Assume that each born child is equally likely to be a boy or a girl. If a family has two children,
Question: Assume that each born child is equally likely to be a boy or a girl. If a family has two children, what is the conditional probability that both are girls given that (i) the youngest is a girl, (ii) at least one is a girl? Solution: Letbandgrepresent the boy and the girl child respectively. If a family has two children, the sample space will be S = {(b,b), (b,g), (g,b), (g,g)} Let A be the event that both children are girls. $\therefore \mathrm{A}=\{(g, g)\}$ (i) Let B be the event tha...
Read More →The sides of a triangle are a = 4, b = 6 and c = 8.
Question: The sides of a triangle are $a=4, b=6$ and $c=8$. Show that $8 \cos A+16 \cos B+4 \cos C=17$. Solution: Given: $a=4, b=6$ and $c=8$. Then, $\cos B=\frac{a^{2}+c^{2}-b^{2}}{2 a c}=\frac{16+64-36}{2 \times 4 \times 8}=\frac{11}{16}$ $\cos A=\frac{b^{2}+c^{2}-a^{2}}{2 b c}=\frac{36+64-16}{2 \times 6 \times 8}=\frac{7}{8}$ $\cos C=\frac{b^{2}+a^{2}-c^{2}}{2 a b}=\frac{16+36-64}{2 \times 4 \times 6}=\frac{-1}{4}$ Now, $8 \cos A+16 \cos B+4 \cos C=8 \times \frac{7}{8}+16 \times \frac{11}{16}...
Read More →Examine whether the following numbers are rational or irrational:
Question: Examine whether the following numbers are rational or irrational: (i) $(5-\sqrt{5})(5+\sqrt{5})$ (ii) $(\sqrt{3}+2)^{2}$ (iii) $\frac{2 \sqrt{13}}{3 \sqrt{52}-4 \sqrt{117}}$ (iv) $\sqrt{8}+4 \sqrt{32}-6 \sqrt{2}$ Solution: (i) $(5-\sqrt{5})(5+\sqrt{5})$ $=5^{2}-(\sqrt{5})^{2} \quad\left[(a-b)(a+b)=a^{2}-b^{2}\right]$ $=25-5$ $=20$, which is an integer Hence, $(5-\sqrt{5})(5+\sqrt{5})$ is rational. (ii) $(\sqrt{3}+2)^{2}$ $=(\sqrt{3})^{2}+2^{2}+2 \times \sqrt{3} \times 2 \quad\left[(a+b...
Read More →If cos θ=23, then 2 sec2 θ + 2 tan2 θ − 7 is equal to
Question: If $\cos \theta=\frac{2}{3}$, then $2 \sec ^{2} \theta+2 \tan ^{2} \theta-7$ is equal to (a) 1(b) 0(c) 3(d) 4 Solution: Given that: $\cos \theta=\frac{2}{3}$ We have to find $2 \sec ^{2} \theta+2 \tan ^{2} \theta-7$ As we are given $\cos \theta=\frac{2}{3}$ $\Rightarrow$ Base $=2$ $\Rightarrow$ Hypotenuse $=3$ $\Rightarrow$ Perpendicular $=\sqrt{(3)^{2}-(2)^{2}}$ $\Rightarrow$ Perpendicular $=\sqrt{5}$ We know that: $\cos \theta=\frac{\text { Base }}{\text { Hypotenuse }}$ $\tan \theta...
Read More →In ΔABC,
Question: In $\Delta A B C$, if $a=\sqrt{2}, b=\sqrt{3}$ and $c=\sqrt{5}$, show that its area is $\frac{1}{2} \sqrt{6}$ sq. units. Solution: Given: $a=\sqrt{2}, b=\sqrt{3}, c=\sqrt{5}$ $\because \cos C=\frac{a^{2}+b^{2}-c^{2}}{2 a b}$ $\Rightarrow \cos C=\frac{2+3-5}{2 \times \sqrt{6}}=0$ $\Rightarrow \cos C=0$ $\Rightarrow \cos C=\cos 90^{\circ}$ $\Rightarrow C=90^{\circ}$ Thus, $\sin C=\sin 90^{\circ}=1$ Hence, Area of $\Delta A B C=\frac{1}{2} a b \sin C=\frac{1}{2} \sqrt{6} \times 1=\frac{\s...
