=5:6\\
Ratio\ of\ volumes&=5^3:6^3=125:216\\
Volume\ req. &= 364\pi \times \frac{125}{216-125} = 500\pi\ cm^3\\
\end{align}$$
b)
$$\begin{align}
Height &= 500\pi \div (5^2\pi) = 20\ cm\\
Wet\ surface &= \pi(5)^2 + 2\pi(5)\times20 = 706.9 > 700\\
\therefore&\ I\ agree
\end{align}$$

12. The following stem-and-leaf diagram shows the ages of the players of a basketball team.

|Stem (tens) | Leaf (units) |
| - | - |
| 1 | b 9 |
| 2 | 1 a 3 3 4 5 5 6 7 9 |
| 3 | 2 4 |

It is given that the mean aand the range of the above distribution are $25$ and $15$ respectively.
a)[5M] Find the inter-quartile range and the standard deviation of the above distribution
b)[2M] If one player is selected randomly, find the probability that the age of the selected player is $23$ or above

a)
$$\begin{align}
b&=34 -15 - 10 = 9\\
a&=2\\
IQR&=27-22 = 5\\
\sigma&=\sqrt{\frac{36+36+16+9+4+4+1+1+4+16+49+81}{14}}=4.28
\end{align}$$
b)
$$\begin{align}
Probability = \frac{10}{14} = \frac{5}{7}
\end{align}$$

13. In quadrilateral $ABCD$, $AB // DC$ and $\angle ADB = \angle BCD$
a)[2M] Prove that $\Delta ABD ~ \Delta BDC$
b)[4M] It is given that $CD = 75\ cm$, $BD = 60\ cm$ and $AD=36\ cm$
i) Find $AB$
ii) Is $\Delta BCD$ a right-angled triangle? Explain your answer

a)
$$\begin{align}
In&\ \Delta ABD\ and\ \Delta BDC\\
\angle ABD &= \angle BDC (alt. \angle s, \sim \Delta s)\\
\angle ADB &= \angle BCD (given)\\
\therefore &\ \Delta ABD\sim\Delta BDC (AA)\\
\end{align}$$
b)i)
$$\begin{align}
\because &\ \Delta ABD\sim\Delta BDC (proved)\\
\frac{AB}{BD}&=\frac{BD}{DC} (corr. \angle s, \sim \Delta s)\\
AB&=\frac{60}{75}\times 60\\
&=48
\end{align}$$
ii)
$$\begin{align}
AB^2&=2304\\
AD^2&=1296\\
BD^2&=3600=AB^2+AD^2\\
\angle BAD &= 90^\circ (inv.\ of\ Pyth\ Thm)\\
\therefore \angle DBC &= 90^\circ (corr. \angle s, \sim \Delta s)
\end{align}$$

14. The equation of the straight line $L_1$ is $3x+4y=0$
a)[6M] Let $L_2$ be the straight line passing through point $A(-7,3)$ and perpendicular to $L_1$
i) Find the equation of $L_2$
ii) Suppose that $G$ is a point lying on $L_2$ different from $A$. Denote the y-coordinate of $G$ by $k$, where $k \neq 3$. Let $C$ be the circle passing through $A$ with centre $G$. Prove that the equation of $C$ is $2x^2+2y^2-(3k-37)x-4ky-9k+143=0$
b)[3M] The coordinates of the point B are $(1, -1)$. Using (a)(ii), or otherwise, find the radius of the circle which passes through $A$ and $B$ with centre $G$.

a)i)
$$\begin{align}
Eq\ of\ L_2\ is:\\
\frac{-3}{4} \times \frac{y-3}{x+7}=-1\\
4x-3y+37=0
\end{align}$$
ii)
$$\begin{align}
G&=(\frac{3k-37}{4}, k)\\
Eq\ of\ C\ is&:\\
(x-\frac{3k-37}{4})^2+(y-k)^2&=(-7-\frac{3k-37}{4})^2+(3-k)^2\\
[4x-(3k-37)]^2+16(y-k)^2&=[-28-(3k-37)]^2+16(3-k)^2\\
16x^2-8(3k-37)x+16y^2-32ky&=784+56(3k-37)+144-96k\\
2x^2+2y^2-(3k-37)x-4ky-9k+143&=0\\
\end{align}$$

b)
$$\begin{align}
&\because\ (1, -1)\ is\ on\ C \\
0&=2+2-(3k-37)+4k-9k+143&\\
k &= 23\\
0&=2x^2+2y^2+32x-92y-64\\
0&=x^2+y^2+16x-46y-32\\
625&=(x+8)^2+(y-23)^2\\
\therefore radius &= \sqrt{625}=25\\
\end{align}$$

15. [3M] The mean score of a class of students in a test is 60. The scores of Patrick and Linda in the test are 75 and 52 respectively. The standard score of Patrick in the test is 2.4. Find the standard score of Linda.
$$\begin{align}
\frac{75-60}{\sigma}&=2.4\\
\sigma&=6.25\\
z_{Linda}=\frac{52-60}{6.25}&=1.28\\
\end{align}$$

16. [3M] The first pattern consists of 4 dots. The second pattern is formed by adding 3 dots to the first one. The third pattern is formed by adding 3 dots to the second one and so on. Find the largest value of $m$ such that the total number of dots in the first m patterns is less than 6542
$$\begin{align}
m^{th}\ pattern &= 4+3(m-1)\\
4+7+10+...+[4+3(m-1)]&\lt 6542\\
\frac{[4+4+3(m-1)]m}{2}&\lt 6542\\
3m^2-5m&\lt13084\\
m^2-\frac{5}{3}m&\lt4361\frac{1}{3}\\
\therefore largest\ m\ &is\ 66\\
\end{align}$$

