The times (in seconds) taken by 150 atheletes

Question: The times (in seconds) taken by 150 atheletes to run a 110 m hurdle race are tabulated below The number of atheletes who completed the race in less than 14.6 s is (a) 11 (b) 71 (c) 82 (d) 130 Solution: (c)The number of atheletes who completed the race in less than 14.6 = 2 + 4+ 5+71 =82...

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Factorise:

Question: Factorise: (i) 14x3+ 21x4y 28x2y2 (ii) 5 10t+ 20t2 Solution: (i) H.C.F. of $14 x^{3}, 21 x^{4} y$ and $28 x^{2} y^{2}$ is $7 x^{2}$. $\therefore 14 x^{3}+21 x^{4} y-28 x^{2} y^{2}=7 x^{2}\left(2 x+3 x^{2} y-4 y^{2}\right)$ (ii) H.C.F. of $-5,-10 t$ and $20 t^{2}$ is 5 . $\therefore-5-10 t+20 t^{2}=5\left(-1-2 t+4 t^{2}\right)$...

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Differentiate the following functions with respect to x :

Question: Differentiate the following functions with respect to $x$ : $\sin \left(\frac{1+x^{2}}{1-x^{2}}\right)$ Solution: Let $y=\sin \left(\frac{1+x^{2}}{1-x^{2}}\right)$ On differentiating $y$ with respect to $x$, we get $\frac{d y}{d x}=\frac{d}{d x}\left[\sin \left(\frac{1+x^{2}}{1-x^{2}}\right)\right]$ We know $\frac{d}{d x}(\sin x)=\cos x$ $\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\cos \left(\frac{1+\mathrm{x}^{2}}{1-\mathrm{x}^{2}}\right) \frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{1+\...

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consider the data.

Question: consider the data. The difference of the upper limit of the median class and the lower limit of the modal class is (a) 0 (b) 19 (c) 20 (d) 38 Solution: Here, $\frac{N}{2}=\frac{67}{2}=33.5$ which lies in the interval $125-145$. Hence, upper limit of median class is 145. ere, we see that the highest frequency is 20 which lies in 125-145. Hence, the lower limit of modal class is 125. Required difference = Upper limit of median class Lower limit of modal class = 145-125 = 20...

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Factorise:

Question: Factorise: (i) 9x3 6x2+ 12x (ii) 8x3 72xy+ 12x (iii) 18a3b3 27a2b3+ 36a3b2 Solution: (i) H.C.F. of $9 x^{3}, 6 x^{2}$ and $12 x$ is $3 x$. $\therefore 9 x^{3}-6 x^{2}+12 x=3 x\left(3 x^{2}-2 x+4\right)$ (ii) H.C.F. of $8 x^{3}, 72 x y$ and $12 x$ is $4 x$. $\therefore 8 x^{3}-72 x y+12 x=4 x\left(2 x^{2}-18 y+3\right)$ (iii) H.C.F. of $18 a^{3} b^{3}, 27 a^{2} b^{3}$ and $36 a^{3} b^{2}$ is $9 a^{2} b^{2}$. $\therefore 18 a^{3} b^{3}-27 a^{2} b^{3}+36 a^{3} b^{2}=9 a^{2} b^{2}(2 a b-3 ...

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Factorise:

Question: Factorise: (i) 24x3 36x2y (ii) 10x3 15x2 (iii) 36x3y 60x2y3z Solution: (i) H.C.F. of $24 x^{3}$ and $36 x^{2} y$ is $12 x^{2}$. $\therefore 24 x^{3}-36 x^{2} y=12 x^{2}(2 x-3 y)$ (ii) H.C.F. of $10 x^{3}$ and $15 x^{2}$ is $5 x^{3}$ $\therefore 10 x^{3}-15 x^{2}=5 x^{2}(2 x-3)$ (iii) H.C.F. of $36 x^{3} y$ and $60 x^{2} y^{3} z$ is $12 x^{2} y$. $\therefore 36 x^{3} y-60 x^{2} y^{3} z=12 x^{2} y\left(3 x-5 y^{2} z\right)$...

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For the following distribution,

Question: For the following distribution, the modal class is (a) 10-20 (b) 20-30 (c) 30-40 (d) 30-40 Solution: Here,we see that the highest frequency is 30. which lies in the interval 30-40....

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Factorise:

Question: Factorise: (i) 16a2 24ab (ii) 15ab2 20a2b (iii) 12x2y3 21x3y2 Solution: (i) H.C.F. of $16 a^{2}$ and $24 a b$ is $8 a$. $\therefore 16 a^{2}-24 a b=8 a(2 a-3 b)$ (ii) H.C.F. of $15 a b^{2}$ and $20 a^{2} b$ is $5 a b$. $\therefore 15 a b^{2}-20 a^{2} b=5 a b(3 b-4 a)$ (iii) H.C.F. of $12 x^{2} y^{3}$ and $21 x^{3} y^{2}$ is $3 x^{2} y^{2}$. $\therefore 12 x^{2} y^{3}-21 x^{3} y^{2}=3 x^{2} y^{2}(4 y-7 x)$...

