S4 E Maths Textbook Solutions >> think! Mathematics 4A (8th Edition) Chapter 1 Solutions >>
Chapter 1 Practise Now
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Solutions
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(i)
\begin{align} A & = \{ 2, 4, 6, 8 \} \end{align}
(ii)(a)
\begin{align} \text{True} \end{align}
(ii)(b)
\begin{align} \text{True} \end{align}
(ii)(c)
\begin{align} \text{False} \end{align}
(ii)(d)
\begin{align} \text{True} \end{align}
(iii)(a)
\begin{align} 2 & \in A \end{align}
(iii)(b)
\begin{align} 5 \notin A \end{align}
(iii)(c)
\begin{align} 11 \notin A \end{align}
(iii)(d)
\begin{align} 6 & \in A \end{align}
(iv)
\begin{align} n(A) & = 4 \end{align}
\begin{align} B & = \{ 3, 6, 9, 12, 15, 18, 21, 24, 27, 30 \} \\ \\ n(B) & = 10 \end{align}
(i)
\begin{align} C & = \{ 11, 12, 13, 14, 15, 16, 17 \} \\ \\ D & = \{ 10, 11, 12, 13, 14, 15, 16, 17 \} \end{align}
(ii)
\begin{align} n(C) & = 7 \\ \\ n(D) & = 8 \\ \\ \therefore n(C) & \ne n(D) \end{align}
(iii)
\begin{align} \text{No, since set } D \text{ has an additional element 10} \end{align}
(i)
\begin{align} P & = \emptyset \phantom{000000} [\text{Consonants are non-vowels, i.e. letters that are not A E I O U}] \\ \\ Q & = \emptyset \phantom{000000} [\text{Prime numbers: } 2, 3, 5, 7, 11, 13, ...] \\ \\ [R & \text{ is not empty since it contains the symbol } \emptyset ] \end{align}
(ii)
\begin{align} & \text{Yes, since both sets do not contain any element} \end{align}
(iii)
\begin{align} & \text{No, since } Q \text{ is an empty set while } R \text{ contains 1 element} \end{align}
(i)
\begin{align} \xi & = \{ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13 \} \\ \\ B & = \{ 2, 3, 5, 7, 11, 13 \} \end{align}
(ii)
(iii)
\begin{align} B' & = \{ 1, 4, 6, 8, 9, 10, 12 \} \end{align}
(iv)
\begin{align} B' \text{ is the set of integers between 1 and 13 inclusive that are not prime numbers} \end{align}
(v)
\begin{align} n (\xi) & = 13 \\ \\ n (B) & = 6 \\ \\ n (B') & = 7 \end{align}
(vi)
\begin{align} \text{Yes, since elements in } B \text{ and } B' \text{ are in the universal set } \xi \end{align}
(i)
(ii)
\begin{align} \text{Yes, since all elements in } D \text{ are in } C \text{ and there are more elements in } C \text{ than in } D \end{align}
(iii)
\begin{align} \text{Yes, since some elements in } D \text{ (2, 4, 6) are not in } C \end{align}
(a)
\begin{align} \text{True} \end{align}
(b)
\begin{align} \text{True} \phantom{000000} [\text{Since both sets have the same elements}] \end{align}
(c)
\begin{align} \text{True} \phantom{000000} [\text{Since both sets have the same elements}] \end{align}
(d)
\begin{align} \text{False} \end{align}
(i)
\begin{align} P & = \{ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13 \} \\ \\ Q & = \{ 2, 3, 5, 7, 11 \} \end{align}
(ii)
\begin{align} Q & \subset P \\ \\ \text{Every element in } Q \text{ is in } P & \text{ and there are elements in } P \text{ that are not in } Q \end{align}
