The diameter of roller 1.5 m long is 84 cm.
Question: The diameter of roller 1.5 m long is 84 cm.If it takes 100 revolutions to level a playground, find the cost of leveling this ground at the rate of 50 paise per square meter. Solution: Given that Diameter of the roller (d) = 85 cm = 0.84 m Length of the roller = 1.5 m Radius of the roller(r) =d/2 =0.842= 0.42 m The total area of the playground covered by the roller in one revolution=covered area of the roller Curved surface area of the roller =2rh = 2 * 3.14 * 0.42 * 1.5 $=0.12 * 22 * 1...
Read More →If tan x + sec x =
Question: If $\tan x+\sec x=\sqrt{3}, 0x\pi$, then $x$ is equal to (a) $\frac{5 \pi}{6}$ (b) $\frac{2 \pi}{3}$ (c) $\frac{\pi}{6}$ (d) $\frac{\pi}{3}$ Solution: (c) $\frac{\pi}{6}$ We have: $\tan x+\sec x=\sqrt{3} \quad[0x\pi]$ $\Rightarrow \sec x+\tan x=\sqrt{3}$ $\Rightarrow \frac{1}{\cos x}+\frac{\sin x}{\cos x}=\sqrt{3}$ $\Rightarrow 1+\sin x=\sqrt{3} \cos x$ $\Rightarrow(1+\sin x)^{2}=(\sqrt{3} \cos x)^{2}$ $\Rightarrow 1+\sin ^{2} x+2 \sin x=3 \cos ^{2} x$ $\Rightarrow 1+\sin ^{2} x+2 \sin...
Read More →If tan x + sec x =
Question: If $\tan x+\sec x=\sqrt{3}, 0x\pi$, then $x$ is equal to (a) $\frac{5 \pi}{6}$ (b) $\frac{2 \pi}{3}$ (c) $\frac{\pi}{6}$ (d) $\frac{\pi}{3}$ Solution: (c) $\frac{\pi}{6}$ We have: $\tan x+\sec x=\sqrt{3} \quad[0x\pi]$ $\Rightarrow \sec x+\tan x=\sqrt{3}$ $\Rightarrow \frac{1}{\cos x}+\frac{\sin x}{\cos x}=\sqrt{3}$ $\Rightarrow 1+\sin x=\sqrt{3} \cos x$ $\Rightarrow(1+\sin x)^{2}=(\sqrt{3} \cos x)^{2}$ $\Rightarrow 1+\sin ^{2} x+2 \sin x=3 \cos ^{2} x$ $\Rightarrow 1+\sin ^{2} x+2 \sin...
Read More →If ∆ABC and ∆DEF are two triangles such that ABDE=BCEF=CAFD=34,
Question: If ∆ABC and ∆DEF are two triangles such thatABDE=BCEF=CAFD=34, then write Area (∆ABC) : Area (∆DEF) Solution: GIVEN: $\triangle \mathrm{ABC}$ and $\triangle \mathrm{DEF}$ are two triangles such that $\frac{\mathrm{AB}}{\mathrm{DE}}=\frac{\mathrm{BC}}{\mathrm{EF}}=\frac{\mathrm{CA}}{\mathrm{FD}}=\frac{3}{4}$. TO FIND: Area (ABC):Area(DEF) We know that two triangles are similar if their corresponding sides are proportional. Here, $\triangle \mathrm{ABC}$ and $\triangle \mathrm{DEF}$ are ...
Read More →Form the differential equation of the family of hyperbolas having foci on x-axis and centre at origin.
Question: Form the differential equation of the family of hyperbolas having foci onx-axis and centre at origin. Solution: The equation of the family of hyperbolas with the centre at origin and foci along thex-axis is: $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ ...(1) Differentiating both sides of equation (1) with respect tox, we get: $\frac{2 x}{a^{2}}-\frac{2 y y^{\prime}}{b^{2}}=0$ $\Rightarrow \frac{x}{a^{2}}-\frac{y y^{\prime}}{b^{2}}=0$ ...(2) Again, differentiating both sides with respec...