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Consider the following frequency distribution

Question: Consider the following frequency distribution The upper limit of the median class is (a) 7 (b) 17.5 (c) 18 (d) 18.5 Solution:...

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Factorise:

Question: Factorise: (i) 12x+ 15 (ii) 14m 21 (iii) 9n 12n2 Solution: (i) $12 x+15=3(4 x+5)$ (ii) $14 m-21=7(2 m-3)$ (iii) $9 n-12 n^{2}=3 n(3-4 n)$...

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Differentiate the following functions with respect to x :

Question: Differentiate the following functions with respect to $x$ : $\sqrt{\frac{1+x}{1-x}}$ Solution: Let $y=\sqrt{\frac{1+x}{1-x}}$ On differentiating $y$ with respect to $x$, we get $\frac{d y}{d x}=\frac{d}{d x}\left(\sqrt{\frac{1+x}{1-x}}\right)$ $\Rightarrow \frac{d y}{d x}=\frac{d}{d x}\left[\left(\frac{1+x}{1-x}\right)^{\frac{1}{2}}\right]$ We know $\frac{d}{d x}\left(x^{n}\right)=n x^{n-1}$ $\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{2}\left(\frac{1+\mathrm{x}}{1-\mathrm{x}}...

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For the following distribution,

Question: For the following distribution, the sum of lower limits of the median class and modal class is (a) 15 (b) 25 (c) 30 (d) 35 Solution: Now, $\frac{N}{2}=\frac{66}{2}=33$, which lies in the interval $10-15$. Therefore, lower limit of the median class is 10. . ,The highest frequency is 20, which lies in the interval 15-20. Therefore, lower limit of modal class is 15. Hence, required sum is 10 + 15 = 25....

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The abscissa of the point of intersection

Question: The abscissa of the point of intersection of the less than type and of the more than type cumulative frequency curves of a grouped data gives its (a) mean (b) median (c) mode (d) All of these Solution: (b) Since, the intersection point of less than ogive and more than ogive gives the median on the abscissa....

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In the formula

Question: In the formula $\bar{x}=a+h\left(\frac{\Sigma f_{i} u_{i}}{\Sigma f_{i}}\right)$, for finding the mean of grouped frequency distribution $u_{i}$ is equal to (a) $\frac{x_{i}+a}{h}$ (b) $h\left(x_{i}-a\right)$ (c) $\frac{x_{i}-a}{h}$ (d) $\frac{a-x_{i}}{h}$ Solution: (c) Given, $\bar{x}=a+h\left(\frac{\Sigma f_{i} U_{i}}{\Sigma f_{j}}\right)$ Above formula is a step deviation formula. $u_{i}=\frac{x_{i}-a}{h}$...

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Tick (✓) the correct answer:

Question: Tick (✓) the correct answer: Ifx= 10, then the value of (4x2+ 20x+ 25) = ? (a) 256 (b) 425 (c) 625 (d) 575 Solution: (c) 625 $\left(4 x^{2}+20 x+25\right)$ $\Rightarrow(2 x)^{2}+2(2 x)(5)+(5)^{2}$ $\Rightarrow(2 x+5)^{2}$ $\Rightarrow(2(10)+5)^{2}$ $\Rightarrow(20+5)^{2}$ $\Rightarrow(25)^{2}$ $\Rightarrow 625$...

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Differentiate the following functions with respect to x :

Question: Differentiate the following functions with respect to $x$ : $\sqrt{\frac{1+x}{1-x}}$ Solution: Let $y=\sqrt{\frac{1+x}{1-x}}$ On differentiating $y$ with respect to $x$, we get $\frac{d y}{d x}=\frac{d}{d x}\left(\sqrt{\frac{1+x}{1-x}}\right)$ $\Rightarrow \frac{d y}{d x}=\frac{d}{d x}\left[\left(\frac{1+x}{1-x}\right)^{\frac{1}{2}}\right]$ We know $\frac{d}{d x}\left(x^{n}\right)=n x^{n-1}$ $\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{2}\left(\frac{1+\mathrm{x}}{1-\mathrm{x}}...

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Prove the following

Question: If $x_{i}{ }^{\prime}$ s are the mid-points of the class intervals of grouped data, $f_{i}{ }^{\prime}$ s are the corresponding frequencies and $\bar{x}$ is the mean, then $\Sigma\left(f_{i} x_{i}-\bar{x}\right)$ is equal to (a) 0 (b) -1 (c) 1 (d) 2 Solution: (a) $\because$ $\bar{x}=\frac{\Sigma f_{i} x_{i}}{n}$ $\therefore$ $\Sigma\left(f_{i} x_{i}-\bar{x}\right)=\Sigma f_{i} x_{i}-\Sigma \bar{x}$ $=n \bar{x}-n \bar{x}$ $[\because \Sigma \bar{x}=n \bar{x}]$ $=0$...

