Choice D

# 2018 December(B05) Math Question 2

Choice G

$$t=10 \Rightarrow 180(t-2)=180(10-2)=180\times8=1440$$

# 2018 December(B05) Math Question 3

Choice D

$$(-8,-3)\Rightarrow (-8+8,-3+3)=(0,0)$$

Choice F

# 2018 December(B05) Math Question 5

Choice C

$$\frac{1}{3}\pi r^{2}h,\ r=3,\ h=6$$ $$\Rightarrow \frac{1}{3}\pi\times3^{2}\times6=18\pi$$

# 2018 December(B05) Math Question 6

Choice G

$$(x^{4})^{6}=x^{4\times6}=x^{24}$$

# 2018 December(B05) Math Question 7

Choice C

$$12\times40+12\times1\frac{1}{2}\times5$$ $$=480+90$$ $$=570$$

# 2018 December(B05) Math Question 8

Choice F

$$=\begin{bmatrix}5+(-6) & 7+3 \\(-4)+6 & 4+8 \end{bmatrix}$$ $$=\begin{bmatrix}-1 & 10 \\2 & 12 \end{bmatrix}$$

# 2018 December(B05) Math Question 9

Choice E

$$6\times4\times6=144$$

# 2018 December(B05) Math Question 10

Choice G

$$(a,0)=(-1+2,0)=(1,0)$$ $$\Rightarrow a=1$$

# 2018 December(B05) Math Question 11

Choice B

$$3\times8\times7\times1\times2=336$$

# 2018 December(B05) Math Question 12

Choice J

$$\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{4-3}{2-(-7)}=\frac{1}{9}$$

# 2018 December(B05) Math Question 13

Choice A

$$60\times\frac{1}{2}=30$$ $$60\times\frac{1}{4}=15$$ $$60\times\frac{1}{6}=10$$ $$60-30-15-10=5$$

# 2018 December(B05) Math Question 14

Choice J

$$6x-2\leq11.2$$ $$\Rightarrow 6x\leq13.2$$ $$\Rightarrow x\leq2.2$$

2 is the greatest solution.

# 2018 December(B05) Math Question 15

Choice D

$$17.5+0.7x\leq50$$ $$\Rightarrow 0.7x\leq 32.5$$ $$\Rightarrow x\leq46.42...$$ $$Ans:\ x=46$$

# 2018 December(B05) Math Question 16

Choice K

$$2:9=8:x$$ $$\Rightarrow x=\frac{8}{2}\times9=36$$

# 2018 December(B05) Math Question 17

Choice C

$$Total=30\times5=150$$ $$150-50=100$$ $$100\div4=25$$

# 2018 December(B05) Math Question 18

Choice K

$$\because\overline{BE}\equiv\overline{BF}$$ $$\therefore ABE=\angle CBF=32°$$ $$\because\overline{AC}\parallel\overline{DF}$$ $$\therefore\angle ABE=\angle BEF=32°$$ $$\angle BED=180°-32°=148°$$

# 2018 December(B05) Math Question 19

Choice C

$$x=-1, y=2,$$ $$x^{3}-2x^{2}y-4xy^{2}+8$$ $$=(-1)^{3}-2(-1)^{2}\times2-4(-1)\times2^{2}+8$$ $$=-1-2\times2+4\times4+8$$ $$=-1-4+16+8$$ $$=19$$

# 2018 December(B05) Math Question 20

Choice H

$$A: 10\times5+12\times10+5\times6=200$$ $$B: 10\times5.5+12\times9.5+5\times5.75$$ $$=55+114+28.75$$ $$=197.75$$ $$200-197.75=2.25$$

# 2018 December(B05) Math Question 21

Choice E

$$Slope\ l=1$$ $$\Rightarrow Slope\ p=1\times\frac{1}{2}=\frac{1}{2}$$ …

# 2018 December(B05) Math Question 22

Choice H

$$\sin\theta=\frac{a}{\sqrt{a^{2}+b^{2}}}$$

# 2018 December(B05) Math Question 23

Choice E

$$50\times50=2500$$ $$10\times x=2500$$ $$\Rightarrow x=250$$

# 2018 December(B05) Math Question 24

Choice J

$$(fg)(x)=(5x^{2}-6x+1)(x^{2}-2)$$ $$=(5x^{2}-6x+1)(x^{2})-(5x^{2}-6x+1)2$$ $$=(5x^{4}-6x^{3}+x^{2})-10x^{2}+12x-2$$ $$=5x^{4}-6x^{3}-9x^{2}+12x-2$$

# 2018 December(B05) Math Question 25

Choice B

$$Total\ marbles:\ 12+14+8=34$$ $$p(red)=\frac{12+x}{34+x}=\frac{3}{5}$$ $$5(12+x)=3(34+x)$$ …

# 2018 December(B05) Math Question 26

Choice H

$$Megan:\frac{120}{40}=3$$ $$Louisa:\frac{120}{70}=1\frac{5}{7}\approx1.71$$

# 2018 December(B05) Math Question 27

Choice E

$$\require{cancel}$$ $$3x-(x+6)+8=2x+14$$ $$\Rightarrow3x-x-6+8=2x+14$$ $$\Rightarrow\cancel{2x}+2=\cancel{2x}+14$$ …

# 2018 December(B05) Math Question 28

Choice F

$$Mean:(\frac{1+2+3+4+5+6+7}{2})\div7$$ $$=\frac{[(1+7)7]\div2}{2}\div7$$ $$=\frac{56\div2}{2}\div7$$ $$=\frac{28}{2}\div7$$ $$=14\div7$$ …

Choice C

# 2018 December(B05) Math Question 36

Choice H

$$\frac{2.4}{16}=0.15\ /ounces$$ $$0.15\times10=1.5$$

Choice D

# 2018 December(B05) Math Question 38

Choice G

The denominator of any fraction …

# 2018 December(B05) Math Question 39

Choice E

$$0.1\times3+0.7\times4+0.2\times5$$ $$=0.3+2.8+1=4.1$$

# 2018 December(B05) Math Question 40

Choice K

Cannot be determined from the …

Choice B

Choice F