Find the measure of an angle which is
Question: Find the measure of an angle which is $30^{\circ}$ less than its supplement. Solution: Let the measure of the angle be $x^{\circ}$. $\therefore$ Supplement of $x^{\circ}=180^{\circ}-x^{\circ}$ It is given that, $\left(180^{\circ}-x^{\circ}\right)-x^{\circ}=30^{\circ}$ $\Rightarrow 180^{\circ}-2 x^{\circ}=30^{\circ}$ $\Rightarrow 2 x^{\circ}=180^{\circ}-30^{\circ}=150^{\circ}$ $\Rightarrow x^{\circ}=75^{\circ}$ Thus, the measure of the angle is $75^{\circ}$....
Read More →Total number of words formed by 2 vowels and 3 consonants taken from 4 vowels and 5 consonants is equal to
Question: Total number of words formed by 2 vowels and 3 consonants taken from 4 vowels and 5 consonants is equal to (a) 60 (b) 120 (c) 7200 (d) 720 Solution: Number of vowels given is 4 Number of consonants given is 5 We have to form words by 2 vowels and 3 consonants number of ways of selecting 2 vowels out of 4 is4C2number ofways of selecting 3 consonants out of 5 is5C3. Hence number of ways of selection is4C25C3 $=\frac{4 !}{2 ! 2 !} \times \frac{5 !}{3 ! 2 !}$ $=6 \times 10=60$ Now, there s...
Read More →The sum of 4th and 8th terms of an A.P. is 24 and the sum of the 6th and 10th terms is 34
Question: The sum of 4thand 8thterms of an A.P. is 24 and the sum of the 6thand 10thterms is 34. Find the first term and the common difference of the A.P. Solution: In the given problem, the sum of 4thand 8thterm is 24 and the sum of 6thand 10thterm is 34. We can write this as, $a_{4}+a_{8}=24$.........(1) $a_{6}+a_{10}=34$......(2) We need to findaandd For the given A.P., let us take the first term asaand the common difference asd As we know, $a_{n}=a+(n-1) d$ For 4thterm (n= 4), $a_{4}=a+(4-1)...
Read More →The number of committees of five persons with a chairperson that can be formed from 12 persons,
Question: The number of committees of five persons with a chairperson that can be formed from 12 persons, is (a)12C5 (b)12C4 (c) 12 11C4 (d)11C4 Solution: For committee consisting of 5 persons with are chair person, from 12 persons. Chair persons can be selected in 12 ways. Remaining 4 can be selected from remaining 11 Hence, number of committee of five person with a chair person is 12 11C4 Hence, the correct answer is option C....
Read More →Find the measure of an angle which is 36° more than its complement.
Question: Find the measure of an angle which is 36 more than its complement. Solution: Let the measure of the required angle be $x^{\circ}$ Then, measure of its complement $=(90-x)^{\circ}$. Therefore, $x-\left(90^{\circ}-x\right)=36^{\circ}$ $\Rightarrow 2 x=126^{\circ}$ $\Rightarrow x=63^{\circ}$ Hence, the measure of the required angle is $63^{\circ}$....
Read More →The straight lines
Question: The straight linesl1,l2andl3are parallel and lie in the same plane. A total number ofmpoints are taken onl1,npoints onl2,kpoints onl3. The maximum number of triangles formed with vertices at these points are (a)m+n+kC3 (b)m+n+kC3mC3nC3kC3 (c)mC3+nC3+kC3 (d)mC3nC3kC3 Solution: Here the total number of points arem+n+k. Which must givem + n + kC3numbers of triangles. Sincempoints liel1, taking 3 points at a time ismC3. similarly, fromnpoints onl2, taking 3 points at a time isnC3andkC3from...
Read More →The number of signals that can be sent by 6 flags of different colours taking one or more at a time is
Question: The number of signals that can be sent by 6 flags of different colours taking one or more at a time is (a) 63 (b) 1956 (c) 720 (d) 21 Solution: Number of signals possible with one flag is 6. Number of signals possible with 2 flags is6P2. Number of signals possible with 3 flags is6P3. Similarly with 4 flags6P4, 5 flags is6P5and with all 6 flags6P6. Total number of signals =6P1+6P2+6P3+6P4+6P5+ 6P6 = 6 + 30 + 120 + 360 + 720 + 720 = 1956...
Read More →Find the measure of an angle which is
Question: Find the measure of an angle which is(i) equal to its complement(ii) equal to its supplement Solution: (i) Let the measure of the required angle be $x^{\circ}$. Then, in case of complementary angles: $x+x=90^{\circ}$ $\Rightarrow 2 x=90^{\circ}$ $\Rightarrow x=45^{\circ}$ Hence, measure of the angle that is equal to its complement is $45^{\circ}$. (ii) Let the measure of the required angle be $x^{\circ}$ Then, in case of supplementary angles: $x+x=180^{\circ}$ $\Rightarrow 2 x=180^{\ci...
