One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting
Question: One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting (i) a king of red colour (ii) a face card (iii) a red face card (iv) the jack of hearts (v) a spade (vi) the queen of diamonds Solution: Total number of cards in a well-shuffled deck = 52 (i) Total number of kings of red colour = 2 $P$ (getting a king of red colour) $=\frac{\text { Number of favourable outcomes }}{\text { Number of total possible outcomes }}$ $=\frac{2}{52}=\frac{1}{26}$ $=\frac{2}...
Read More →Find the principal and general solutions of the equation
Question: Find the principal and general solutions of the equation Solution: $\tan x=\sqrt{3}$ It is known that $\tan \frac{\pi}{3}=\sqrt{3}$ and $\tan \left(\frac{4 \pi}{3}\right)=\tan \left(\pi+\frac{\pi}{3}\right)=\tan \frac{\pi}{3}=\sqrt{3}$ Therefore, the principal solutions are $x=\frac{\pi}{3}$ and $\frac{4 \pi}{3}$. Now, $\tan x=\tan \frac{\pi}{3}$ $\Rightarrow \mathrm{x}=\mathrm{n} \pi+\frac{\pi}{3}$, where $\mathrm{n} \in \mathrm{Z}$ Therefore, the general solution is $\mathrm{x}=\math...
Read More →The value of Kc for the reaction
Question: The value of $K_{c}$ for the reaction $3 \mathrm{O}_{2}(\mathrm{~g}) \longleftrightarrow 2 \mathrm{O}_{3}(\mathrm{~g})$ is $2.0 \times 10^{-50}$ at $25^{\circ} \mathrm{C}$. If the equilibrium concentration of $\mathrm{O}_{2}$ in air at $25^{\circ} \mathrm{C}$ is $1.6 \times 10^{-2}$, what is the concentration of $\mathrm{O}_{3} ?$ Solution: The given reaction is: $3 \mathrm{O}_{2(\mathrm{~g})} \longleftrightarrow 2 \mathrm{O}_{3(\mathrm{~g})}$ Then, $K_{C}=\frac{\left[\mathrm{O}_{3(\ma...
Read More →A die is thrown once. Find the probability of getting
Question: A die is thrown once. Find the probability of getting (i) a prime number; (ii) a number lying between 2 and 6; (iii) an odd number. Solution: The possible outcomes when a dice is thrown = {1, 2, 3, 4, 5, 6} Number of possible outcomes of a dice = 6 (i) Prime numbers on a dice are 2, 3, and 5. Total prime numbers on a dice = 3 Probability of getting a prime number $=\frac{3}{6}=\frac{1}{2}$ (ii) Numbers lying between 2 and 6 = 3, 4, 5 Total numbers lying between 2 and 6 = 3 Probability ...
Read More →Prove that: cos 6x = 32 cos6 x – 48 cos4 x + 18 cos2 x – 1
Question: Prove that $\cos 6 x=32 \cos ^{6} x-48 \cos ^{4} x+18 \cos ^{2} x-1$ Solution: L.H.S. $=\cos 6 x$ $=\cos 3(2 x)$ $=4 \cos ^{3} 2 x-3 \cos 2 x\left[\cos 3 A=4 \cos ^{3} A-3 \cos A\right]$ $=4\left[\left(2 \cos ^{2} x-1\right)^{3}-3\left(2 \cos ^{2} x-1\right)\left[\cos 2 x=2 \cos ^{2} x-1\right]\right.$ $=4\left[\left(2 \cos ^{2} x\right)^{3}-(1)^{3}-3\left(2 \cos ^{2} x\right)^{2}+3\left(2 \cos ^{2} x\right)\right]-6 \cos ^{2} x+3$ $=4\left[8 \cos ^{6} x-1-12 \cos ^{4} x+6 \cos ^{2} x\...
