Show that x = −3 is a solution of
Question: Show that $x=-3$ is a solution of $x^{2}+6 x+9=0$. Solution: Given that the equation $x^{2}+6 x+9=0$ $x^{2}+3 x+3 x+9=0$ $x(x+3)+3(x+3)=0$ $(x+3)(x+3)=0$ $(x+3)^{2}=0$ Square root both side, we get $(x+3)=0$ $x=-3$ Therefore, $x=-3$ is the solution of given equation. Hence, proved....
Read More →Find the value
Question: 305 308 = ?(a) 94940(b) 93840(c) 93940(d) 94840 Solution: (c) 93940 $305 \times 308=(300+5) \times(300+8)$ $=(300)^{2}+300 \times(5+8)+5 \times 8$ $=90000+3900+40$ $=93940$...
Read More →If a denotes the number of permutations of (x + 2) things taken all at a time,
Question: Ifadenotes the number of permutations of (x+ 2) things taken all at a time, b the number of permutations ofxthings taken 11 at a time andcthe number of permutations ofx 11 things taken all at a time such thata= 182bc, find the value ofx. Solution: $a={ }^{x+2} P_{x+2}=(x+2) !$ $b={ }^{x} P_{11}=\frac{x !}{(x-11) !}$ $c={ }^{x-11} P_{x-11}=(x-11) !$ $a=182 b c$ $\Rightarrow(x+2) !=182 \frac{x !}{(x-11) !} \times(x-11) !$ $\Rightarrow(x+2) !=182(x !)$ $\Rightarrow \frac{(x+2) !}{x !}=182...
Read More →Write a quadratic polynomial, sum of whose zeros is
Question: Write a quadratic polynomial, sum of whose zeros is $2 \sqrt{3}$ and their product is 2 . Solution: As we know that the quadratic polynomial $f(x)=k\left[x^{2}-(\right.$ sum of their roots $) x+($ product of their roots $\left.)\right]$ According to question, (sum of their roots) $=2 \sqrt{3}$ And (product of their roots) $=2$ Thus putting the value in above, $f(x)=k\left[x^{2}-2 \sqrt{3} x+2\right]$ where $k$ is real number. Therefore, the quadratic polynomial be $f(x)=k\left[x^{2}-2 ...
Read More →In how many ways can 6 boys and 5 girls be arranged for a group photograph if the girls are to sit on chairs in a row and the boys are to stand in a row behind them?
Question: In how many ways can 6 boys and 5 girls be arranged for a group photograph if the girls are to sit on chairs in a row and the boys are to stand in a row behind them? Solution: Number of arrangements of the boys = Number of arrangements of the 6 boys taken 6 at a time = 6! Number of arrangements of the girls = Number of arrangements of the 5 girls taken 5 at a time = 5! Total number of arrangements = 6!5! = 86400...
Read More →How many 6-digit telephone numbers can be constructed with digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
Question: How many 6-digit telephone numbers can be constructed with digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 if each number starts with 35 and no digit appears more than once? Solution: Total available digits = 10 Out of these, 3 and 5 have already been used to make the first two digits. Number of available digits = 8 The telephone number consists of 6 digits. The initial numbers have already been fixed as 35. Since repetition is not allowed, the number of telephone numbers that can be formed is equ...
Read More →How many three-digit numbers are there, with no digit repeated?
Question: How many three-digit numbers are there, with no digit repeated? Solution: Total number of 3-digit numbers $=$ Number of arrangements of 10 numbers, taken 3 at a time $={ }^{10} P_{3}=\frac{10 !}{7 !}=10 \times 9 \times 8=720$ Total number of 3-digit numbers, having 0 at its hundred's place $={ }^{9} \mathrm{P}_{2}=\frac{9 !}{7 !}=9 \times 8=72$ Total number of 3-digit numbers with distinct digits $={ }^{10} P_{3}-{ }^{9} P_{2}=720-72=648$...
Read More →There are 6 items in column A and 6 items in column B. A student is asked to match each item in column A with an item in column B.
Question: There are 6 items in column A and 6 items in column B. A student is asked to match each item in column A with an item in column B. How many possible, correct or incorrect, answers are there to this question? Solution: Each answer is an arrangement of the 6 items of column B keeping the order of column A fixed. Total number of answers = Number of arrangements of items in column B =6P6= 6! = 720...
