If x + y + z = 9 and xy + yz + zx = 23,
Question: If $x+y+z=9$ and $x y+y z+z x=23$, then the value of $\left(x^{3}+y^{3}+z^{3}-3 x y z\right)=?$ (a) 108(b) 207(c) 669(d) 729 Solution: (a) 108 $x^{3}+y^{3}+z^{3}-3 x y z=(x+y+z)\left(x^{2}+y^{2}+z^{2}-x y-y z-z x\right)$ $=(x+y+z)\left[(x+y+z)^{2}-3(x y+y z+z x)\right]$ $=9 \times(81-3 \times 23)$ $=9 \times 12$ $=108$...
Read More →In how many ways can a lawn tennis mixed double be made up from seven married couples if no husband and wife play in the same set?
Question: In how many ways can a lawn tennis mixed double be made up from seven married couples if no husband and wife play in the same set? Solution: We arrange any 2 men in ${ }^{\prime} P_{2}$ ways and then the wives of the remaining 5 men can be arranged in ${ }^{5} P_{2}$ ways. This is because these two men should not be with their respective wives. $\therefore$ By fundamental principle of counting, the required number of ways $={ }^{7} P_{2} \times{ }^{5} P_{2}$ $=\frac{7 !}{5 !} \times \f...
Read More →If a + b + c = 0, then
Question: If $a+b+c=0$, then $\left(\frac{a^{2}}{b c}+\frac{b^{2}}{c a}+\frac{c^{2}}{a b}\right)=?$ Solution: (d) 3$a+b+c=0 \Rightarrow a^{3}+b^{3}+c^{3}=3 a b c$ Thus, we have: $\left(\frac{a^{2}}{b c}+\frac{b^{2}}{c a}+\frac{c^{2}}{a b}\right)=\frac{a^{3}+b^{3}+c^{3}}{a b c}$ $=\frac{3 a b c}{a b c}$ $=3$...
Read More →How many words can be formed out of the letters of the word 'ARTICLE', so that vowels occupy even places?
Question: How many words can be formed out of the letters of the word 'ARTICLE', so that vowels occupy even places? Solution: The word ARTICLE consists of 3 vowels, which have to be arranged in 3 even places. This can be done in 3! ways. Now, the remaining 4 consonants can be arranged in the remaining 4 places in 4! ways $\therefore$ Total number of words in which the vowels occupy only even places $=3 ! \times 4 !=144$...
Read More →How many permutations can be formed by the letters of the word, 'VOWELS', when
Question: How many permutations can be formed by the letters of the word, 'VOWELS', when (i) there is no restriction on letters? (ii) each word begins with E? (iii) each word begins with O and ends with L? (iv) all vowels come together? (v) all consonants come together? Solution: (i) The word VOWELS consists of 6 distinct letters that can be arranged amongst themselves in 6! ways. Number of words that can be formed with the letters of the word VOWELS, without any restriction = 6! = 720 (ii) If w...
Read More →Find the value
Question: $3 x^{3}+2 x^{2}+3 x+2=?$ (a) $(3 x-2)\left(x^{2}-1\right)$ (b) $(3 x-2)\left(x^{2}+1\right)$ (c) $(3 x+2)\left(x^{2}-1\right)$ (d) $(3 x+2)\left(x^{2}+1\right)$ Solution: (d) $(3 x+2)\left(x^{2}+1\right)$ $3 x^{3}+2 x^{2}+3 x+2=x^{2}(3 x+2)+1(3 x+2)$ $=(3 x+2)\left(x^{2}+1\right)$...
Read More →How many different words can be formed from the letters of the word 'GANESHPURI'?
Question: How many different words can be formed from the letters of the word 'GANESHPURI'? In how many of these words: (i) the letter G always occupies the first place? (ii) the letters P and I respectively occupy first and last place? (iii) the vowels are always together? (iv) the vowels always occupy even places? Solution: The word GANESHPURI consists of 10 distinct letters. Number of letters = 10! (i) If we fix the first letter as G, the remaining 9 letters can be arranged in 9! ways to form...
Read More →(x + 1) is a factor of the polynomial
Question: (x+ 1) is a factor of the polynomial (a) $x^{3}-2 x^{2}+x+2$ (b) $x^{3}+2 x^{2}+x-2$ (c) $x^{3}-2 x^{2}-x-2$ (d) $x^{3}-2 x^{2}-x+2$ Solution: (c) $x^{3}-2 x^{2}-x-2$ Let: $f(x)=x^{3}-2 x^{2}+x+2$ By the factor theorem, (x+ 1) will be a factor off(x) iff(1) = 0.We have: $f(-1)=(-1)^{3}-2 \times(-1)^{2}+(-1)+2$ $=-1-2-1+2$ $=-2 \neq 0$ Hence, $(x+1)$ is not a factor of $f(x)=x^{3}-2 x^{2}+x+2$. Now,Let: $f(x)=x^{3}+2 x^{2}+x-2$ By the factor theorem, (x+ 1) will be a factor off(x) iff(-...
