If (n + 3)! = 56 [(n + 1)!], find n.
Question: If (n+ 3)! = 56 [(n+ 1)!], findn. Solution: (n+ 3)! = 56 [(n+ 1)!] $\Rightarrow(n+3) \times(n+2) \times(n+1) !=56[(n+1) !]$ $\Rightarrow(n+3) \times(n+2)=56$ $\Rightarrow(n+3) \times(n+2)=8 \times 7$ $\Rightarrow n+3=8$ $\therefore n=5$...
Read More →x = 5, y = 2 is a solution of the linear equation
Question: x= 5,y= 2 is a solution of the linear equation(a)x +2y =7(b) 5x +2y =7(c)x +y =7(d) 5x +y =7 Solution: Substituting the valuesx= 5,y = 2 in (a)x +2y =7, we get LHS $=5+2(2)=5+4=9 \neq 7=$ RHS i.e. LHS $\neq$ RHS (b) 5x +2y =7, we get LHS $=5(5)+2(2)=25+4=29 \neq 7=$ RHS i.e. LHS $\neq$ RHS (c)x +y =7, we get LHS $=5+2=7=$ RHS i.e. $\mathrm{LHS}=\mathrm{RHS}$ (d) 5x +y =7, we get $\mathrm{LHS}=5(5)+2=25+2=27 \neq 7=\mathrm{RHS}$ i.e. $\mathrm{LHS} \neq \mathrm{RHS}$ Hence, the correct o...
Read More →Write the number of numbers that can be formed using all for digits 1, 2, 3, 4.
Question: Write the number of numbers that can be formed using all for digits 1, 2, 3, 4. Solution: Disclaimer:- (1) Here, we can form 4 digits, 5 digits , 6 digits numbers and so on.... using the given digits. Thus, infinite numbers can be formed. (2) Taking into account only four digit numbers. We have to find all the numbers that can be formed by using the digits 1, 2, 3 and 4. This means that repetition of digits is not allowed as all the digits have to be used. Total numbers that can be for...
Read More →Write the remainder obtained when 1! + 2! + 3! + ... + 200! is divided by 14.
Question: Write the remainder obtained when 1! + 2! + 3! + ... + 200! is divided by 14. Solution: Every number after 6! (i.e. 7! onwards) till 200! will consist a power of 2 and 7, which will be exactly divisible by 14. So, we need to divide only the sum till 6!. 1! + 2! + 3! + 4! + 5! + 6! = 1 + 2 + 6 + 24 + 120 + 720 = 873 When 873 is divided, the remainder would be same as when 1! + 2! + 3! + ... + 200! is divided by 14. Remainder obtained when 1! + 2! + 3! + ... + 200! is divided by 14 = Rem...
Read More →Write the sequence with nth term:
Question: Write the sequence withnth term: (i) $a_{n}=3+4 n$ (ii) $a_{n}=5+2 n$ (iii) $a_{n}=6-n$ (iv) $a_{n}=9-5 n$ Show that all of the above sequences form A.P. Solution: In the given problem, we are given the sequence with the $n^{\text {th }}$ term $\left(a_{n}\right)$. We need to show that these sequences form an A.P (i) $a_{n}=3+4 n$ Now, to show that it is an A.P, we will first find its few terms by substituting $n=1,2,3$ So, Substituting $n=1$, we get $a_{1}=3+4(1)$ $a_{1}=7$ Substituti...
Read More →Write the sequence with nth term:
Question: Write the sequence withnth term: (i) $a_{n}=3+4 n$ (ii) $a_{n}=5+2 n$ (iii) $a_{n}=6-n$ (iv) $a_{n}=9-5 n$ Show that all of the above sequences form A.P. Solution: In the given problem, we are given the sequence with the $n^{\text {th }}$ term $\left(a_{n}\right)$. We need to show that these sequences form an A.P (i) $a_{n}=3+4 n$ Now, to show that it is an A.P, we will first find its few terms by substituting $n=1,2,3$ So, Substituting $n=1$, we get $a_{1}=3+4(1)$ $a_{1}=7$ Substituti...
Read More →Let f : R−{n}→R be a function defined by
Question: Let $f: R-\{n\} \rightarrow R$ be a function defined by $f(x)=\frac{x-m}{x-n}$, where $m \neq n .$ Then, (a)fis one-one onto(b)fis one-one into(c)fis many one onto(d)fis many one into Solution: Injectivity:Letxandybe two elements in the domainR-{n}, such that $f(x)=f(y)$ $\Rightarrow \frac{x-m}{x-n}=\frac{y-m}{y-n}$ $\Rightarrow(x-m)(y-n)=(x-n)(y-m)$ $\Rightarrow x y-n x-m y+m n=x y-m x-n y+m n$ $\Rightarrow(m-n) x=(m-n) y$ $\Rightarrow x=y$ So,fis one-one. Surjectivity:Letybe an eleme...
