In the adjoining figure, AOB is a straight line. Find the value of x.

Question: In the adjoining figure,AOBis a straight line. Find the value ofx. Solution: We know that the sum of angles in a linear pair is $180^{\circ}$. Therefore, $\angle A O C+\angle B O C=180^{\circ}$ $\Rightarrow 62^{\circ}+x^{\circ}=180^{\circ}$ $\Rightarrow x^{\circ}=\left(180^{\circ}-62^{\circ}\right)$ $\Rightarrow x=118^{\circ}$ Hence, the value of $x$ is $118^{\circ}$....

The number of ways in which a team of eleven players can be selected from

Question: The number of ways in which a team of eleven players can be selected from 22 players always including 2 of them and excluding 4 of them, is (a)16C11 (b)16C5 (c)16C6 (d)20C9 Solution: Total number of players = 22 We need to select 11 players Given condition says, Exclude 4 particular player gives us 18 players available and including 2 particular players. Implies only 9 need to be selected out of remaining 16 . Hence, number of ways of selection is16C9. Hence, the correct answer is opti...

Let f : A → B and g : B → C be the bijective functions

Question: Let $f: A \rightarrow B$ and $g: B \rightarrow C$ be the bijective functions. Then, $(g o f)^{-1}=$ (a) $f^{-1} \circ g^{-1}$ (b) $f \circ g$ (c) $g^{-1} o r^{-1}$ (d) gof Solution: Given:f:ABandg:BCbe the bijective functions Since, $f: A \rightarrow B$ Thus, $f^{-1}: B \rightarrow A$ $\ldots(1)$ Since, $g: B \rightarrow C$ Thus, $g^{-1}: C \rightarrow B$ $\ldots(2)$ From (1) and (2), we get $f^{-1} \mathrm{og}^{-1}: C \rightarrow A$ $\ldots(3)$ Also, $g o f: A \rightarrow C$ $\Rightar...

The number of ways in which we can choose a committee from four men and six women so that

Question: The number of ways in which we can choose a committee from four men and six women so that the committee includes at least two man and exactly twice as many women as men, is (a) 94 (b) 126 (C) 128 (d) none of these Solution: Number of men = 4 Number of women = 6 (given) Given condition says, committee include at-least two man and exactly twice as many women as men . So, we can select either 2 men and 4 women or 3 men and 6 women. $\therefore$ Required number of ways of forming committee...

Find the value of x for which the angles

Question: Find the value ofxfor which the angles (2x 5) and (x 10) are the complementary angles. Solution: Two angles whose sum is $90^{\circ}$ are called complementary angles. It is given that the angles $(2 x-5)^{\circ}$ and $(x-10)^{\circ}$ are the complementary angles. $\therefore(2 x-5)^{\circ}+(x-10)^{\circ}=90^{\circ}$ $\Rightarrow 3 x^{\circ}-15^{\circ}=90^{\circ}$ $\Rightarrow 3 x^{\circ}=90^{\circ}+15^{\circ}=105^{\circ}$ $\Rightarrow x^{\circ}=\frac{105^{\circ}}{3}=35^{\circ}$ Thus, t...

Find a30 − a20 for the A.P.

Question: Find $a_{30}-a_{20}$ for the A.P. (i) $-9,-14,-19,-24, \ldots$ (ii) $a, a+d, a+2 d, a+3 d, \ldots$ Solution: In this problem, we are given different A.P. and we need to find $a_{30}-a_{20}$. (i) A.P. $-9,-14,-19,-24, \ldots$ Here, First term (a) = -9 Common difference of the A.P. $(d)=-14-(-9)$ $=-14+9$ $=-5$ Now, as we know, $a_{n}=a+(n-1) d$ Here, we find $a_{30}$ and $a_{20}$ So, for $30^{\text {th }}$ term, $a_{30}=a+(30-1) d$ $=-9+(29)(-5)$ $=-9-145$ $=-154$ Also, for $20^{\text {...

