Find:
Question: Find: (i) Is 68 a term of the A.P. 7, 10, 13, ...?(ii) Is 302 a term of the A.P. 3, 8, 13, ...?(iii) Is 150 a term of the A.P. 11, 8, 5, 2, ...? Solution: In the given problem, we are given an A.P and the value of one of its term. We need to find whether it is a term of the A.P or not. So here we will use the formula, $a_{n}=a+(n-1) d$ (i) Here, A.P is $7,10,13, \ldots$ $a_{e}=68$ $a=7$ Now, Common difference $(d)=a_{1}-a$ $=10-7$ $=3$ Thus, using the above mentioned formula, we get, $...
Read More →How many words each of 3 vowels and 2 consonants can be formed from the letters of the word INVOLUTE?
Question: How many words each of 3 vowels and 2 consonants can be formed from the letters of the word INVOLUTE? Solution: There are 4 vowels and 4 consonants in the word INVOLUTE. Out of these, 3 vowels and 2 consonants can be chosen in $\left({ }^{4} C_{3} \times{ }^{4} C_{2}\right)$ ways. The 5 letters that have been selected can be arranged in $5 !$ ways. $\therefore$ Required number of words $=\left({ }^{4} C_{3} \times{ }^{4} C_{2}\right) \times 5 !=4 \times 6 \times 120=2880$...
Read More →Find the number of permutations of n distinct things taken r together,
Question: Find the number of permutations ofndistinct things takenrtogether, in which 3 particular things must occur together. Solution: Givenrplaces, we first fill up 3 places by 3 particular things. This can be done inrP3ways. Now, we have to fill remaining r 3 places with remainingn 3 things. This can be done inn 3Pr 3ways. Thus, the required number of permutations will berP3n 3Pr 3 ways....
Read More →How many words, with or without meaning can be formed from the letters of the word 'MONDAY',
Question: How many words, with or without meaning can be formed from the letters of the word 'MONDAY', assuming that no letter is repeated, if (i) 4 letters are used at a time (ii) all letters are used at a time (iii) all letters are used but first letter is a vowel? Solution: There are six letters in the word MONDAY. (i) 4 letters are used at a time: Four letters can be chosen out of six letters in6C4ways. So, there are ${ }^{6} \mathrm{C}_{4}$ groups containing four letters that can be arrange...
Read More →How many words, with or without meaning can be formed from the letters of the word 'MONDAY',
Question: How many words, with or without meaning can be formed from the letters of the word 'MONDAY', assuming that no letter is repeated, if (i) 4 letters are used at a time (ii) all letters are used at a time (iii) all letters are used but first letter is a vowel? Solution: There are six letters in the word MONDAY. (i) 4 letters are used at a time: Four letters can be chosen out of six letters in6C4ways. So, there are ${ }^{6} \mathrm{C}_{4}$ groups containing four letters that can be arrange...
Read More →Find:
Question: Find: (i) Which term of the A.P. $3,8,13, \ldots$ is $248 ?$ (ii) Which term of the A.P. $84,80,76, \ldots$ is $248 ?$ (iii) Which term of the A.P. $4,9,14, \ldots$ is 254 ? (iv) Which term of the A.P. $21,42,63,84, \ldots$ is 420 ? (v) Which term of the A.P. $121,117,113, \ldots$ is its first negative term? Solution: In the given problem, we are given an A.P and the value of one of its term. We need to find which term it is (n) So here we will find the value of $n$ using the formula, ...
Read More →There are 10 persons named
Question: There are 10 persons named $P_{1}, P_{2}, P_{3}, \ldots, P_{10}$. Out of 10 persons, 5 persons are to be arranged in a line such that in each arrangement $P_{1}$ must occur whereas $P_{4}$ and $P_{5}$ do not occur. Find the number of such possible arrangements. Solution: We need to arrange 5 persons in a line out of 10 persons, such thatin each arrangementP1must occur whereasP4andP5 do not occur. First we choose 5 persons out of 10 persons, such that in each arrangementP1must occur whe...