Read More →If x = r sin θ cos ϕ,
Question: Ifx=rsin cos ϕ,y=rsin sinϕ andz=rcos , thenx2+y2+z2is independent of (a) , ϕ (b)r, (c)r, ϕ (d)r. Solution: (a) , ϕ We have: x=rsin cos ϕ ,y=rsin sinϕ andz=rcos , $\therefore x^{2}+y^{2}+z^{2}$ $=(r \sin \theta \cos \phi)^{2}+(r \sin \theta \sin \phi)^{2}+(r \cos \theta)^{2}$ $=r^{2} \sin ^{2} \theta \cos ^{2} \phi+r^{2} \sin ^{2} \theta \sin ^{2} \phi+r^{2} \cos ^{2} \theta$ $=r^{2} \sin ^{2} \theta\left(\cos ^{2} \phi+\sin ^{2} \phi\right)+r^{2} \cos ^{2} \theta$ $=r^{2} \sin ^{2} \th...
Read More →If x = r sin θ cos ϕ,
Question: Ifx=rsin cos ϕ,y=rsin sinϕ andz=rcos , thenx2+y2+z2is independent of (a) , ϕ (b)r, (c)r, ϕ (d)r. Solution: (a) , ϕ We have: x=rsin cos ϕ ,y=rsin sinϕ andz=rcos , $\therefore x^{2}+y^{2}+z^{2}$ $=(r \sin \theta \cos \phi)^{2}+(r \sin \theta \sin \phi)^{2}+(r \cos \theta)^{2}$ $=r^{2} \sin ^{2} \theta \cos ^{2} \phi+r^{2} \sin ^{2} \theta \sin ^{2} \phi+r^{2} \cos ^{2} \theta$ $=r^{2} \sin ^{2} \theta\left(\cos ^{2} \phi+\sin ^{2} \phi\right)+r^{2} \cos ^{2} \theta$ $=r^{2} \sin ^{2} \th...
Read More →The students of Vidyalaya were asked to participate in a competition for making and decorating penholders in the shape of a cylinder with a base using cardboard. Each penholder was to be of radius 3 cm and height 10.5 cm.
Question: The students of Vidyalaya were asked to participate in a competition for making and decorating penholders in the shape of a cylinder with a base using cardboard. Each penholder was to be of radius 3 cm and height 10.5 cm. The Vidyalaya was to supply the competitors with cardboard. If there were 35 competitors, how much cardboard was required to be bought for the competition? (Take = 3.14). Solution: It is given that Radius of the circular part of the penholder (r) = 3 cm The height of ...
Read More →If the altitude of two similar triangles are in the ratio 2 : 3, what is the ratio of their areas?
Question: If the altitude of two similar triangles are in the ratio 2 : 3, what is the ratio of their areas? Solution: GIVEN: Altitudes of two similar triangles are in ratio 2:3. TO FIND: Ratio of the areas of two similar triangles. Let first triangle be ΔABC and the second triangle be ΔPQR We know that the areas of two similar triangles are in the ratio of the squares of the corresponding altitudes. $\Rightarrow \frac{\text { Area }(\mathrm{ABC})}{\text { Area }(\mathrm{PQR})}=\frac{2^{2}}{3^{2...
Read More →If ABC and DEF are similar triangles such that ∠A = 57° and ∠E = 73°,
Question: If $\mathrm{ABC}$ and $\mathrm{DEF}$ are similar triangles such that $\angle \mathrm{A}=57^{\circ}$ and $\angle \mathrm{E}=73^{\circ}$, what is the measure of $\angle \mathrm{C}$ ? Solution: GIVEN: There are two similar triangles ΔABC and ΔDEF. $\angle \mathrm{A}=57^{\circ} \cdot \angle \mathrm{E}=73^{\circ}$ TO FIND: measure of $\angle \mathrm{C}$ SAS Similarity Criterion: If in two triangles, one pair of corresponding sides are proportional and the included angles are equal then two ...
