List all the elements of each of the sets given below.
Question: List all the elements of each of the sets given below. $E=\left\{x: x \in Z\right.$ and $\left.x^{2}=x\right\}$ Solution: Given: $x \in Z$ and $x^{2}=x$ Z is a set of integers Integers are -2 , -1, 0, 1, 2, Now, if we take x = -2 then we have to check that it satisfies the given condition x2 = x $(-2)^{2}=4 \neq 2$ So, $-2 \notin E$ If $x=-1$ then $(-1)^{2}=1 \neq-1$ [not satisfying $\left.x^{2}=x\right]$ So, $-1 \notin E$ If $x=0$ then $(0)^{2}=0$ [satisfying $\left.x^{2}=x\right]$ $\...
Read More →The area of the base of a right circular cylinder is 616 cm
Question: The area of the base of a right circular cylinder is 616 cm2and its height is 2.5 cm. Find the curved surface area of the cylinder. Solution: Given: Area of the base of a right circular cylinder $=616 \mathrm{~cm}^{2}$ Height $=2.5 \mathrm{~cm}$ Let $r$ be the radius of the base of a right circular cylinder. $\pi \mathrm{r}^{2}=616$ $\Rightarrow \mathrm{r}^{2}=616 \times \frac{7}{22}$ $\Rightarrow \mathrm{r}^{2}=196$ $\Rightarrow \mathrm{r}=14 \mathrm{~cm}$ Curved surface area of the r...
Read More →List all the elements of each of the sets given below
Question: List all the elements of each of the sets given below $D=\left\{x: x=n^{2}, n \in N\right.$ and $\left.2 \leq n \leq 5\right\}$ Solution: Here, $x=n^{2}$ and $2 \leq n \leq 5$ n = 2, 3, 4, 5 [it is given that n is less than equal to 2 and greater than equal to 5] If $n=2, x=(2)^{2}=4$ $n=3, x=(3)^{2}=9$ $n=4, x=(4)^{2}=16$ $n=5, x=(5)^{2}=25$ So, D = {4, 9, 16, 25}...
Read More →The curved surface area of a cylindrical road is 132
Question: The curved surface area of a cylindrical road is 132 cm2. Find its length if the radius is 0.35 cm. Solution: Consider $h$ to be the height of the cylindrical rod. Given: Radius, $r=0.35 \mathrm{~cm}$ Curved surface area $=132 \mathrm{~cm}^{2}$ We know : Curved surface area $=2 \times \pi \times \mathrm{r} \times \mathrm{h}$ $132=2 \times \frac{22}{7} \times 0.35 \times \mathrm{h}$ $h=\frac{132 \times 7}{2 \times 22 \times 0.35}$ $h=60$ Therefore, the length of the cylindrical rod is $...
Read More →The angles of a triangle are in AP.
Question: The angles of a triangle are in AP. The greatest angle is twice the least. Find all the angles of the triangle. Solution: Given that, the angles of a triangle are in AR Let A, B and C are angles of a Δ ABC. $\therefore$$B=\frac{A+C}{2}$ $\Rightarrow \quad 2 B=A+C$ ...(i) We know that, sum of all interior angles of a $\triangle A B C=180^{\circ}$ $A+B+C=180^{\circ}$ $\Rightarrow \quad 2 B+B=180^{\circ}$ [from Eq. (i)] $\Rightarrow$ $3 B=180^{\circ} \Rightarrow B=60^{\circ}$ Let the grea...
Read More →List all the elements of each of the sets given below
Question: List all the elements of each of the sets given below $\mathrm{C}=\left\{\mathrm{x}: \mathrm{x}=\frac{1}{\mathrm{n}}, \mathrm{n} \in \mathrm{N}\right.$ and $\left.\mathrm{n} \leq 6\right\}$ Solution: Here, $x=\frac{1}{n}$ and $\mathrm{n} \leq 6$ So, $n=1,2,3,4,5$ and $6[\because n \in N]$ Given: $x=\frac{1}{n}$ $n=1, x=\frac{1}{1}=1$ $n=2, x=\frac{1}{2}$ $n=3, x=\frac{1}{3}$ $n=4, x=\frac{1}{4}$ $n=5, x=\frac{1}{5}$ $n=6, x=\frac{1}{6}$ So $C=\left\{1, \frac{1}{2}, \frac{1}{3}, \frac{1...