Read More →Write the set of value of k for which the quadratic
Question: Write the set of value of $k$ for which the quadratic equations has $2 x^{2}+k x-8=0$ has real roots. Solution: The given quadric equation is $2 x^{2}+k x-8=0$, and roots are real. Then find the value ofk. Here, $a=2, b=k$ and, $c=-8$ As we know that $D=b^{2}-4 a c$ Putting the value of $a=2, b=k$ and,$c=-8$ $=(k)^{2}-4 \times 2 \times(-8)$ $=k^{2}+64$ The given equation will have real roots, if $D0$ I.e., $k^{2}+640$ which is true for all real values of $k$ Therefore, for all real val...
Read More →There are 6 items in column A and 6 items in column B. A student is asked to match each item in column A with an item in column B.
Question: There are 6 items in column A and 6 items in column B. A student is asked to match each item in column A with an item in column B. How many possible, correct or incorrect, answers are there to this question? Solution: Each answer is an arrangement of the 6 items of column B keeping the order of column A fixed. Total number of answers = Number of arrangements of items in column B =6P6= 6! = 720...
Read More →There are two works each of 3 volumes and two works each of 2 volumes;
Question: There are two works each of 3 volumes and two works each of 2 volumes; In how many ways can the 10 books be placed on a shelf so that the volumes of the same work are not separated? Solution: There are 4 different types of works. Number of arrangements of these 4 works, taken 4 at a time = 4! Of these 4 works, two of the works with 3 volumes each can be arranged in 3! ways each and two of the works with 2 volumes each can be arranged in 2! ways. Total number of arrangements $=4 ! \time...
Read More →Find the value
Question: 104 96 = ?(a) 9864(b) 9984(c) 9684(d) 9884 Solution: (b) 9984 $104 \times 96=(100+4)(100-4)$ $=100^{2}-4^{2}$ $=(10000-16)$ $=9984$...
Read More →How many words, with or without meaning,
Question: How many words, with or without meaning, can be formed by using the letters of the word 'TRIANGLE'? Solution: There are 8 letters in the word TRIANGLE. Number of 8 letter words = Number of arrangements of 8 letters, taken 8 at a time =8P8= 8!...
Read More →Write the condition to be satisfied for which equations
Question: Write the condition to be satisfied for which equations $a x^{2}+2 b x+c=0$ and $b x^{2}-2 \sqrt{a c} x+b=0$ have equal roots. Solution: The given equations are $a x^{2}+2 b x+c=0 \ldots \ldots$ (1) And, $b x^{2}-2 \sqrt{a c} x+b=0$......(2) roots are equal. Let $D_{1}$ and $D_{2}$ be the discriminants of equation (1) and (2) respectively, Then, $D_{1}=(2 b)^{2}-4 a c$ $=4 b^{2}-4 a c$ And $D_{2}=(-2 \sqrt{a c})^{2}-4 \times b \times b$ $=4 a c-4 b^{2}$ Both the given equation will hav...
Read More →How many words, with or without meaning, can be formed by using all the letters of the word 'DELHI',
Question: How many words, with or without meaning, can be formed by using all the letters of the word 'DELHI', using each letter exactly once? Solution: There are 5 letters in the word DELHI. Number of 5 letter words = Number of arrangements of 5 letters, taken 5 at a time =5P5= 120...
Read More →How many three-digit numbers are there, with distinct digits, with each digit odd?
Question: How many three-digit numbers are there, with distinct digits, with each digit odd? Solution: The odd digits are 1, 3, 5, 7 and 9. Required number of ways = Number of arrangements of five digits ( 1, 3, 5, 7 and 9), taken three at a time =5P3 $=\frac{5 !}{(5-3) !}$ $=\frac{5 !}{2 !}$ $=\frac{5 \times 4 \times 3 \times 2 !}{2 !}$ $=5 \times 4 \times 3$ = 60...
Read More →If (x + 2) and (x − 1) are factors of
Question: If $(x+2)$ and $(x-1)$ are factors of $\left(x^{3}+10 x^{2}+m x+n\right)$, then (a) $m=5, n=-3$ (b) $m=7, n=-18$ (c) $m=17, n=-8$ (d) $m=23, n=-19$ Solution: (b) $m=7, n=-18$ Let: $p(x)=x^{3}+10 x^{2}+m x+n$ Now, $x+2=0 \Rightarrow x=-2$ $(x+2)$ is a factor of $p(x)$ So, we have $p(-2)=0$ $\Rightarrow(-2)^{3}+10 \times(-2)^{2}+m \times(-2)+n=0$ $\Rightarrow-8+40-2 m+n=0$ $\Rightarrow 32-2 m+n=0$ $\Rightarrow 2 m-n=32$ ........(i) Now, $x-1=0 \Rightarrow x=1$ Also, $(x-1)$ is a factor o...