Read More →The number of ways in which a committee consisting of 3 men and 2 women,
Question: The number of ways in which a committee consisting of 3 men and 2 women, can be chose from 7 men and 5 women, is (a) 45 (b) 350 (c) 4200 (d) 230 Solution: The selection of 5 members, consisting of 3 men and 2 women from 7 men and 5 women, can be made by selecting 3 men from 7 men and 2 women from 5 available $\therefore$ This can be done in ${ }^{7} C_{3} \times{ }^{5} C_{2}$ $=\frac{7 !}{4 ! 3 !} \times \frac{5 !}{3 ! 2 !}$ $=\frac{7 \times 6 \times 5 \times 4 !}{4 ! \times 3 !} r \fr...
Read More →Mark the correct alternative in the following question:
Question: Mark the correct alternative in the following question:If the setAcontains 7 elements and the setBcontains 10 elements, then the number one-onefunctions fromAtoBis (a) ${ }^{10} \mathrm{C}_{7}$ (b) ${ }^{10} \mathrm{C}_{7} \times 7 !$ (c) $7^{10}$ (d) $10^{7}$ Solution: As, the number of one-one functions from $A$ to $B$ with $m$ and $n$ elements, respectively $={ }^{n} P_{m}={ }^{n} C_{m} \times m !$ So, the number of one-one functions from $A$ to $B$ with 7 and 10 elements, respectiv...
Read More →An A.P. consists of 60 terms.
Question: An A.P. consists of 60 terms. If the first and the last terms be 7 and 125 respectively, find the 32nd term. Solution: In the given problem, we need to find the 32ndterm of an A.P. which contains a total of 60 terms. Here we are given the following, First term (a) = 7 Last term (an) = 125 Number of terms (n) = 60 So, let us take the common difference asd Now, as we know, $a_{n}=a+(n-1) d$ So, for the last term, $125=7+(60-1) d$ $125=7+(59) d$ $125-7=59 d$ $118=59 d$ Further simplifying...
Read More →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) 9 (c) 12 (d) 18 Solution: (d) 18 A parallelogram can be formed by choosing two parallel lines from the set of four parallel lines and two parallel lines from the set of three parallel lines. Two parallel lines from the set of four parallel lines can be chosen in4C2ways and two parallel lines from the set of 3 parallel lines can be chosen in3C2w...
Read More →Find the supplement of each of the following angles.
Question: Find the supplement of each of the following angles. (i) $42^{\circ}$ (ii) $90^{\circ}$ (iii) $124^{\circ}$ (iv) $\frac{3}{5}$ of a right angle Solution: Two angles whose sum is $180^{\circ}$ are called supplementary angles. (i) Supplement of $42^{\circ}=180^{\circ}-42^{\circ}=138^{\circ}$ (ii) Supplement of $90^{\circ}=180^{\circ}-90^{\circ}=90^{\circ}$ (iii) Supplement of $124^{\circ}=180^{\circ}-124^{\circ}=56^{\circ}$ (iv) $\frac{3}{5}$ of a right angle $\Rightarrow\left(\frac{3}{5...
Read More →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) 9 (c) 12 (d) 18 Solution: (d) 18 (d) 18 A parallelogram can be formed by choosing two parallel lines from the set of four parallel lines and two parallel lines from the set of three parallel lines. Two parallel lines from the set of four parallel lines can be chosen in4C2ways and two parallel lines from the set of 3 parallel lines can be chosen...
Read More →Mark the correct alternative in the following question:
Question: Mark the correct alternative in the following question:If the setAcontains 5 elements and the setBcontains 6 elements, then the number ofone-one and onto mappings fromAtoBis(a) 720 (b) 120 (c) 0 (d) none of these Solution: As, the number of bijection from $A$ into $B$ can only be possible when provided $n(A) \geq n(B)$ But here $n(A)n(B)$ So, the number of bijection i.e. one $-$ one and onto mappings from $A$ to $B=0$ Hence, the correct alternative is option (c)....
Read More →Solve the following
Question: Ifn+1C3= 2 nC2, thenn= (a) 3 (b) 4 (c) 5 (d) 6 Solution: (c) 5 ${ }^{n+1} C_{3}=2 \times{ }^{n} C_{2}$ $\Rightarrow \frac{(n+1) !}{3 !(n-2) !}=2 \times \frac{n !}{2 !(n-2) !}$ $\Rightarrow \frac{(n+1) n !}{3 \times 2 !(n-2) !}=2 \times \frac{n !}{2 !(n-2) !}$ $\Rightarrow n+1=6$ $\Rightarrow n=5$...
Read More →How many numbers of two digit are divisible by 3?
Question: How many numbers of two digit are divisible by 3? Solution: In this problem, we need to find out how many numbers of two digits are divisible by 3. So, we know that the first two digit number that is divisible by 3 is 12 and the last two digit number divisible by 3 is 99. Also, all the terms which are divisible by 3 will form an A.P. with the common difference of 3. So here, First term (a) = 12 Last term (an) = 99 Common difference (d) = 3 So, let us take the number of terms asn Now, a...
Read More →Find the complement of each of the following angles.
