Find the mean deviation about median for the following data:
Question: Find the mean deviation about median for the following data:/spanbr data-mce-bogus="1"/ppimg src="https://www.esaral.com/qdb/uploads/2021/12/23/image83305.png" alt="" Solution: The following table is formed. The class interval containing the $\left(\frac{\mathrm{N}}{2}\right)$ or $25^{\text {th }}$ item is $20-30$. Therefore, 20 30 is the median class. It is known that, Median $=l+\frac{\frac{\mathrm{N}}{2}-\mathrm{C}}{f} \times h$ Here, $I=20, C=14, f=14, h=10$, and $N=50$ $\therefore...
Read More →Find the mean deviation about the mean for the data
Question: Find the mean deviation about the mean for the data Solution: The following table is formed. Here, $N=\sum_{i=1}^{6} f_{i}=100, \sum_{i=1}^{6} f_{i} x_{i}=12530$ $\therefore \overline{\mathrm{x}}=\frac{1}{\mathrm{~N}} \sum_{\mathrm{i}=1}^{6} \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}=\frac{1}{100} \times 12530=125.3$ M.D. $(\bar{x})=\frac{1}{N} \sum_{i=1}^{6} f_{i}\left|x_{i}-\bar{x}\right|=\frac{1}{100} \times 1128.8=11.28$...
Read More →Find the mean deviation about the mean for the data.
Question: Find the mean deviation about the mean for the data. Solution: The following table is formed. Here, $N=\sum_{i=1}^{8} f_{i}=50, \sum_{i=1}^{8} f_{i} x_{i}=17900$ $\therefore \bar{x}=\frac{1}{N} \sum_{i=1}^{s} f_{i} x_{i}=\frac{1}{50} \times 17900=358$ M.D. $(\bar{x})=\frac{1}{N} \sum_{i=1}^{8} \mathrm{f}_{i}\left|\mathrm{x}_{i}-\bar{x}\right|=\frac{1}{50} \times 7896=157.92$...
Read More →Find the mean deviation about the median for the data
Question: Find the mean deviation about the median for the data Solution: The given observations are already in ascending order. Adding a column corresponding to cumulative frequencies of the given data, we obtain the following table. Here, $N=29$, which is odd. $\therefore$ Median $=\left(\frac{29+1}{2}\right)^{\text {th }}$ observation $=15^{\text {th }}$ observation This observation lies in the cumulative frequency 21, for which the corresponding observation is 30. $\therefore$ Median $=30$ T...
Read More →Find the mean deviation about the median for the data.
Question: Find the mean deviation about the median for the data. Solution: The given observations are already in ascending order. Adding a column corresponding to cumulative frequencies of the given data, we obtain the following table. Here, N = 26, which is even. Median is the mean of $13^{\text {th }}$ and $14^{\text {th }}$ observations. Both of these observations lie in the cumulative frequency 14 , for which the corresponding observation is 7 . $\therefore$ Median $=\frac{13^{\text {bh }} \...
Read More →Find the mean deviation about the mean for the data
Question: Find the mean deviation about the mean for the data Solution: $N=\sum_{i=1}^{5} f_{i}=80, \sum_{i=1}^{5} f_{i} x_{i}=4000$ $\therefore \overline{\mathrm{x}}=\frac{1}{\mathrm{~N}} \sum_{\mathrm{i}=1}^{5} \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}=\frac{1}{80} \times 4000=50$ $\operatorname{MD}(\bar{x}) \frac{1}{N} \sum_{i=1}^{5} f_{i}\left|x_{i}-\bar{x}\right|=\frac{1}{80} \times 1280=16$...
Read More →Find the mean deviation about the mean for the data.