Read More →If
Question: If $\frac{\pi}{2}x\pi$, then $\sqrt{\frac{1-\sin x}{1+\sin x}}+\sqrt{\frac{1+\sin x}{1-\sin x}}$ is equal to (a) 2 secx (b) 2 secx (c) secx (d) secx Solution: (b) 2 secx $\sqrt{\frac{1-\sin x}{1+\sin x}}+\sqrt{\frac{1+\sin x}{1-\sin x}}$ $=\sqrt{\frac{(1-\sin x)(1-\sin x)}{(1+\sin x)(1-\sin x)}}+\sqrt{\frac{(1+\sin x)(1+\sin x)}{(1-\sin x)(1+\sin x)}}$ $=\sqrt{\frac{(1-\sin x)^{2}}{1-\sin ^{2} x}}+\sqrt{\frac{(1+\sin x)^{2}}{1-\sin ^{2} x}}$ $=\sqrt{\frac{(1-\sin x)^{2}}{\cos ^{2} x}}+...
Read More →If
Question: If $\frac{\pi}{2}x\pi$, then $\sqrt{\frac{1-\sin x}{1+\sin x}}+\sqrt{\frac{1+\sin x}{1-\sin x}}$ is equal to (a) 2 secx (b) 2 secx (c) secx (d) secx Solution: (b) 2 secx $\sqrt{\frac{1-\sin x}{1+\sin x}}+\sqrt{\frac{1+\sin x}{1-\sin x}}$ $=\sqrt{\frac{(1-\sin x)(1-\sin x)}{(1+\sin x)(1-\sin x)}}+\sqrt{\frac{(1+\sin x)(1+\sin x)}{(1-\sin x)(1+\sin x)}}$ $=\sqrt{\frac{(1-\sin x)^{2}}{1-\sin ^{2} x}}+\sqrt{\frac{(1+\sin x)^{2}}{1-\sin ^{2} x}}$ $=\sqrt{\frac{(1-\sin x)^{2}}{\cos ^{2} x}}+...
Read More →If
Question: If $\frac{\pi}{2}x\pi$, then $\sqrt{\frac{1-\sin x}{1+\sin x}}+\sqrt{\frac{1+\sin x}{1-\sin x}}$ is equal to (a) 2 secx (b) 2 secx (c) secx (d) secx Solution: (b) 2 secx $\sqrt{\frac{1-\sin x}{1+\sin x}}+\sqrt{\frac{1+\sin x}{1-\sin x}}$ $=\sqrt{\frac{(1-\sin x)(1-\sin x)}{(1+\sin x)(1-\sin x)}}+\sqrt{\frac{(1+\sin x)(1+\sin x)}{(1-\sin x)(1+\sin x)}}$ $=\sqrt{\frac{(1-\sin x)^{2}}{1-\sin ^{2} x}}+\sqrt{\frac{(1+\sin x)^{2}}{1-\sin ^{2} x}}$ $=\sqrt{\frac{(1-\sin x)^{2}}{\cos ^{2} x}}+...
Read More →Find the lateral surface area of a petrol storage tank is 4.2 m in diameter and 4.5 m high.
Question: Find the lateral surface area of a petrol storage tank is $4.2 \mathrm{~m}$ in diameter and $4.5 \mathrm{~m}$ high. How much steel was actually used, if $1 / 12^{\text {th }}$ of the steel actually used was wasted in making the closed tank? Solution: It is given that Diameter of cylinder = 4.2 m Radius of cylinder = 4.22 m = 2.1 m Height of cylinder = 4.5 m Therefore, Lateral or Curved surface area = 2rh $=2 * 3.14 * 2.1 * 4.5=59.4 \mathrm{~m}^{2}$ Total surface area of tank $=2^{\star...
Read More →Form the differential equation of the family of ellipses having foci on y-axis and centre at origin.
Question: Form the differential equation of the family of ellipses having foci ony-axis and centre at origin. Solution: The equation of the family of ellipses having foci on they-axis and the centre at origin is as follows: $\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1$ ...(1) Differentiating equation (1) with respect tox, we get: $\frac{2 x}{b^{2}}+\frac{2 y y^{\prime}}{b^{2}}=0$ $\Rightarrow \frac{x}{b^{2}}+\frac{y y^{\prime}}{a^{2}}=0$ ...(2) Again, differentiating with respect tox, we get: $\fr...
Read More →The areas of two similar triangles are 169 cm2 and 121 cm2 respectively.
