R2

Revision Exercise 2

1. CH 4 Q1. Simplify:
   (a) $-2(3m + 4n - 7) + 5(2m - 6n + 3) = $ 检查
   (b) $\dfrac{5(2x+1)}{3} - \dfrac{4x-7}{2} = $ 检查
   💡 思路

   (a) Expand: $-6m - 8n + 14 + 10m - 30n + 15 = 4m - 38n + 29$.
   (b) Common denominator 6: $\dfrac{10(2x+1) - 3(4x-7)}{6} = \dfrac{20x+10-12x+21}{6} = \dfrac{8x+31}{6}$.

2. CH 4 Q2. Factorise completely:
   (a) $3ax - 6ay + 21a = $ 检查
   (b) $6(kt - 3d) - 2(2ky - 9d) = $ 检查
   💡 思路

   (a) GCF $3a$: $3a(x - 2y + 7)$.
   (b) Expand first: $6kt - 18d - 4ky + 18d = 6kt - 4ky = 2k(3t - 2y)$.

3. CH 5 Q3. Solve:
   (a) $5(2x - 3) = 3(x + 2)$: $x = $ 检查
   (b) $\dfrac{2x}{5} + \dfrac{7(x-4)}{2} = 25$: $x = $ 检查
   💡 思路

   (a) $10x - 15 = 3x + 6 \Rightarrow 7x = 21 \Rightarrow x = 3$.
   (b) ×10 both sides: $4x + 35(x-4) = 250 \Rightarrow 39x = 390 \Rightarrow x = 10$.

4. CH 4 Q4. A charging cable's unit price is $x$ cents per cm. Lucas bought a cable for $\$y$. Find the length of the cable in metres, in terms of $x$ and $y$.
   Length $= $ metres 检查
   💡 思路

   $\$y = 100y$ cents. Length in cm $= 100y/x$. In metres $= 100y/x \div 100 = y/x$ metres.

5. CH 5 Q5. Tom works $x$ hours a week at $\$14$/h normally. Overtime rate is $1.5 \times 14 = \$21$/h. In a certain week he worked $54$ h total and earned $\$819$. Find $x$.
   $x = $ hours 检查
   💡 思路

   $14x + 21(54 - x) = 819 \Rightarrow 14x + 1134 - 21x = 819 \Rightarrow -7x = -315 \Rightarrow x = 45$.

6. CH 4 Q6. (a) Simplify $\dfrac{5(3n - 4.5)}{3} - \dfrac{-(2n + 9)}{2}$.
   $= $ 检查
   (b) Hence show that the above expression is a multiple of $3$ for all positive integers $n$.
   Factored form: $6n - 3 = 3(2n - 1)$, which is $3 \times $ (integer), so always a multiple of 3.
   💡 思路

   Common denominator 6: $\dfrac{10(3n - 4.5)}{6} + \dfrac{3(2n+9)}{6} = \dfrac{30n - 45 + 6n + 27}{6} = \dfrac{36n - 18}{6} = 6n - 3 = 3(2n - 1)$.

7. CH 6+7 Q7. In the diagram, $AB \parallel DE$, $\angle BCD = 59°$ and $\angle CDE = 105°$. Find $\angle ABC$.
   $\angle ABC = $ $°$ 检查
   💡 思路

   Draw a line through $C$ parallel to $AB$ (and to $DE$). Using co-interior angles with $DE$: the part of the angle at $C$ on the $D$ side $= 180 - 105 = 75°$. Then the remaining part to $\angle BCD = 59°$ means the part toward $B$ $= 75 + 59 = 134°$. By alt angles with $AB$, $\angle ABC = 134°$.

8. CH 7 Q8. $EABF$ is a straight line. $\angle EAD = 100°$, $\angle FBC = 115°$, $\angle ADC = 113°$. Find $\angle BCD$.
   $\angle BCD = $ $°$ 检查
   💡 思路

   $\angle DAB = 180 - 100 = 80°$ (linear pair). $\angle ABC = 180 - 115 = 65°$. Sum of quad $ABCD$: $80 + 65 + 113 + \angle BCD = 360 \Rightarrow \angle BCD = 102°$.

9. CH 7 Q9. A pentagon has three angles of $118°$ each. The other two are $x°$ and $(x + 16)°$. Find $x$.
   $x = $ 检查
   💡 思路

   Sum $= 540°$. $3(118) + x + (x+16) = 540 \Rightarrow 2x = 540 - 354 - 16 = 170 \Rightarrow x = 85$.