Read More →How many different words can be formed with the letters of word 'SUNDAY'?
Question: How many different words can be formed with the letters of word 'SUNDAY'? How many of the words begin with N? How many begin with N and end in Y? Solution: Total number of words that can be formed with the letters of the word SUNDAY = 6! = 720 Now, if we fix the first letter as N, the remaining 5 places can be filled with the remaining 5 letters in 5! ways, i.e. 120. If we fix the first letter as N and the last word as Y: Remaining 4 places can be filled with 4 letters in 4! ways = 24...
Read More →How many words can be formed out of the letters of the word, 'ORIENTAL',
Question: How many words can be formed out of the letters of the word, 'ORIENTAL', so that the vowels always occupy the odd places? Solution: There are 8 letters in the word ORIENTAL. We wish to find the total number of arrangements of these 8 letters so that the vowels occupy only odd positions. There are 4 vowels and 4 odd positions. These 4 vowels can be arranged in the 4 positions in 4! ways. Now, the remaining 4 consonants can be arranged in the remaining 4 positions in 4! ways. By fundamen...
Read More →How many words can be formed from the letters of the word 'SUNDAY'?
Question: How many words can be formed from the letters of the word 'SUNDAY'? How many of these begin with D? Solution: Total number of words that can be formed with the letters of the word SUNDAY = 6! = 720 Fixing the first letter as D: Number of arrangements of the remaining 5 letters, taken 5 at a time = 5! = 120 Number of words with the starting letter D = 120...
Read More →In how many ways can the letters of the word 'STRANGE' be arranged so that
Question: In how many ways can the letters of the word 'STRANGE' be arranged so that (i) the vowels come together? (ii) the vowels never come together? and (iii) the vowels occupy only the odd places? Solution: (i) Number of vowels = 2 Number of consonants = 5 Considering the two vowels as a single entity, we are now to arrange 6 entities taken all at a time. Total number of ways = 6! Also, the two vowels can be mutually arranged amongst themselves in 2! ways. By fundamental principle of countin...
Read More →Find the discriminant of the quadratic equation
Question: Find the discriminant of the quadratic equation $3 \sqrt{3} x^{2}+10 x+\sqrt{3}=0$. Solution: Given that quadric equation is $3 \sqrt{3} x^{2}+10 x+\sqrt{3}=0$. Then, find the value of discrimenant. Here, $a=3 \sqrt{3}, b=10$ and, $c=\sqrt{3}$ As we know that discrimenant $D=b^{2}-4 a c$ Putting the value of $a=3 \sqrt{3}, b=10$ and,$c=\sqrt{3}$ $=(10)^{2}-4 \times 3 \sqrt{3} \times \sqrt{3}$ $=100-36$ $=64$ Thus, the value of discrimenant be $\mathrm{D}=64$....
Read More →Find the value
Question: $6 x^{2}+17 x+5=?$ (a) (2x+ 1)(3x+ 5)(b) (2x+ 5)(3x+ 1)(c) (6x+ 5)(x+ 1)(d) none of these Solution: (b) (2x+ 5)(3x+ 1) $6 x^{2}+17 x+5=6 x^{2}+15 x+2 x+5$ $=3 x(2 x+5)+1(2 x+5)$ $=(2 x+5)(3 x+1)$...
Read More →Find the value
Question: $\left(4 x^{2}+4 x-3\right)=?$ (a) (2x 1) (2x 3)(b) (2x+ 1) (2x 3)(c) (2x+ 3) (2x 1)(d) none of these Solution: (c) $(2 x+3)(2 x-1)$ $4 x^{2}+4 x-3=4 x^{2}+6 x-2 x-3$ $=2 x(2 x+3)-1(2 x+3)$ $=(2 x+3)(2 x-1)$...
Read More →In how many ways can the letters of the word 'FAILURE' be arranged so that the consonants may occupy only odd positions?
Question: In how many ways can the letters of the word 'FAILURE' be arranged so that the consonants may occupy only odd positions? Solution: There are 7 letters in the word FAILURE. We wish to find the total number of arrangements of these 7 letters so that the consonants occupy only odd positions. There are 3 consonants and 4 odd positions. These 3 consonants can be arranged in the 4 positions in 4! ways. Now, the remaining 4 vowels can be arranged in the remaining 4 positions in 4! ways. By fu...