Read More →In ΔABC,
Question: In $\Delta A B C$, if $a=5, b=6$ and $C=60^{\circ}$, show that its area is $\frac{15 \sqrt{3}}{2}$ sq. units. Solution: Given : $a=5, b=6, c=60^{\circ}$ Area of a triangle $=\frac{1}{2} a b \sin C$ $=\frac{1}{2} \times 5 \times 6 \times \sin 60^{\circ}=15 \times \frac{\sqrt{3}}{2}$ sq. units hence proved....
Read More →A fair die is rolled. Consider events
Question: A fair die is rolled. Consider events E = {1, 3, 5}, F = {2, 3} and G = {2, 3, 4, 5} Find Solution: When a fair die is rolled, the sample space S will be S = {1, 2, 3, 4, 5, 6} It is given that E = {1, 3, 5}, F = {2, 3}, and G = {2, 3, 4, 5} $\therefore P(E)=\frac{3}{6}=\frac{1}{2}$ $\mathrm{P}(\mathrm{F})=\frac{2}{6}=\frac{1}{3}$ $P(G)=\frac{4}{6}=\frac{2}{3}$ (i) $E \cap F=\{3\}$ $\therefore P(E \cap F)=\frac{1}{6}$ $\therefore P(E \mid F)=\frac{P(E \cap F)}{P(F)}=\frac{\frac{1}{6}}{...
Read More →If the sides a, b and c of ∆ABC are in H.P.,
Question: If the sides $a, b$ and $c$ of $\triangle A B C$ are in H.P., prove that $\sin ^{2} \frac{A}{2}, \sin ^{2} \frac{B}{2}$ and $\sin ^{2} \frac{C}{2}$ are in H.P. Solution: $\sin ^{2} \frac{A}{2}, \sin ^{2} \frac{B}{2}$ and $\sin ^{2} \frac{C}{2}$ is a H.P. $\Leftrightarrow \frac{1}{\sin ^{2} \frac{A}{2}}, \frac{1}{\sin ^{2} \frac{B}{2}}$ and $\frac{1}{\sin ^{2} \frac{C}{2}}$ is an A.P. $\Leftrightarrow \frac{1}{\sin ^{2} \frac{B}{2}}-\frac{1}{\sin ^{2} \frac{A}{2}}=\frac{1}{\sin ^{2} \fr...
Read More →If A, B and C are interior angles of a triangle ABC, then sin (B+C2)=
Question: If $A, B$ and $C$ are interior angles of a triangle $A B C$, then $\sin \left(\frac{B+C}{2}\right)=$ (a) $\sin \frac{A}{2}$ (b) $\cos \frac{A}{2}$ (c) $-\sin \frac{A}{2}$ (d) $-\cos \frac{A}{2}$ Solution: We know that in triangle $A B C$ $A+B+C=180^{\circ}$ $\Rightarrow B+C=180^{\circ}-A$ $\Rightarrow \frac{B+C}{2}=\frac{90^{\circ}}{2}-\frac{A}{2}$ $\Rightarrow \sin \left(\frac{B+C}{2}\right)=\sin \left(90^{\circ}-\frac{A}{2}\right)$ Since $\sin \left(90^{\circ}-A\right)=\cos A$ So $\s...
Read More →Simplify
Question: Simplify $(3+\sqrt{3})(2+\sqrt{2})^{2}$. Solution: $(3+\sqrt{3})(2+\sqrt{2})^{2}$ $=(3+\sqrt{3})\left[2^{2}+(\sqrt{2})^{2}+2 \times 2 \sqrt{2}\right]$ $=(3+\sqrt{3})[4+2+4 \sqrt{2}]$ $=(3+\sqrt{3})[6+4 \sqrt{2}]$ $=3 \times 6+3 \times 4 \sqrt{2}+\sqrt{3} \times 6+\sqrt{3} \times 4 \sqrt{2}$ $=18+12 \sqrt{2}+6 \sqrt{3}+4 \sqrt{6}$...
