Mathematics Guide

Maths Demo (I) ​

S1 Estimation ​

Rounding ​

Consider the numbers 6.1 , 6.4 , 6.5 , 6.9.

  1. Round off the numbers to the nearest integer.

  2. Round up the numbers to the nearest integer.

  3. Round down the numbers to the nearest integer.

Solution

  1. 6 , 6 , 7 , 7 四捨五入

  2. 7 , 7 , 7 , 7 無論如何，進位

  3. 6 , 6 , 6 , 6 無論如何，捨去

S1 Basic Coordinate Geometry ​

Translate / Reflect / Rotate a Point ​

Given P(2,1)P(2, 1).

  1. Translate PP upwards by 2 units to get QQ.

  2. Reflect PP with respect to the yy-axis to get RR.

  3. Rotate PP about the origin by 90° clockwise to get SS.

Solution

figure

  1. Q=(2,3)Q=(2, 3)

  2. R=(−2,1)R=(-2, 1)

  3. S=(1,−2)S=(1, -2)

S2 Identity ​

Find Unknown Constants in Identity ​

Given (x−2)(x+a)−3bx+b≡x(x−18)−3(x-2)(x+a)-3bx+b\equiv x(x-18)-3. Find aa and bb.

Method 1: Compare Terms

(x−2)(x+a)−3bx+b≡x(x−18)−3x2−2x+ax−2a−3bx+b≡x2−18x−3←expand both sidesx2+(a−3b−2)x+(b−2a)≡x2−18x−3\begin{aligned} (x-2)(x+a)-3bx+b &\equiv x(x-18)-3 \\ x^{2}-2x+ax-2a-3bx+b &\equiv x^{2}-18x-3 && \leftarrow \text{expand both sides} \\ x^{2}+(a-3b-2)x+(b-2a) &\equiv x^{2}-18x-3 \end{aligned}

Compare xx term: a−3b−2=−18a-3b-2=-18 ...(1)

Compare constant term: b−2a=−3b-2a=-3 ...(2)

Solving, a=5a=5 and b=7b=7

Method 2: Kill x

Put x=2x=2:

(2−2)(2+a)−3b(2)+b=2(2−18)−3−5b=−35b=7\begin{aligned} (2-2)(2+a)-3b(2)+b &= 2(2-18)-3 \\ -5b &= -35 \\ b &= 7 \end{aligned}

Put x=0x=0:

(0−2)(0+a)−3b(0)+b=0(0−18)−3−2a+b=−3a=5\begin{aligned} (0-2)(0+a)-3b(0)+b &= 0(0-18)-3 \\ -2a+b &= -3 \\ a &= 5 \end{aligned}

S3 Factorization ​

Cross Method by Calculator ​

Factorize

  1. x2+x−6x^{2}+x-6

  2. x2+xy−6y2x^{2}+xy-6y^{2}

Solution

  1. ∵     The roots of x2+x−6=0x^{2}+x-6=0 are 2 and −3.
    ∴     x2+x−6=(x−2)(x+3)x^{2}+x-6=(x-2)(x+3)

  2. x2+xy−6y2=(x−2y)(x+3y)x^{2}+xy-6y^{2}=(x-2y)(x+3y) ← just put yy back in place

Factorize

  1. 2x2+5x−122x^{2}+5x-12

  2. 2x2+5xy−12y22x^{2}+5xy-12y^{2}

Solution

  1. ∵     The roots of 2x2+5x−12=02x^{2}+5x-12=0 are 32\dfrac{3}{2} and −4.
    ∴     2x2+5x−12=2(x−32)(x+4)=(2x−3)(x+4)\begin{aligned} 2x^{2}+5x-12 &= 2(x-\dfrac{3}{2})(x+4) \\ &= (2x-3)(x+4) \end{aligned}
    ← don’t forget the leading coefficient is 2!

  2. 2x2+5xy−12y2=(2x−3y)(x+4y)2x^{2}+5xy-12y^{2}=(2x-3y)(x+4y)

S2 Ratio ​

Read a Ratio ​

  1. If a2=b3\dfrac{a}{2}=\dfrac{b}{3}, find a:ba:b.

  2. If 3a=7b3a=7b, find a:ba:b.

  3. If a2=b3=c5\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{5}, find a:b:ca:b:c.

Solution

分子：分子 = 分母：分母

  1. a:b=2:3a:b=2:3

  2. 3a=7ba7=b3\begin{aligned} 3a &= 7b \\ \dfrac{a}{7} &= \dfrac{b}{3} \end{aligned} ← write as fraction first
    ∴     a:b=7:3a:b=7:3

  3. a:b:c=2:3:5a:b:c=2:3:5

k-method ​

If a:b=2:3a:b=2:3 and a+2b=56a+2b=56, find aa and bb.

Solution

Let a=2ka=2k and b=3kb=3k

2k+2(3k)=56k=7\begin{aligned} 2k+2(3k) &= 56 \\ k &= 7 \end{aligned}

∴     a=2k=14b=3k=21\begin{aligned} a &= 2k \\ &= 14 \\ b &= 3k \\ &= 21 \end{aligned}

S3 Number System ​

Convert to Base 10 ​

  1. Convert 11001211001_{2} to a decimal number (base 10).

  2. Convert AB8516\text{AB85}_{16} to a decimal number (base 10).

Solution

  1. 110012=1−×24+1−×23+0−×22+0−×21+1−×20=25\begin{aligned} 11001_{2} &= \underset{\scriptscriptstyle-}{1}\times 2^{4}+\underset{\scriptscriptstyle-}{1}\times 2^{3}+\underset{\scriptscriptstyle-}{0}\times 2^{2}+\underset{\scriptscriptstyle-}{0}\times 2^{1}+\underset{\scriptscriptstyle-}{1}\times 2^{0} \\ &= 25 \end{aligned}

