The table below shows the salaries of 280 persons.
Question: The table below shows the salaries of 280 persons. caluculate the median and mode of the data. Solution: First, we construct a cumulative frequency table. $\therefore$$\frac{N}{2}=\frac{280}{2}=140$ (i) Here, median class is $10-15$, because 140 lies in it. Lower limit $(l)=10$, Frequency $(f)=133$, Cumulative frequency $(c f)=49$ and class width $(h)=5$ $\therefore$ Median $=l+\frac{\left(\frac{N}{2}-c f\right)}{f} \times h$ $=10+\frac{(140-49)}{133} \times 5$ $=10+\frac{91 \times 5}{...
Read More →Factorise:
Question: Factorise:x2 22x+ 117 Solution: The given expression is $x^{2}-22 x+117$. Find two numbers that follow the conditions given below: Sum $=-22$ Product $=117$ Clearly, the numbers are $-13$ and $-9$. $x^{2}-22 x+117=x^{2}-13 x-9 x+117$ $=x(x-13)-9(x-13)$ $=(x-13)(x-9)$...
Read More →Differentiate the following functions with respect to x :
Question: Differentiate the following functions with respect to $x$ : $\frac{e^{x} \sin x}{\left(x^{2}+2\right)^{3}}$ Solution: Let $y=\frac{e^{x} \sin x}{\left(x^{2}+2\right)^{2}}$ On differentiating y with respect to $x$, we get $\frac{d y}{d x}=\frac{d}{d x}\left[\frac{e^{x} \sin x}{\left(x^{2}+2\right)^{3}}\right]$ Recall that $\left(\frac{\mathrm{u}}{\mathrm{v}}\right)^{\prime}=\frac{\mathrm{vu}^{\prime}-\mathrm{uv}^{\prime}}{\mathrm{v}^{2}}$ (quotient rule) $\Rightarrow \frac{d y}{d x}=\fr...
Read More →Differentiate the following functions with respect to x :
Question: Differentiate the following functions with respect to $x$ : $\frac{e^{x} \sin x}{\left(x^{2}+2\right)^{3}}$ Solution: Let $y=\frac{e^{x} \sin x}{\left(x^{2}+2\right)^{2}}$ On differentiating y with respect to $x$, we get $\frac{d y}{d x}=\frac{d}{d x}\left[\frac{e^{x} \sin x}{\left(x^{2}+2\right)^{3}}\right]$ Recall that $\left(\frac{\mathrm{u}}{\mathrm{v}}\right)^{\prime}=\frac{\mathrm{vu}^{\prime}-\mathrm{uv}^{\prime}}{\mathrm{v}^{2}}$ (quotient rule) $\Rightarrow \frac{d y}{d x}=\fr...
Read More →Refer to Q.5 above. Draw the less than type
Question: Refer to Q.5 above. Draw the less than type and more than type ogives for the data and use them to find the meadian weight. Solution: For less than type table we follow the Q.5. Here, we observe that, the weight of all 70 packets is more than or equal to 200. Since, 13 packets lie in the interval 200-201. So, the weight of 70 -13 = 57 packets is more than or equal to 201. Continuing in this manner we will get remaining more than or equal to 202, 203, 204, 205 and 206. To draw the less ...
Read More →Factorise:
Question: Factorise:y2 21y+ 90 Solution: The given expression is $y^{2}-21 y+90$. Find two numbers that follow the conditions given below: Sum $=-21$ Product $=90$ Clearly, the numbers are $-15$ and $-6$. $y^{2}-21 y+90=y^{2}-15 y-6 y+90$ $=y(y-15)-6(y-15)$ $=(y-15)(y-6)$...
Read More →Refer to Q.4 above. Draw the less than type
Question: Refer to Q.4 above. Draw the less than type ogive for this data and use it to find the median weight. Solution: We observe that, the number of packets less than 200 is 0, Similarly, less than 201 include the number of packets from 0-200 as well as the number of packets from 200-201. So, the total number of packets less than 201 is 0 + 13 = 13. We say that, the cumulative frequency of the class 200-201 is 13. Similarly, for other class. To draw the less than type ogive, we plot the poin...
