There are 200 individuals with a skin disorder, 120 had been exposed to the chemical $C_1, 50$ to chemical $C_2$ and 30 to both the chemicals $C _1$ and $C _2$. Find the number of individuals exposed to (i) chemical $C _1$ but not chemical $C _2$ (ii) Chemical $C _2$ but not chemical $C _1$ (iii) Chemical $C _2$ or chemical $C _1$.
Answer
Let S denote the universal set consisting of individuals suffering from the skin disorder, A denote the set of individuals exposed to chemical $C _1$ and B denote the set of individuals exposed to chemical $C _2$. Now, n(S) = 200 n(A) = 120 n(B) = 50 and $n(A \cap B)=30$ i. Chemical $C _1$ but not chemical $C _2$ Number of individuals exposed to chemical $C _1$ but not chemical $C _2$ is $\begin{array}{l}=n\left(A \cap B^{\prime}\right) \\ = n ( A )-n(A \cap B) \\ =120-30=90\end{array}$ ii. Number of individuals exposed to chemical $C _2$ but not chemical $C _1$ $\begin{array}{l}=n\left(A^{\prime} \cap B\right) \\ = n ( B )- n ( A \cap B ) \\ =50-30=20\end{array}$ iii. Number of individuals exposed to chemical $C_1$ or chemical $C_2$ $\begin{array}{l}=n(A \cup B) \\ = n ( A )+ n ( B )- n ( A \cap B ) \\ =120+50-30 \\ =140\end{array}$
Let $x + yi =\sqrt{3-4 i}$ Squaring both sides, we get $x^2-y^2+2 x y i=3-4 i$ Equating the real and imaginary parts $x^2-y^2=3 \ldots . \text { (i) }$ and $2 x y=-4 \Rightarrow x y=-2$ Now from the identity, we know $\begin{array}{l}\left(x^2+y^2\right)=\left(x^2-y^2\right)^2+4 x^2 y^2 \\ \left(x^2+y^2\right)^2=(3)^2+4(-2)^2 \\ =9+16=25\end{array}$ $\therefore x^2+y^2=5 \ldots$ (ii) [Neglecting (-) sign as $\left.x^2+y^2>0\right]$ Solving (i) and (ii) we get $\begin{array}{l} x ^2=4 \text { and } y ^2=1 \\ x= \pm 2 \text { and } y= \pm 1\end{array}$ Since the sign of xy is negative $\begin{array}{l}\therefore \text { if, } x =2, y =-1 \\ \text { and if } x =-2, y =1 \\ \therefore \sqrt{-5+12 i}= \pm(2-i)\end{array}$
Find the point on the y-axis which is equidistant from the points A(3, 1, 2) and B(5, 5, 2).
Answer
Consider, C(0, y, 0) point which lies on the y-axis and is equidistant from points A(3, 1, 2) and B(5, 5, 2). $\begin{array}{l}\therefore AC = BC \\ \sqrt{(0-3)^2+(y-1)^2+(0-2)^2}=\sqrt{(0-5)^2+(y-5)^2+(0-2)^2}\end{array}$ Squaring both sides, $\begin{array}{l}\Rightarrow(0-3)^2+(y-1)^2+(0-2)^2=(0-5)^2+(y-5)^2+(0-2)^2 \\ \Rightarrow 9+y^2-2 y+1+4=25+y^2-10 y+25+4 \\ \Rightarrow 8 y=40 \\ \Rightarrow y=5\end{array}$ The coordinate of C is (0, 5, 0).
To receive Grade A, in a mathematics course, one must obtain an average of 90 marks or more in five examinations (each of 100 marks). If Ragini's marks in first four examinations are 87, 92, 94 and 95, find minimum marks that Ragini must obtain in fifth examination to get Grade A in the course.
Answer
Let the marks obtained by Ragini in fifth examination be x. Then average of five examinations $=\frac{87+92+94+95+x}{5}$ Now $\frac{87+92+94+95+x}{5} \geq 90 \Rightarrow \frac{368+x}{5} \geq 90$ Multiplying both sides by 5, we have $\begin{array}{l}368+x \geq 450 \\ \Rightarrow x \geq 450-368 \\ \Rightarrow x \geq 82\end{array}$ Thus the minimum marks needed to be obtained by Ragini = 82.