10. CH 7 Q10. Carol measured an exterior angle of a regular polygon as $50°$.
    (a) Why is Carol's measurement not exactly correct? Hint: $360 / 50 = $ (not an integer) 检查
    (b) If Carol's measurement is very close, find the number of sides.
    $n = $ 检查
    💡 思路

    (a) For a regular $n$-gon, $n = 360/$ext must be an integer. $360/50 = 7.2$ — not integer, so $50°$ exactly is impossible.
    (b) Nearest integer giving close to $50°$: $n=7 \Rightarrow$ ext $= 360/7 \approx 51.4°$. Diff $|51.4 - 50| = 1.4$. $n=8 \Rightarrow 45°$, diff $5°$. So $n = 7$.

11. CH 7 Q11. Construct $\triangle ABC$ with $AB = 9$ cm, $BC = 7$ cm, $\angle BAC = 40°$.
    (a) Measure $\angle ABC$. Possible values: $°$ 检查
    (b) Number of distinct triangles satisfying the given measurements: 检查
    💡 思路

    SSA ambiguous case. By law of sines: $\sin(\angle C)/9 = \sin(40°)/7 \Rightarrow \sin(\angle C) = 9\sin(40°)/7 \approx 0.826$. So $\angle C \approx 55.7°$ or $124.3°$. Then $\angle B = 180 - 40 - 55.7 = 84.3°$ or $180 - 40 - 124.3 = 15.7°$. Both yield valid triangles.

12. CH 4 Q12. Paul's monthly salary is $\$3x$. Queenie's is $\$600$ more than Paul's. Robert's is two-thirds of Queenie's.
    (a) Queenie's salary (in terms of $x$): 检查
    (b) Robert's salary: 检查
    (c) If sum of their salaries $= \$10{,}600$, Paul's salary = dollars 检查
    💡 思路

    Queenie $= 3x + 600$. Robert $= (2/3)(3x+600) = 2x + 400$.
    Sum: $3x + (3x+600) + (2x+400) = 8x + 1000 = 10600 \Rightarrow 8x = 9600 \Rightarrow x = 1200$. Paul $= 3 \cdot 1200 = \$3600$.

13. CH 6+7 Q13. $BCD$ is a straight line, $BA \parallel DE$, $AC = AD$, $\angle ABC = 40°$ and $\angle CAD = 66°$.
    (a) $\angle CDA = $ $°$ 检查
    (b) $\angle BAC = $ $°$ 检查
    (c) $\angle ADE = $ $°$ 检查
    💡 思路

    (a) $\triangle ACD$ isos ($AC=AD$): $\angle ACD = \angle ADC = (180-66)/2 = 57°$.
    (b) Tri $ABC$: $\angle ACB = 180 - 57 = 123°$ (linear pair on $BCD$). $\angle BAC = 180 - 40 - 123 = 17°$.
    (c) $\angle BAD = \angle BAC + \angle CAD = 17 + 66 = 83°$. With $BA \parallel DE$ and transversal $AD$, $\angle ADE = \angle BAD = 83°$ (alt angles).

14. CH 7 Q14. A triangle has sides $x$, $y$, $z$ cm. Given $x$ is an odd integer, $y$ is a whole number that is neither prime nor composite (so $y = 1$), and $5(z + 1) - 3(z - 2) = 21$. Find the perimeter of the triangle.
    Perimeter $= $ cm 检查
    💡 思路

    Solve $z$: $5z + 5 - 3z + 6 = 21 \Rightarrow 2z = 10 \Rightarrow z = 5$. $y = 1$ (not prime, not composite). For triangle inequality with $y = 1, z = 5$: need $|5-1| < x < 5+1$, i.e. $4 < x < 6$. Only odd integer in (4,6) is $x = 5$. Perimeter $= 5 + 1 + 5 = 11$ cm.

15. CH 7 Q15. The diagram shows part of two regular polygons with $n$ sides and $2n$ sides sharing a vertex. The angle at one vertex is $78.75°$.
    (a) $n = $ 检查
    (b) Exterior angle of a regular $3n$-gon = $°$ 检查
    💡 思路

    (a) 图中 $78.75°$ 是正 $2n$ 边形某个内角被一条对角线（或对称轴）二等分后的一半。
    正 $2n$ 边形每个内角 $= \dfrac{(2n-2) \cdot 180°}{2n} = 180° - \dfrac{180°}{n}$。
    二等分后：$\dfrac{1}{2}\left(180° - \dfrac{180°}{n}\right) = 78.75°$
    $\Rightarrow 90° - \dfrac{90°}{n} = 78.75°$
    $\Rightarrow \dfrac{90°}{n} = 11.25°$
    $\Rightarrow n = 8$。
    验证：正 $16$ 边形内角 $= (16-2) \cdot 180°/16 = 157.5°$，一半 $= 78.75°$ ✓。
    (b) $3n = 24$，正 $24$ 边形每个外角 $= 360°/24 = 15°$。