Question: Find the mean deviation about the mean for the data. Solution: $\mathrm{N}=\sum_{i=1}^{5} \mathrm{f}_{i}=25$ $\sum_{i=1}^{5} f_{i} x_{i}=350$ $\therefore \overline{\mathrm{X}}=\frac{1}{\mathrm{~N}} \sum_{\mathrm{i}=1}^{5} \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}=\frac{1}{25} \times 350=14$ $\therefore \mathrm{MD}(\overline{\mathrm{x}})=\frac{1}{\mathrm{~N}} \sum_{\mathrm{i}=1}^{3} \mathrm{f}_{\mathrm{i}}\left|\mathrm{x}_{\mathrm{i}}-\overline{\mathrm{x}}\right|=\frac{1}{25} \tim...
Read More →Find the mean deviation about the median for the data
Question: Find the mean deviation about the median for the data 36, 72, 46, 42, 60, 45, 53, 46, 51, 49 Solution: The given data is 36, 72, 46, 42, 60, 45, 53, 46, 51, 49 Here, the number of observations is 10, which is even. Arranging the data in ascending order, we obtain 36, 42, 45, 46, 46, 49, 51, 53, 60, 72 Median $\mathrm{M}=\frac{\left(\frac{10}{2}\right)^{t h} \text { observation }+\left(\frac{10}{2}+1\right)^{m} \text { observation }}{2}$ $=\frac{5^{\text {th }} \text { observation }+6^{...
Read More →The line $y=x+1$ is a tangent to the curve $y^{2}=4 x$ at the point
Question: The line $y=x+1$ is a tangent to the curve $y^{2}=4 x$ at the point (A) (1, 2) (B) (2, 1) (C) (1, 2) (D) (1, 2) Solution: The equation of the given curve is $y^{2}=4 x$. Differentiating with respect tox, we have: $2 y \frac{d y}{d x}=4 \Rightarrow \frac{d y}{d x}=\frac{2}{y}$ Therefore, the slope of the tangent to the given curve at any point (x,y) is given by, $\frac{d y}{d x}=\frac{2}{y}$ The given line is $y=x+1$ (which is of the form $y=m x+c$ ) Slope of the line = 1 The liney=x+ 1...
Read More →Find the mean deviation about the median for the data.
Question: Find the mean deviation about the median for the data. 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17 Solution: The given data is 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17 Here, the numbers of observations are 12, which is even. Arranging the data in ascending order, we obtain 10, 11, 11, 12, 13, 13, 14, 16, 16, 17, 17, 18 Median, $\mathrm{M}=\frac{\left(\frac{12}{2}\right)^{\text {min }} \text { observation }+\left(\frac{12}{2}+1\right)^{\text {mi }} \text { observation }}{2}$ $=...
Read More →Find the mean deviation about the mean for the data
Question: Find the mean deviation about the mean for the data 38, 70, 48, 40, 42, 55, 63, 46, 54, 44 Solution: The given data is 38, 70, 48, 40, 42, 55, 63, 46, 54, 44 Mean of the given data, $\bar{x}=\frac{38+70+48+40+42+55+63+46+54+44}{10}=\frac{500}{10}=50$ The deviations of the respective observations from the mean $\bar{x}$, i.e. $x_{i}-\bar{x}$, are $-12,20,-2,-10,-8,5,13,-4,4,-6$ The absolute values of the deviations, i.e. $\left|x_{i}-\bar{x}\right|$, are 12, 20, 2, 10, 8, 5, 13, 4, 4, 6...
Read More →Find the mean deviation about the mean for the data
Question: Find the mean deviation about the mean for the data 4, 7, 8, 9, 10, 12, 13, 17 Solution: The given data is 4, 7, 8, 9, 10, 12, 13, 17 Mean of the data, $\bar{x}=\frac{4+7+8+9+10+12+13+17}{8}=\frac{80}{8}=10$ The deviations of the respective observations from the mean $\bar{x}$, i.e. $x_{i}-\bar{x}$, are $-6,-3,-2,-1,0,2,3,7$ The absolute values of the deviations, i.e. $\left|x_{i}-\bar{x}\right|$, are 6, 3, 2, 1, 0, 2, 3, 7 The required mean deviation about the mean is M.D. $(\bar{x})=...