Read More →How many different words, each containing 2 vowels and 3 consonants can be formed with 5 vowels and 17 consonants?
Question: How many different words, each containing 2 vowels and 3 consonants can be formed with 5 vowels and 17 consonants? Solution: 2 out of 5 vowels and 3 out of 17 consonants can be chosen in ${ }^{5} C_{2} \times{ }^{17} C_{3}$ ways. Thus, there are ${ }^{5} C_{2} \times{ }^{17} C_{3}$ groups, each containing 2 vowels and 3 consonants. Each group contains 5 letters, which can be arranged in $5 !$ ways. $\therefore$ Required number of words $=\left({ }^{5} C_{2} \times{ }^{17} C_{3}\right) ...
Read More →What is the difference between a theorem and an axiom?
Question: What is the difference between a theorem and an axiom? Solution: An axiom is a basic fact that is taken for granted without proof. Examples: i) Halves of equals are equal. ii) The whole is greater than each of its parts. Theorem: A statement that requires proof is called theorem. Examples: i) The sum of all the angles around a point is $360^{\circ} .$ ii) The sum of all the angles of triangle is $180^{\circ}$....
Read More →Out of 18 points in a plane, no three are in the same straight line except five points which are collinear.
Question: Out of 18 points in a plane, no three are in the same straight line except five points which are collinear. How many (i) straight lines (ii) triangles can be formed by joining them? Solution: (i) Number of straight lines formed joining the 18 points, taking 2 points at a time $={ }^{18} C_{2}=\frac{18}{2} \times \frac{17}{1}=153$ Number of straight lines formed joining the 5 points, taking 2 points at a time $={ }^{5} C_{2}=\frac{5}{2} \times \frac{4}{1}=10$ But, when 5 collinear point...
Read More →A parallelogram is cut by two sets of m lines parallel to its sides.
Question: A parallelogram is cut by two sets of m lines parallel to its sides. Find the number of parallelograms thus formed. Solution: Each set of parallel lines consists of $(m+2)$ lines. Each parallelogram is formed by choosing two lines from the first set and two straight lines from the second set. $\therefore$ Total number of parallelograms $={ }^{m+2} C_{2} \times{ }^{m+2} C_{2}=\left({ }^{m+2} C_{2}\right)^{2}$...
Read More →In an examination, a question paper consists of 12 questions divided into two parts i.e.,
Question: In an examination, a question paper consists of 12 questions divided into two parts i.e., Part I and Part II, containing 5 and 7 questions, respectively. A student is required to attempt 8 questions in all, selecting at least 3 from each part. In how many ways can a student select the questions? Solution: A question paper consists of 12 questions divided into 2 parts, one with 5 and the other with 7 questions. A student has to attempt 8 questions out of the 12 questions by selecting at...
Read More →A committee of 7 has to be formed from 9 boys and 4 girls.
Question: A committee of 7 has to be formed from 9 boys and 4 girls. In how many ways can this be done when the committee consists of: (i) exactly 3 girls? (ii) at least 3 girls? (iii) at most 3 girls? Solution: A committee of 7 has to be formed from 9 boys and 4 girls. (i) When the committee consists of exactly 3 girls: Required number of ways $={ }^{4} C_{3} \times{ }^{9} C_{4}=\frac{4}{3} \times \frac{3}{2} \times \frac{2}{1} \times \frac{9}{4} \times \frac{8}{3} \times \frac{7}{2} \times \fr...