(iii)
\begin{align} R & = \{ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13 \} \end{align}
(iv)
\begin{align} P \subseteq R \text{ and } R & \subseteq P \\ \\ \text{Every element in } P \text{ are in } R & \text{ (and vice versa)} \end{align}
(a)
\begin{align} \{ \}, \{7 \} , \{ 8 \}, \{ 7, 8 \} \end{align}
(b)
\begin{align} \{ \}, \{ \text{a} \}, \{ \text{b} \}, \{ \text{c} \}, \{ \text{a, b} \}, \{ \text{b, c} \}, \{ \text{a, c} \}, \{ \text{a, b, c} \} \end{align}
(i)
\begin{align} C \cap D & = \{ \text{e, g} \} \end{align}
(ii)
(i)
\begin{align} E & = \{ 6, 12, 18 \} \\ \\ F & = \{ 3, 6, 9, 12, 15, 18 \} \end{align}
(ii)
\begin{align} E \cap F & = \{ 6, 12, 18 \} \end{align}
(iii)
(iv)
\begin{align} \text{Yes, since } E \text{ is a proper subset of } F \end{align}
(i)
\begin{align} G & = \{ 1, 2, 3, 4, 6, 12 \} \\ \\ H & = \{ 5, 7, 11, 13 \} \end{align}
(ii)
\begin{align} G \cap H & = \emptyset \\ \\ \text{Prime number only has } & \text{two factors, 1 and itself} \end{align}
(iii)
(i)
(ii)
\begin{align} C \cup D & = \{ \text{p, q, r, s, t, u, v} \} \end{align}
(i)
\begin{align} E & = \{ 1, 2, 4, 8 \} \\ \\ F & = \{ 1, 2, 4, 8, 16 \} \end{align}
(ii)
(iii)
\begin{align} E \cup F & = \{ 1, 2, 4, 8, 16 \} \end{align}
(iv)
\begin{align} \text{Yes, since } E \text{ is a proper subset of } F \end{align}
(i)
\begin{align} G & = \{ 7, 14, 21, 28, 35, 42, 49, 56 \} \\ \\ H & = \{ 9, 18, 27, 36, 45, 54 \} \end{align}
(ii)
(iii)
\begin{align} G \cup H & = \{ 7, 9, 14, 18, 21, 27, 28, 35, 36, 42, 45, 49, 54, 56 \} \end{align}
(i)
\begin{align} \xi & = \{ 1, 2, 3, 4, 5, 6, 7, 8, 9 \} \\ \\ A & = \{ 1, 3, 5, 7, 9 \} \\ \\ B & = \{ 3, 6, 9 \} \end{align}
(ii)
(iii)(a)
\begin{align} (A \cup B)' & = \{ 2, 4, 8 \} \end{align}
(iii)(b)
\begin{align} A \cap B' & = \{ 1, 5, 7 \} \end{align}
\begin{align} 3^2 + 4^2 & = 25 \\ 5^2 & = 25 \\ \\ \text{By the converse of Pytha} & \text{goras theorem, }E \text{ is a right-angled triangle} \end{align}
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(a)
$$ X' \cap Y $$
(b)
$$ (X' \cap Y) \cup (X \cap Y') $$
(c)
$$ Y' $$
(d)
$$ X \cup Y' $$
\begin{align} \text{No. of students} & = 38 - 3 - 8 - 12 \\ & = 15 \end{align}
\begin{align} \text{No. of students} & = 117 - 65 - 23 \\ & = 29 \end{align}
(i)
\begin{align} \text{Largest value of } n(A \cup B)' & = 23 - 15 \\ & = 8 \end{align}
(ii)
\begin{align} 15 + 11 & = 26 \\ \\ 26 - 23 & = 3 \end{align}
\begin{align} \text{Smallest value of } n(A \cup B)' & = 0 \end{align}
(i)
\begin{align} \text{Largest possible no. of students} & = 8 + 11 \\ & = 19 \end{align}
(ii)
\begin{align} \text{Smallest possible no. of students} & = 36 - 19 \phantom{000000} [\text{Refer to Venn diagram in (i)}] \\ & = 17 \end{align}
(iii)
\begin{align} \text{Largest possible no. of students} & = 8 \end{align}
(iv)
\begin{align} \text{Smallest possible no. of students} & = 11 - 8 \phantom{000000} [\text{Refer to Venn diagram in (iii)}] \\ & = 3 \end{align}