Read More →The slope of the normal to the curve
Question: The slope of the normal to the curve $y=2 x^{2}+3 \sin x$ at $x=0$ is (A) 3 (B) $\frac{1}{3}$ (C) $-3$ (D) $-\frac{1}{3}$ Solution: The equation of the given curve is $y=2 x^{2}+3 \sin x$. Slope of the tangent to the given curve atx= 0 is given by, $\left.\left.\frac{d y}{d x}\right]_{x-0}=4 x+3 \cos x\right]_{x=0}=0+3 \cos 0=3$ Hence, the slope of the normal to the given curve atx= 0 is $\frac{-1}{\text { Slope of the tangent at } x=0}=\frac{-1}{3}$ The correct answer is D....
Read More →Find the equation of the tangent to the curve
Question: Find the equation of the tangent to the curve $y=\sqrt{3 x-2}$ which is parallel to the line $4 x-2 y+5=0$. Solution: The equation of the given curve is $y=\sqrt{3 x-2}$. The slope of the tangent to the given curve at any point (x,y) is given by, $\frac{d y}{d x}=\frac{3}{2 \sqrt{3 x-2}}$ The equation of the given line is $4 x-2 y+5=0$. $4 x-2 y+5=0 \Rightarrow y=2 x+\frac{5}{2}($ which is of the form $y=m x+c)$ Slope of the line = 2 Now, the tangent to the given curve is parallel to t...
Read More →Find the equations of the tangent and normal to the hyperbola
Question: Find the equations of the tangent and normal to the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ at the point $\left(x_{0}, y_{0}\right)$. Solution: Differentiating $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ with respect to $x$, we have: $\frac{2 x}{a^{2}}-\frac{2 y}{b^{2}} \frac{d y}{d x}=0$ $\Rightarrow \frac{2 y}{b^{2}} \frac{d y}{d x}=\frac{2 x}{a^{2}}$ $\Rightarrow \frac{d y}{d x}=\frac{b^{2} x}{a^{2} y}$ Therefore, the slope of the tangent at $\left(x_{0}, y_{0}\right)$...
Read More →Find the values of x in each of the following:
Question: Find the values of x in each of the following: (i) $2^{5 x} \div 2 x=\sqrt[5]{2^{20}}$ (ii) $\left(2^{3}\right)^{4}=\left(2^{2}\right)^{x}$ (iii) $\left(\frac{3}{5}\right)^{x}\left(\frac{5}{3}\right)^{2 x}=\frac{125}{27}$ (iv) $5^{x-2} \times 3^{2 x-3}=135$ (v) $2^{x-7} \times 5^{x-4}=1250$ (vi) $(\sqrt[3]{4})^{2 x+\frac{1}{2}}=\frac{1}{32}$ (vii) $5^{2 x+3}=1$ (viii) $(13)^{\sqrt{x}}=4^{4}-3^{4}-6$ (ix) $\left(\sqrt{\frac{3}{5}}\right)^{x+1}=\frac{125}{27}$ Solution: From the followin...
Read More →Prove that the curves
Question: Prove that the curves $x=y^{2}$ and $x y=k$ cut at right angles if $8 k^{2}=1$. [Hint: Two curves intersect at right angle if the tangents to the curves at the point of intersection are perpendicular to each other.] Solution: The equations of the given curves are given as $x=y^{2}$ and $x y=k$. Putting $x=y^{2}$ in $x y=k$, we get: $y^{3}=k \Rightarrow y=k^{\frac{1}{3}}$ $\therefore x=k^{\frac{2}{3}}$ Thus, the point of intersection of the given curves is $\left(k^{\frac{2}{3}}, k^{\fr...
