The number lock of a suitcase has 4 wheels, each labelled with ten digits i.e.,
Question: The number lock of a suitcase has 4 wheels, each labelled with ten digits i.e., from 0 to 9. The lock opens with a sequence of four digits with no repeats. What is the probability of a person getting the right sequence to open the suitcase? Solution: The number lock has 4 wheels, each labelled with ten digits i.e., from 0 to 9. Number of ways of selecting 4 different digits out of the 10 digits $={ }^{10} \mathrm{C}_{4}$ $\therefore$ Number of four digits with no repetitions $={ }^{10}...
Read More →If 4-digit numbers greater than 5,000 are randomly formed from the digits 0, 1, 3, 5, and 7,
Question: If 4-digit numbers greater than 5,000 are randomly formed from the digits 0, 1, 3, 5, and 7, what is the probability of forming a number divisible by 5 when, (i) the digits are repeated? (ii) the repetition of digits is not allowed? Solution: (i)When the digits are repeated Since four-digit numbers greater than 5000 are formed, the leftmost digit is either 7 or 5. The remaining 3 places can be filled by any of the digits 0, 1, 3, 5, or 7 as repetition of digits is allowed. $\therefore$...
Read More →If a and b are distinct positive primes such that
Question: If a and b are distinct positive primes such that $\sqrt[3]{a^{6} b^{-4}}=a^{x} b^{2} y$, , find $x$ and $y$ Solution: $\sqrt[3]{a^{6} b^{-4}}=a^{x} b^{2} y$ $\left(a^{6} b^{-4}\right)^{1 / 3}=a^{x} b^{2 y}$ $a^{6 / 3} b^{-4 / 3}=a^{x} b^{2 y}$ $a^{2} b^{-4 / 3}=a^{x} b^{2 y}$ x = 2, 2y = 4/3 $y=\frac{\frac{-4}{3}}{2}$ $y=-\frac{2}{3}$...
Read More →From the employees of a company, 5 persons are selected to represent them in the managing committee of the company.
Question: From the employees of a company, 5 persons are selected to represent them in the managing committee of the company. Particulars of five persons are as follows: A person is selected at random from this group to act as a spokesperson. What is the probability that the spokesperson will be either male or over 35 years? Solution: Let E be the event in which the spokesperson will be a male and F be the event in which the spokesperson will be over 35 years of age. Accordingly, $\mathrm{P}(\ma...
Read More →A and B are two events such that
Question: A and B are two events such that P(A) = 0.54, P(B) = 0.69 and P(A B) = 0.35. Find (i) P(A B) (ii) P(A B) (iii) P(A B) (iv) P(B A) Solution: It is given that $P(A)=0.54, P(B)=0.69, P(A \cap B)=0.35$ (i) We know that $P(A \cup B)=P(A)+P(B)-P(A \cap B)$ $\therefore \mathrm{P}(\mathrm{A} \cup \mathrm{B})=0.54+0.69-0.35=0.88$ (ii) $A^{\prime} \cap B^{\prime}=(A \cup B)^{\prime}[$ by De Morgan's law] $\therefore P\left(A^{\prime} \cap B^{\prime}\right)=P(A \cup B)^{\prime}=1-P(A \cup B)=1-0....
Read More →Solve the following equations
Question: Solve the following equations (i) $3^{x+1}=27 \times 3^{4}$ (ii) $4^{2 x}=(\sqrt[3]{16})^{-\frac{6}{y}}=(\sqrt{8})^{2}$ (iii) $3^{x-1} \times 5^{2 y-3}=225$ (iv) $8^{x+1}=16^{y+2}$ and $(1 / 2)^{3+x}=(1 / 4)^{3 y}$ Solution: (i) $3^{x+1}=27 \times 3^{4}$ $3^{x+1}=3^{3} \times 3^{4}$ $3^{x+1}=3^{3+4}$ x + 1 = 3 + 4 [By equating exponents] x + 1 = 7 x = 7 1 x = 6 (ii) $4^{2 x}=(\sqrt[3]{16})^{-\frac{6}{y}}=(\sqrt{8})^{2}$ $\left(2^{2}\right)^{2 x}=\left(16^{\frac{1}{3}}\right)^{-\frac{6}...