Read More →Factorise:
Question: Factorise:x2 17x+ 16 Solution: The given expression is $x^{2}-17 x+16$. Find two numbers that follow the conditions given below: Sum $=-17$ Product $=16$ Clearly, the numbers are $-16$ and $-1$. $x^{2}-17 x+16=x^{2}-16 x-x+16$ $=x(x-16)-1(x-16)$ $=(x-16)(x-1)$...
Read More →The weights of tea in 70 packets are shown
Question: The weights of tea in 70 packets are shown in the following table Find the mean weight of packets. Solution: First,we find the class marks of the given data as follows, Here, (assume mean) $a=203.5$ and (class width) $\quad h=1$ By assumed mean method, $\operatorname{Mean}(\bar{x})=a+\frac{\Sigma f d_{i}}{\sum f_{i}}$ $=203.5-\frac{108}{70}$ $=203.5-1.54=201.96$ Hence, the required mean weight is 201.96 g...
Read More →Factorise:
Question: Factorise:x2 23x+ 42 Solution: The given expression is $x^{2}-23 x+42$. Find two numbers that follow the conditions given below: Sum $=-23$ Product $=42$ Clearly, the numbers are $-21$ and $-2$. $x^{2}-23 x+42=x^{2}-21 x-2 x+42$ $=x(x-21)-2(x-21)$ $=(x-21)(x-2)$...
Read More →Differentiate the following functions with respect to x :
Question: Differentiate the following functions with respect to $x$ : $\sin ^{-1}\left(\frac{x}{\sqrt{x^{2}+a^{2}}}\right)$ Solution: Let $y=\sin ^{-1}\left(\frac{x}{\sqrt{x^{2}+a^{2}}}\right)$ On differentiating $y$ with respect to $x$, we get $\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left[\sin ^{-1}\left(\frac{\mathrm{x}}{\sqrt{\mathrm{x}^{2}+\mathrm{a}^{2}}}\right)\right]$ We have $\frac{\mathrm{d}}{\mathrm{dx}}\left(\sin ^{-1} \mathrm{x}\right)=\frac{1}{\sqrt{1-\mathrm{...
Read More →Factorise:
Question: Factorise:x2 10x+ 24 Solution: The given expression is $x^{2}-10 x+24$. Find two numbers that follow the conditions given below: Sum $=-10$ Product $=24$ Clearly, the numbers are $-6$ and $-4$. $x^{2}-10 x+24=x^{2}-6 x-4 x+24$ $=x(x-6)-4(x-6)$ $=(x-6)(x-4)$...
Read More →Find the mean age of 100 residents
Question: Find the mean age of 100 residents of a town from the following data. Solution: Here, we observe that, all 100 residents of a town have age equal and above 0. Since, 90 residents of a town have age equal and above 10. So, 100 90 = 10 residents lies in the interval 0-10 and so on. Continue in this manner, we get frequency of all class intervals. Now, we construct the frequency distribution table. Here, (assumed mean) $a=35$ and (class width) $h=10$ By step deviation method, $\operatorna...
Read More →Factorise:
Question: Factorise:p2+ 6p 16 Solution: The given expression is $p^{2}+6 p-16$. Find two numbers that follow the conditions given below: Sum $=6$ Product $=-16$ Clearly, the numbers are 8 and $-2$. $p^{2}+6 p-16=p^{2}+8 p-2 p-16$ $=p(p+8)-2(p+8)$ $=(p+8)(p-2)$...
Read More →Prove the following
Question: Solution: Here, we observe that, 5 students have scored marks below 10, i.e. it lies between class interval 0-10 and 9 students have scored marks below 20, So, (9 5) = 4 students lies in the class interval 10-20. Continuing in the same manner, we get the complete frequency distribution table for given data. Here, (assumed mean) $a=45$ and (class width) $h=10$ By step deviation method, $\operatorname{Mean}(\bar{x})=a+\frac{\Sigma f_{i} u_{i}}{\Sigma f_{i}} \times h=45+\frac{29}{85} \tim...
Read More →Differentiate the following functions with respect to x :
Question: Differentiate the following functions with respect to $x$ : $\left(\sin ^{-1} x^{4}\right)^{4}$ Solution: Let $y=\left(\sin ^{-1} x^{4}\right)^{4}$ On differentiating $y$ with respect to $x$, we get $\frac{d y}{d x}=\frac{d}{d x}\left[\left(\sin ^{-1} x^{4}\right)^{4}\right]$ We know $\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{\mathrm{n}}\right)=\mathrm{nx}^{\mathrm{n}-1}$ $\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=4\left(\sin ^{-1} \mathrm{x}^{4}\right)^{4-1} \frac{\mathrm{d}}{...
