S4 E Maths Textbook Solutions >> think! Mathematics 4A (8th Edition) Chapter 1 Solutions >>
Review Ex 1
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Solutions
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(i)
\begin{align} 5 \in B \end{align}
(ii)
\begin{align} \{ 6 , 9 \} \not\subset A \end{align}
(iii)
\begin{align} \{ 4, 11, 16 \} \subset C' \end{align}
(iv)
\begin{align} B' \cup C' & = \xi \end{align}
(i)
\begin{align} A & = \{ 2, 4, 6, 8, 10, 12 \} \end{align}
(ii)(a)
$$ \text{False} $$
(ii)(b)
$$ \text{False} $$
(ii)(c)
$$ \text{True} $$
(ii)(d)
$$ \text{True} $$
(iii)(a)
\begin{align} -4 \notin A \end{align}
(iii)(b)
\begin{align} 6 \in A \end{align}
(iii)(c)
\begin{align} 0 \notin A \end{align}
(iii)(d)
\begin{align} 9 \notin A \end{align}
(a)
\begin{align} A & = \{ 2 \} \end{align}
(b)
\begin{align} B & = \{ \text{Saturday, Sunday} \} \end{align}
(c)
\begin{align} C & = \emptyset \phantom{000000} [\text{Factors of 24: 1, 2, 3, 4, 6, 8, 12, 24 (None multiples of 9)}] \end{align}
(d)
\begin{align} D & = \emptyset \phantom{000000} [360^\circ \div 4 = 90^\circ \implies \text{Not possible for four obtuse angles}] \end{align}
(i)
\begin{align} A' & = \{ -4, -2, 0, 1, 3 \} \end{align}
(ii)
\begin{align} B' & = \emptyset \end{align}
(iii)
\begin{align} C & = \{ 2, 3 \} \\ \\ C' & = \{ -5, -4, -3, -2, -1, 0, 1 \} \end{align}
(iv)
\begin{align} D & = \{ 3 \} \\ \\ D' & = \{ -5, -4, -3, -2, -1, 0, 1, 2 \} \end{align}
(a)
\begin{align} \text{True (since every element in } X \text{ is in } Y) \end{align}
(b)
\begin{align} \text{False} \end{align}
(c)
\begin{align} \text{True} \end{align}
(d)
\begin{align} \text{True} \end{align}
(i)
(ii)
\begin{align} A \cup B' & = \{ 4, 5, 7, 8, 10, 11, 12, 13, 14, 15, 16, 17, 19, 20, 21, 22, 23 \} \end{align}
(i)
\begin{align} \xi & = \{ -7, -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7 \} \\ \\ A & = \{ -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6 \} \\ \\ A' & = \{ -7, 7 \} \end{align}
(ii)
\begin{align} B & = \{ 1, 2, 3, 4, 5, 6, 7 \} \\ \\ A \cap B & = \{ 1, 2, 3, 4, 5, 6 \} \end{align}
(iii)
\begin{align} B' & = \{ -7, -6, -5, -4, -3, -2, -1, 0 \} \\ \\ A \cup B' & = \{ -7, -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6 \} \end{align}
(i)
(ii)
\begin{align} A' \cap B' & = \{ 3, 5, 6, 7, 8, 10, 11, 12, 14, 15 \} \end{align}
(i)
(ii)
\begin{align} A \cap B' & = \{ 5 \} \end{align}
\begin{align} S & = \{ \}, \{ \text{s} \}, \{ \text{i} \}, \{ \text{t} \}, \{ \text{s, i} \}, \{ \text{s, t} \}, \{ \text{i, t} \} \phantom{000000} [\{ \text{s, i, t} \} \text{ is not a proper subset}] \end{align}
Note: Rational number is a number that can be written as a fraction. Thus, all integers are rational numbers, i.e. $ 5 = {5 \over 1} = {10 \over 2} $
(i)
(ii)
(iii)
(a)
\begin{align} (P \cup Q)' \end{align}
(b)
\begin{align} P' \cup Q \end{align}
(c)
\begin{align} P \cup Q' \end{align}
(d)
\begin{align} P \cap Q \end{align}
\begin{align} \text{Let no. of people with both types of cars} & = x \\ \\ 6 - x + x + 19 - x + 30 & = 53 \\ -x + x - x & = 53 - 6 - 19 - 30 \\ -x & = - 2 \\ x & = 2 \\ \\ \text{No. of people with either electric or petrol cars, but not both} & = 6 - x + 19 - x \\ & = 25 - 2x \\ & = 25 - 2(2) \\ & = 21 \end{align}
(i)
\begin{align} \text{No. of students} & = 13 + 5 \\ & = 18 \end{align}
(ii)
\begin{align} \text{No. of students} & = 21 \phantom{000000} [\text{Refer to Venn diagram in (i)}] \end{align}
(iii)
\begin{align} \text{No. of students} & = 5 \end{align}
(iv)
\begin{align} \text{No. of students} & = 0 \phantom{000000} [\text{Refer to Venn diagram in (iii) - Every student who plays piano} \\ & \phantom{0000000000} \text{also plays guitar}] \end{align}