Read More →If the zeros of the polynomial f(x)
Question: If the zeros of the polynomialf(x) =ax3+ 3bx2+ 3cx+dare in A.P., prove that 2b3 3abc+a2d= 0. Solution: Let $a-d, a$ and $a+d$ be the zeros of the polynomial $f(x)$. Then, Sum of the zeros $=\frac{\text { Coefficient of } x^{2}}{\text { Coefficient of } x^{3}}$ $(a+d)+a+(a-d)=-\frac{3 b}{a}$ $3 a=\frac{-3 b}{a}$ $a=\frac{-b}{a}$ Sinceais a zero of the polynomialf(x). Therefore, $f(x)=a x^{3}+3 b x^{2}+3 c x+d$ $f(a)=0$ $f(a)=a a^{3}+3 b a^{2}+3 c a+d$ $a a^{3}+3 b a^{2}+3 c a+d=0$ $a\le...
Read More →Three letters are dictated to three persons and an envelope is addressed to each of them,
Question: Three letters are dictated to three persons and an envelope is addressed to each of them, the letters are inserted into the envelopes at random so that each envelope contains exactly one letter. Find the probability that at least one letter is in its proper envelope. Solution: Let $L_{1}, L_{2}, L_{3}$ be three letters and $E_{1}, E_{2}$, and $E_{3}$ be their corresponding envelops respectively. There are 6 ways of inserting 3 letters in 3 envelops. These are as follows: There are 4 wa...
Read More →Out of 100 students, two sections of 40 and 60 are formed.
Question: Out of 100 students, two sections of 40 and 60 are formed. If you and your friend are among the 100 students, what is the probability that (a) you both enter the same sections? (b) you both enter the different sections? Solution: My friend and I are among the 100 students. Total number of ways of selecting 2 students out of 100 students $={ }^{100} \mathrm{C}_{2}$ (a)The two of us will enter the same section if both of us are among 40 students or among 60 students. $\therefore$ Number ...
Read More →Find the condition that the zeros of the polynomial f(x)
Question: Find the condition that the zeros of the polynomialf(x) =x3+ 3px2+ 3qx+rmay be in A.P. Solution: Let $a-d, a$ and $a+d$ be the zeros of the polynomials $f(x)$. Then, Sum of the zeros $=\frac{\text { Coefficient of } x^{2}}{\text { Coefficient of } x^{3}}$ $a-d+a+a+d=\frac{-3 p}{1}$ $3 a=-3 p$ $a=\frac{-3 \times p}{3}$ $a=-p$ Since $a$ is a zero of the polynomial $f(x)$. Therefore, $f(x)=x^{3}+3 p x^{2}+3 q x+r$ $f(a)=0$ $f(a)=a^{3}+3 p a^{2}+3 q a+r$ $a^{3}+3 p a^{2}+3 q a+r=0$ Substit...
Read More →In a certain lottery, 10,000 tickets are sold and ten equal prizes are awarded.
Question: In a certain lottery, 10,000 tickets are sold and ten equal prizes are awarded. What is the probability of not getting a prize if you buy (a) one ticket (b) two tickets (c) 10 tickets? Solution: Total number of tickets sold = 10,000 Number prizes awarded = 10 (i)If we buy one ticket, then $P$ (getting a prize) $=\frac{10}{10000}=\frac{1}{1000}$ $\therefore P($ not getting a prize $)=1-\frac{1}{1000}=\frac{999}{1000}$ (ii)If we buy two tickets, then Number of tickets not awarded = 10,00...
Read More →Show that semi-vertical angle of right circular cone of given
Question: Show that semi-vertical angle of right circular cone of given surface area and maximum volume is $\operatorname{Sin}^{-1}\left(\frac{1}{3}\right)$. Solution: Letrbe the radius,lbe the slant height andhbe the height of the cone of given surface area,S. Also, letbe the semi-vertical angle of the cone. Then $S=\pi r l+\pi r^{2}$ $\Rightarrow l=\frac{S-\pi r^{2}}{\pi r} \ldots .(1)$ Let $V$ be the volume of the cone. $\begin{aligned} \Rightarrow V^{2} =\frac{1}{9} \pi^{2} r^{4} h^{2} \\ =\...