Read More →Find the number of different 4-letter words, with or without meanings,
Question: Find the number of different 4-letter words, with or without meanings, that can be formed from the letters of the word 'NUMBER'. Solution: Here, we need to permute four of the letters from the available 6 letters of the word NUMBER. Number of different four letter words = Number of arrangements of 6 letters, taken 4 at a time =6P4 $=\frac{6 !}{(6-4) !}$ $=\frac{6 !}{2 !}$ $=\frac{6 \times 5 \times 4 \times 3 \times 2 !}{2 !}$ $=6 \times 5 \times 4 \times 3$ = 360...
Read More →Four books, one each in Chemistry, Physics, Biology and Mathematics,
Question: Four books, one each in Chemistry, Physics, Biology and Mathematics, are to be arranged in a shelf. In how many ways can this be done? Solution: Here, all the four books are to be arranged on a shelf. This means that we have to find the number of arrangements of the books, taken all at a time. ⇒4P4 Now,nPn= n! Similarly,4P4= 4! = 24...
Read More →Four letters E, K, S and V, one in each, were purchased from a plastic warehouse.
Question: Four letters E, K, S and V, one in each, were purchased from a plastic warehouse. How many ordered pairs of letters, to be used as initials, can be formed from them? Solution: Here, we need to find out the number of pairs of the letters that can be formed with the 4 letters. Required number of ordered pairs = Number of arrangements of four letters, taken two at a time =4P2 $=\frac{4 !}{(4-2) !}$ $=\frac{4 !}{2 !}$ $=\frac{4 \times 3 \times 2 !}{2 !}$ $=4 \times 3$ $=12$...
Read More →From among the 36 teachers in a school, one principal and one vice-principal are to be appointed.
Question: From among the 36 teachers in a school, one principal and one vice-principal are to be appointed. In how many ways can this be done? Solution: Here, we need to permute 2 teachers out of the 36 available teachers. It can also be understood as the arrangement of 36 teachers, taken two at a time. Required number of ways =36P2 $=\frac{36 !}{(36-2) !}$ $=\frac{36 !}{34 !}$ $=\frac{36 \times 35 \times 34 !}{34 !}$ $=36 \times 35$ $=1260$...
Read More →Write the value of λ for which
Question: Write the value of $\lambda$ for which $x^{2}+4 x+\lambda$ is a perfect square. Solution: The given quadric equation is $x^{2}+4 x+\lambda=0$ Then find the value of $k$. Here, $a=1, b=4$ and,$c=\lambda$ As we know that $D=b^{2}-4 a c$ Putting the value of $a=1, b=4$ and, $c=\lambda$ $=(4)^{2}-4 \times 1 \times \lambda$ $=16-4 \lambda$ The given equation are perfect square, if $D=0$ $14-4 \lambda=0$ $4 \lambda=16$ $\lambda=\frac{16}{4}$ $=4$ Therefore, the value of $\lambda=4$...
Read More →From among the 36 teachers in a school, one principal and one vice-principal are to be appointed.
Question: From among the 36 teachers in a school, one principal and one vice-principal are to be appointed. In how many ways can this be done? Solution: Here, we need to permute 2 teachers out of the 36 available teachers. It can also be understood as the arrangement of 36 teachers, taken two at a time. Required number of ways =36P2 $=\frac{36 !}{(36-2) !}$ $=\frac{36 !}{34 !}$ $=\frac{36 \times 35 \times 34 !}{34 !}$ $=36 \times 35$ $=1260$...
Read More →In how many ways can five children stand in a queue?
Question: In how many ways can five children stand in a queue? Solution: Required number of ways = Number of arrangements of all the children =5P5= 5! We know: nPn= n! 5P5= 5! = 120...
Read More →Solve the following
Question: If $n+5 P_{n+1}=\frac{11(n-1)}{2} n+3 P_{n}$, find $n$. Solution: ${ }^{n+5} \mathrm{P}_{n+1}=\frac{11(n-1)}{2} n+3 \mathrm{P}_{n}$ $\Rightarrow \frac{(n+5) !}{(n+5-n-1) !}=\frac{11(n-1)}{2} \times \frac{(n+3) !}{(n+3-n) !}$ $\Rightarrow \frac{(n+5) !}{4 !}=\frac{11(n-1)}{2} \times \frac{(n+3) !}{3 !}$ $\Rightarrow \frac{(n+5) !}{(n+3) !}=\frac{11(n-1)}{2} \times \frac{4 !}{3 !}$ $\Rightarrow \frac{(n+5)(n+4)(n+3) !}{(n+3) !}=\frac{11(n-1)}{2} \times \frac{4 \times 3 !}{3 !}$ $\Rightar...
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