Read More →Factorise:
Question: Factorise:q2 10q+ 21 Solution: The given expression is $q^{2}-10 q+21$. Find two numbers that follow the conditions given below: Sum $=-10$ Product $=21$ Clearly, the numbers are $-7$ and $-3$. $q^{2}-10 q+21=q^{2}-7 q-3 q+21$ $=q(q-7)-3(q-7)$ $=(q-7)(q-3)$...
Read More →Find the mean marks of students for the following distribution
Question: Find the mean marks of students for the following distribution Solution:...
Read More →Factorise:
Question: Factorise:x2+ 13x+ 40 Solution: The given expression is $x^{2}+13 x+40$. Find two numbers that follow the conditions given below: Sum $=13$ Product $=40$ Clearly, the numbers are 8 and 5 . $x^{2}+13 x+40=x^{2}+8 x+5 x+40$ $=x(x+8)+5(x+8)$ $=(x+8)(x+5)$...
Read More →At a fete, cards bearing numbers 1 to 1000,
Question: At a fete, cards bearing numbers 1 to 1000, one number on one card, are put in a box. Each player selects one card at random and that card is not replaced. If the selected card has a perfect square greater than 500, the player wins a prize. What is the probability that (i) the first player wins a prize? (ii) the second player wins a prize, if the first has won? Solution: Given that,, at a fete, cards bearing numbers 1 to 1000 one number on one card, are put in a box. Each player select...
Read More →Factorise:
Question: Factorise:y2+ 19y+ 60 Solution: The given expression is $y^{2}+19 y+60$. Find two numbers that follow the conditions given below: Sum $=19$ Product $=60$ Clearly, the numbers are 15 and $4 .$ $y^{2}+19 y+60=y^{2}+15 y+4 y+60$ $=y(y+15)+4(y+15)$ $=(y+15)(y+4)$...
Read More →Differentiate the following functions with respect to x :
Question: Differentiate the following functions with respect to $x$ : $\log \left\{x+2+\sqrt{x^{2}+4 x+1}\right\}$ Solution: Let $y=\log \left(x+2+\sqrt{x^{2}+4 x+1}\right)$ On differentiating y with respect to $x$, we get $\frac{d y}{d x}=\frac{d}{d x}\left[\log \left(x+2+\sqrt{x^{2}+4 x+1}\right)\right]$ We know $\frac{d}{d x}(\log x)=\frac{1}{x}$ $\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{x+2+\sqrt{\mathrm{x}^{2}+4 \mathrm{x}+1}} \frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}+2+\sq...
Read More →A bag contains 24 balls of which x are red,
Question: A bag contains 24 balls of which x are red, 2x are white and 3x am are. A ball is selected at random. What is the probability that it (i) not red? (ii) white Solution: Given that, A bag contains total number of balls = 24 A bag contains number of red bails = 24 A bag contains number of white balls = 2x and a bag contains number of blue balls = x By condition, x + 2x + 3x = 24 ⇒ 6x = 24 x = 4 Number of red balls = x = 4 Number of white balls = 2x = 2 x 4 = 8 and number of blue balls = 3...
Read More →A lot consists of 48 mobile phones of which 42 are good,
Question: A lot consists of 48 mobile phones of which 42 are good, 3 have only minor defects and 3 have major defects. Varnika will buy a phone, if it is good but the trader will only buy a mobile, if it has no major defect. One phone is selected at random from the lot. What is the probability that it is (i) acceptable to Varnika? (ii) acceptable to the trader? Solution: Given, total number of mobile phones $n(S)=48$ (I) Let $E_{1}=$ Event that Varnika will buy a mobile phone $=$ Varnika buy onl...
Read More →Solve the following
Question: Factorise:x2+ 15x+ 56 Solution: The given expression is $x^{2}+15 x+56$. Find two numbers that follow the conditions given below: Sum $=15$ Product $=56$ Clearly, the numbers are 8 and 7 . $x^{2}+15 x+56=x^{2}+8 x+7 x+56$ $=x(x+8)+7(x+8)$ $=(x+8)(x+7)$...
Read More →