Read More →A die has two faces each with number ‘1’,
Question: A die has two faces each with number 1, three faces each with number 2 and one face with number 3. If die is rolled once, determine (i) P(2) (ii) P(1 or 3) (iii) P(not 3) Solution: Total number of faces = 6 (i) Number faces with number 2 = 3 $\therefore P(2)=\frac{3}{6}=\frac{1}{2}$ (ii) $P(1$ or 3$)=P(\operatorname{not} 2)=1-P(2)=1-\frac{1}{2}=\frac{1}{2}$ (iii) Number of faces with number 3 = 1 $\therefore P(3)=\frac{1}{6}$ Thus, $\mathrm{P}($ not 3$)=1-\mathrm{P}(3)=1-\frac{1}{6}=\f...
Read More →4 cards are drawn from a well-shuffled deck of 52 cards.
Question: 4 cards are drawn from a well-shuffled deck of 52 cards. What is the probability of obtaining 3 diamonds and one spade? Solution: Number of ways of drawing 4 cards from 52 cards $={ }^{52} \mathrm{C}_{4}$ In a deck of 52 cards, there are 13 diamonds and 13 spades. $\therefore$ Number of ways of drawing 3 diamonds and one spade $={ }^{13} \mathrm{C}_{3} \times{ }^{13} \mathrm{C}_{1}$ Thus, the probability of obtaining 3 diamonds and one spade $=\frac{{ }^{13} \mathrm{C}_{3} \times{ }^{1...
Read More →If the zeros of the polynomial f(x)
Question: If the zeros of the polynomialf(x) = 2x3 15x2+ 37x 30 are in A.P., find them. Solution: Let $\alpha=a-d, \beta=a$ and $\gamma=a+d$ be the zeros of the polynomial $f(x)=2 x^{3}-15 x^{2}+37 x-30$ Therefore $\alpha+\beta+\gamma=\frac{\text { Coefficient of } x^{2}}{\text { Coefficient of } x^{3}}$ $=-\left(\frac{-15}{2}\right)$ $=\frac{15}{2}$ $\alpha \beta \gamma=\frac{-\text { Constant term }}{\text { Coefficient of } x^{2}}$ $=-\left(\frac{-30}{2}\right)$ = 15 Sum of the zeros $=\frac{...
Read More →A box contains 10 red marbles, 20 blue marbles and 30 green marbles.
Question: A box contains 10 red marbles, 20 blue marbles and 30 green marbles. 5 marbles are drawn from the box, what is the probability that (i) all will be blue? (ii) atleast one will be green? Solution: Total number of marbles = 10 + 20 + 30 = 60 Number of ways of drawing 5 marbles from 60 marbles $={ }^{60} C_{5}$ (i) All the drawn marbles will be blue if we draw 5 marbles out of 20 blue marbles. 5 blue marbles can be drawn from 20 blue marbles in ${ }^{20} C_{5}$ ways. $\therefore$ Probabil...
Read More →Show that the semi-vertical angle of the cone of the maximum volume and of given slant height is.
Question: Show that the semi-vertical angle of the cone of the maximum volume and of given slant height is $\tan ^{-1} \sqrt{2}$. Solution: Let $\theta$ be the semi-vertical angle of the cone. It is clear that $\theta \in\left[0, \frac{\pi}{2}\right]$. Let $r, h$, and $/$ be the radius, height, and the slant height of the cone respectively. The slant height of the cone is given as constant. Now, $r=I \sin \theta$ and $h=I \cos \theta$ The volume (V)of the cone is given by, $\begin{aligned} V =\f...
Read More →Show that the semi-vertical angle of the cone of the maximum volume and of given slant height is.
Question: Show that the semi-vertical angle of the cone of the maximum volume and of given slant height is $\tan ^{-1} \sqrt{2}$. Solution: Let $\theta$ be the semi-vertical angle of the cone. It is clear that $\theta \in\left[0, \frac{\pi}{2}\right]$. Let $r, h$, and $/$ be the radius, height, and the slant height of the cone respectively. The slant height of the cone is given as constant. Now, $r=I \sin \theta$ and $h=I \cos \theta$ The volume (V)of the cone is given by, $\begin{aligned} V =\f...