Read More →A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers
Question: A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8 (see the given figure), and these are equally likely outcomes. What is the probability that it will point at (i) 8? (ii) an odd number? (iii) a number greater than 2? (iv) a number less than 9? Solution: Total number of possible outcomes = 8 (i) Probability of getting $8=\frac{\text { Number of favourable outcomes }}{\text { Number of total possible outcomes }}=\frac...
Read More →Prove that cos 4x = 1 – 8sin2 x cos2 x
Question: Prove that $\cos 4 x=1-8 \sin ^{2} x \cos ^{2} x$ Solution: L.H.S. $=\cos 4 x$ $=\cos 2(2 x)$ $=1-12 \sin ^{2} 2 x\left[\cos 2 A=1-2 \sin ^{2} A\right]$ $=1-2(2 \sin x \cos x)^{2}[\sin 2 A=2 \sin A \cos A]$ $=1-8 \sin ^{2} x \cos ^{2} x$ $=$ R.H.S....
Read More →Predict which of the following reaction will have appreciable concentration of reactants and products:
Question: Predict which of the following reaction will have appreciable concentration of reactants and products: Solution: If the value of $K_{c}$ lies between $10^{-3}$ and $10^{3}$, a reaction has appreciable concentration of reactants and products. Thus, the reaction given in (c) will have appreciable concentration of reactants and products....
Read More →Prove that the square of any positive integer is of the form 3m or, 3m + 1 but not of the form 3m + 2.
Question: Prove that the square of any positive integer is of the form 3mor, 3m+ 1 but not of the form 3m+ 2. Solution: To Prove: that the square of an positive integer is of the form 3mor 3m+ 1 but not of the form 3m+ 2. Proof: Since positive integernis of the form of 3q, 3q+ 1 and 3q+ 2 Ifn= 3q $\Rightarrow n^{2}=(3 q)^{2}$ $\Rightarrow n^{2}=9 q^{2}$ $\Rightarrow n^{2}=3\left(3 q^{2}\right)$ $\Rightarrow n^{2}=3 m\left(m=3 q^{2}\right)$ Ifn= 3q+ 1 Then, $n^{2}=(3 q+1)^{2}$ $\Rightarrow n^{2}=...
Read More →Determine whether or not each of the definition of given below gives a binary operation.
Question: Determine whether or not each of the definition of given below gives a binary operation. In the event that * is not a binary operation, give justification for this. (i) On $\mathbf{Z}^{+}$, define ${ }^{*}$ by $a{ }^{*} b=a-b$ (ii) $\mathrm{On} \mathbf{Z}^{+}$, define ${ }^{*}$ by $a^{*} b=a b$ (iii) On $\mathbf{R}$, define * ${ }^{*}$ by $a^{*} b=a b^{2}$ (iv) On $\mathbf{Z}^{+}$, define ${ }^{*}$ by $a^{*} b=|a-b|$ (v) On $\mathbf{Z}^{+}$, define ${ }^{*}$ by $a^{*} b=a$ Solution: (i...
Read More →Gopi buys a fish from a shop for his aquarium.
Question: Gopi buys a fish from a shop for his aquarium. The shopkeeper takes out one fish at random from a tank containing 5 male fish and 8 female fish (see the given figure). What is the probability that the fish taken out is a male fish? Solution: Total number of fishes in a tank = Number of male fishes + Number of female fishes = 5 + 8 = 13 Probability of getting a male fish $=\frac{\text { Number of favourable outcomes }}{\text { Number of total possible outcomes }}$ $=\frac{5}{13}$...
Read More →Dihydrogen gas used in Haber’s process is produced by reacting methane from natural gas with high temperature steam.
Question: Dihydrogen gas used in Habers process is produced by reacting methane from natural gas with high temperature steam. The first stage of two stage reaction involves the formation of CO and H2. In second stage, CO formed in first stage is reacted with more steam in water gas shift reaction, $\mathrm{CO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \longleftrightarrow \mathrm{CO}_{2}(\mathrm{~g})+\mathrm{H}_{2}(\mathrm{~g})$ If a reaction vessel at $400^{\circ} \mathrm{C}$ is charged ...