Question: The areas of two similar triangles are $169 \mathrm{~cm}^{2}$ and $121 \mathrm{~cm}^{2}$ respectively. If the longest side of the larger triangle is $26 \mathrm{~cm}$, what is the length of the longest side of the smaller triangle? Solution: Let $\triangle \mathrm{ABC}$ and $\triangle \mathrm{PQR}$ are similar triangles. The area of triangles is $169 \mathrm{~cm}^{2}$ and $121 \mathrm{~cm}^{2}$, respectively. Longest side of the larger triangle is 26cm TO FIND: length of longest side o...
Read More →If 0
Question: If $0x\frac{\pi}{2}$, and if $\frac{y+1}{1-y}=\sqrt{\frac{1+\sin x}{1-\sin x}}$, then $y$ is equal to (a) $\cot \frac{x}{2}$ (b) $\tan \frac{x}{2}$ (c) $\cot \frac{x}{2}+\tan \frac{x}{2}$ (d) $\cot \frac{x}{2}-\tan \frac{x}{2}$ Solution: (b) $\tan \frac{x}{2}$ We have: $\frac{y+1}{1-y}=\sqrt{\frac{1+\sin x}{1-\sin x}}$ $\Rightarrow \frac{y+1}{1-y}=\sqrt{\frac{\cos ^{2} \frac{x}{2}+\sin ^{2} \frac{x}{2}+2 \sin \frac{x}{2} \cos \frac{x}{2}}{2}}$ $\Rightarrow \frac{y+1}{1-y}=\sqrt{\frac{\...
Read More →If 0
Question: If $0x\frac{\pi}{2}$, and if $\frac{y+1}{1-y}=\sqrt{\frac{1+\sin x}{1-\sin x}}$, then $y$ is equal to (a) $\cot \frac{x}{2}$ (b) $\tan \frac{x}{2}$ (c) $\cot \frac{x}{2}+\tan \frac{x}{2}$ (d) $\cot \frac{x}{2}-\tan \frac{x}{2}$ Solution: (b) $\tan \frac{x}{2}$ We have: $\frac{y+1}{1-y}=\sqrt{\frac{1+\sin x}{1-\sin x}}$ $\Rightarrow \frac{y+1}{1-y}=\sqrt{\frac{\cos ^{2} \frac{x}{2}+\sin ^{2} \frac{x}{2}+2 \sin \frac{x}{2} \cos \frac{x}{2}}{2}}$ $\Rightarrow \frac{y+1}{1-y}=\sqrt{\frac{\...
Read More →The inner diameter of a circular well is 3.5 m. It is 10 m deep. Find:
Question: The inner diameter of a circular well is 3.5 m. It is 10 m deep. Find: (a) Inner curved surface area (b) The cost of plastering this curved surface at the rate of Rs 40 per $m^{2}$. Solution: The inner diameter of the well = 3.5 m Inner radius = 3.52 = 1.75 m Height of the well = 10 m (a) Inner curved surface area = 2rh = 2 * 3.14 * 1.75 * 10 $=110 m^{2}$ (b) Cost of painting $1 \mathrm{~m}^{2}$ area of the well = Rs 40 Cost of painting $110 \mathrm{~m}^{2}$ area of the well = Rs (50*1...
Read More →If the areas of two similar triangles ABC and PQR are in the ratio 9 : 16 and BC = 4.5 cm,
Question: If the areas of two similar triangles ABC and PQR are in the ratio 9 : 16 and BC = 4.5 cm, what is the length of QR? Solution: Given: ΔABC and ΔPQR are similar triangles. Area of ΔABC: Area of ΔPQR = 9:16 and BC = 4.5cm. To find: Length of QR We know that the ratio of the areas of two similar triangles is equal to the ratio of the squares of any two corresponding sides. Hence, $\frac{\operatorname{ar}(\Delta \mathrm{ABC})}{\operatorname{ar}(\Delta \mathrm{PQR})}=\frac{\mathrm{BC}^{2}}{...
