A circular pond is 17.5 m is of diameter.

Question: A circular pond is 17.5 m is of diameter. It is surrounded by a 2m wide path. Find the cost of constructing the path at the rate of ₹ 25 Per m2? Solution: Given that, a circular pond is surrounded by a wide path.The diameter of circular pond = 17.5 m $\therefore \quad$ Radius of circular pond $\left(r_{i}\right)=\frac{\text { Diameter }}{2}$ i.e.,$O A=r_{i}=\frac{17.5}{2}=8.75 \mathrm{~m}$ and the width of the path $=2 \mathrm{~m}$ i.e. $A B=2 \mathrm{~m}$ Now, $\quad$ length of $O B=O...

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Mark (✓) against the correct answer

Question: Mark (✓) against the correct answer $\sqrt[3]{\frac{-343}{729}}=?$ (a) $\frac{7}{9}$ (b) $\frac{-7}{9}$ (C) $\frac{-9}{7}$ (d) $\frac{9}{7}$ Solution: (b) $\frac{-7}{9}$ By prime factorisation: $\sqrt[3]{\frac{-343}{729}}=\frac{\sqrt[3]{-343}}{\sqrt[3]{729}}=\frac{\sqrt[3]{(-7) \times(-7) \times(-7)}}{\sqrt[3]{3 \times 3 \times 3 \times 3 \times 3 \times 3}}=\frac{\sqrt[3]{(-7)^{3}}}{\sqrt[3]{(3)^{3} \times(3)^{3}}}$ $\sqrt[3]{\frac{-343}{729}}=\frac{\sqrt[3]{(-7)^{3}}}{\sqrt[3]{(9)^{3...

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Solve this

Question: If $x=2 \cos \theta-\cos 2 \theta$ and $y=2 \sin \theta-\sin 2 \theta$, prove that $\frac{d y}{d x}=\tan \left(\frac{3 \theta}{2}\right)$ Solution: We have, $x=2 \cos \theta-\cos 2 \theta$ $\Rightarrow \frac{d x}{d \theta}=2(-\sin \theta)-(-\sin 2 \theta) \frac{d}{d \theta}(2 \theta)$ $\Rightarrow \frac{d x}{d \theta}=-2 \sin \theta+2 \sin 2 \theta$ $\Rightarrow \frac{d x}{d \theta}=2(\sin 2 \theta-\sin \theta)$ .....(1) and, $y=2 \sin \theta-\sin 2 \theta$ $\Rightarrow \frac{d y}{d \t...

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Mark (✓) against the correct answer

Question: Mark (✓) against the correct answer $\sqrt[3]{216 \times 64}=?$ (a) 64 (b) 32 (c) 24 (d) 36 Solution: (c) 24 $\sqrt[3]{216 \times 64}=\sqrt[3]{216} \times \sqrt[3]{64}=\sqrt[3]{2 \times 2 \times 2 \times 3 \times 3 \times 3} \times \sqrt[3]{2 \times 2 \times 2 \times 2 \times 2 \times 2}$ $\sqrt[3]{216 \times 64}=\sqrt[3]{(2)^{3} \times(3)^{3}} \times \sqrt[3]{(2)^{3} \times(2)^{3}}=\sqrt[3]{(6)^{3}} \times \sqrt[3]{(4)^{3}}$ $\sqrt[3]{216 \times 64}=6 \times 4=24$ $\therefore \sqrt[3]...

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Mark (✓) against the correct answer

Question: Mark (✓) against the correct answer Which of the following numbers is a perfect cube? (a) 121 (b) 169 (c) 196 (d) 216 Solution: (d) 216 $121=11 \times 11$ $169=13 \times 13$ $196=7 \times 7 \times 2 \times 2$ $216=2 \times 2 \times 2 \times 3 \times 3 \times 3=(2)^{3} \times(3)^{3}=(6)^{3}$ $216=6^{3}$ Hence, 216 is a perfect cube....

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Find the area of the segment of a circle

Question: Find the area of the segment of a circle of radius 12 cm whose corresponding sector has a centrel angle of 60. (use = 3.14) Solution: Given that, radius of a circle (r) = 12 cm and central angle of sector OBCA () = 60 $\therefore \quad$ Area of sector $O B C A=\frac{\pi r^{2}}{360} \times \theta$[here, $O B C A=$ sector and $A B C A=$ segment ] $=3.14 \times 2 \times 12$ $=3.14 \times 24=75.36 \mathrm{~cm}^{2}$ Since, $\triangle O A B$ is an isosceles triangle. Let $\angle O A B=\angle...

