If A lies in second quadrant 3tanA + 4 = 0,
Question: IfAlies in second quadrant 3tanA+ 4 = 0, then the value of 2cotA 5cosA+ sinAis equal to (a) $-\frac{53}{10}$ (b) $\frac{23}{10}$ (c) $\frac{37}{10}$ (d) $\frac{7}{10}$ Solution: It is given that $\frac{\pi}{2}A\pi$. $3 \tan A+4=0$ $\Rightarrow \tan A=-\frac{4}{3}$ $\Rightarrow \cot A=-\frac{3}{4}$ Now, $\sec A=\pm \sqrt{1+\tan ^{2} A}=\pm \sqrt{1+\frac{16}{9}}=\pm \sqrt{\frac{25}{9}}=\pm \frac{5}{3}$ $\therefore \sec A=-\frac{5}{3} \quad$ (A lies in 2 nd quadrant) $\Rightarrow \cos A=-...
Read More →If ∆ABC and ∆DEF are similar such that 2AB = DE and BC = 8 cm, then EF =
Question: If ∆ABC and ∆DEF are similar such that 2AB = DE and BC = 8 cm, then EF = (a) 16 cm(b) 12 cm(c) 8 cm(d) 4 cm Solution: Given: ΔABC and ΔDEF are similar triangles such that 2AB = DE and BC = 8 cm. To find: EF We know that if two triangles are similar then there sides are proportional. Hence, for similar triangles ΔABC and ΔDEF $\frac{\mathrm{AB}}{\mathrm{DE}}=\frac{\mathrm{BC}}{\mathrm{EF}}=\frac{\mathrm{CA}}{\mathrm{FD}}$ $\frac{\mathrm{AB}}{\mathrm{DE}}=\frac{\mathrm{BC}}{\mathrm{EF}}$...
Read More →If x sin 45
Question: If $x \sin 45^{\circ} \cos ^{2} 60^{\circ}=\frac{\tan ^{2} 60^{\circ} \operatorname{cosec} 30^{\circ}}{\sec 45^{\circ} \cot ^{2} 30^{\circ}}$, then $x=$ (a) 2 (b) 4 (c) 8 (d) 16 Solution: (c) 8 We have: $x \sin 45^{\circ} \cos ^{2} 60^{\circ}=\frac{\tan ^{2} 60^{\circ} \operatorname{cosec} 30^{\circ}}{\sec 45^{\circ} \cot ^{2} 30^{\circ}}$ $\Rightarrow x \times\left(\frac{1}{\sqrt{2}}\right) \times\left(\frac{1}{2}\right)^{2}=\frac{(\sqrt{3})^{2} \times(2)}{(\sqrt{2}) \times(\sqrt{3})^...
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Question: $\frac{d y}{d x}+y=1(y \neq 1)$ Solution: The given differential equation is: $\frac{d y}{d x}+y=1$ $\Rightarrow d y+y d x=d x$ $\Rightarrow d y=(1-y) d x$\ Separating the variables, we get: $\Rightarrow \frac{d y}{1-y}=d x$ Now, integrating both sides, we get: $\int \frac{d y}{1-y}=\int d x$ $\Rightarrow \log (1-y)=x+\log \mathrm{C}$ $\Rightarrow-\log \mathrm{C}-\log (1-y)=x$ $\Rightarrow \log \mathrm{C}(1-y)=-x$ $\Rightarrow \mathrm{C}(1-y)=e^{-x}$ $\Rightarrow 1-y=\frac{1}{\mathrm{C...
Read More →If x sin 45
Question: If $x \sin 45^{\circ} \cos ^{2} 60^{\circ}=\frac{\tan ^{2} 60^{\circ} \operatorname{cosec} 30^{\circ}}{\sec 45^{\circ} \cot ^{2} 30^{\circ}}$, then $x=$ (a) 2 (b) 4 (c) 8 (d) 16 Solution: (c) 8 We have: $x \sin 45^{\circ} \cos ^{2} 60^{\circ}=\frac{\tan ^{2} 60^{\circ} \operatorname{cosec} 30^{\circ}}{\sec 45^{\circ} \cot ^{2} 30^{\circ}}$ $\Rightarrow x \times\left(\frac{1}{\sqrt{2}}\right) \times\left(\frac{1}{2}\right)^{2}=\frac{(\sqrt{3})^{2} \times(2)}{(\sqrt{2}) \times(\sqrt{3})^...