Read More →Write the number of ways in which 5 boys and 3 girls can be seated in a row so that each girl is between 2 boys.
Question: Write the number of ways in which 5 boys and 3 girls can be seated in a row so that each girl is between 2 boys. Solution: Five boys can be arranged amongst themselves in 5! ways, at the places shown above. The three girls are now to be arranged in the remaining four places taken three at a time =4P3= 4! By fundamental principle of counting, total number of ways = 5!4! = 12024 = 2880...
Read More →Any point on the y-axis is of the form
Question: Any point on they-axis is of the form(a) (x, y)(b) (0, y)(c) (x,0)(d) (y, y) Solution: (b) $(0, y)$, Any point on the $y$-axis is of the form $(0, y)$, where $y \neq 0$....
Read More →Write the number of ways in which 6 men and 5 women can dine at a round table if no two women sit together.
Question: Write the number of ways in which 6 men and 5 women can dine at a round table if no two women sit together. Solution: Each of the six men can be arranged amongst themselves in 6! ways. The five women can be arranged amongst themselves in the six places in 5! ways. $\therefore$ By fundamental principle of counting, total number of ways $=6 ! \times 5 !$...
Read More →Any point on the x-axis is of the form (a) (x, y)
Question: Any point on thex-axis is of the form(a) (x, y)(b) (0, y)(c) (x,0)(d) (x, x) Solution: (c) $(x, 0)$, Any point on the $\mathrm{x}$-axis is of the form $(\mathrm{x}, 0)$, where $\mathrm{x} \neq 0$....
Read More →Write the number of all possible words that can be formed using the letters of the word 'MATHEMATICS'.
Question: Write the number of all possible words that can be formed using the letters of the word 'MATHEMATICS'. Solution: The word 'MATHEMATICS' consists of 11 letters including two Ms, two Ts and two As Number of words that can be formed out of the letters of the word MATHEMATICS = Number of arrangements of 11 things of which 2 are similar to the first kind, 2 are similar to the second kind and 2 are similar to the third kind $=\frac{11 !}{2 ! 2 ! 2 !}$...
Read More →Write the number of words that can be formed out of the letters of the word 'COMMITTEE'.
Question: Write the number of words that can be formed out of the letters of the word 'COMMITTEE'. Solution: The word COMMITTEE consists of 9 letters including two Ms, two Ts and two Es. Number of words that can be formed out of the letters of the word COMMITTEE = Number of arrangements of 9 things of which 2 are similar to the first kind, 2 are similar to the second kind and 2 are similar to the third kind $=\frac{9 !}{2 ! 2 ! 2 !}=\frac{9 !}{(2 !)^{3}}$...
Read More →If (2, 0) is a solution of the linear equation 2x + 3y = k then the value of k is
Question: If (2, 0) is a solution of the linear equation 2x+ 3y=kthen the value ofkis(a) 6(b) 5(c) 2(d) 4 Solution: Since, (2, 0) is a solution of the linear equation 2x+ 3y=k.Substituting the valuesx= 2 andy= 0 in the equation2x+ 3y=k, we get $2(2)+3(0)=k$ $\Rightarrow 4=k$ or, $k=4$ Hence, the correct option is (d)....
Read More →Write the number of ways in which 7 men and 7 women can sit on a round table such that no two women sit together.
Question: Write the number of ways in which 7 men and 7 women can sit on a round table such that no two women sit together. Solution: Each of the seven men can be arranged amongst themselves in 7! ways. The women can be arranged amongst themselves in seven places, in 6! ways (i.e.nthings can be arranged in (n-1)! ways around a round table). By fundamental principle of counting, total number of ways $=7 ! \times 6 !$...
Read More →A linear equation in two variables x and y is of the form ax + by + c = 0, where
Question: A linear equation in two variablesxandyis of the formax+by+c= 0, where(a)a 0, b 0(b)a 0,b= 0(c)a= 0,b 0(d)a= 0,c= 0 Solution: (a) $a \neq 0, b \neq 0$ A linear equation in two variables $\mathrm{x}$ and $y$ is of the form $a x+b y+c=0$, where $a \neq 0$ and $b \neq 0$....
Read More →Write the number of arrangements of the letters of the word BANANA in which two N's come together.
Question: Write the number of arrangements of the letters of the word BANANA in which two N's come together. Solution: The word BANANA consists of 6 letters including three As and two Ns. Considering both Ns together or as a single letter, we are left with 5 letters including three As. $\therefore$ Number of arrangements of 5 things in which 3 are similar to one kind $=\frac{5 !}{3 !}=20$...
