The areas of two similar triangles are 121 cm2 and 64 cm2 respectively.
Question: The areas of two similar triangles are $121 \mathrm{~cm}^{2}$ and $64 \mathrm{~cm}^{2}$ respectively. If the median of the first triangle is $12.1 \mathrm{~cm}$, then the corresponding median of the other triangle is (a) 11 cm(b) 8.8 cm(c) 11.1 cm(d) 8.1 cm Solution: Given: The area of two similar triangles is $121 \mathrm{~cm}^{2}$ and $64 \mathrm{~cm}^{2}$ respectively. The median of the first triangle is $2.1 \mathrm{~cm}$. To find: Corresponding medians of the other triangle We kno...
Read More →The sum of the radius of the base and height of a solid cylinder is 37 m.
Question: The sum of the radius of the base and height of a solid cylinder is $37 \mathrm{~m}$. If the total surface area of the solid cylinder is $1628 \mathrm{~cm}$. Find the volume of the cylinder. Solution: Given data is as follows: h + r = 37 cm Total surface area of the cylinder $=1628 \mathrm{~cm}^{2}$ That is, $2 \pi r h+2 \pi r^{2}=1628$ 2r(h + 2r)= 1628 But it is already given in the problem that, h + r = 37 cm Therefore,2r 37= 1628 2 22/7 r 37= 1628 r = 7 cm Since, h + r = 37 cm We ha...
Read More →If tan x + cot x = 4,
Question: If $\tan x+\cot x=4$, then $\tan ^{4} x+\cot ^{4} x=$ ___________. Solution: Given tanx+ cotx = 4 Since $\left(\tan ^{2} x+\cot ^{2} x\right)^{2}=\tan ^{4} x+\cot ^{4} x+2 \cot ^{2} x \tan ^{2} x$ i.e $\left(\tan ^{2} x+\cot ^{2} x\right)^{2}=\tan ^{4} x+\cot ^{4} x+2$ i. e $\tan ^{4} x+\cot ^{4} x=\left(\tan ^{2} x+\cot ^{2} x\right)^{2}-2$ ....(1) also, $(\tan x+\cot x)^{2}=(4)^{2}$ i. e $\tan ^{2} x+\cot ^{2} x+2 \cot x \tan x=16$ $\tan ^{2} x+\cot ^{2} x+2=16$ $\tan ^{2} x+\cot ^{2...
Read More →If tan x + cot x = 4,
Question: If $\tan x+\cot x=4$, then $\tan ^{4} x+\cot ^{4} x=$ ___________. Solution: Given tanx+ cotx = 4 Since $\left(\tan ^{2} x+\cot ^{2} x\right)^{2}=\tan ^{4} x+\cot ^{4} x+2 \cot ^{2} x \tan ^{2} x$ i.e $\left(\tan ^{2} x+\cot ^{2} x\right)^{2}=\tan ^{4} x+\cot ^{4} x+2$ i. e $\tan ^{4} x+\cot ^{4} x=\left(\tan ^{2} x+\cot ^{2} x\right)^{2}-2$ ....(1) also, $(\tan x+\cot x)^{2}=(4)^{2}$ i. e $\tan ^{2} x+\cot ^{2} x+2 \cot x \tan x=16$ $\tan ^{2} x+\cot ^{2} x+2=16$ $\tan ^{2} x+\cot ^{2...
Read More →How many litres of water flow out of a pipe having an area of cross-section of 5 cm2
Question: How many litres of water flow out of a pipe having an area of cross-section of $5 \mathrm{~cm}^{2}$ in one minute, if the speed of water in the pipe is $30 \mathrm{~cm} / \mathrm{sec}$ ? Solution: Given data is as follows: Area of cross section of the pipe $=5 \mathrm{~cm}^{2}$ Speed of water = 30 cm/sec We have to find the volume of water that flows through the pipe in 1 minute. Volume of water that flows through the pipe in one second $=\pi r^{2} h$ Here, $\pi r^{2}$ is nothing but t...