Read More →2 tan 30°1−tan2 30° is equal to
Question: $\frac{2 \tan 30^{\circ}}{1-\tan ^{2} 30^{\circ}}$ is equal to (a) $\cos 60^{\circ}$ (b) $\sin 60^{\circ}$ (c) $\tan 60^{\circ}$ (d) $\sin 30^{\circ}$ Solution: We are asked to find the value of the following $\frac{2 \tan 30^{\circ}}{1-\tan ^{2} 30^{\circ}}$ $=\frac{2 \tan 30^{\circ}}{1-\tan ^{2} 30^{\circ}}$ $=\frac{2 \times \frac{1}{\sqrt{3}}}{1-\left(\frac{1}{\sqrt{3}}\right)^{2}}$ $=\frac{\frac{2}{\sqrt{3}}}{1-\frac{1}{3}}$ $=\frac{\frac{2}{\sqrt{3}}}{\frac{2}{3}}$ We know that $\...
Read More →A person observes the angle of elevation of the peak of a hill from a station to be α.
Question: A person observes the angle of elevation of the peak of a hill from a station to be $\alpha$. He walks $c$ metres along a slope inclined at an angle $\beta$ and finds the angle of elevation of the peak of the hill to be $Y$. Show that the height of the peak above the ground is $\frac{c \sin \alpha \sin (\gamma-\beta)}{(\sin \gamma-\alpha)}$. Solution: Suppose, AB is a peak whose height above the ground ist+x. In $\triangle D F C$, $\sin \beta=\frac{x}{c}$ $\Rightarrow x=c \sin \beta$ a...
Read More →A black and a red dice are rolled.
Question: A black and a red dice are rolled. (a) Find the conditional probability of obtaining a sum greater than 9, given that the black die resulted in a 5. (b) Find the conditional probability of obtaining the sum 8, given that the red die resulted in a number less than 4. Solution: Let the first observation be from the black die and second from the red die. When two dice (one black and another red) are rolled, the sample space S has 6 6 = 36 number of elements. 1. Let A: Obtaining a sum grea...
Read More →Sin 2A = 2 sin A is true when A =
Question: Sin 2A = 2 sin A is true when A = (a) $0^{\circ}$ (b) $30^{\circ}$ (c) $45^{\circ}$ (d) $60^{\circ}$ Solution: We are given $\sin 2 A=2 \sin A \cdot \cos A$ So $\Rightarrow \sin 2 A=2 \sin A$ $\Rightarrow 2 \sin A \cdot \cos A=2 \sin A$ $\Rightarrow \cos A=1$ $\Rightarrow \cos A=\cos 0^{\circ}$ As $A=0^{\circ}$ Hence the correct option is (a)...
Read More →Simplify
Question: Simplify (i) $(3-\sqrt{11})(3+\sqrt{11})$ (ii) $(-3+\sqrt{5})(-3-\sqrt{5})$ (iii) $(3-\sqrt{3})^{2}$ (iv) $(\sqrt{5}-\sqrt{3})^{2}$ (v) $(5+\sqrt{7})(2+\sqrt{5})$ (vi) $(\sqrt{5}-\sqrt{2})(\sqrt{2}-\sqrt{3})$ Solution: (i) $(3-\sqrt{11})(3+\sqrt{11})$ $=3^{2}-(\sqrt{11})^{2} \quad\left[(a-b)(a+b)=a^{2}-b^{2}\right]$ $=9-11$ $=-2$ (ii) $(-3+\sqrt{5})(-3-\sqrt{5})$ $=(-3)^{2}-(\sqrt{5})^{2} \quad\left[(a+b)(a-b)=a^{2}-b^{2}\right]$ $=9-5$ $=4$ (iii) $(3-\sqrt{3})^{2}$ $=3^{2}+(\sqrt{3})^...