Read More →How many linear equations in x and y can be satisfied by x = 2, y = 3?
Question: How many linear equations inxandycan be satisfied byx= 2,y= 3?(a) Only one(b) Only two(c) Infinitely many(d) None of these Solution: (c) Infinitely many Infinite linear equations are satisfied by $x=2, y=3$....
Read More →Let f : R→R be a function defined by
Question: Let $f: R \rightarrow R$ be a function defined by $f(x)=\frac{e^{|x|}-e^{-x}}{e^{x}+e^{-x}}$. Then, (a)fis a bijection(b)fis an injection only(c)fis surjection on only(d)fis neither an injection nor a surjection Solution: (d)fis neither an injection nor a surjection $f: R \rightarrow R$ $f(x)=\frac{e^{|x|}-e^{-x}}{e^{x}+e^{-x}}$ For $x=-2$ and $-3 \in R$ $f(-2)=\frac{e^{|-2|}-e^{2}}{e^{-2}+e^{2}}$ $=\frac{e^{2}-e^{2}}{e^{-2}+e^{2}}$ $=0$ $\ f(-3)=\frac{e^{|-3|-e^{3}}}{e^{-3}+e^{3}}$' $...
Read More →Write the total number of possible outcomes in a throw of 3 dice in which at least one of the dice shows an even number.
Question: Write the total number of possible outcomes in a throw of 3 dice in which at least one of the dice shows an even number. Solution: Total number of outcomes when 3 dice are thrown $=6 \times 6 \times 6=216$ Number of outcomes in which there is an odd number on all the three dice $=3 \times 3 \times 3=27$ $\therefore$ Number of outcomes in which there is an even number at least on one dice $=\{$ Total possible outcomes $\}-\{$ Number of outcomes in which there is an odd number on all the...
Read More →Each of the points (−2, 2), (0, 0), (2, −2) satisfies the linear equation
Question: Each of the points (2, 2), (0, 0), (2, 2) satisfies the linear equation(a)xy= 0(b)x+y= 0(c) x+ 2y= 0(d)x 2y= 0 Solution: (b) $x+y=0$ Given points: $(-2,2),(0,0)$ and $(2,-2)$ We have to check which equation satisfies the given points. Let us check for $(a) x-y=0$ Substituting $(-2,2)$ in the equation, we get: $x-y=-2-2=-4$ Substituting $(0,0)$ in the equation, we get: $x-y=0-0=0$ Substituting $(2,-2)$ in the equation, we get: $x-y=2+2=4$ So, the given points do not satisfy the equation...
Read More →In how many ways 4 women draw water from 4 taps,
Question: In how many ways 4 women draw water from 4 taps, if no tap remains unused? Solution: Number of ways to draw water from the 1st tap = Number of women available to draw water = 4 Number of ways to draw water from the 2nd tap = Number of women available to draw water = 3 Number of ways to draw water from the 3rd tap = Number of women available to draw water = 2 Number of ways to draw water from the 4th tap = Number of women available to draw water = 1 $\therefore$ Total number of ways $=4...
Read More →Prove that no matter what the real numbers a and b are,
Question: Prove that no matter what the real numbersaandbare, the sequence withnth terma+nbis always an A.P. What is the common difference? Solution: In the given problem, we are given the sequence with the $n^{\text {th }}$ term $\left(a_{n}\right)$ as $a+n b$ where $a$ and $b$ are real numbers. We need to show that this sequence is an A.P and then find its common difference $(d)$ Here, $a_{n}=a+n b$ Now, to show that it is an A.P, we will find its few terms by substituting $n=1,2,3$ So, Substi...
Read More →Write the number of 5 digit numbers that can be formed using digits 0, 1 and 2.
Question: Write the number of 5 digit numbers that can be formed using digits 0, 1 and 2. Solution: Number of ways in which the first digit can be filled = Number of digits available for filling it = 2 {1,2} (Since the first one cannot be 0) Number of ways of filling the remaining four palaces = 3 each (as each place can be filled with either 1, 2 or 0) By fundamental principle of counting, number of five digit numbers that can be formed $=2 \times 3 \times 3 \times 3 \times 3=2 \times 3^{4}$...
Read More →In how many ways can 4 letters be posted in 5 letter boxes?
Question: In how many ways can 4 letters be posted in 5 letter boxes? Solution: Each of the letter can be posted in anyone of the letter boxes. This means that every letter can be posted in 5 ways. $\therefore$ Total number of ways of posting 4 letters $=5 \times 5 \times 5 \times 5=5^{4}$...
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