Read More →A rectangular sheet of paper 30 cm × 18 cm can be transformed into the curved surface of a right circular
Question: A rectangular sheet of paper 30 cm 18 cm can be transformed into the curved surface of a right circular cylinder in two ways i.e. either by rolling the paper along its length or by rolling it along its breadth. Find the ratio of the volumes of the two cylinders thus formed. Solution: Given data is as follows: Dimensions of the rectangular sheet of paper = 30 cm18 cm We have to find the ratio of the volumes of the cylinders formed by rolling the sheet along its length and along its brea...
Read More →If sec x=t
Question: If $\sec x=t+\frac{1}{4 t}$, then the value of $\sec x+\tan x$ is ______________ . Solution: Given sec $x=t+\frac{1}{4 t}$ Since $\tan ^{2} x=\sec ^{2} x-1$ $=\left(t+\frac{1}{4 t}\right)^{2}-1$ $=t^{2}+\frac{1}{16 t^{2}}+\frac{1}{2}-1$ $=t^{2}+\frac{1}{16 t^{2}}-\frac{1}{2}$ $\tan ^{2} x=\left(t-\frac{1}{4 t}\right)^{2}$ i. e $\tan x=\pm\left(t-\frac{1}{4 t}\right)$ $\therefore \sec x+\tan x=\left(t+\frac{1}{4 t}\right) \pm\left(t-\frac{1}{4 t}\right)$ Hence, sec $x+\tan x=2 t$ or $\f...
Read More →If sec x=t
Question: If $\sec x=t+\frac{1}{4 t}$, then the value of $\sec x+\tan x$ is ______________ . Solution: Given sec $x=t+\frac{1}{4 t}$ Since $\tan ^{2} x=\sec ^{2} x-1$ $=\left(t+\frac{1}{4 t}\right)^{2}-1$ $=t^{2}+\frac{1}{16 t^{2}}+\frac{1}{2}-1$ $=t^{2}+\frac{1}{16 t^{2}}-\frac{1}{2}$ $\tan ^{2} x=\left(t-\frac{1}{4 t}\right)^{2}$ i. e $\tan x=\pm\left(t-\frac{1}{4 t}\right)$ $\therefore \sec x+\tan x=\left(t+\frac{1}{4 t}\right) \pm\left(t-\frac{1}{4 t}\right)$ Hence, sec $x+\tan x=2 t$ or $\f...
Read More →If sin x=
Question: If $\sin x=\frac{2 t}{1+t^{2}}$ and $x$ lies in the second quadrant, then $\cos x=$ ______________ . Solution: Given $\sin x=\frac{2 t}{1+t^{2}}$ and $x$ lies in 2 nd quadrant Since sin2x+ cos2x= 1 cos2x= 1 sin2x Sincexlies in II Quadrant ⇒ cosx 0 $\therefore \cos x=-\sqrt{1-\sin ^{2} \theta}$ $=-\sqrt{1-\left(\frac{2 t}{1+t^{2}}\right)^{2}}$ $=-\sqrt{1-\frac{4 t^{2}}{\left(1+t^{2}\right)^{2}}}$ $=-\sqrt{\frac{\left(1+t^{2}\right)^{2}-4 t^{2}}{\left(1+t^{2}\right)^{2}}}$ $=-\sqrt{\frac...
Read More →∆ABC ∼ ∆PQR such that ar(∆ABC) = 4 ar(∆PQR). If BC = 12 cm, then QR =
Question: $\triangle \mathrm{ABC} \sim \triangle \mathrm{PQR}$ such that $\operatorname{ar}(\triangle \mathrm{ABC})=4 \operatorname{ar}(\triangle \mathrm{PQR})$. If $\mathrm{BC}=12 \mathrm{~cm}$, then $\mathrm{QR}=$ (a) 9 cm(b) 10 cm(c) 6 cm(d) 8 cm Solution: Given: In Δ ABC and ΔPQR $\triangle \mathrm{ABC} \sim \triangle \mathrm{PQR}$ $\operatorname{Ar}(\triangle \mathrm{ABC})=4 \mathrm{Ar}(\triangle \mathrm{PQR})$ $\mathrm{BC}=12 \mathrm{~cm}$ To find: Measure of QR We know that the ratio of a...
