第 5 章总复习。共 19 题，覆盖 §5.1–5.4。试着不翻书独立完成，再对答案。

1. Q1. Solve the following equations.
   (a) $13x + 22 = 3x - 8.5$ → $x = $ 检查
   (b) $25 - 4(9 - x) = 6x$ → $x = $ 检查
   (c) $2(5x - 8) + 6 = -11(-2 - x)$ → $x = $ 检查
2. Q2. Solve the following equations.
   (a) $\dfrac{2x}{3} + \dfrac{x}{5} = 13$ → $x = $ 检查
   (b) $1 - \dfrac{4}{7}x = 23 + x$ → $x = $ 检查
   (c) $\dfrac{3x - 5}{2} = \dfrac{7x - 3}{9}$ → $x = $ 检查
   (d) $\dfrac{x - 4}{3} - \dfrac{2x + 1}{6} = \dfrac{5x - 1}{2}$ → $x = $ 检查
3. Q3. Solve the following equations.
   (a) $\dfrac{2x}{x - 7} = -6$ → $x = $ 检查
   (b) $\dfrac{4x + 1}{5x - 2} = \dfrac{5}{7}$ → $x = $ 检查
   (c) $\dfrac{3}{x + 1} - \dfrac{7}{2x} = 0$ → $x = $ 检查
   (d) $\dfrac{5}{x - 1} = \dfrac{-2}{1 + 2x}$ → $x = $ 检查
4. Q4. Given the formula $D = b^2 - 4ac$, find
   (a) the value of $D$ when $a = -1$, $b = -5$ and $c = 3$ → $D = $ 检查
   (b) the value of $c$ when $a = 2$, $b = 3$ and $D = 49$ → $c = $ 检查
   💡 思路

   (a) $(-5)^2 - 4(-1)(3) = 25 + 12 = 37$. (b) $49 = 9 - 8c \Rightarrow 8c = -40 \Rightarrow c = -5$.

5. Q5. Given the formula $S = \dfrac{n(a + b)}{2}$, find
   (a) the value of $S$ when $a = 1$, $b = 25$ and $n = 12$ → $S = $ 检查
   (b) the value of $a$ when $b = 41$, $n = 15$ and $S = 330$ → $a = $ 检查
6. Q6. The sides of a triangle are $(2x + 1)$ cm, $(3x + 2)$ cm and $(4x - 1)$ cm.
   (a) Find the perimeter of the triangle in terms of $x$. → cm 检查
   (b) If the perimeter of the triangle is 47 cm, find the value of $x$. → $x = $ 检查
7. Q7. In a quiz, 5 marks are awarded for each correct answer and 1 mark is deducted for each wrong answer. The quiz has 20 questions. Reena attempted all the questions and scored 76 marks. Form an equation and solve it to find the number of correct answers Reena had.
   Correct answers = 检查
   💡 思路

   Let $c$ = correct. Wrong = $20 - c$. Score: $5c - (20-c) = 76 \Rightarrow 6c - 20 = 76 \Rightarrow c = 16$.

8. Q8. Peter has 96 stamps and Ajit has 63 stamps. How many stamps should Ajit give Peter so that Peter will have twice as many stamps as Ajit?
   Stamps to give = 检查
   💡 思路

   Let $x$ = stamps Ajit gives Peter. After: Peter = $96 + x$, Ajit = $63 - x$. Peter = 2 × Ajit: $96 + x = 2(63 - x) \Rightarrow 96 + x = 126 - 2x \Rightarrow 3x = 30 \Rightarrow x = 10$.

9. Q9. Mohan wants to buy some buns at $2 each. The cashier informs Mohan that if he buys 2 more buns, the price for all the buns will be $1.50 each and the total amount Mohan needs to pay is the same as what he will have to pay if he buys some buns at $2 each. Form an equation and solve it to find the total amount Mohan needs to pay for the buns.
   Total amount = $ 检查
   💡 思路

   Let $n$ = original bun count. $2n = 1.50(n + 2) \Rightarrow 2n = 1.5n + 3 \Rightarrow 0.5n = 3 \Rightarrow n = 6$. Total = $2 \times 6 = \$12$.

10. Q10. A boy is 26 years younger than his father. In 3 years' time, his age will be $\dfrac{1}{3}$ of his father's age. Find the present age of the boy.
    Present age = years 检查
    💡 思路

    Let father now = $f$; boy = $f - 26$. In 3 yrs: $(f - 26 + 3) = \tfrac{1}{3}(f + 3) \Rightarrow f - 23 = \tfrac{f+3}{3} \Rightarrow 3f - 69 = f + 3 \Rightarrow 2f = 72 \Rightarrow f = 36$. Boy now = $36 - 26 = 10$.

11. Q11. The price of a skirt is $25 more than the price of a T-shirt. The total price of 3 skirts and 8 T-shirts is $339. Find the price of a skirt.
    Skirt price = $ 检查
    💡 思路

    Let T-shirt = $t$; skirt = $t + 25$. $3(t+25) + 8t = 339 \Rightarrow 11t + 75 = 339 \Rightarrow t = 24$. Skirt = $49$.