Read More →The area of ∆AOB having vertices A(0, 6), O(0, 0) and B(6, 0) is
Question: The area of ∆AOBhaving verticesA(0, 6),O(0, 0) andB(6, 0) is(a) 12 sq units(b) 36 sq units(c) 18 sq units(d) 24 sq units Solution: The points A(0, 6), O(0, 0) and B(6, 0) can be plotted on the Cartesian plane as follows: Here,∆AOB is a right triangle right angled at O.OA = 6 units and OB = 6 units $\therefore$ Area of $\Delta \mathrm{AOB}=\frac{1}{2} \times \mathrm{OA} \times \mathrm{OB}=\frac{1}{2} \times 6 \times 6=18$ square units Hence, the correct answer is option (c)....
Read More →Let A={x ∈ R : x ≤1} and f : A→A be defined as f(x)
Question: Let $A=\{x \in R: x \leq 1\}$ and $f: A \rightarrow A$ be defined as $f(x)=x(2-x)$. Then, $f^{-1}(x)$ is (a) $1+\sqrt{1-x}$ (b) $1-\sqrt{1-x}$ (c) $\sqrt{1-x}$ (d) $1 \pm \sqrt{1-x}$ Solution: LetybetheelementinthecodomainR suchthat $f^{-1}(x)=y \ldots(1)$ $\Rightarrow f(y)=x$ and $\mathrm{y} \leq 1$ $\Rightarrow y(2-y)=x$ $\Rightarrow 2 y-y^{2}=x$ $\Rightarrow y^{2}-2 y+x=0$ $\Rightarrow y^{2}-2 y=-x$ $\Rightarrow y^{2}-2 y+1=1-x$ $\Rightarrow(y-1)^{2}=1-x$ $\Rightarrow y-1=\pm \sqrt{...
Read More →In how many ways can a student choose a programme of 5 courses if 9 courses are available and 2 specific courses are compulsory for every student?
Question: In how many ways can a student choose a programme of 5 courses if 9 courses are available and 2 specific courses are compulsory for every student? Solution: 2 courses are compulsory out of the 9 available courses. There are 7 more courses. So, we need to choose 3 courses out of 7 courses. $\therefore$ Required number of ways $={ }^{7} C_{3}=\frac{7}{3} \times \frac{6}{2} \times \frac{5}{1}=35$...
Read More →A bag contains 5 black and 6 red balls.
Question: A bag contains 5 black and 6 red balls. Determine the number of ways in which 2 black and 3 red balls can be selected. Solution: 2 black and 3 red balls are to be selected from 5 black and 6 red balls. Required number of ways $={ }^{5} C_{2} \times{ }^{6} C_{3}=\frac{5}{2} \times \frac{4}{1} \times \frac{6}{3} \times \frac{5}{2} \times \frac{4}{1}=200$...
Read More →In how many ways can one select a cricket team of eleven from 17 players in which only 5 persons can bowl if each cricket team of 11 must include exactly 4 bowlers?
Question: In how many ways can one select a cricket team of eleven from 17 players in which only 5 persons can bowl if each cricket team of 11 must include exactly 4 bowlers? Solution: Out of 17 players, 11 need to be selected. There are 5 bowlers, of which four must be selected in the team. So, we have to choose 7 players from the remaining 12 players. Required number of ways $={ }^{5} C_{4} \times{ }^{12} C_{7}=5 \times \frac{12 !}{7 ! 5 !}=5 \times 792=3960$...
Read More →Which of the following points does not lie in any quadrant?
Question: Which of the following points does not lie in any quadrant?(a) (3, 6)(b) (3, 4)(c) (5, 7)(d) (0, 3) Solution: The point (3, 6) lies in the fourth quadrant.The point(3, 4) lies in the second quadrant.The point (5, 7) lies in the first quadrant.The point (0, 3) lies on the positive direction ofy-axis. Thus, the point (0, 3) does not lie in any quadrant.Hence, the correct answer is option (d)....