Read More →Find the equation of a curve passing through the point
Question: Find the equation of a curve passing through the point (0, 2) given that at any pointon the curve, the product of the slope of its tangent andy-coordinate of the point is equal to thex-coordinate of the point. Solution: Letxandybe thex-coordinate andy-coordinate of the curve respectively. We know that the slope of a tangent to the curve in the coordinate axis is given by the relation, $\frac{d y}{d x}$ According to the given information, we get: $y \cdot \frac{d y}{d x}=x$ $\Rightarrow...
Read More →If cos
Question: If cos2x+ sinx+ 1 = 0, and 0 x 2 thenx= _________. Solution: Given, $\cos ^{2} x+\sin x+1=0$ and $0x2 \pi$ $1-\sin ^{2} x+\sin x+1=0$ i. e $(1-\sin x)(1+\sin x)+(1+\sin x)=0$ $(1+\sin x)[1-\sin x+1]=0$ $(1+\sin x)[2-\sin x]=0$ i. e $\sin x=-1$ or $\sin x=2$ since $\sin x \neq 2$ $\Rightarrow \sin x=-1 \quad$ (only possibility) i. e $x=\frac{3 \pi}{2}$...
Read More →∆ABC ∼ ∆DEF. If BC = 3 cm, EF = 4 cm and ar(∆ABC) = 54 cm2, then ar(∆DEF) =
Question: $\triangle \mathrm{ABC} \sim \triangle \mathrm{DEF}$. If $\mathrm{BC}=3 \mathrm{~cm}, \mathrm{EF}=4 \mathrm{~cm}$ and $\operatorname{ar}(\triangle \mathrm{ABC})=54 \mathrm{~cm}^{2}$, then $\operatorname{ar}(\triangle \mathrm{DEF})=$ (a) $108 \mathrm{~cm}^{2}$ (b) $96 \mathrm{~cm}^{2}$ (c) $48 \mathrm{~cm}^{2}$ (d) $100 \mathrm{~cm}^{2}$ Solution: Given: In Δ ABC and Δ DEF $\triangle \mathrm{ABC} \sim \triangle \mathrm{DEF}$ $\operatorname{Ar}(\triangle \mathrm{ABC})=54 \mathrm{~cm}^{2}...
Read More →If sec x = m and tan x = n,
Question: If $\sec x=m$ and $\tan x=n$, then $\frac{1}{m}\left\{(m+n)+\frac{1}{m+n}\right\}$ is equal to ____________. Solution: If secx=mand tanx=n $\frac{1}{m}\left\{(m+n)+\frac{1}{m+n}\right\}$ i. e $\frac{1}{\sec x}\left\{\sec x+\tan x+\frac{1}{\sec x+\tan x}\right\}$ $=\frac{1}{\sec x}\left\{\sec x+\tan x+\frac{1}{\sec x+\tan x} \times\left(\frac{\sec x-\tan x}{\sec x-\tan x}\right)\right\}$ $=\frac{1}{\sec x}\left\{\sec x+\tan x+\frac{\sec x-\tan x}{\sec ^{2} x-\tan ^{2} x}\right\}$ $=\fra...
Read More →For the differential equation
Question: For the differential equation $x y \frac{d y}{d x}=(x+2)(y+2)$, find the solution curve passing through the point $(1,-1)$. Solution: The differential equation of the given curve is: $x y \frac{d y}{d x}=(x+2)(y+2)$ $\Rightarrow\left(\frac{y}{y+2}\right) d y=\left(\frac{x+2}{x}\right) d x$ $\Rightarrow\left(1-\frac{2}{y+2}\right) d y=\left(1+\frac{2}{x}\right) d x$ Integrating both sides, we get: $\int\left(1-\frac{2}{y+2}\right) d y=\int\left(1+\frac{2}{x}\right) d x$ $\Rightarrow \in...