12. Q12. In a certain week, the time Lisa spent on sport was 3 hours less than twice the time she spent doing her school homework. The time she spent on reading was $\dfrac{2}{3}$ of the time she spent doing her school homework. The total time she spent on these three activities was 30 hours in that week. Form an equation and solve it to find the amount of time Lisa spent doing her school homework.
    Homework time = hours 检查
    💡 思路

    Let $h$ = homework hr; sport = $2h - 3$; reading = $\tfrac{2h}{3}$. $h + (2h-3) + \tfrac{2h}{3} = 30 \Rightarrow \tfrac{11h}{3} = 33 \Rightarrow h = 9$.

13. Q13. 144 coins are divided equally among some children. If there were 3 fewer children, each child would have 16 coins. How many children are there?
    Children = 检查
    💡 思路

    Let $n$ = original children. If $(n-3)$ children, each gets 16 coins → total = $16(n-3) = 144 \Rightarrow n - 3 = 9 \Rightarrow n = 12$. Verify: 144 / 12 = 12 coins each normally; 144 / 9 = 16 ✓.

14. Q14. The numerator of a fraction is 4 less than its denominator. If 1 is subtracted from both the numerator and the denominator, the new fraction obtained is $\dfrac{3}{5}$. Find the original fraction.
    Original = 检查
    💡 思路

    Let numerator = $x$; denominator = $x + 4$. $\dfrac{x - 1}{x + 3} = \dfrac{3}{5} \Rightarrow 5(x-1) = 3(x+3) \Rightarrow 5x - 5 = 3x + 9 \Rightarrow 2x = 14 \Rightarrow x = 7$. Original = $\tfrac{7}{11}$.

15. Q15. The number of books in a class library is 17 more than 3 times the number of students in the class. If 5 students are absent, each student can borrow exactly 4 books from the library. Find the number of students in the class.
    Students = 检查
    💡 思路

    Let $s$ = students; books = $3s + 17$. $4(s - 5) = 3s + 17 \Rightarrow 4s - 20 = 3s + 17 \Rightarrow s = 37$.

16. Q16. In a flower fertilisation experiment, the number of pink flowers grown is twice that of red flowers. The number of white flowers grown is 3 more than that of red flowers. The number of white flowers grown is 37 fewer than that of pink flowers. Find the total number of flowers grown.
    Total = 检查
    💡 思路

    Let red = $r$; pink = $2r$; white = $r + 3$. Also white = pink - 37 → $r + 3 = 2r - 37 \Rightarrow r = 40$. Pink = 80, white = 43. Total = $40 + 80 + 43 = \mathbf{163}$.

17. Q17. The average travelling times from HarbourFront MRT station to Serangoon MRT station on the North East Line and on the Circle Line are $(2x + 5)$ minutes and $(5x - 4)$ minutes respectively. The difference between the two average travelling times is $\dfrac{1}{4}$ hour.
    (a) Form two equations in $x$ and solve them. Two possible values of $x$: 检查
    (b) Hence find the average travelling time from HarbourFront MRT station to Serangoon MRT station on the North East Line. → minutes 检查
    💡 思路

    $\tfrac{1}{4}$ hour = 15 min. So either $(5x-4) - (2x+5) = 15$ → $3x = 24 \Rightarrow x = 8$;
    or $(2x+5) - (5x-4) = 15$ → $-3x = 6 \Rightarrow x = -2$.
    Since $x$ represents a time-related coefficient, reject $x = -2$ (would make CL time negative: $5(-2)-4 = -14 < 0$).
    Take $x = 8$: NEL = $2(8)+5 = \mathbf{21}$ min, CL = $5(8)-4 = 36$ min.

18. Q18. There is a 2-digit number. The sum of its ones digit and its tens digit is 5. When the digits are reversed, the new number formed is 9 more than the original number. Find the original 2-digit number.
    Original number = 检查
    💡 思路

    Let tens digit = $x$, ones = $y$. $x + y = 5$. Original = $10x + y$; reversed = $10y + x$.
    $10y + x = 10x + y + 9 \Rightarrow 9y - 9x = 9 \Rightarrow y - x = 1$.
    From two equations: $x + y = 5$, $y - x = 1$ → $2y = 6$ → $y = 3, x = 2$. Original = $\mathbf{23}$.

19. Q19. The price of each type of ticket at an amusement park:

    	Single ticket	Group tickets (buy 4 tickets or more)
    Adult	$30	$25
    Child	$15	$12

    Mrs Lee and her friends brought their children to the amusement park. Altogether, there were 8 people and the total cost of the tickets was $170. Is each of the following cases possible?
    (a) There are more children than adults. → 检查
    (b) There are more adults than children. → 检查
    If yes, find the number of adults and children. → adults: ; children: 检查

    💡 系统解析

    Let $a$ = adults, $c$ = children; $a + c = 8$. Group price applies when that category has $\geq 4$ tickets.
    (a) more children: Try $(0, 8), (1, 7), (2, 6), (3, 5)$:
    $(0, 8)$: $8 \times 12 = 96$.
    $(1, 7)$: $30 + 7 \times 12 = 114$.
    $(2, 6)$: $60 + 72 = 132$.
    $(3, 5)$: $90 + 60 = 150$.
    都 ≠ $170 → 不可能。
    (b) more adults:
    $(4, 4)$: $4 \times 25 + 4 \times 15 = 160$.
    $(5, 3)$: $5 \times 25 + 3 \times 15 = 125 + 45 = \mathbf{170}$ ✓
    $(6, 2)$: $150 + 30 = 180$.
    $(7, 1)$: $175 + 15 = 190$.
    $(8, 0)$: $200$.
    Only $(5, 3)$ matches. 5 adults + 3 children.