Read More →Let A={x ∈ R : x ≥ 1}. The inverse of the function, f : A→A given by
Question: Let $A=\{x \in R: x \geq 1\}$. The inverse of the function, $f: A \rightarrow A$ given by $f(x)=2^{x(x-1)}$, is (a) $\left(\frac{1}{2}\right)^{x(x-1)}$ (b) $\frac{1}{2}\left\{1+\sqrt{1+4 \log _{2} x}\right\}$ (c) $\frac{1}{2}\left\{1-\sqrt{1+4 \log _{2} x}\right\}$ (d) not defined Solution: Let $f^{-1}(x)=y \ldots(1)$ $\Rightarrow f(y)=x$ $\Rightarrow 2^{y(y-1)}=x$ $\Rightarrow 2^{y^{2}-y}=x$ $\Rightarrow y^{2}-y=\log _{2} x$ $\Rightarrow y^{2}-y+\frac{1}{4}=\log _{2} x+\frac{1}{4}$ $\...
Read More →Determine the number of 5 cards combinations out of a deck of 52 cards if there is exactly one ace in each combination.
Question: Determine the number of 5 cards combinations out of a deck of 52 cards if there is exactly one ace in each combination. Solution: There are total 4 aces in the deck of 52 cards. So, we are left with 48 cards. $\therefore$ Required ways $={ }^{4} C_{1} \times{ }^{48} C_{4}=\frac{4}{1} \times \frac{48}{4} \times \frac{47}{3} \times \frac{46}{2} \times \frac{45}{1}=778320$...
Read More →Which of the following points does not lie on the line y = 3x + 4?
Question: Which of the following points does not lie on the liney= 3x+ 4?(a) (1, 7)(b) (2, 0)(c) (1, 1)(d) (4, 12) Solution: (d) (4,12)Explanation:(a) Point (1,7) satisfy the equationy= 3x+ 4. (∵y= 3 1 + 4 = 7)(b) Point (2,10) satisfy the equationy= 3x+ 4. (∵y= 3 2 + 4 = 10)(c) Point (-1,1) satisfy the equationy= 3x+ 4. (∵y= 3 -1 + 4 = 1)(d) Point (4,12) does not satisfy the equationy= 3x+ 4. (∵y= 3 4 + 4 = 16 12)Hence, the point (4,12) donot lie on the liney=3x +4....
Read More →Find the number of ways of selecting 9 balls from 6 red balls,
Question: Find the number of ways of selecting 9 balls from 6 red balls, 5 white balls and 5 blue balls if each selection consists of 3 balls of each colour. Solution: Required number of ways $={ }^{6} C_{3} \times{ }^{5} C_{3} \times{ }^{5} C_{3}=\frac{6 !}{3 ! 3 !} \times \frac{5 !}{3 ! 2 !} \times \frac{5 !}{3 ! 2 !}=2000$...
Read More →Which of the following points lies on the line y = 2x + 3?
Question: Which of the following points lies on the liney= 2x+ 3?(a) (2, 8)(b) (3, 9)(c) (4, 12)(d) (5, 15) Solution: (b) (3,9) Explanation:Point (2,8) does not satisfy the equationy= 2x+ 3. (∵y= 2 2 + 8 = 128)Point (3,9) satisfy the equationy= 2x+ 3. (∵y=2 3 + 3 = 9)Point (4,12) does not satisfy the equationy= 2x+ 3. (∵y= 2 4 + 3 = 1112)Point (5,15) does not satisfy the equationy= 2x+3. (∵y= 2 5 + 3 = 1315)Hence, the point (3,9) lies on the liney=2x +3....
Read More →In how many ways can a team of 3 boys and 3 girls be selected from 5 boys and 4 girls?
Question: In how many ways can a team of 3 boys and 3 girls be selected from 5 boys and 4 girls? Solution: 3 boys and 3 girls are to be selected from 5 boys and 4 girls. $\therefore$ Required ways $={ }^{5} C_{3} \times{ }^{4} C_{3}=\frac{5}{3} \times \frac{4}{2} \times \frac{3}{1} \times \frac{4}{3} \times \frac{3}{2} \times \frac{2}{1}=40$...
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