Question: Let $f: A \rightarrow B$ and $g: B \rightarrow C$ be the bijective functions. Then, $(g o f)^{-1}=$ (a) $f^{-1} \circ g^{-1}$ (b) fog (c) $g^{-1} o r^{-1}$ (d) gof Solution: Given:f:ABandg:BC be the bijective functions Since, $f: A \rightarrow B$ Thus, $f^{-1}: B \rightarrow A$ $\ldots(1)$ Since, $g: B \rightarrow C$ Thus, $g^{-1}: C \rightarrow B$ $\ldots(2)$ From (1) and (2), we get $f^{-1} \mathrm{og}^{-1}: C \rightarrow A$ $\ldots(3)$ Also, $g o f: A \rightarrow C$ $\Rightarrow(g o...

The number of 5-digit telephone numbers having at least one of their digits repeated is

Question: The number of 5-digit telephone numbers having at least one of their digits repeated is (a) 90,000 (b) 10,000 (c) 30240 (d) 69760 Solution: Total number of five digit telephone number possible without any restriction is 105and number of 5 digit telephone number having all digits different is10P5 Required number of ways i.e having at-least one digit repeated is =10510P5 i.e 1000000 10 9 8 7 6 = 105 30240 = 69760 Hence, the correct answer is option D....

Two complementary angles are in the ratio 4 : 5. Find the angles.

Question: Two complementary angles are in the ratio 4 : 5. Find the angles. Solution: Let the two angles be 4xand 5x,respectively.Then, $4 x+5 x=90$ $\Rightarrow 9 x=90$ $\Rightarrow x=10^{\circ}$ Hence, the two angles are $4 x=4 \times 10^{\circ}=40^{\circ}$ and $5 x=5 \times 10^{\circ}=50^{\circ}$....

The number of parallelograms that can be formed from a set of four parallel lines intersecting another set of three parallel lines is

Question: The number of parallelograms that can be formed from a set of four parallel lines intersecting another set of three parallel lines is (a) 6 (b) 18 (c) 12 (d) 9 Solution: Since parallelogram needs two set of parallel lines i.e selecting two parallel lines from a set of four can be done is ${ }^{4} C_{2}=\frac{4 \times 3 \times 2 !}{2 ! \times 2 !}=6$ and selecting two parallel lines from set of three parallel lines can be done in ${ }^{3} C_{2}=\frac{3 !}{2 !}=3$ Number of parallelogram...

Find the angle whose complement is one-third of its supplement.

Question: Find the angle whose complement is one-third of its supplement. Solution: Let the measure of the required angle be $x^{\circ}$ Then, the measure of its complement $=(90-x)^{\circ}$. And the measure of its supplement $=(180-x)^{\circ}$. Therefore, $(90-x)=\frac{1}{3}(180-x)$ $\Rightarrow 3(90-x)=(180-x)$ $\Rightarrow 270-3 x=180-x$ $\Rightarrow 2 x=90$ $\Rightarrow x=45$ Hence, the measure of the required angle is $45^{\circ}$....

The total number of 9 digit numbers which have all different digits is

Question: The total number of 9 digit numbers which have all different digits is (a) 10! (b) 9! (c) 9 9! (d) 10 10! Solution: Out of 10 digit 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 _ _ _ _ _ _ _ _ First place has 9 options since 0 can not be place for number to be 9 digit. Second place has 9 options since out of 10 are is already choosen. Similarly third place has 8 options and so on... Number of 9 digit number with different digit is $9 \times 9 \times 8 \times 7 \times 6 \times \frac{5 \times 4 \times 3...

Which of the following functions from Z to Z are bijections?

Question: Which of the following functions fromZtoZare bijections? (a) $f(x)=x^{3}$ (b) $f(x)=x+2$ (c) $f(x)=2 x+1$ (d) $f(x)=x^{2}+1$ Solution: Given: $f: Z \rightarrow Z$ (a) $f(x)=x^{3}$ It is one-one but not onto. Thus, it is not bijective. (b) $f(x)=x+2$ It is one-one and onto. Thus, it is bijective. (c) $f(x)=2 x+1$ It is one-one but not onto. Thus, it is not bijective. (d) $f(x)=x^{2}+1$ It is neither one-one nor onto. Thus, it is not bijective. Hence, the correct option is (b)....