Read More →Find the curved surface area and total surface area of a cylinder,
Question: Find the curved surface area and total surface area of a cylinder, the diameter of whose base is 7 cm and height is 60 cm. Solution: Let $r$ and $h$ be the radius and the height of the cylinder. Given : $\mathrm{r}=\frac{7}{2} \mathrm{~cm}$ $\mathrm{~h}=60 \mathrm{~cm}$ Curved surface area of the cylinder $=2 \pi \times \mathrm{r} \times \mathrm{h}$ $=2 \times \frac{22}{7} \times \frac{7}{2} \times 60$ $=22 \times 60=1320 \mathrm{~cm}^{2}$ Total surface area of the cylinder $=2 \pi \ti...
Read More →List all the elements of each of the sets given below.
Question: List all the elements of each of the sets given below. B = {x : x = 2n + 1, n ϵ W and n 5}. Solution: Given: x = 2n + 1 and n 5 ⇒ n = 0, 1, 2, 3, 4 and 5 [∵ n W] Given $x=2 n+1$ $n=0, x=2 \times 0+1=1$ $n=1, x=2 \times 1+1=3$ $n=2, x=2 \times 2+1=5$ $n=3, x=2 \times 3+1=7$ $n=4, x=2 \times 4+1=9$ $n=5, x=2 \times 5+1=11$ So, the elements of B are 1, 3, 5, 7, 9 and 11 , B = {1, 3, 5, 7, 9, 11}...
Read More →Split 207 into three parts such that these
Question: Split 207 into three parts such that these are in AP and the product of the two smaller parts is 4623. Solution: Let the three parts of the number 207 are (a d), a and (a + d), which are in AP Now, by given condition, $\Rightarrow \quad$ Sum of these parts $=207$ $\Rightarrow \quad a-d+a+a+d=207$ $\Rightarrow \quad 3 a=207$ $a=69$ Given that, product of the two smaller parts $=4623$ $\Rightarrow \quad a(a-d)=4623$ $\Rightarrow \quad 69 \cdot(69-d)=4623$ $\Rightarrow \quad 69-d=67$ $\Ri...
Read More →List all the elements of each of the sets given below.
Question: List all the elements of each of the sets given below. A = {x : x = 2n, n ϵ N and n 5}. Solution: Given: x = 2n and n 5 $\Rightarrow \mathrm{n}=1,2,3,4$ and $5[\because \mathrm{n} \in \mathrm{N}]$ Given $x=2 n$ $n=1, x=2 \times 1=2$ $n=2, x=2 \times 2=4$ $n=3, x=2 \times 3=6$ $n=4, x=2 \times 4=8$ $n=5, x=2 \times 5=10$ So, the elements of A are 2, 4, 6, 8 and 10 , A = {2, 4, 6, 8, 10}...
Read More →Determine k, so that k2 + 4k + 8,
Question: Determine k, so that k2+ 4k + 8, 2k2+ 3k + 6 and 3k2+ 4k + 4 are three consecutive terms of an AP. Solution: Since, $k^{2}+4 k+8,2 k^{2}+3 k+6$ and $3 k^{2}+4 k+4$ are consecutive terms of an AP $2 k^{2}+3 k+6-\left(k^{2}+4 k+8\right)=3 k^{2}+4 k+4-\left(2 k^{2}+3 k+6\right)=$ Common difference $\Rightarrow \quad 2 k^{2}+3 k+6-k^{2}-4 k-8=3 k^{2}+4 k+4-2 k^{2}-3 k-6$ $\Rightarrow \quad k^{2}-k-2=k^{2}+k-2$ $\Rightarrow \quad-k=k \Rightarrow 2 k=0 \Rightarrow k=0$...
Read More →Write the following sets iroster from:
Question: Write the following sets iroster from: K = {x : ϵ N, x is a multiple of 5 and x2 400} Solution: Multiple of 5 are 5, 10, 15, 20, 25, 30, So, $5^{2}=25$ $10^{2}=100$ $15^{2}=225$ $20^{2}=400$ $25^{2}=625400$ The elements which are multiple of 5 and x2 400 are 5, 10, 15 So, K = {5, 10, 15}...