Read More →1−tan2 45°1+tan2 45° is equal to
Question: $\frac{1-\tan ^{2} 45^{\circ}}{1+\tan ^{2} 45^{\circ}}$ is equal to (a) $\tan 90^{\circ}$ (b) 1 (c) $\sin 45^{\circ}$ (d) $\sin 0^{\circ}$ Solution: We have to find the value of the following $\frac{1-\tan ^{2} 45^{\circ}}{1+\tan ^{2} 45^{\circ}}$ So $\frac{1-\tan ^{2} 45^{\circ}}{1+\tan ^{2} 45^{\circ}}$ $=\frac{1-(1)^{2}}{1+(1)^{2}}$ $=\frac{0}{1}$ $=0$ We know that $\left[\begin{array}{r}\tan 45^{\circ}=1 \\ \sin 0^{\circ}=0\end{array}\right]$ $=\sin 0^{\circ}$ Hence the correct opt...
Read More →Mother, father and son line up at random for a family picture
Question: Mother, father and son line up at random for a family picture E: son on one end, F: father in middle Solution: If mother (M), father (F), and son (S) line up for the family picture, then the sample space will be S = {MFS, MSF, FMS, FSM, SMF, SFM} $\Rightarrow \mathrm{E}=\{\mathrm{MFS}, \mathrm{FMS}, \mathrm{SMF}, \mathrm{SFM}\}$ F = {MFS, SFM} $\therefore \mathrm{E} \cap \mathrm{F}=\{\mathrm{MFS}, \mathrm{SFM}\}$ $\therefore E \cap F=\{M F S, S F M\}$ $P(E \cap F)=\frac{2}{6}=\frac{1}{...
Read More →At the foot of a mountain, the elevation of it summit is 45°;
Question: At the foot of a mountain, the elevation of it summit is 45; after ascending 1000 m towards the mountain up a slope of 30 inclination, the elevation is found to be 60. Find the height of the mountain. Solution: Suppose, AB is a mountain of heightt + x. In $\triangle D F C$, $\sin 30^{\circ}=\frac{x}{1000}$ $\Rightarrow x=1000 \times\left(\frac{1}{2}\right)=500 \mathrm{~m}$ and, $\tan 30^{\circ}=\frac{x}{y}$ $\Rightarrow y=\frac{x}{\tan 30^{\circ}}=500 \sqrt{3}$ In $\Delta A B C$, $\tan...
Read More →2 tan 30°1+tan2 30° is equal to
Question: $\frac{2 \tan 30^{\circ}}{1+\tan ^{2} 30^{\circ}}$ is equal to (a) $\sin 60^{\circ}$ (b) $\cos 60^{\circ}$ (c) $\tan 60^{\circ}$ (d) $\sin 30^{\circ}$ Solution: We have to find the value of the following expression $\frac{2 \tan 30^{\circ}}{1+\tan ^{2} 30^{\circ}}$ $\frac{2 \tan 30^{\circ}}{1+\tan ^{2} 30^{\circ}}$ $=\frac{2 \times \frac{1}{\sqrt{3}}}{1+\left(\frac{1}{\sqrt{3}}\right)^{2}}$ $=\frac{\frac{2}{\sqrt{3}}}{1+\frac{1}{3}}$ $=\frac{\frac{2}{\sqrt{3}}}{\frac{4}{3}}$ $\left[\be...
Read More →A die is thrown three times,
Question: A die is thrown three times, E: 4 appears on the third toss, F: 6 and 5 appears respectively on first two tosses Solution: If a die is thrown three times, then the number of elements in the sample space will be 6 6 6 = 216 $\mathrm{E}=\left\{\begin{array}{l}(1,1,4),(1,2,4), \ldots(1,6,4) \\ (2,1,4),(2,2,4), \ldots(2,6,4) \\ (3,1,4),(3,2,4), \ldots(3,6,4) \\ (4,1,4),(4,2,4), \ldots(4,6,4) \\ (5,1,4),(5,2,4), \ldots(5,6,4) \\ (6,1,4),(6,2,4), \ldots(6,6,4)\end{array}\right\}$ $\mathrm{F}...