Read More →In a class of 60 students, 30 opted for NCC, 32 opted for NSS and 24 opted for both NCC and NSS.
Question: In a class of 60 students, 30 opted for NCC, 32 opted for NSS and 24 opted for both NCC and NSS. If one of these students is selected at random, find the probability that (i) The student opted for NCC or NSS. (ii) The student has opted neither NCC nor NSS. (iii) The student has opted NSS but not NCC. Solution: Let A be the event in which the selected student has opted for NCC and B be the event in which the selected student has opted for NSS. Total number of students = 60 Number of stu...
Read More →If 53x = 125 and 10y
Question: If $5^{3 x}=125$ and $10^{y}=0.001 .$ Find $x$ and $y .$ Solution: $5^{3 x}=125$ and $10^{y}=0.001$ $5^{3 x}=5^{3}$ $x=1$ Now, $10^{\mathrm{y}}=0.001$ $10^{\mathrm{y}}=10^{-3}$ $\mathrm{y}=-3$ Therefore, the value of x = 1 and the value of y = 3...
Read More →If 34x = (81)-1 and (10)1/y
Question: If $3^{4 x}=(81)^{-1}$ and $(10)^{1 / y}=0.0001$, find the value of $2^{-x+4 y}$. Solution: $3^{4 x}=(81)^{-1}$ and $(10)^{1 / y}=0.0001$ $3^{4 x}=(3)^{-4}$ x = -1 And, $(10)^{1 / y}=0.0001$ $(10)^{1 / y}=(10)^{-4}$ $1 / y=-4$ $y=1 /-4$...
Read More →Show that the right circular cone of least curved surface
Question: Show that the right circular cone of least curved surface and given volume has an altitude equal to $\sqrt{2}$ time the radius of the base. Solution: Letrandhbe the radius and the height (altitude) of the cone respectively. Then, the volume (V)of the cone is given as: $V=\frac{1}{3} \pi r^{2} h \Rightarrow h=\frac{3 V}{\pi r^{2}}$ The surface area (S)of the cone is given by, S= rl(wherelis the slant height) $=\pi r \sqrt{r^{2}+h^{2}}$ $=\pi r \sqrt{r^{2}+\frac{9 V^{2}}{\pi^{2} r^{4}}}=...
Read More →Show that the right circular cone of least curved surface
Question: Show that the right circular cone of least curved surface and given volume has an altitude equal to $\sqrt{2}$ time the radius of the base. Solution: Letrandhbe the radius and the height (altitude) of the cone respectively. Then, the volume (V)of the cone is given as: $V=\frac{1}{3} \pi r^{2} h \Rightarrow h=\frac{3 V}{\pi r^{2}}$ The surface area (S)of the cone is given by, S= rl(wherelis the slant height) $=\pi r \sqrt{r^{2}+h^{2}}$ $=\pi r \sqrt{r^{2}+\frac{9 V^{2}}{\pi^{2} r^{4}}}=...
Read More →Show that the right circular cone of least curved surface
Question: Show that the right circular cone of least curved surface and given volume has an altitude equal to $\sqrt{2}$ time the radius of the base. Solution: Letrandhbe the radius and the height (altitude) of the cone respectively. Then, the volume (V)of the cone is given as: $V=\frac{1}{3} \pi r^{2} h \Rightarrow h=\frac{3 V}{\pi r^{2}}$ The surface area (S)of the cone is given by, S= rl(wherelis the slant height) $=\pi r \sqrt{r^{2}+h^{2}}$ $=\pi r \sqrt{r^{2}+\frac{9 V^{2}}{\pi^{2} r^{4}}}=...
Read More →The probability that a student will pass the final examination in both English and Hindi is 0.5 and the probability of passing neither is 0.1.
Question: The probability that a student will pass the final examination in both English and Hindi is 0.5 and the probability of passing neither is 0.1. If the probability of passing the English examination is 0.75, what is the probability of passing the Hindi examination? Solution: Let A and B be the events of passing English and Hindi examinations respectively. Accordingly, $P(A$ and $B)=0.5, P($ not $A$ and not $B)=0.1$, i.e., $P\left(A^{\prime} \cap B^{\prime}\right)=0.1$ P(A) = 0.75 Now, $(...
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