4 Marks Questions

Question 14 Marks

$A \cap(B-C)=(A \cap B)-(A \cap C)$

Answer

$\begin{array}{l}\text { Let } x \in\{A \cap(B-C)\} \\ x \in A \text { and } x \in B-C \\ x \in A \text { and } x \in B \text { and } x \notin C \\ (x \in A \text { and } x \in B) \text { and }(x \in A \text { and } x \notin C) \\ x \in A \cap B \text { and } x \notin A \cap C \\ x \in(A \cap B)-(A \cap C)\end{array}$
$A \cap(B-C) \subseteq(A \cap B)-(A \cap C) \ldots(i) $
Again, let
$
\begin{array}{l}
y \in(A \cap B) \cap(A-C) \\
y \in A \text { and }(y \in B \text { and } y \notin C) \\
y \in A \text { and } y \in B-C \\
y \in\{A \cap(B-C)\} \\
(A \cap B)-(A \cap C) \subseteq A \cap(B-C) \ldots(ii)
\end{array}
$
From eqs. (i) and (ii), we get
$
A \cap(B-C)=(A \cap B)-(A \cap C)
$
Hence Proved

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Question 24 Marks

Out of $2 5$ members in a family, 12 like to take tea, 15 like to take coffee and 7 like to take coffee and tea both. How many like (i) at least one of the two drinks (ii) only tea but not coffee (iii) only coffee but not tea (iv) neither tea nor coffee.

Answer

Given that,
$
\begin{aligned}
n(T) & =12 \\
n(C) & =15 \\
n(T \cap C) & =7
\end{aligned}
$

(i)
$
\begin{aligned}
n(T \cup C) & =n(T)+n(C)-n(T \cap C) \\
& =12+15-7 \\
n(T \cup C) & =20
\end{aligned}
$
20 members like at least one of the two drinks.

(ii) Only tea but not coffee
$
\begin{array}{l}
=n(T)-n(T \cap C) \\
=12-7 \\
=5
\end{array}
$

(iii) Only coffee but not tea
$
\begin{array}{l}
=n(C)-n(T \cup C) \\
=15-7 \\
=8
\end{array}
$

(iv) Neither tea nor coffee
$
\begin{array}{l}
=n(U)-n(T \cup C) \\
=25-20 \\
=5
\end{array}
$

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Question 34 Marks

Using properties of sets and their complements prove that:
(i) $(A \cup B) \cap\left(A \cap B^{\prime}\right)=A$
(ii) $A-(A \cap B)=A-B$

Answer

(i) $(A \cup B) \cap\left(A \cap B^{\prime}\right)=A$
$
\begin{array}{rlr}
\text { L. H. S. } & =(A \cup B) \cap\left(A \cap B^{\prime}\right) \\
& =A \cup\left(B \cap B^{\prime} \right)\quad(\text { By distributive law }) \\
& =A \cup \phi \ \left(B \cap B^{\prime}=\phi\right)\\
& =A \\
& =\text { R.H.S. } \text { Hence proved. }
\end{array}
$

(ii)
$
\begin{aligned}
A-(A \cap B)= & A-B \\
\text { L.H.S. }= & A-(A \cap B) \\
= & A \cap(A \cap B)^{\prime} \\
& \quad\left[\therefore A-B=A \cap B^{\prime}\right] \\
= & A \cap\left(A^{\prime} \cup B^{\prime}\right) \\
& (B y \text { Demorgan's law }) \\
= & \left(A \cap A^{\prime}\right) \cup\left(A \cap B^{\prime}\right) \\
& \quad(\text { By distributive law }) \\
= & \phi \cup A \cap B^{\prime}\left(\therefore A \cap A^{\prime}=\phi\right) \\
= & A \cap B^{\prime} \\
= & A-B \\
= & \text { R.H.S } \quad \text { Hence Proved }
\end{aligned}
$

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Question 44 Marks

If $(2 a+b, a-b)=(8,3)$, find $a$ and $b$.

Answer

We know that, two ordered pairs are equal, if their corresponding first and second elements are equal We have, $(2 a+b, a-b)=(8,3)$
$\therefore 2 a+b=8\quad \ldots(i)$
and $a-b=3\quad \ldots(ii)$
From eq. (ii), we get $b=a-3\quad \ldots(iii) $
On substituting, $b=a-3$, in eq. (i), we get
$
\begin{aligned}
\Rightarrow & & 2 a+(a-3) & =8 \\
\Rightarrow & & 3 a & =11 \\
\Rightarrow & & a & =\frac{11}{3}
\end{aligned}
$
Again, on substituting $a=\frac{11}{3}$ in eq. (iii), we get
$
\begin{array}{ll}
\Rightarrow & b=\frac{11}{3}-3 \\
\Rightarrow & b=\frac{2}{3} \\
\therefore & a=\frac{11}{3} \text { and } b=\frac{2}{3}
\end{array}
$

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Question 54 Marks

What is represented by the shaded regions in each of the following Venn-diagrams

Answer

(i)

(ii)

$\begin{array}{lr}\therefore & (A \cap B) \cup(A \cap C) \\ \text { or } & A \cap(B \cup C)\end{array}$

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Applied Maths 4 Marks Questions

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