Read More →A piggy bank contains hundred 50 p coins
Question: A piggy bank contains hundred 50 p coins, fifty Rs 1 coins, twenty Rs 2 coins and ten Rs 5 coins. If it is equally likely that one of the coins will fall out when the bank is turned upside down, what is the probability that the coin (i) Will be a 50 p coin? (ii) Will not be a Rs.5 coin? Solution: Total number of coins in a piggy bank = 100 + 50 + 20 + 10 = 180 (i) Number of 50 p coins = 100 Probability of getting a $50 \mathrm{pcoin}=\frac{\text { Number of favourable outcomes }}{\text...
Read More →Prove that cot x cot 2x – cot 2x cot 3x – cot 3x cot x = 1
Question: Prove that $\cot x \cot 2 x-\cot 2 x \cot 3 x-\cot 3 x \cot x=1$ Solution: L.H.S. $=\cot x \cot 2 x-\cot 2 x \cot 3 x-\cot 3 x \cot x$ $=\cot x \cot 2 x-\cot 3 x(\cot 2 x+\cot x)$ $=\cot x \cot 2 x-\cot (2 x+x)(\cot 2 x+\cot x)$ $=\cot x \cot 2 x-\left[\frac{\cot 2 x \cot x-1}{\cot x+\cot 2 x}\right](\cot 2 x+\cot x)$ $\left[\because \cot (A+B)=\frac{\cot A \cot B-1}{\cot A+\cot B}\right]$ $=\cot x \cot 2 x-(\cot 2 x \cot x-1)$ $=1=\mathrm{R} \cdot \mathrm{H} \cdot \mathrm{S}$...
Read More →Letbe a function defined as. The inverse of f is map g:
Question: Let $f: \mathbf{R}-\left\{-\frac{4}{3}\right\} \rightarrow \mathbf{R}$ be a function defined as $f(x)=\frac{4 x}{3 x+4}$. The inverse of $f$ is map $g$ : Range $f \rightarrow \mathbf{R}-\left\{-\frac{4}{3}\right\}$ given by (A) $g(y)=\frac{3 y}{3-4 y}$ (B) $g(y)=\frac{4 y}{4-3 y}$ (C) $g(y)=\frac{4 y}{3-4 v}$ (D) $g(y)=\frac{3 y}{4-3 y}$ Solution: Let $y$ be an arbitrary element of Range $f$. Then, there exists $x \in \mathbf{R}-\left\{-\frac{4}{3}\right\}$ such that $y=f(x)$. $\Righta...
Read More →A box contains 5 red marbles, 8 white marbles and 4 green marbles.
Question: A box contains 5 red marbles, 8 white marbles and 4 green marbles. One marble is taken out of the box at random. What is the probability that the marble taken out will be (i) red? (ii) white? (iii) not green? Solution: Total number of marbles = 5 + 8 + 4 = 17 (i)Number of red marbles = 5 Probability of getting a red marble $=\frac{\text { Number of favourable outcomes }}{\text { Number of total possible outcomes }}$ $=\frac{5}{17}$ (ii)Number of white marbles = 8 Probability of getting...
Read More →At 473 K, equilibrium constant Kc for decomposition of phosphorus pentachloride,
Question: At $473 \mathrm{~K}$, equilibrium constant $K_{c}$ for decomposition of phosphorus pentachloride, $\mathrm{PCl}_{5}$ is $8.3 \times 10^{-3} .$ If decomposition is depicted as, $\mathrm{PCl}_{5}(\mathrm{~g}) \longleftrightarrow \mathrm{PCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g}) \Delta_{r} H^{\circ}=124.0 \mathrm{kJmol}^{-1}$ a) Write an expression for $K_{c}$ for the reaction. b) What is the value of $K_{c}$ for the reverse reaction at the same temperature? c) What would be the ...
Read More →A bag contains 3 red balls and 5 black balls.