Given 5 different green dyes, four different blue dyes and three different red dyes,

Question: Given 5 different green dyes, four different blue dyes and three different red dyes, the number of combinations of dyes which can be chosen taking at least one green and one blue dye is (a) 3600 (b) 3720 (c) 3800 (d) 3600 Solution: Number of ways of selecting one green dye is5C1+5C2+5C3+5C4+5C5 = 251ways. Number of ways of selecting one blue dye can be choosen is4C1+4C2+4C3+4C4= 241ways. and Number of ways of selecting red dye can be choosen in3C0+3C1+3C2+3C3= 23ways. So, total number ...

Let f : R → R be defined by

Question: Let $f: R \rightarrow R$ be defined by $f(x)=3 x^{2}-5$ and $g: R \rightarrow R$ by $g(x)=\frac{x}{x^{2}+1} .$ Then $(g$ of $)(x)$ is (a) $\frac{3 x^{2}-5}{9 x^{4}-30 x^{2}+26}$ (b) $\frac{3 x^{2}-5}{9 x^{4}-6 x^{2}+26}$ (c) $\frac{3 x^{2}}{x^{4}+2 x^{2}-4}$ (d) $\frac{3 x^{2}}{9 x^{4}+30 x^{2}-2}$ Solution: Given: $f(x)=3 x^{2}-5$ and $g(x)=\frac{x}{x^{2}+1}$ $(g o f)(x)=g(f(x))$ $=g\left(3 x^{2}-5\right)$ $=\frac{3 x^{2}-5}{\left(3 x^{2}-5\right)^{2}+1}$ $=\frac{3 x^{2}-5}{\left(3 x^...

Find the angle whose supplement is four times its complement.

Question: Find the angle whose supplement is four times its complement. Solution: Let the measure of the required angle be $x^{\circ}$ Then measure of its complement $=(90-x)^{\circ}$ And, measure of its supplement $=(180-x)^{\circ}$. Therefore, $(180-x)=4(90-x)$ $\Rightarrow 180-x=360-4 x$ $\Rightarrow 3 x=180$ $\Rightarrow x=60$ Hence, the measure of the required angle is $60^{\circ}$....

The number of triangles that are formed by choosing the vertices from a set of 12 points,

Question: The number of triangles that are formed by choosing the vertices from a set of 12 points, seven of which lie on the same line is (a) 105 (b) 15 (c) 175 (d) 185 Solution: Number of ways of selecting 3 points from given 12 points =12C3. But any three points selected from given seven collinear points does not from triangle number ways of selecting three points team seven collinear points =7C3 Required number of triangle =12C37C3 $=\frac{12 !}{3 ! 9 !}-\frac{7 !}{3 ! 4 !}$ $=\frac{12 \time...

Everybody in a room shakes hands with everybody else.

Question: Everybody in a room shakes hands with everybody else. The total number of hand shakes is 66. The total number of persons in the room is (a) 11 (b) 12 (c) 13 (d) 14 Solution: Let us suppose there arenpersons in a room for a handshake, 2 people are requried who will shake hand with each other. So, number of ways to make a pair out ofnpersons isnC2= 66 (given) i. $\mathrm{e} \frac{n !}{2 !(n-2) !}=66$ i. $\mathrm{e} \frac{n(n-1)(n-2) !}{2 \times(n-2) !}=66$ i.e $n^{2}-n=132$ i.e $n^{2}-n-...

Find the angle which is five times its supplement.

Question: Find the angle which is five times its supplement. Solution: Let the measure of the required angle be $x$. Then, measure of its supplement $=\left(180^{\circ}-x\right)$. Therefore, $x=\left(180^{\circ}-x\right) 5$ $\Rightarrow x=900^{\circ}-5 x$ $\Rightarrow 6 x=900^{\circ}$ $\Rightarrow x=150^{\circ}$ Hence, the measure of the required angle is $150^{\circ}$....