Read More →Two isosceles triangles have equal angles and their areas are in the ratio 16 : 25. The ratio of their corresponding heights is
Question: Two isosceles triangles have equal angles and their areas are in the ratio 16 : 25. The ratio of their corresponding heights is (a) 4 : 5(b) 5 : 4(c) 3 : 2(d) 5 : 7 Solution: Given: Two isosceles triangles have equal vertical angles and their areas are in the ratio of 16:25. To find: Ratio of their corresponding heights. Let ∆ABC and ∆PQR be two isosceles triangles such thatA=P. Suppose AD BC and PS QR .In ∆ABC and ∆PQR,ABPQ=ACPRA=P∆ABC~∆PQRSASsimilarity We know that the ratio of areas...
Read More →If tan A + cot A = 4,
Question: If tanA+ cotA= 4, then tan4A+ cot4Ais equal to (a) 110 (b) 191 (c) 80 (d) 194 Solution: (d) 194 We have: $\tan \mathrm{A}+\cot \mathrm{A}=4$ Squaring both the sides: $(\tan \mathrm{A}+\cot \mathrm{A})^{2}=4^{2}$ $\Rightarrow \tan ^{2} \mathrm{~A}+\cot ^{2} \mathrm{~A}+2(\tan \mathrm{A})(\cot \mathrm{A})=16$ $\Rightarrow \tan ^{2} \mathrm{~A}+\cot ^{2} \mathrm{~A}+2=16$ $\Rightarrow \tan ^{2} \mathrm{~A}+\cot ^{2} \mathrm{~A}=14$ Squaring both the sides again: $\left(\tan ^{2} \mathrm{~...
Read More →∆ABC and ∆BDE are two equilateral triangles such that D is the mid-point of BC. The ratio of the areas of triangle ABC and BDE is
Question: ∆ABC and ∆BDE are two equilateral triangles such that D is the mid-point of BC. The ratio of the areas of triangle ABC and BDE is (a) 2 : 1(b) 1 : 2(c) 4 : 1(d) 1 : 4 Solution: Given: ΔABC and ΔBDE are two equilateral triangles such that D is the midpoint of BC. To find: Ratio of areas of ΔABC and ΔBDE. ΔABC and ΔBDE are equilateral triangles; hence they are similar triangles. Since D is the midpoint of BC, BD = DC. We know that the ratio of areas of two similar triangles is equal to t...
Read More →If tan A + cot A = 4,
Question: If tanA+ cotA= 4, then tan4A+ cot4Ais equal to (a) 110 (b) 191 (c) 80 (d) 194 Solution: (d) 194 We have: $\tan \mathrm{A}+\cot \mathrm{A}=4$ Squaring both the sides: $(\tan \mathrm{A}+\cot \mathrm{A})^{2}=4^{2}$ $\Rightarrow \tan ^{2} \mathrm{~A}+\cot ^{2} \mathrm{~A}+2(\tan \mathrm{A})(\cot \mathrm{A})=16$ $\Rightarrow \tan ^{2} \mathrm{~A}+\cot ^{2} \mathrm{~A}+2=16$ $\Rightarrow \tan ^{2} \mathrm{~A}+\cot ^{2} \mathrm{~A}=14$ Squaring both the sides again: $\left(\tan ^{2} \mathrm{~...
Read More →The areas of two similar triangles ∆ABC and ∆DEF are 144 cm2 and 81 cm2 respectively. If the longest side of larger ∆ABC be 36 cm, then the longest side of the smaller triangle ∆DEF is
Question: The areas of two similar triangles ∆ABC and ∆DEF are 144 cm2and 81 cm2 respectively. If the longest side of larger ∆ABC be 36 cm, then the longest side of the smaller triangle ∆DEF is (a) 20 cm(b) 26 cm(c) 27 cm(d) 30 cm Solution: Given: Areas of two similar triangles $\triangle \mathrm{ABC}$ and $\triangle \mathrm{DEF}$ are $144 \mathrm{~cm}^{2}$ and $81 \mathrm{~cm}^{2}$. If the longest side of larger ΔABC is 36cm To find: the longest side of the smaller triangle ΔDEF We know that th...