  2. AB8516=10‾×163+11‾×162+8‾×161+5‾×160=43909\begin{aligned} \text{AB85}_{16} &= \underline{10}\times 16^{3}+\underline{11}\times 16^{2}+\underline{8}\times 16^{1}+\underline{5}\times 16^{0} \\ &= 43909 \end{aligned}

Convert from Base 10 ​

  1. Convert 25 to binary base (base 2).

  2. Convert 43909 to hexadecimal base (base 16).

Solution

短除法，由下至上讀result

  1. 25=11001225=11001_{2}

    225‾25225‾12  … 1225‾6  … 0225‾3  … 0251  … 1\begin{array}{r|r} 2 & \underline{\phantom{25}}\llap{25} \\ 2 & \underline{\phantom{25}}\llap{12}\rlap{\;\ldots\,1} \\ 2 & \underline{\phantom{25}}\llap{6}\rlap{\;\ldots\,0} \\ 2 & \underline{\phantom{25}}\llap{3}\rlap{\;\ldots\,0} \\ & \phantom{25}\llap{1}\rlap{\;\ldots\,1} \end{array}

  2. 43909=AB851643909=\text{AB85}_{16}

    1643909‾439091643909‾2744  … 51643909‾171  … 84390910  … 11\begin{array}{r|r} 16 & \underline{\phantom{43909}}\llap{43909} \\ 16 & \underline{\phantom{43909}}\llap{2744}\rlap{\;\ldots\,5} \\ 16 & \underline{\phantom{43909}}\llap{171}\rlap{\;\ldots\,8} \\ & \phantom{43909}\llap{10}\rlap{\;\ldots\,11} \end{array}

    10 = AA, 11 = BB

S3 Area and Volume ​

Volume of Frustum ​

Find the volume of this frustum.

figure

Solution

A frustum is formed by cutting off a smaller cone from a larger cone.

figure

hh+18=26h=9\begin{aligned} \dfrac{h}{h+18} &= \dfrac{2}{6} \\ h &= 9 \end{aligned} ← similar triangle

Volume of smaller cone=13π⋅22⋅9=12π\begin{aligned} \text{Volume of smaller cone} &= \dfrac{1}{3}π \cdot 2^{2}\cdot 9 \\ &= 12π \end{aligned}

Volume of larger cone=13π⋅62⋅(9+18)=324π\begin{aligned} \text{Volume of larger cone} &= \dfrac{1}{3}π \cdot 6^{2}\cdot (9+18) \\ &= 324π \end{aligned}

∴     Volume of frustum=324π−12π=312π cm3\begin{aligned} \text{Volume of frustum} &= 324π-12π \\ &= 312π \text{ cm}^{3} \end{aligned}

S3 Basic Coordinate Geometry ​

Length Ratio of Segment (影比例法) ​

Given A(2,1)A(2, 1), B(5,2)B(5, 2) and C(11,4)C(11, 4). ABCABC is a straight line. Find AB:BCAB:BC.

Method 1: Distance Formula (SLOW!)

AB:BC=(5−2)2+(2−1)2:(11−5)2+(4−2)2=10:40=10:210=1:2\begin{aligned} AB:BC &= \sqrt{(5-2)^{2}+(2-1)^{2}}:\sqrt{(11-5)^{2}+(4-2)^{2}} \\ &= \sqrt{10}:\sqrt{40} \\ &= \sqrt{10}:2\sqrt{10} \\ &= 1:2 \end{aligned}

figure

Method 2: Shadow on x-axis

AB:BC=X1:X2=(5−2):(11−5)=3:6=1:2\begin{aligned} AB:BC &= X_{1}:X_{2} \\ &= (5-2):(11-5) \\ &= 3:6 \\ &= 1:2 \end{aligned} ← ratio = shadows' ratio on xx-axis

figure

Method 3: Shadow on y-axis

AB:BC=Y1:Y2=(2−1):(4−2)=1:2\begin{aligned} AB:BC &= Y_{1}:Y_{2} \\ &= (2-1):(4-2) \\ &= 1:2 \end{aligned}

figure

Area Ratio of Adjacent Triangles (同高異底) ​

Given A(2,1)A(2, 1), B(5,2)B(5, 2), C(11,4)C(11, 4) and DD. ABCABC is a straight line. Find the area ratio of △ABD\triangle ABD to △BCD\triangle BCD.

figure

Solution

If ABAB and BCBC are treated as their bases, the two triangle have the same height hh.

Area of ΔABD:ΔBCD=12×AB×h:12×BC×h=AB:BC=1:2\begin{aligned} \text{Area of }Δ ABD:Δ BCD &= \dfrac{1}{2}\times AB\times h:\dfrac{1}{2}\times BC\times h \\ &= AB:BC \\ &= 1:2 \end{aligned}

figure

S3 Statistics ​

Composite Mean Problem ​

There are 40 boys and 60 girls in a school. The mean height of the boys is 150 cm. The mean height of all the students is 144 cm. Find the mean height of the girls.

Solution

Let the mean height of the girls be mm.

sum of height of boys+sum of height of girls100=14440×150+60×m100=144m=140 cm\begin{aligned} \dfrac{\text{sum of height of boys}+\text{sum of height of girls}}{100} &= 144 \\ \dfrac{40\times 150+60\times m}{100} &= 144 \\ m &= 140\text{ cm} \end{aligned}

S3 Probability ​

Expected Value ​

In a game, there is a 50% chance of winning $1, a 30% chance of winning $5 and a 20% chance of winning $10.

Find the expected value of the prize.

Solution

(0.5)(1)+(0.3)(5)+(0.2)(10)=$4(0.5)(1)+(0.3)(5)+(0.2)(10)=\$4