Question: Find the complement of each of the following angles. (i) $55^{\circ}$ (ii) $16^{\circ}$ (iii) $90^{\circ}$ (iv) $\frac{2}{3}$ of a right angle Solution: Two angles whose sum is $90^{\circ}$ are called complementary angles. (i) Complement of $55^{\circ}=90^{\circ}-55^{\circ}=35^{\circ}$ (ii) Complement of $16^{\circ}=(90-16)^{\circ}=74^{\circ}$ (iii) Complement of $90^{\circ}=90^{\circ}-90^{\circ}=0^{\circ}$ (iv) $\frac{2}{3}$ of a right angle $\Rightarrow\left(90 \times \frac{2}{3}\rig...
Read More →Mark the correct alternative in the following question:
Question: Mark the correct alternative in the following question: Let $A=\{1,2, \ldots, n\}$ and $B=\{a, b\}$. Then the number of subjections from $A$ into $B$ is (a) ${ }^{n} P_{2}$ (b) $2^{n}-2$ (c) $2^{n}-1$ (d) ${ }^{n} \mathrm{C}_{2}$ Solution: As, the number of surjections from $A$ to $B$ is equal to the number of functions from $A$ to $B$ minus the number of functions from $A$ to $B$ whose images are proper subsets of $B$. And, the number of functions from a set with $n$ number of element...
Read More →A lady gives a dinner party for six guests.
Question: A lady gives a dinner party for six guests. The number of ways in which they may be selected from among ten friends if two of the friends will not attend the party together is (a) 112 (b) 140 (c) 164 (d) none of these Solution: (b) 140 Suppose there are two friends, A and B, who do not attend the party together. If both of them do not attend the party, then the number of ways of selecting 6 guests =8C6= 28 If one of them attends the party, then the number of ways of selecting 6 guests ...
Read More →Find the second term and nth term of an A.P.
Question: Find the second term andnthterm of an A.P. whose 6thterm is 12 and 8thterm is 22. Solution: In the given problem, we are given 6thand 8thterm of an A.P. We need to find the 2ndandnthterm Here, let us take the first term asaand the common difference asd We are given, $a_{6}=12$ $a_{5}=22$ Now, we will find $a_{6}$ and $a_{8}$ using the formula $a_{e}=a+(n-1) d$ So, $a_{6}=a+(6-1) d$ $12=a+5 d$ .......(1) Also, $a_{8}=a+(8-1) d$ $22=a+7 d$.........(2) So, to solve foraandd On subtracting...
Read More →Define the following terms:
Question: Define the following terms:(i) Angle(ii) Interior of an angle(iii) Obtuse angle(iv) Reflex angle(v) Complementary angles(vi) Supplementary angles Solution: (i) Two rays OA and OB, with a common end-point O, form an angle AOB that is represented asAOBAOB. (ii) The interior of an angle is the set of all points in its plane, which lie on the same side of OA as B and also on the same side of OB as A. (iii) An angle greater than $90^{\circ}$ but less than $180^{\circ}$ is called an obtuse a...
Read More →Among 14 players, 5 are bowlers. In how many ways a team of 11 may be formed with at least 4 bowlers?
Question: Among 14 players, 5 are bowlers. In how many ways a team of 11 may be formed with at least 4 bowlers? (a) 265 (b) 263 (c) 264 (d) 275 Solution: (c) 264 Among 14 players, 5 are bowlers. A team of 11 players has to be selected such that at least 4 bowlers are included in the team. $\therefore$ Required number of ways $={ }^{5} C_{4} \times{ }^{9} C_{7}+{ }^{5} C_{5} \times{ }^{9} C_{6}$ $=180+84$ $=180+84$ $=264$...
Read More →The value of
Question: The value of $\left({ }^{7} C_{0}+{ }^{7} C_{1}\right)+\left({ }^{7} C_{1}+{ }^{7} C_{2}\right)+\ldots+\left({ }^{7} C_{6}+{ }^{7} C_{7}\right)$ is (a) 27 1 (b) 28 2 (c) 28 1 (d) 28 Solution: (b) $2^{8}-2$ $\left({ }^{7} C_{0}+{ }^{7} C_{1}\right)+\left({ }^{7} C_{1}+{ }^{7} C_{2}\right)+\left({ }^{7} C_{2}+{ }^{7} C_{3}\right)+\left({ }^{7} C_{3}+{ }^{7} C_{4}\right)+\left({ }^{7} C_{4}+{ }^{7} C_{5}\right)+\left({ }^{7} C_{5}+{ }^{7} C_{6}\right)+\left({ }^{7} C_{6}+{ }^{7} C_{7}\rig...
Read More →Mark the correct alternative in the following question:
Question: Mark the correct alternative in the following question: Let $f: \mathrm{R} \rightarrow \mathrm{R}$ be defined as $f(x)=\left\{\begin{array}{c}2 x, \text { if } x3 \\ x^{2}, \text { if } 1 Then, findf(-1) +f(2) +f(4) (a) 9 (b) 14 (c) 5 (d) none of these Solution: We have, $f(x)=\left\{\begin{array}{c}2 x, \text { if } x3 \\ x^{2}, \text { if } 1 Now, $f(-1)+f(2)+f(4)$ $=3(-1)+2^{2}+2(4)$ $=-3+4+8$ $=9$ Hence, the correct alternative is option (a)....
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