Read More →sin
Question: sin2/18 + sin2/9 + sin27/18 + sin24/9 = (a) 1 (b) 4 (c) 2 (d) 0 Solution: (c) 2 We have: $\sin ^{2} \frac{\pi}{18}+\sin ^{2} \frac{\pi}{9}+\sin ^{2} \frac{7 \pi}{18}+\sin ^{2} \frac{4 \pi}{9}$ $=\sin ^{2} \frac{\pi}{18}+\sin ^{2} \frac{2 \pi}{18}+\sin ^{2} \frac{7 \pi}{18}+\sin ^{2} \frac{8 \pi}{18}$ $=\sin ^{2} \frac{\pi}{18}+\sin ^{2} \frac{2 \pi}{18}+\sin ^{2}\left(\frac{7 \pi}{18}\right)+\sin ^{2}\left(\frac{8 \pi}{18}\right)$ $=\sin ^{2} \frac{\pi}{18}+\sin ^{2} \frac{2 \pi}{18}+\...
Read More →A hollow garden roller, 63 cm wide with a girth of 440 cm,
Question: A hollow garden roller, 63 cm wide with a girth of 440 cm, is made of 4 cm thick iron. Find the volume of the iron. Solution: Given, h = 63 cm The outer circumference of the roller = 440 cm Thickness of the roller = 4 cm Let,R be the external radius We know that, 22/7 * R = 440 2 * 22/7 * R = 440 R = 70 The thickness is given as 4cm, so the inner radius 'r' is given as ⟹ r = R - 4 ⟹70 - 4 ⟹ 66cm Here, we know the inner and outer radii So, the volume is given as $\Rightarrow V=\pi\left(...
Read More →sin
Question: sin2/18 + sin2/9 + sin27/18 + sin24/9 = (a) 1 (b) 4 (c) 2 (d) 0 Solution: (c) 2 We have: $\sin ^{2} \frac{\pi}{18}+\sin ^{2} \frac{\pi}{9}+\sin ^{2} \frac{7 \pi}{18}+\sin ^{2} \frac{4 \pi}{9}$ $=\sin ^{2} \frac{\pi}{18}+\sin ^{2} \frac{2 \pi}{18}+\sin ^{2} \frac{7 \pi}{18}+\sin ^{2} \frac{8 \pi}{18}$ $=\sin ^{2} \frac{\pi}{18}+\sin ^{2} \frac{2 \pi}{18}+\sin ^{2}\left(\frac{7 \pi}{18}\right)+\sin ^{2}\left(\frac{8 \pi}{18}\right)$ $=\sin ^{2} \frac{\pi}{18}+\sin ^{2} \frac{2 \pi}{18}+\...
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Question: $\frac{d y}{d x}=\sqrt{4-y^{2}}(-2y2)$ Solution: The given differential equation is: $\frac{d y}{d x}=\sqrt{4-y^{2}}$ Separating the variables, we get: $\Rightarrow \frac{d y}{\sqrt{4-y^{2}}}=d x$ Now, integrating both sides of this equation, we get: $\int \frac{d y}{\sqrt{4-y^{2}}}=\int d x$ $\Rightarrow \sin ^{-1} \frac{y}{2}=x+\mathrm{C}$ $\Rightarrow \frac{y}{2}=\sin (x+\mathrm{C})$ $\Rightarrow y=2 \sin (x+\mathrm{C})$ This is the required general solution of the given differentia...
Read More →The areas of two similar triangles are in respectively $9 mathrm{~cm}^{2}$ and $16 mathrm{~cm}^{2}$. The ratio of their corresponding sides is
Question: The areas of two similar triangles are in respectively $9 \mathrm{~cm}^{2}$ and $16 \mathrm{~cm}^{2}$. The ratio of their corresponding sides is (a) 3 : 4(b) 4 : 3(c) 2 : 3(d) 4 : 5 Solution: Given: Areas of two similar triangles are 9cm2and 16cm2. To find: Ratio of their corresponding sides. We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides. $\frac{\operatorname{ar}(\text { triangle } 1)}{\operatorname{ar}(\text { tr...