Read More →Show that x = −2 is a solution of
Question: Show that $x=-2$ is a solution of $3 x^{2}+13 x+14=0$. Solution: Given that the equation $3 x^{2}+13 x+14=0$ $3 x^{2}+7 x+6 x+14=0$ $x(3 x+7)+2(3 x+7)=0$ $(3 x+7)(x+2)=0$ $(3 x+7)=0$ $x=\frac{-7}{3}$ $x=\frac{-7}{3}$ or $(x+2)=0$ $x=-2$ Therefore, $x=-2$ is the solution of given equation. Hence, proved....
Read More →All the letters of the word 'EAMCOT' are arranged in different possible ways.
Question: All the letters of the word 'EAMCOT' are arranged in different possible ways. Find the number of arrangements in which no two vowels are adjacent to each other. Solution: We note that, there are 3 consonants M, C, T and 3 vowels E, A, O. Since, no two vowels have to be together, the possible choice for volwels are the blank spaces, ${ }_{-} \mathrm{M}_{-} \mathrm{C}_{-} \mathrm{T}_{-}$ These vowels can be arranged in4P3ways. 3 consonants can be arranged in 3! ways. Hence, the required ...
Read More →Find the value
Question: (x2 4x 21) = ?(a) (x 7)(x 3)(b) (x + 7)(x 3)(c) (x 7)(x + 3)(d) none of these Solution: (c) $(x-7)(x+3)$ $x^{2}-4 x-21=x^{2}-7 x+3 x-21$ $=x(x-7)+3(x-7)$ $=(x-7)(x+3)$...
Read More →All the letters of the word 'EAMCOT' are arranged in different possible ways.
Question: All the letters of the word 'EAMCOT' are arranged in different possible ways. Find the number of arrangements in which no two vowels are adjacent to each other. Solution: We note that, there are 3 consonants M, C, T and 3 vowels E, A, O.Since, no two vowels have to be together, the possible choice for volwels are the blank spaces,_M_C_T_These vowels can be arranged in4P3ways.3consonants can be arranged in 3! ways.Hence, the required numbers of ways = 3!4P3 = 144 ways....
Read More →Find the value
Question: $4 a^{2}+b^{2}+4 a b+8 a+4 b+4=?$ (a) $(2 a+b+2)^{2}$ (b) $(2 a-b+2)^{2}$ (c) $(a+2 b+2)^{2}$ (d) none of these Solution: (a) $(2 a+b+2)^{2}$ $4 a^{2}+b^{2}+4 a b+8 a+4 b+4$ $=4 a^{2}+b^{2}+4+4 a b+4 b+8 a$ $=(2 a)^{2}+b^{2}+2^{2}+2 \times 2 a \times b+2 \times b \times 2+2 \times 2 a \times 2$ $=(2 a+b+2)^{2}$...
Read More →Find the number of 4-digit numbers that can be formed using the digits
Question: Find the number of 4-digit numbers that can be formed using the digits 1, 2, 3, 4, 5, if no digit is repeated? How many of these will be even? Solution: Number of 4-digit numbers that can be formed using the digits 1, 2, 3, 4, 5 = Number of arrangements of 5 digits taken 4 at a time =5P4= 5! = 120 Now, these numbers also consist of numbers in which the last digit is an odd digit. So, in order to find the number of even digits, we subtract the cases in which the unit's digit have been f...
Read More →Find the value
Question: 207 193 = ?(a) 39851(b) 39951(c) 39961(d) 38951 Solution: (b) 39951 $207 \times 193$ $=(200+7)(200-7)$ $=(200)^{2}-(7)^{2}$ $=40000-49$ $=39951$...
Read More →How many 3-digit even number can be made using the digits
Question: How many 3-digit even number can be made using the digits 1, 2, 3, 4, 5, 6, 7, if no digits is repeated? Solution: In order to find the number of even digits, we fix the unit's digit as an even digit.Fixing the unit's digit as 2: Number of arrangements possible =6P2= 65 = 30 Similarly, fixing the unit's digit as 4: Number of arrangements possible =6P2= 65 = 30 Fixing the unit's digit as 6: Number of arrangements possible =6P2= 65 = 30 Number of 3-digit even numbers that can be formed =...
Read More →How many 3-digit numbers can be formed by using the digits 1 to 9 if no digit is repeated?
Question: How many 3-digit numbers can be formed by using the digits 1 to 9 if no digit is repeated? Solution: Total number of arrangements of 9 digits, taken 3 at a time =9P3 Total 3-digit numbers that can be formed by using the digits 1 to 9, if no digit is repeated =9P3= 987 = 504...
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