Read More →Find the equations of the tanqent and normal to the parabola
Question: Find the equations of the tanqent and normal to the parabola $y^{2}=4 a x$ at the point ( $a t^{2}, 2 a t$ ). Solution: The equation of the given parabola is $y^{2}=4 a x$. On differentiating $y^{2}=4 a x$ with respect to $x$, we have: $2 y \frac{d y}{d x}=4 a$ $\Rightarrow \frac{d y}{d x}=\frac{2 a}{y}$ $\therefore$ The slope of the tangent at $\left(a t^{2}, 2 a t\right)$ is $\left.\frac{d y}{d x}\right]_{\left(a t^{2}, 2 a t\right)}=\frac{2 a}{2 a t}=\frac{1}{t}$. Then, the equation...
Read More →Write the following statement in five different ways, conveying the same meaning.
Question: Write the following statement in five different ways, conveying the same meaning. p: If triangle is equiangular, then it is an obtuse angled triangle. Solution: The given statement can be written in five different ways as follows. (i) A triangle is equiangular implies that it is an obtuse-angled triangle. (ii) A triangle is equiangular only if it is an obtuse-angled triangle. (iii) For a triangle to be equiangular, it is necessary that the triangle is an obtuse-angled triangle. (iv) Fo...
Read More →Check the validity of the statements given below by the method given against it.
Question: Check the validity of the statements given below by the method given against it. Check the validity of the statements given below by the method given against it. (i)p: The sum of an irrational number and a rational number is irrational (by contradiction method). (ii) $q$ : If $n$ is a real number with $n3$, then $n^{2}9$ (by contradiction method). Solution: (i) The given statement is as follows.p: the sum of an irrational number and a rational number is irrational. Let us assume that t...
Read More →Find the equation of the normals to the curve
Question: Find the equation of the normals to the curve $y=x^{3}+2 x+6$ which are parallel to the line $x+14 y+4=0$. Solution: The equation of the given curve is $y=x^{3}+2 x+6$. The slope of the tangent to the given curve at any point (x,y) is given by, $\frac{d y}{d x}=3 x^{2}+2$ Slope of the normal to the given curve at any point (x,y) = $=\frac{-1}{3 x^{2}+2}$ The equation of the given line isx+ 14y+ 4 = 0. $x+14 y+4=0 \Rightarrow y=-\frac{1}{14} x-\frac{4}{14}($ which is of the form $y=m x+...
Read More →Given below are two statements
Question: Given below are two statements p:25 is a multiple of 5. q: 25 is a multiple of 8. Write the compound statements connecting these two statements with And and Or. In both cases check the validity of the compound statement. Solution: The compound statement with And is 25 is a multiple of 5 and 8. This is a false statement, since 25 is not a multiple of 8. The compound statement with Or is 25 is a multiple of 5 or 8. This is a true statement, since 25 is not a multiple of 8 but it is a mul...
Read More →Re write each of the following statements in the form “p if and only if q”.
Question: Re write each of the following statements in the form pif and only ifq. (i)p: If you watch television, then your mind is free and if your mind is free, then you watch television. (ii)q: For you to get an A grade, it is necessary and sufficient that you do all the homework regularly. (iii)r: If a quadrilateral is equiangular, then it is a rectangle and if a quadrilateral is a rectangle, then it is equiangular. Solution: (i) You watch television if and only if your mind is free. (ii) You...
Read More →Write each of the statements in the form “if p, then q”.
Question: Write each of the statements in the form ifp, thenq. (i)p: It is necessary to have a password to log on to the server. (ii)q: There is traffic jam whenever it rains. (iii)r: You can access the website only if you pay a subscription fee. Solution: (i) Statementpcan be written as follows. If you log on to the server, then you have a password. (ii) Statementqcan be written as follows. If it rains, then there is a traffic jam. (iii) Statementrcan be written as follows. If you can access th...
Read More →State the converse and contrapositive of each of the following statements:
Question: State the converse and contrapositive of each of the following statements: (i)p: A positive integer is prime only if it has no divisors other than 1 and itself. (ii)q: I go to a beach whenever it is a sunny day. (iii)r: If it is hot outside, then you feel thirsty. Solution: (i) Statementpcan be written as follows. If a positive integer is prime, then it has no divisors other than 1 and itself. The converse of the statement is as follows. If a positive integer has no divisors other than...
Read More →