Read More →The value of sin
Question: The value of sin25 + sin210 + sin215 + ... + sin285 + sin290 is (a) 7 (b) 8 (c) 9.5 (d) 10 Solution: (c) 9.5 We have: $\sin ^{2} 5^{\circ}+\sin ^{2} 10^{\circ}+\sin ^{2} 15^{\circ}+\ldots+\sin ^{2} 85^{\circ}+\sin ^{2} 90^{\circ}$ $=\sin ^{2} 5^{\circ}+\sin ^{2} 10^{\circ}+\sin ^{2} 15^{\circ}+\ldots+\sin ^{2}\left(90^{\circ}-10^{\circ}\right)+\sin ^{2}\left(90^{\circ}-5^{\circ}\right)+\sin ^{2} 90^{\circ}$ $=\sin ^{2} 5^{\circ}+\sin ^{2} 10^{\circ}+\sin ^{2} 15^{\circ}+\ldots+\cos ^{2...
Read More →The value of sin
Question: The value of sin25 + sin210 + sin215 + ... + sin285 + sin290 is (a) 7 (b) 8 (c) 9.5 (d) 10 Solution: (c) 9.5 wqe have: $\sin ^{2} 5^{\circ}+\sin ^{2} 10^{\circ}+\sin ^{2} 15^{\circ}+\ldots+\sin ^{2} 85^{\circ}+\sin ^{2} 90^{\circ}$ $=\sin ^{2} 5^{\circ}+\sin ^{2} 10^{\circ}+\sin ^{2} 15^{\circ}+\ldots+\sin ^{2}\left(90^{\circ}-10^{\circ}\right)+\sin ^{2}\left(90^{\circ}-5^{\circ}\right)+\sin ^{2} 90^{\circ}$ $=\sin ^{2} 5^{\circ}+\sin ^{2} 10^{\circ}+\sin ^{2} 15^{\circ}+\ldots+\cos ^{...
Read More →A patient in a hospital is given soup daily in a cylindrical bowl of diameter 7 cm.
Question: A patient in a hospital is given soup daily in a cylindrical bowl of diameter 7 cm. If the bowl is filled with soup to a height of 4 cm, how much soup the hospital has to prepare daily to serve 250 patients? Solution: Given, Radius (R) of cylindrical bowl =7/2cm = 3.5 cm [Diameter = 7 cm] Height up to which the bowl is filled with soup = 4 cm Volume of soup in 1 bowl $=\pi r^{2} h$ $=22 / 7 \times 3.52 \times 4=154 \mathrm{~cm}^{3}$ Volume of soup $n 250$ bowls $=(250 \times 154) \math...
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Question: $\frac{d y}{d x}=\frac{1-\cos x}{1+\cos x}$ Solution: The given differential equation is: $\frac{d y}{d x}=\frac{1-\cos x}{1+\cos x}$ $\Rightarrow \frac{d y}{d x}=\frac{2 \sin ^{2} \frac{x}{2}}{2 \cos ^{2} \frac{x}{2}}=\tan ^{2} \frac{x}{2}$ $\Rightarrow \frac{d y}{d x}=\left(\sec ^{2} \frac{x}{2}-1\right)$ Separating the variables,we get: $d y=\left(\sec ^{2} \frac{x}{2}-1\right) d x$ Now, integrating both sides of this equation, we get: $\int d y=\int\left(\sec ^{2} \frac{x}{2}-1\rig...
Read More →Sides of two similar triangles are in the ratio 4 : 9. Areas of these triangles are in the ratio.
Question: Sides of two similar triangles are in the ratio 4 : 9. Areas of these triangles are in the ratio. (a) $2: 3$ (b) $4: 9$ (c) $81: 16$ (d) $16: 81$ Solution: Given: Sides of two similar triangles are in the ratio 4:9 To find: Ratio of area of these triangles We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides. $\frac{\operatorname{ar}(\text { triangle } 1)}{\operatorname{ar}(\text { triangle } 2)}=\left(\frac{\operatornam...