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Tick (✓) the correct answer:

Question: Tick (✓) the correct answer: If (ab) = 7 andab= 9, then (a2+b2) = ? (a) 67 (b) 31 (c) 40 (d) 58 Solution: (a) 67 $(a-b)=7$ $\Rightarrow$ Squaring both the sides : $\Rightarrow(a-b)^{2}=(7)^{2}$ $\Rightarrow\left(a^{2}+b^{2}-2 a b\right)=49$ $\Rightarrow\left(a^{2}+b^{2}\right)=49+2 a b$ $\Rightarrow\left(a^{2}+b^{2}\right)=49+2(9)$ $\Rightarrow\left(a^{2}+b^{2}\right)=49+18$ $\Rightarrow\left(a^{2}+b^{2}\right)=67$...

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While computing mean of grouped data,

Question: While computing mean of grouped data, we assume that the frequencies are (a) evenly distributed over all the classes (b) centred at the class marks of the classes (c) centred at the upper limits of the classes (d) centred at the lower limits of the classes Solution: (b)In computing the mean of grouped data, the frequencies are centred at the class marks of the classes....

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Tick (✓) the correct answer:

Question: Tick (✓) the correct answer: If (a+b) = 12 andab= 14, then (a2+b2) = ? (a) 172 (b) 116 (c) 165 (d) 126 Solution: (b) 116 $(a+b)=12$ $\Rightarrow$ Squaring both the sides: $\Rightarrow(a+b)^{2}=(12)^{2}$ $\Rightarrow\left(a^{2}+b^{2}+2 a b\right)=144$ $\Rightarrow\left(a^{2}+b^{2}\right)=144-2 a b$ $\Rightarrow\left(a^{2}+b^{2}\right)=144-2(14)$ $\Rightarrow\left(a^{2}+b^{2}\right)=144-28$ $\Rightarrow\left(a^{2}+b^{2}\right)=116$...

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In the formula

Question: In the formula $\bar{x}=a+\frac{\Sigma f_{i} d_{i}}{\Sigma f_{i}}$, for finding the mean of grouped data $d_{i}$ 's are deviation from $a$ of (a) lower limits of the classes (b) upper limits of the classes (c) mid-points of the classes (d) frequencies of the class marks Solution: (c) We know that, $d_{i}=x_{i}-\mathrm{a}$ i.e., $d_{i}$ 's are the deviation from a of mid-points of the classes....

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Tick (✓) the correct answer:

Question: Tick (✓) the correct answer: (197 203) = ? (a) 39991 (b) 39999 (c) 40009 (d) 40001 Solution: (a) 39991 $(197) \times(203)$ [using the identity (a+b) (a-b) =a2-b2] $\Rightarrow(200-3)(200+3)$ $\Rightarrow(200)^{2}-(3)^{2}$ $\Rightarrow 40000-9$ $\Rightarrow 39991$...

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Differentiate the following functions with respect to x :

Question: Differentiate the following functions with respect to $x$ : $(\log \sin x)^{2}$ Solution: Let $y=(\log \sin x)^{2}$ On differentiating y with respect to $x$, we get $\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left[(\log (\sin \mathrm{x}))^{2}\right]$ We know $\frac{d}{d x}\left(x^{n}\right)=n x^{n-1}$ $\Rightarrow \frac{d y}{d x}=2(\log (\sin x))^{2-1} \frac{d}{d x}[\log (\sin x)]$ [using chain rule] $\Rightarrow \frac{d y}{d x}=2 \log (\sin x) \frac{d}{d x}[\log (\...

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A pen stand made of wood is in the shap

Question: A pen stand made of wood is in the shape of a cuboid with four conical depressions and a cubical depression to hold the pens and pins, respectively. The dimensions of cubiod are 10 cm, 5 cm and 4 cm. The radius of each of the conical depressions is 0.5 cm and the depth is 2.1 cm. The edge of the cubical depression is 3 cm. Find the volume of the wood in the entire stand. Solution: Given that, length of cuboid pen stand (l) = 10 cm Breadth of cubiod pen stand (b) = 5 c and height of cub...

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Tick (✓) the correct answer:

Question: Tick (✓) the correct answer: (82)2 (18)2= ? (a) 8218 (b) 6418 (c) 6400 (d) 7204 Solution: (c) 6400 $(82)^{2}-(18)^{2}$ [using the identity(a-b)(a+b)=a2-b2] $=(82+18)(82-18)$ $=(100)(64)$ $=6400$...

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