Read More →A man goes 24 m due west and then 7 m due north. How far is he from the starting point?
Question: A man goes 24 m due west and then 7 m due north. How far is he from the starting point? (a) 31 m(b) 17 m(c) 25 m(d) 26 m Solution: A man goes 24m due to west and then 7m due north. Let the man starts from point B and goes 24 m due to west and reaches point A, then walked 7m north and reaches point C. Now we have to find the distance between the starting point and the end point i.e. BC. In right triangle ABC, applying Pythagoras theorem, we get $\mathrm{BC}^{2}=\mathrm{AB}^{2}+\mathrm{A...
Read More →The minimum value of
Question: The minimum value of 9 tan2+ 4 cot2is ____________. Solution: 9 tan2+ 4 cot2 Since Arithmetic mass Geometric mean for 2 tans. i. e $\frac{a+b}{2} \geq \sqrt{a b}$ Let $a=9 \tan ^{2} \theta$ $b=4 \cot ^{2} \theta$ $\Rightarrow \frac{9 \tan ^{2} \theta+4 \cot ^{2} \theta}{2} \geq \sqrt{9 \tan ^{2} \theta \times 4 \cot ^{2} \theta}$ $=\sqrt{9 \times 4}$ $=3 \times 2$ i. e $9 \tan ^{2} \theta+4 \cot ^{2} \theta \geq 2 \times 6=12$ i.e minimum value of $9 \tan ^{2} \theta+4 \cot ^{2} \theta...
Read More →The minimum value of
Question: The minimum value of 9 tan2+ 4 cot2is ____________. Solution: 9 tan2+ 4 cot2 Since Arithmetic mass Geometric mean for 2 tans. i. e $\frac{a+b}{2} \geq \sqrt{a b}$ Let $a=9 \tan ^{2} \theta$ $b=4 \cot ^{2} \theta$ $\Rightarrow \frac{9 \tan ^{2} \theta+4 \cot ^{2} \theta}{2} \geq \sqrt{9 \tan ^{2} \theta \times 4 \cot ^{2} \theta}$ $=\sqrt{9 \times 4}$ $=3 \times 2$ i. e $9 \tan ^{2} \theta+4 \cot ^{2} \theta \geq 2 \times 6=12$ i.e minimum value of $9 \tan ^{2} \theta+4 \cot ^{2} \theta...
Read More →A cylindrical water tank of diameter 1.4 m and height 2.1 m is being fed by a pipe of diameter 3.5 cm
Question: A cylindrical water tank of diameter 1.4 m and height 2.1 m is being fed by a pipe of diameter 3.5 cm through which water flows at the rate of 2 metre per second. In how much time the tank will be filled? Solution: Given data is as follows: Diameter of the tank = 1.4 m Height of the tank = 2.1 m Diameter of the pipe = 3.5 cm Water flow rate = 2 m/sec We have to find the time required to fill the tank using the pipe. The diameter of the tank is given which is 1.4 m. Let us find the radi...
Read More →The length of the hypotenuse of an isosceles right triangle whose one side is 42 cm is
Question: The length of the hypotenuse of an isosceles right triangle whose one side is42cmis (a) 12 cm(b) 8 cm(c)82cm(d)122cm Solution: Given:One side of isosceles right triangle is 42cm To find:Length of the hypotenuse. We know that in isosceles triangle two sides are equal. In isosceles right triangle $\mathrm{ABC}$, let $\mathrm{AB}$ and $\mathrm{AC}$ be the two equal sides of measure $4 \sqrt{2} \mathrm{~cm}$. Applying Pythagoras theorem, we get $\mathrm{BC}^{2}=\mathrm{AB}^{2}+\mathrm{AC}^...