Read More →The upper part of a tree broken by the wind makes an angle of 30° with the ground and the distance from the root to the point where the top of the tree touches the ground is 15 m.
Question: The upper part of a tree broken by the wind makes an angle of 30 with the ground and the distance from the root to the point where the top of the tree touches the ground is 15 m. Using sine rule, find the height of the tree Solution: Suppose BD be the tree and the upper part of the tree is broken over by the wind at point A. The total height of the tree is $x+y$. In $\triangle A B C$ $\angle C=30^{\circ}$ and $\angle B=90^{\circ} .$ $\therefore \angle A=60^{\circ} .$ So, on using sine ...
Read More →If A + B = 90°, then tan A tan B+tan A cot Bsin A sec B−sin2 Bcos2 A is equal to
Question: If $A+B=90^{\circ}$, then $\frac{\tan A \tan B+\tan A \cot B}{\sin A \sec B}-\frac{\sin ^{2} B}{\cos ^{2} A}$ is equal to (a) $\cot ^{2} \mathrm{~A}$ (b) $\cot ^{2} \mathrm{~B}$ (c) $-\tan ^{2} A$ (d) $-\cot ^{2} A$ Solution: We have: $A+B=90^{\circ}$ $\Rightarrow B=90^{\circ}-A$ We have to find the value of the following expression $\frac{\tan A \tan B+\tan A \cot B}{\sin A \sec B}-\frac{\sin ^{2} B}{\cos ^{2} A}$ So $\frac{\tan A \tan B+\tan A \cot B}{\sin A \sec B}-\frac{\sin ^{2} B...
Read More →Divide:
Question: Divide: (i) $16 \sqrt{6}$ by $4 \sqrt{2}$ (ii) $12 \sqrt{5}$ by $4 \sqrt{3}$ (iii) $18 \sqrt{21}$ by $6 \sqrt{7}$ Solution: (i) $\frac{16 \sqrt{6}}{4 \sqrt{2}}=\frac{16 \sqrt{2} \sqrt{3}}{4 \sqrt{2}}=4 \sqrt{3}$ (ii) $\frac{12 \sqrt{15}}{4 \sqrt{3}}=\frac{12 \sqrt{5} \times \sqrt{3}}{4 \sqrt{3}}=3 \sqrt{5}$ (iii) $\frac{18 \sqrt{21}}{6 \sqrt{7}}=\frac{18 \sqrt{7} \sqrt{3}}{6 \sqrt{7}}=3 \sqrt{3}$...
Read More →If 5θ and 4θ are acute angles satisfying sin 5θ = cos 4θ,
Question: If $5 \theta$ and $4 \theta$ are acute angles satisfying $\sin 5 \theta=\cos 4 \theta$, then $2 \sin 3 \theta-\sqrt{3} \tan 3 \theta$ is equal to (a) 1 (b) 0 (c) $-1$ (d) $1+\sqrt{3}$ Solution: We are given that $5 \theta$ and $4 \theta$ are acute angles satisfying the following condition $\sin 5 \theta=\cos 4 \theta$. We are asked to find $2 \sin 3 \theta-\sqrt{3} \tan 3 \theta$ $\Rightarrow \sin 5 \theta=\cos 4 \theta$ $\Rightarrow \cos \left(90^{\circ}-5 \theta\right)=\cos 4 \theta$...
Read More →Two coins are tossed once, where
Question: Two coins are tossed once, where (i) E: tail appears on one coin, F: one coin shows head (ii) E: not tail appears, F: no head appears Solution: If two coins are tossed once, then the sample space S is S = {HH, HT, TH, TT} (i) E = {HT, TH} F = {HT, TH} $\therefore \mathrm{E} \cap \mathrm{F}=\{\mathrm{HT}, \mathrm{TH}\}$ $\mathrm{P}(\mathrm{F})=\frac{2}{8}=\frac{1}{4}$ $\mathrm{P}(\mathrm{E} \cap \mathrm{F})=\frac{2}{8}=\frac{1}{4}$ $\therefore P(E \mid F)=\frac{P(E \cap F)}{P(F)}=\frac{...
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