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Solve this

Question: Find $\frac{d y}{d x}$, when $x=\frac{1-t^{2}}{1+t^{2}}$ and $y=\frac{2 t}{1+t^{2}}$ Solution: We have, $y=\frac{2 t}{1+t^{2}}$ $\Rightarrow \frac{d y}{d t}=\left[\frac{\left(1+t^{2}\right) \frac{d}{d t}(2 t)-2 t \frac{d}{d t}\left(1+t^{2}\right)}{\left(1+t^{2}\right)^{2}}\right]$ [using quotient rule] $\Rightarrow \frac{d y}{d t}=\left[\frac{\left(1+t^{2}\right)(2)-2 t(2 t)}{\left(1+t^{2}\right)^{2}}\right]$ $\Rightarrow \frac{d y}{d t}=\left[\frac{2+2 t^{2}-4 t^{2}}{\left(1+t^{2}\rig...

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Mark (✓) against the correct answer

Question: Mark (✓) against the correct answer $\left(1 \frac{3}{4}\right)^{3}=?$ (a) $1 \frac{27}{64}$ (b) $2 \frac{27}{64}$ (c) $5 \frac{23}{64}$ (d) none of these Solution: (c) $5 \frac{23}{64}$ $\left(1 \frac{3}{4}\right)^{3}=\left(\frac{7}{4}\right)^{3}=\frac{(7)^{3}}{(4)^{3}}=\frac{7 \times 7 \times 7}{4 \times 4 \times 4}=\frac{343}{64}$ $\left(1 \frac{3}{4}\right)^{3}=\frac{343}{64}=5 \frac{23}{64}$ $\therefore\left(1 \frac{3}{4}\right)^{3}=5 \frac{23}{64}$...

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Evaluate

Question: Evaluate $\sqrt[3]{\frac{-64}{125}}$ Solution: $\sqrt[3]{\frac{-64}{125}}$ By prime factorisation method: $\sqrt[3]{\frac{-64}{125}}=\frac{\sqrt[3]{-64}}{\sqrt[3]{125}}=\frac{\sqrt[3]{(-4) \times(-4) \times(-4)}}{\sqrt[3]{5 \times 5 \times 5}}=\frac{\sqrt[3]{(-4)^{3}}}{\sqrt[3]{(5)^{3}}}$ $\sqrt[3]{\frac{-64}{125}}=\frac{-4}{5}$ $\therefore \sqrt[3]{\frac{-64}{125}}=\frac{-4}{5}$...

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Evaluate

Question: Evaluate $\sqrt[3]{4096}$ Solution: $\sqrt[3]{4096}$ By prime factorisation method: $\sqrt[3]{4096}=\sqrt[3]{2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2}=\sqrt[3]{(2)^{3} \times(2)^{3} \times(2)^{3} \times(2)^{3}}$ $\sqrt[3]{4096}=(2) \times(2) \times(2) \times(2)=16$ $\therefore \sqrt[3]{4096}=16$...

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Solve this

Question: Find $\frac{d y}{d x}$, when $x=\cos ^{-1} \frac{1}{\sqrt{1+t^{2}}}$ and $y=\sin ^{-1} \frac{t}{\sqrt{1+t^{2}}}, t \in R$ Solution: We have, $x=\cos ^{-1}\left(\frac{1}{\sqrt{1+t^{2}}}\right)$ $\Rightarrow \frac{d x}{d t}=\frac{-1}{\sqrt{1-\left(\frac{1}{\sqrt{1+t^{2}}}\right)^{2}}} \frac{d}{d t}\left(\frac{1}{\sqrt{1+t^{2}}}\right)$ $\Rightarrow \frac{d x}{d t}=\frac{-1}{\sqrt{1-\frac{1}{\left(1+t^{2}\right)}}}\left\{\frac{-1}{2\left(1+t^{2}\right)^{\frac{3}{2}}}\right\} \frac{d}{d t}...