Read More →If π < x <2π,
Question: If $\pix2 \pi$, then $\sqrt{\frac{1+\cos x}{1-\cos x}}$ is equal to (a) cosecx+ cotx (b) cosecx cotx (c) cosecx+ cotx (d) cosecx cotx Solution: (d) cosecx cotx $\sqrt{\frac{1+\cos x}{1-\cos x}}$ $=\sqrt{\frac{(1+\cos x)(1+\cos x)}{(1-\cos x)(1+\cos x)}}$ $=\sqrt{\frac{(1+\cos x)^{2}}{1-\cos ^{2} x}}$ $=\sqrt{\frac{(1+\cos x)^{2}}{\sin ^{2} x}}$ $=\frac{(1+\cos x)}{-\sin x} \quad[\operatorname{as}, \pix2 \pi$, so $\sin x$ will be negative $]$ $=-(\operatorname{cosec} x+\cot x)$ $=-\oper...
Read More →In the figure given below DE || BC. If AD = 2.4 cm,
Question: In the figure given below DE || BC. If AD = 2.4 cm, DB = 3.6 cm, AC = 5 cm. Find AE. Solution: GIVEN:AD = 2.4cm, BD = 3.6cm and AC = 5cm. TO FIND:AE According toBASIC PROPORTIONALITY THEOREM If a line is drawn parallel to one side of a triangle intersecting the other two sides, then it divides the two sides in the same ratio. In ∆ABC, DE || BC. $\frac{\mathrm{AD}}{\mathrm{AB}}=\frac{\mathrm{AE}}{\mathrm{AC}}$ $\frac{2.4}{2.4+3.6}=\frac{\mathrm{AE}}{\mathrm{AC}}$ $\frac{2.4}{6}=\frac{\m...
Read More →A cylindrical vessel, without lid, has to be tin-coated on its both sides. If the radius of the base is 70 cm
Question: A cylindrical vessel, without lid, has to be tin-coated on its both sides. If the radius of the base is $70 \mathrm{~cm}$ and its height is $1.4 \mathrm{~m}$, calculate the cost of tin-coating at the rate of Rs $3.50$ per $1000 \mathrm{~cm}$. Solution: Given that Radius of the vessel (r) = 70 cm Height ofthe vessel (h) = 1.4 m = 140 cm The area to be tin coated $=2 *\left(2 \pi r h+\pi r^{2}\right)$ = 2r (2h + r) = 2 * 3.14 * 70[(2 * 140) + 70] $=154000 \mathrm{~cm}^{2}$ Required cost ...
Read More →If π < x <2π,
Question: If $\pix2 \pi$, then $\sqrt{\frac{1+\cos x}{1-\cos x}}$ is equal to (a) cosecx+ cotx (b) cosecx cotx (c) cosecx+ cotx (d) cosecx cotx Solution: (d) cosecx cotx $\sqrt{\frac{1+\cos x}{1-\cos x}}$ $=\sqrt{\frac{(1+\cos x)(1+\cos x)}{(1-\cos x)(1+\cos x)}}$ $=\sqrt{\frac{(1+\cos x)^{2}}{1-\cos ^{2} x}}$ $=\sqrt{\frac{(1+\cos x)^{2}}{\sin ^{2} x}}$ $=\frac{(1+\cos x)}{-\sin x} \quad[\operatorname{as}, \pix2 \pi$, so $\sin x$ will be negative $]$ $=-(\operatorname{cosec} x+\cot x)$ $=-\oper...
Read More →If π < x <2π,
Question: If $\pix2 \pi$, then $\sqrt{\frac{1+\cos x}{1-\cos x}}$ is equal to (a) cosecx+ cotx (b) cosecx cotx (c) cosecx+ cotx (d) cosecx cotx Solution: (d) cosecx cotx $\sqrt{\frac{1+\cos x}{1-\cos x}}$ $=\sqrt{\frac{(1+\cos x)(1+\cos x)}{(1-\cos x)(1+\cos x)}}$ $=\sqrt{\frac{(1+\cos x)^{2}}{1-\cos ^{2} x}}$ $=\sqrt{\frac{(1+\cos x)^{2}}{\sin ^{2} x}}$ $=\frac{(1+\cos x)}{-\sin x} \quad[\operatorname{as}, \pix2 \pi$, so $\sin x$ will be negative $]$ $=-(\operatorname{cosec} x+\cot x)$ $=-\oper...
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