17. The equations of $L_1$ and $L_2$ are $4x-5y+6=0$ and $9x-y-32=0$ respectively. $R$ is the region (including the boundary) bounded by $L_1$, $L_2$ and $L_3$ wich passes through $(0, 4)$ and $(2, 0)$
a)[3M] If $R$ represents the solutions of a system of inequalities, find the system of inequalities.
b)[2M] Let $x$ and $y$ be integers such that $(x, y)$ lies in $R$. Someone claims that the maximum value of the function $F=2x-y+2$ is $10$. Do you agree? Explain your answer

a)
$$\begin{align}
\left\{ \begin{matrix}
4x-5y+6 \ge 0 \\
9x-y-32 \le 0 \\
\frac{x}{2}+\frac{y}{4} \ge 1 \\
\end{matrix} \right.
\end{align}$$

b)
$$\begin{align}
F &= 2x-y+2\ is\ larger\ to\ the\ bottom\ right\\
\therefore\ &At\ (3, -2),\ F=2(3)-(-2)+2=10\ is\ maximum\\
\end{align}$$

18. Let $f(x)=x^2+4kx+7k^2-4$, where $k$ is a positive real constant
a)[2M] Using the method of completing the square, express the coordinates of the vertex of the graph of $y=f(x)$ in terms of $k$
b)[4M] On the same rectangular coordinate system, let $U$ and $V$ be the vertices of the graphs of $y=f(x+4)$ and $y=6-f(x)$ respectively. $W$ is a point on the rectangular coordinate system such that the circumcentre of $\Delta UVW$ is $(2,0)$. Find the equation of the circumcircle of $\Delta UVW$.

a)
$$\begin{align}
f(x)&=(x+2k)^2+3k^2-4\\
\therefore Vertex &= (-2k, 3k^2-4)\\
\end{align}$$

b)
$$\begin{align}
U &= (-2k-4,3k^2-4)\\
V &= (-2k, 10-3k^2)\\
(-2k-4-2)^2+(3k^2-4)^2 &= (-2k-2)^2+(10-3k^2)^2\\
0 &= 36k^2+16k-52\\
0 &= (18k+26)(x-1)\\
k &= 1\ or\ -\frac{13}{9} (rej)\\
U &= (-6,-1),\ V=(-2,7)\\
Eq.\ req.\ is:\\
(x-2)^2+y^2&=(-6-2)^2+(-1-0)^2=65\\
\end{align}$$

19. a)[2M] Let $r$ be the radius of the inscribed circle of $\Delta ABC$, $AB=a$, $BC=b$ and $AC=c$. Prove that the area of $\Delta ABC$ is $\frac{(a+b+c)r}{2}$
b)[5M] The coordinates of points $O$, $P$ and $Q$ are $(0,0)$, $(7,0)$ and $(0,24)$ respectively. Using (a), find the equation of the inscribed circle of $\Delta OPQ$

a)
$$\begin{align}
Area &= \frac{1}{2}(a)(r) + \frac{1}{2}(b)(r) + \frac{1}{2}(c)(r) \\
&= \frac{(a+b+c)r}{2}\\
\end{align}$$

b)
$$\begin{align}
OP &= 7\\
OQ &= 24\\
PQ &= \sqrt{7^2+24^2} = 25\\
\frac{1}{2}(7)(24) &= \frac{(7+24+25)r}{2}\\
r &= 3\\
Eq.\ req.\ is:\\
(x-3)^2+(y-3)^2 &= 9\\
\end{align}$$

20. $ABCDB'E$ is a hexagonal paper card. It is given that $AB=EB' = 15\ cm$, $BC=B'D=25\ cm$. $ACDE$ is a rectangle with $AE=10\ cm$. Let $\angle ABC=\theta$, where $55^\circ \le \theta \le 85^\circ$
a)[4M] Suppose that $\theta = 55^\circ$
i) Find the length of $AC$
ii) Find the area of the paper card $ABCDB'E$
b)[2M] Describe how the area of the paper card $ABCDB'E$ varies when $\theta$ increase from $55^\circ$ to $85^\circ$. Explain your answer
c)[5M] Suppose that $\theta = 73^\circ$. The paper card is folded along $AC$ and $ED$ such that $B$ and $B'$ join together to form a pyramid $ACDEB$. Find the volume of the pyramid $ACDEB$

a)i)
$$\begin{align}
AC^2 &= 15^2 + 25^2 -2(15)(25)\cos{55^\circ}\\
AC &= 20.49\ cm\\
\end{align}$$
ii)
$$\begin{align}
Area &= 2\times\frac{1}{2}(25)(15)\sin{55^\circ} + (10)(20.49)\\
&= 512.1\ cm^2\\
\end{align}$$

b)
As $\theta$ increases, $\cos{\theta}$ decreases, so $AC$ increases and the area increases

c)
Denote $H$ on $AC$ such that $BH\perp AC$
and $H'$ on $$DE$ such taht $B'H'\perp DE$

$$\begin{align}
\frac{1}{2}BH(20.49)&=\frac{1}{2}(15)(25)\sin{73^\circ}\\
BH&=17.5\ cm\\
17.5^2+10^2-2(17.5)(10)\cos{\angle BHH'}&=17.5^2\\
\angle BHH'&=73.4^\circ\\
Volume &= \frac{1}{3}(10)(20.49)(17.5\sin{73.4^\circ})\\
&= 1145\ cm^3\\
\end{align}$$