Question: A bag contains 3 red balls and 5 black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is (i) red? (ii) not red? Solution: (i) Total number of balls in the bag = 8 Probability of getting a red ball $=\frac{\text { Number of favourable outcomes }}{\text { Number of total possible outcomes }}$ $=\frac{3}{8}$ (ii) Probability of not getting red ball = 1 Probability of getting a red ball $=1-\frac{3}{8}$ $=\frac{5}{8}$...
Read More →Prove that the square of any positive integer of the form 5q + 1 is of the same form.
Question: Prove that the square of any positive integer of the form 5q+ 1 is of the same form. Solution: To Prove: that the square of a positive integer of the form 5q+ 1 is of the same form Proof: Since positive integernis of the form 5q+ 1 Ifn= 5q+ 1 Then $n^{2}=(5 q+1)^{2}$ $\Rightarrow n^{2}=(5 q)^{2}+(1)^{2}+2(5 q)(1)$ $\Rightarrow n^{2}=25 q^{2}+1+10 q$ $\Rightarrow n^{2}=25 q^{2}+10 q+1$ $\Rightarrow n^{2}=5\left(5 q^{2}+2 q\right)+1$ $\Rightarrow n^{2}=5 m+1$ (where $m=\left(5 q^{2}+2 q\...
Read More →It is given that in a group of 3 students, the probability of 2 students not having the same birthday is 0.992.
Question: It is given that in a group of 3 students, the probability of 2 students not having the same birthday is 0.992. What is the probability that the 2 students have the same birthday? Solution: Probability that two students are not having same birthday $P(\overline{\mathrm{E}})=0.992$ Probability that two students are having same birthday $P(E)=1-P(\bar{E})$ $=1-0.992$ $=0.008$...
Read More →A bag contains lemon flavoured candies only.
Question: A bag contains lemon flavoured candies only. Malini takes out one candy without looking into the bag. What is the probability that she takes out (i) an orange flavoured candy? (ii) a lemon flavoured candy? Solution: (i) The bag contains lemon flavoured candies only. It does not contain any orange flavoured candies. This implies that every time, she will take out only lemon flavoured candies. Therefore, event that Malini will take out an orange flavoured candy is an impossible event. He...
Read More →If P(E) = 0.05, what is the probability of ‘not E’?
Question: If P(E) = 0.05, what is the probability of not E? Solution: We know that, $P(\bar{E})=1-P(E)$ $P(\bar{E})=1-0.05$ $=0.95$ Therefore, the probability of not E is 0.95....
Read More →Which of the following cannot be the probability of an event?
Question: Which of the following cannot be the probability of an event? (A) $\frac{2}{3}$ (B) $-1.5$ (C) $15 \%$ (D) $0.7$ Solution: Probability of an event (E) is always greater than or equal to 0. Also, it is always less than or equal to one. This implies that the probability of an event cannot be negative or greater than 1. Therefore, out of these alternatives, 1.5 cannot be a probability of an event. Hence, (B)...
Read More →If f: R → R be given by, then fof(x) is
Question: If $f: \mathbf{R} \rightarrow \mathbf{R}$ be given by $f(x)=\left(3-x^{3}\right)^{\frac{1}{3}}$, then $f \circ f(x)$ is (A) $\frac{1}{x^{3}}$ (B) $x^{3}$ (C) $x$ (D) $\left(3-x^{3}\right)$ Solution: $f \mathbf{R} \rightarrow \mathbf{R}$ is given as $f(x)=\left(3-x^{3}\right)^{\frac{1}{3}}$ $f(x)=\left(3-x^{3}\right)^{\frac{1}{3}}$ The correct answer is C....
Read More →Why is tossing a coin considered to be a fair way of deciding
Question: Why is tossing a coin considered to be a fair way of deciding which team should get the ball at the beginning of a football game? Solution: When we toss a coin, the possible outcomes are only two, head or tail, which are equally likely outcomes. Therefore, the result of an individual toss is completely unpredictable....
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