Read More →If
Question: If $\pix2 \pi$ and $\sqrt{\frac{1+\cos x}{1-\cos x}}+\sqrt{\frac{1-\cos x}{1+\cos x}}=k \operatorname{cosec} x$, then $k=$ __________ Solution: If $\pix2 \pi$ Given $\sqrt{\frac{1+\cos x}{1-\cos x}}+\sqrt{\frac{1-\cos x}{1+\cos x}}=k \operatorname{cosec} x$ L.H.S is $\sqrt{\frac{1+\cos x}{1-\cos x}}+\sqrt{\frac{1-\cos x}{1+\cos x}}$ $=\sqrt{\frac{1+\cos x}{1-\cos x} \times \frac{1+\cos x}{1+\cos x}}+\sqrt{\frac{1-\cos x}{1+\cos x} \times \frac{1-\cos x}{1-\cos x}}$ $=\sqrt{\frac{(1+\co...
Read More →From a tap of inner radius 0.75 cm, water flows at the rate of 7 m per second.
Question: From a tap of inner radius 0.75 cm, water flows at the rate of 7 m per second. Find the volume in litres of water delivered by the pipe in one hour. Solution: Given data is as follows: r = 0.75 cm Water flow rate = 7 m/sec Time = 1 hour We have to find the volume of water that flows through the pipe for 1 hour. Let us first convert water flow from m/sec to cm/sec, since radius of the pipe is in centimeters We have, Waterflow rate =7 m/sec = 700 cm/sec Volume of water delivered by the p...
Read More →Find the equation of a curve passing through the point
Question: Find the equation of a curve passing through the point $(0,0)$ and whose differential equation is $y^{\prime}=e^{x} \sin x$. Solution: The differential equation of the curve is: $y^{\prime}=e^{x} \sin x$ $\Rightarrow \frac{d y}{d x}=e^{x} \sin x$ $\Rightarrow d y=e^{x} \sin x$ Integrating both sides, we get: $\int d y=\int e^{x} \sin x d x$ ...(1) $\int d y=\int e^{x} \sin x d x$ $\Rightarrow I=\sin x \int e^{x} d x-\int\left(\frac{d}{d x}(\sin x) \cdot \int e^{x} d x\right) d x$ $\Rig...
Read More →If
Question: If $\frac{\pi}{2}x\pi$ and $\sqrt{\frac{1+\sin x}{1-\sin x}}+\sqrt{\frac{1-\sin x}{1+\sin x}}=k$ sec $x$, then $k=$ Solution: If $\frac{\pi}{2}x\pi$ $\sqrt{\frac{1+\sin x}{1-\sin x}}+\sqrt{\frac{1-\sin x}{1+\sin x}}=k \sec x \quad$ (given) L.H. $\mathrm{S}=\sqrt{\frac{1+\sin x}{1-\sin x}}+\sqrt{\frac{1-\sin x}{1+\sin x}}$ $=\sqrt{\frac{1+\sin x}{1-\sin x} \times \frac{1+\sin x}{1+\sin x}}+\sqrt{\frac{1-\sin x}{1+\sin x} \times \frac{1-\sin x}{1-\sin x}}$ $=\sqrt{\left[\frac{(1+\sin x)^...
Read More →A cylindrical tube, open at both ends, is made of metal.
Question: A cylindrical tube, open at both ends, is made of metal. The internal diameter of the tube is 10.4 cm and its length is 25 cm. The thickness of the metal is 8 mm everywhere. Calculate the volume of the metal. Solution: Given data is as follows: Internal diameter = 10.4 cm Thickness of the metal = 8 mm Length of the pipe = 25 cm We have to find the volume of the metal used in the pipe. We know that, Volume of the hollow pipe $=\pi\left(R^{2}-r^{2}\right) h$ Given is the internal diamete...
Read More →