Read More →If x is an acute angle and tan x
Question: If $x$ is an acute angle and $\tan x=\frac{1}{\sqrt{7}}$, then the value of $\frac{\operatorname{cosec}^{2} x-\sec ^{2} x}{\operatorname{cosec}^{2} x+\sec ^{2} x}$ is (a) 3/4 (b) 1/2 (c) 2 (d) 5/4 Solution: (a) 3/4 We have: $\tan x=\frac{1}{\sqrt{7}}$ $\therefore \tan ^{2} x=\frac{1}{7}$ Now, dividing the numerator and the denominator of $\frac{\operatorname{cosec}^{2} x-\sec ^{2} x}{\operatorname{cosec}^{2} x+\sec ^{2} x}$ by $\operatorname{cosec}^{2} x$ : $\frac{1-\tan ^{2} x}{1+\tan...
Read More →The capacity of a closed cylindrical vessel of height 1 m is 15.4 liters
Question: The capacity of a closed cylindrical vessel of height 1 m is 15.4 liters. How many square meters of the metal sheet would be needed to make it? Solution: Given, Height of the cylindrical vessel $=15.4$ litres $=0.0154 \mathrm{~m}^{3}\left[1 \mathrm{~m}^{3}=1000\right.$ litres $]$ Let the radius of the circular ends of the cylinders be 'r' $\pi r^{2} h=0.0154 \mathrm{~m}^{3}$ $r=0.07 \mathrm{~m} \quad[\pi=31.4, h=1 \mathrm{~m}]$ Total surface area of a vessel $=2 \pi r(r+h)$ $=2(3.14(0....
Read More →A vertical stick 20 m long casts a shadow 10 m long on the ground. At the same time, a tower casts a shadow 50 m long on the ground. The height of the tower is
Question: A vertical stick 20 m long casts a shadow 10 m long on the ground. At the same time, a tower casts a shadow 50 m long on the ground. The height of the tower is (a) $100 \mathrm{~m}$ (b) $120 \mathrm{~m}$ (c) $25 \mathrm{~m}$ (d) $200 \mathrm{~m}$ Solution: Given: Vertical stick 20m long casts a shadow 10m long on the ground. At the same time a tower casts the shadow 50 m long on the ground. To determine: Height of the tower Let AB be the vertical stick and AC be its shadow. Also, let D...
Read More →sec
Question: $\sec ^{2} x=\frac{4 x y}{(x+y)^{2}}$ is true if and only if (a)x+y 0 (b)x=y,x 0 (c)x=y (d)x0,y 0 Solution: (b)x=y,x 0 We hsve: $\sec ^{2} \mathrm{x}=\frac{4 x y}{(x+y)^{2}}$ $\Rightarrow \frac{4 x y}{(x+y)^{2}} \geq 1 \quad\left[\because \sec ^{2} \mathrm{x} \geq 1\right]$ $\Rightarrow 4 x y \geq(x+y)^{2}$ $\Rightarrow 4 x y \geq x^{2}+y^{2}+2 x y$ $\Rightarrow 2 x y \geq x^{2}+y^{2}$ $\Rightarrow(x-y)^{2} \leq 0$ $\Rightarrow(x-y) \leq 0$ $\Rightarrow x=y$ For $x=0, \sec ^{2} x$ will...
Read More →In the given figure, DE || BC and AD=12BD. If BC = 4.5 cm, find DE.
Question: In the given figure, DE || BC andAD=12BD. If BC = 4.5 cm, find DE. Solution: Given: In $\triangle A B C, D E \| B C . A D=\frac{1}{2} B D$ and $B C=4.5 \mathrm{~cm}$. To find: DE In ∆ABC and ∆ADEB=ADECorrespondinganglesA=ACommon∆ABC~∆ADEAASimilarity $\frac{\mathrm{AD}}{\mathrm{AB}}=\frac{\mathrm{DE}}{\mathrm{BC}}$ $\frac{\mathrm{AD}}{\mathrm{AD}+\mathrm{BD}}=\frac{\mathrm{DE}}{\mathrm{BC}}$ $\frac{\frac{1}{2} \mathrm{BD}}{\frac{1}{2} \mathrm{BD}+\mathrm{BD}}=\frac{\mathrm{DE}}{\mathrm{...
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