Read More →Show that
Question: Show that $f(x)=\left\{\begin{array}{rrr}\frac{|x-a|}{x-a}, \text { when } x \neq a \\ 1 , \text { when } x=a\end{array}\right.$ is discontinuous atx=a. Solution: The given function can be rewritten as: $f(x)=\left\{\begin{array}{l}\frac{x-a}{x-a}, \text { when } xa \\ \frac{a-x}{x-a}, \text { when } xa \\ 1, \quad \text { when } x=a\end{array}\right.$ $\Rightarrow f(x)=\left\{\begin{array}{c}1, \text { when } xa \\ -1, \text { when } xa \\ 1, \text { when } x=a\end{array}\right.$ $\Ri...
Read More →Write the following sets in roster from:
Question: Write the following sets in roster from: J = {x : ϵ R and x2 + x 12 = 0}. Solution: The given equation is: $x^{2}+x-12=0$ $\Rightarrow x^{2}+4 x-3 x-12=0$ $\Rightarrow x(x+4)-3(x+4)=0$ $\Rightarrow(x-3)(x+4)=0$ $\Rightarrow x-3=0$ or $x+4=0$ $\Rightarrow x=3$ or $x=-4$ therefore, the solution set of the given equation can be written in roaster form as {3, -4} So, J = {3, -4}...
Read More →Show that
Question: Show that $f(x)=\left\{\begin{array}{rrr}\frac{|x-a|}{x-a}, \text { when } x \neq a \\ 1 , \text { when } x=a\end{array}\right.$ is discontinuous atx=a. Solution: The given function can be rewritten as: $f(x)=\left\{\begin{array}{l}\frac{x-a}{x-a}, \text { when } xa \\ \frac{a-x}{x-a}, \text { when } xa \\ 1, \quad \text { when } x=a\end{array}\right.$ $\Rightarrow f(x)=\left\{\begin{array}{c}1, \text { when } xa \\ -1, \text { when } xa \\ 1, \text { when } x=a\end{array}\right.$ $\Ri...
Read More →The volume of a cuboidal box is 48 cm
Question: The volume of a cuboidal box is 48 cm3. If its height and length are 3 cm and 4 cm respectively, find its breadth. Solution: Suppose that the breadth of the box is $b \mathrm{~cm}$. Volume of the cuboidal box $=48 \mathrm{~cm}^{3}$ Height of the box $=3 \mathrm{~cm}$ Length of the box $=4 \mathrm{~cm}$ Now, volume of box $=$ length $\times$ breadth $\times$ height $\Rightarrow 48=4 \times b \times 3$ $\Rightarrow 48=12 \times b$ $\Rightarrow b=\frac{48}{12}=4 \mathrm{~cm}$ $\therefore$...
Read More →Find whether 55 is a term of
Question: Find whether 55 is a term of the AP 7, 10, 13, or not. If yes, find which term it is. Solution: Yes, let the first term, common difference and the number of terms of an AP are a, d and n respectively. Let the nth term of an AP be 55. i.e., Tn= 55. Let the $n$th term of an AP be 55. i.e.. $T_{n}=55$. We know that, the $n$th term of an AP, $T_{n}=a+(n-1) d$...(i) Given that, first term $(a)=7$ and common difference $(d)=10-7=3$ From Eq. (i), $\quad 55=7+(n-1) \times 3$ $\Rightarrow \quad...
Read More →Write the following sets in roster from:
Question: Write the following sets in roster from: H = {x : x is a perfect square and x 50}. Solution: Perfect squares are: $0^{2}=0$ $1^{2}=1$ $2^{2}=4$ $3^{2}=9$ $4^{2}=16$ $5^{2}=25$ $6^{2}=36$ $7^{2}=49$ $8^{2}=6450$ The elements which are perfect square and x 50 are 0, 1, 2, 3, 4, 5, 6, 7...