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Evaluate

Question: Evaluate $\sqrt[3]{216 \times 343}$ Solution: $\sqrt[3]{216 \times 343}$ By prime factorisation: $\sqrt[3]{216 \times 343}=\sqrt[3]{216} \times \sqrt[3]{343}=\sqrt[3]{2 \times 2 \times 2 \times 3 \times 3 \times 3} \times \sqrt[3]{7 \times 7 \times 7}=\sqrt[3]{(2)^{3} \times(3)^{3}} \times \sqrt[3]{(7)^{3}}$ $\sqrt[3]{216 \times 343}=(2) \times(3) \times(7)=42$ $\therefore \sqrt[3]{216 \times 343}=42$...

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Sides of a triangular field are 15 m, 16m and 17m.

Question: Sides of a triangular field are 15 m, 16m and 17m. with the three corners of the field a cow, a buffalo and a horse are tied separately with ropes of length 7m each to graze in the field. Find the area of the field which cannot be grazed by the three animals. Solution: Given that, a triangular field with the three corners of the field a cow, a buffalo and a horse are tied separately with ropes. So, each animal grazed the field in each corner of triangular field as a sectorial form. Giv...

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Evaluate:

Question: Evaluate: $\sqrt[3]{4096}$ Solution: $\sqrt[3]{4096}$ By prime factorisation method: $\sqrt[3]{4096}=\sqrt[3]{2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2}=\sqrt[3]{(2)^{3} \times(2)^{3} \times(2)^{3} \times(2)^{3}}$ $\sqrt[3]{4096}=(2) \times(2) \times(2) \times(2)=16$ $\therefore \sqrt[3]{4096}=16$...

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Solve this

Question: Find $\frac{d y}{d x}$, when $x=\frac{2 t}{1+t^{2}}$ and $y=\frac{1-t^{2}}{1+t^{2}}$ Solution: We have, $x=\frac{2 t}{1+t^{2}}$ $\Rightarrow \frac{d x}{d t}=\left[\frac{\left(1+t^{2}\right) \frac{d}{d t}(2 t)-2 t \frac{d}{d t}\left(1+t^{2}\right)}{\left(1+t^{2}\right)^{2}}\right]$ [using quotient rule] $\Rightarrow \frac{d x}{d t}=\left[\frac{\left(1+t^{2}\right)(2)-2 t(2 t)}{\left(1+t^{2}\right)^{2}}\right]$ $\Rightarrow \frac{d x}{d t}=\left[\frac{2+2 t^{2}-4 t^{2}}{\left(1+t^{2}\rig...

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The diameters of front and rear wheels of a tractor

Question: The diameters of front and rear wheels of a tractor are 80 cm and 2m, respectively. Find the number of revolutions that rear wheel will make in covering a distance in which the front wheel makes 1400 revolutions. Solution: Given, diameter of front wheels, d1= 80 cm and diameter of rear wheels, d2= 2 m = 200 cm $\therefore \quad$ Radius of front wheel $\left(r_{1}\right)=\frac{80}{2}=40 \mathrm{~cm}$ $\therefore \quad$ Circumference of the front wheel $=2 \pi r_{1}=\frac{2 \times 22}{7}...

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Tick (✓) the correct answer

Question: Tick (✓) the correct answer (0.8)3= ? (a) 51.2 (b) 5.12 (c) 0.512 (d) none of these Solution: (c) 0.512 $(0.8)^{3}=(0.8) \times(0.8) \times(0.8)=0.512$ $\therefore(0.8)^{3}=0.512$...

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Solve this

Question: Find $\frac{d y}{d x}$, when $x=e^{\theta}\left(\theta+\frac{1}{\theta}\right)$ and $y=e^{-\theta}\left(\theta-\frac{1}{\theta}\right)$ Solution: We have, $x=e^{\theta}\left(\theta+\frac{1}{\theta}\right)$ Differentiating it with respect to $\theta$, $\frac{d x}{d \theta}=e^{\theta} \frac{d}{d \theta}\left(\theta+\frac{1}{\theta}\right)+\left(\theta+\frac{1}{\theta}\right) \frac{d}{d \theta}\left(e^{\theta}\right)$ [using product rule] $\Rightarrow \frac{d x}{d \theta}=e^{\theta}\left(...