Let f : R → R be defined by

Question: Let $f: R \rightarrow R$ be defined by $f(x)=\frac{1}{x}$. Then, $f$ is (a) one-one(b) onto(e) bijective(d) not defined Solution: Given: The function $f: R \rightarrow R$ be defined by $f(x)=\frac{1}{x}$. To checkf is one-one: Let $f\left(x_{1}\right)=f\left(x_{2}\right)$ $\Rightarrow \frac{1}{x_{1}}=\frac{1}{x_{2}}$ $\Rightarrow x_{1}=x_{2}$ Hence, $f$ is one-one. To check $f$ is onto:\ Since, $y=\frac{1}{x}$ $\Rightarrow x=\frac{1}{y}$ $\Rightarrow y \in R-\{0\} \neq R$ There is no p...

Let f : R → R be defined by

Question: Let $f: R \rightarrow R$ be defined by $f(x)=\frac{1}{x}$. Then, $f$ is (a) one-one(b) onto(e) bijective(d) not defined Solution: Given: The function $f: R \rightarrow R$ be defined by $f(x)=\frac{1}{x}$. To checkf is one-one: Let $f\left(x_{1}\right)=f\left(x_{2}\right)$ $\Rightarrow \frac{1}{x_{1}}=\frac{1}{x_{2}}$ $\Rightarrow x_{1}=x_{2}$ Hence, $f$ is one-one. To check $f$ is onto:\ Since, $y=\frac{1}{x}$ $\Rightarrow x=\frac{1}{y}$ $\Rightarrow y \in R-\{0\} \neq R$ There is no p...

Find the angle which is four times its complement.

Question: Find the angle which is four times its complement. Solution: Let the measure of the required angle be $x$. Then, measure of its complement $=\left(90^{\circ}-x\right)$. Therefore, $x=\left(90^{\circ}-x\right) 4$ $\Rightarrow x=360^{\circ}-4 x$ $\Rightarrow 5 x=360^{\circ}$ $\Rightarrow x=72^{\circ}$ Hence, the measure of the required angle is $72^{\circ}$....

The first term of an A.P. is 5 and its 100th term is −292

Question: The first term of an A.P. is 5 and its $100^{\text {th }}$ term is $-292$. Find the $50^{\text {th }}$ term of this A.P. Solution: In the given problem, we are given 1stand 100thterm of an A.P. We need to find the 50thterm Here, $a=5$ $a_{100}=-292$ Now, we will finddusing the formula So, Also, $a_{100}=a+(100-1) d$ $-292=a+99 d$ So, to solve ford Substitutinga= 5, we get $-292=5+99 d$ $-292-5=99 d$ $\frac{-297}{99}=d$ $d=-3$ Thus, $a=5$ $d=-3$ $n=50$ Substituting the above values in t...

A five digit number divisible by 3 is to be formed using the numbers 0, 1, 2, 3, 4 and 5 without repetitions.

Question: A five digit number divisible by 3 is to be formed using the numbers 0, 1, 2, 3, 4 and 5 without repetitions. The total number of ways this can be done is (a) 216 (b) 600 (c) 240 (d) 3125 Solution: Five digit number is to be formed from 0,1,2,3,4 and 5. Such that number is divisible by 3. Any number is divisible by 3 if sum of its digits out of these digits, 1, 2, 3, 4, 5 and 0, 1, 2, 4, 5 sum up to be a multiple of 3 for 1,2,3,4,5, The number of ways a five digit number which is divis...

Mark the correct alternative in the following question:

Question: Mark the correct alternative in the following question: Let $f: \mathrm{R}-\left\{\frac{3}{5}\right\} \rightarrow \mathrm{R}$ be defined by $f(x)=\frac{3 x+2}{5 x-3} .$ Then, (a) $f^{-1}(x)=f(x)$ (b) $f^{-1}(x)=-f(x)$ (c) fof $f(x)=-x$ (d) $f^{-1}(x)=\frac{1}{19} f(x)$ Solution: $f: \mathbf{R}-\left\{\frac{3}{5}\right\} \rightarrow \mathbf{R}$ is defined by $f(x)=\frac{3 x+2}{5 x-3}$ $f \circ f(x)=f(f(x))$ $=f\left(\frac{3 x+2}{5 x-3}\right)$ $=\frac{3\left(\frac{3 x+2}{5 x-3}\right)+2...