Read More →How many soap cakes can be placed in a box of size 56 cm × 0.4 m × 0.25 m,
Question: How many soap cakes can be placed in a box of size 56 cm 0.4 m 0.25 m, if the size of a soap cake is 7 cm 5 cm 2.5 cm? Solution: Dimension of a soap cake $=7 \mathrm{~cm} \times 5 \mathrm{~cm} \times 2.5 \mathrm{~cm}$ Its volum $e=$ length $\times$ breadth $\times$ height $=(7 \times 5 \times 2.5) \mathrm{cm}^{3}=87.5 \mathrm{~cm}^{3}$ Also, the dimension of the box that contains the soap cakes is $56 \mathrm{~cm} \times 0.4 \mathrm{~m} \times 0.25 \mathrm{~m}$, i.e., $56 \mathrm{~cm} ...
Read More →Show that
Question: Show that $f(x)=\left\{\begin{array}{ccc}\frac{x-|x|}{2}, \text { when } x \neq 0 \\ 2, \text { when } x=0\end{array}\right.$ is discontinuous atx= 0. Solution: The given function can be rewritten as: $f(x)=\left\{\begin{aligned} \frac{x-x}{2}, \text { when } x0 \\ \frac{x+x}{2}, \text { when } x0 \\ 2, \text { when } x=0 \end{aligned}\right.$ $\Rightarrow f(x)=\left\{\begin{array}{l}0, \text { when } x0 \\ x, \text { when } x0 \\ 2, \text { when } x=0\end{array}\right.$ We observe $(\...
Read More →Write the following sets in roster from
Question: Write the following sets in roster from G = {x : x is a prime number and 80 x 100}. Solution: Prime number = Those number which is divisible by 1 and the number itself. Prime numbers are 2, 3, 5, 7, 11, 13, The numbers 80 x 100 are 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99, 100 The elements which are prime and lies between 80 and 100 are 83, 89, 97 So, G = {83, 89, 97}...
Read More →The weight of a metal block of size 5 cm by 4 cm by 3 cm is 1 kg.
Question: The weight of a metal block of size 5 cm by 4 cm by 3 cm is 1 kg. Find the weight of a block of the same metal of size 15 cm by 8 cm by 3 cm. Solution: The weight of the metal block of dimension $5 \mathrm{~cm} \times 4 \mathrm{~cm} \times 3 \mathrm{~cm}$ is $1 \mathrm{~kg}$. Its volume $=$ length $\times$ breadth $\times$ height $=(5 \times 4 \times 3) \mathrm{cm}^{3}=60 \mathrm{~cm}^{3}$ i. e., the weight of $60 \mathrm{~cm}^{3}$ of the metal is $1 \mathrm{~kg}$ Again, the dimension ...
Read More →Write the following sets in roster from:
Question: Write the following sets in roster from: F = {x : x is a letter in the word MATHEMATICS}. Solution: There are 11 letters in the word MATHEMATICS, out of which M, A and T are repeated. So, F = {M, A, T, H, E, I, C, S}...
Read More →A tea-packet measures 10 cm × 6 cm × 4 cm.
Question: A tea-packet measures 10 cm 6 cm 4 cm. How many such tea-packets can be placed in a cardboard box of dimensions 50 cm 30 cm 0.2 m? Solution: Dimension of a tea packet is $10 \mathrm{~cm} \times 6 \mathrm{~cm} \times 4 \mathrm{~cm}$. Volume of a tea packet $=$ length $\times$ breadth $\times$ height $=(10 \times 6 \times 4) \mathrm{cm}^{3}=240 \mathrm{~cm}^{3}$ Also, it is given that the dimension of the cardboard box is $50 \mathrm{~cm} \times 30 \mathrm{~cm} \times 0.2 \mathrm{~m}$, i...
Read More →If the 9th term of an AP is zero,
Question: If the 9th term of an AP is zero, then prove that its 29th term is twice its 19th term. Solution: Let the first term, common difference and number of terms of an AP are a, d and n respectively. Given that, $\quad$ 9th term of an AP, $T_{9}=0 \quad\left[\because n\right.$th term of an AP, $\left.T_{n}=a+(n-1) d\right]$ $\Rightarrow \quad a+(9-1) d=0$ $\Rightarrow \quad a+8 d=0 \Rightarrow a=-8 d$ $\ldots$ (i) Now, its 19th term, $T_{19}=a+(19-1) d$ $=-8 d+18 d \quad$ [from Eq. (i)] $=10...
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