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Tick (✓) the correct answer

Question: Tick (✓) the correct answer $\left(1 \frac{3}{10}\right)^{3}=?$ (a) $1 \frac{27}{1000}$ (b) $2 \frac{27}{1000}$ (c) $2 \frac{197}{1000}$ (d) none of these Solution: (c) $2 \frac{197}{1000}$ $\left(1 \frac{3}{10}\right)^{3}=\left(\frac{13}{10}\right)^{3}=\frac{(13)^{3}}{(10)^{3}}=\frac{(13 \times 13 \times 13)}{(10 \times 10 \times 10)}$ $\left(1 \frac{3}{10}\right)^{3}=\frac{2197}{1000}=2 \frac{197}{1000}$ $\therefore\left(1 \frac{3}{10}\right)^{3}=2 \frac{197}{1000}$...

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The area of a circular playground is

Question: The area of a circular playground is $22176 \mathrm{~m}^{2}$. Find the cost of fencing this ground at the rate of $₹ 50$ per $\mathrm{m}$. Solution: Given, area of a circular playaround $=22176 \mathrm{~m}^{2}$ $\therefore \quad \pi r^{2}=22176 \quad\left[\because\right.$ area of circle $\left.=\pi r^{2}\right]$ $\Rightarrow$ $\frac{22}{7} r^{2}=22176 \Rightarrow r^{2}=1008 \times 7$ $\Rightarrow$ $r^{2}=7056 \Rightarrow r=84 \mathrm{~m}$ $\therefore$ Circumference of a circle $=2 \pi ...

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Tick (✓) the correct answer

Question: Tick (✓) the correct answer By what least number should 1536 be divided to get a perfect cube? (a) 3 (b) 4 (c) 6 (d) 8 Solution: (a) 3 $1536=2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3=(2)^{3} \times(2)^{3} \times(2)^{3} \times 3$ Therefore, to get a perfect cube, we need to divide 1536 by 3....

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Tick (✓) the correct answer

Question: Tick (✓) the correct answer By what least number should 648 be multiplied to get a perfect cube? (a) 3 (b) 6 (c) 9 (d) 8 Solution: (c) 9 $648=2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 3=(2)^{3} \times(3)^{3} \times 3$ Therefore, to get a perfect cube, we need to multiply 648 by 9 , i.e. $(3 \times 3)$....

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Solve this

Question: Find $\frac{d y}{d x}$, when $x=a(\cos \theta+\theta \sin \theta)$ and $y=a(\sin \theta-\theta \cos \theta)$ Solution: We have, $x=a(\cos \theta+\theta \sin \theta)$ and $y=a(\sin \theta-\theta \cos \theta)$ $\Rightarrow \frac{d x}{d \theta}=a\left[\frac{d}{d \theta} \cos \theta+\frac{d}{d \theta}(\theta \sin \theta)\right]$ and $\frac{d y}{d \theta}=a\left[\frac{d}{d \theta}(\sin \theta)-\frac{d}{d \theta}(\theta \cos \theta)\right]$ $\Rightarrow \frac{d x}{d \theta}=a\left[-\sin \the...

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A piece of wire 20 cm long is bent into

Question: A piece of wire 20 cm long is bent into the from of an arc of a circle, subtending an angle of 60 at its centre. Find the radius of the circle. Solution: Length of arc of circle = 20 cm Here, $\quad$ central angle $\theta=60^{\circ}$ $\therefore$Length of $\operatorname{arc}=\frac{\theta}{360^{\circ}} \times 2 \pi r$ $\Rightarrow$ $20=\frac{60^{\circ}}{360^{\circ}} \times 2 \pi r \Rightarrow \frac{20 \times 6}{2 \pi}=r$ $\therefore$ $r=\frac{60}{\pi} \mathrm{cm}$ Hence, the radius of c...

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Tick (✓) the correct answer

Question: Tick (✓) the correct answer $\sqrt[3]{\frac{-512}{729}}=?$ (a) $\frac{-7}{9}$ (b) $\frac{-8}{9}$ (C) $\frac{7}{9}$ (d) $\frac{8}{9}$ Solution: (b) $\frac{-8}{9}$ $\sqrt[3]{\frac{-512}{729}}=\frac{\sqrt[3]{-512}}{\sqrt[3]{729}}=\frac{\sqrt[3]{(-8) \times(-8) \times(-8)}}{\sqrt[3]{9 \times 9 \times 9}}=\frac{\sqrt[3]{(-8)^{3}}}{\sqrt[3]{(9)^{3}}}$ $\sqrt[3]{\frac{-512}{729}}=\frac{-8}{9}$ $\therefore \sqrt[3]{\frac{-512}{729}}=\frac{-8}{9}$...

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