Determine order and degree(if defined) of differential equation
Question: Determine order and degree(if defined) of differential equation $\frac{d^{4} y}{d x^{4}}+\sin \left(y^{\prime \prime}\right)=0$ Solution: $\frac{d^{4} y}{d x^{4}}+\sin \left(y^{\prime \prime \prime}\right)=0$ $\Rightarrow y^{\prime \prime \prime \prime}+\sin \left(y^{\prime \prime \prime}\right)=0$ The highest order derivative present in the differential equation is $y^{\prime \prime \prime}$. Therefore, its order is four. The highest order derivative present in the differential equati...
Read More →The dimensions of a room are 12.5 m by 9 m by 7 m.
Question: The dimensions of a room are 12.5 m by 9 m by 7 m. There are 2 doors and 4 windows in the room; each door measures 2.5 m by 1.2 m and each window 1.5 m by 1 m. Find the cost of painting the walls at Rs 3.50 per square meter. Solution: Given Length of the room = 12.5 m Breadth of the room = 9 m Height of the room = 7 m Therefore, total surface area of the four walls = 2(l + b) * h = 2(12.5 + 9) * 7 $=301 \mathrm{~m}^{2}$ Area of 2 doors = 2(2.5 *1.2) $=6 \mathrm{~m}^{2}$ Area of 4 windo...
Read More →In ∆ABC, ray AD bisects ∠A and intersects BC in D. If BC = a, AC = b and AC = c, prove that
Question: In ∆ABC, ray AD bisects A and intersects BC in D. If BC =a, AC =band AC =c, prove that (i) $\mathrm{BD}=\frac{a c}{b+c}$ (ii) $\mathrm{DC}=\frac{a b}{b+c}$ Solution: Given: In $\triangle A B C$ ray $A D$ bisects angle $A$ and intersects $B C$ in $D$, If $B C=a, A C=b$ and $A B=c$ To Prove: (i) $B D=\frac{a c}{b+c}$ (ii) $D C=\frac{a b}{b+c}$ (i) The corresponding figure is as follows Proof. In triangle $A B C, A D$ is the bisector of $\angle A$ Therefore $\frac{A B}{A C}=\frac{B D}{C D...
Read More →The cost of preparing the walls of a room 12 m long at the rate of Rs 1.35 per square meter is Rs 340.20 and
Question: The cost of preparing the walls of a room 12 m long at the rate of Rs 1.35 per square meter is Rs 340.20 and the cost of matting the floor at 85 paise per square meter is Rs 91.80. Find the height of the room. Solution: Given that, Length of the room = 12m Let the height of the room be 'h' Area of 4 walls = 2(l + b)*h According to the question 2(l + b) * h * 1.35 = 340.20 2(12 + b0 * h * 1.35 = 340.20 $(12+\mathrm{b}) * \mathrm{~h}=\frac{170.10}{1.35}=126 \ldots$ Also Area of the Floor...
Read More →The area bounded by the
Question: The area bounded by the $y$-axis, $y=\cos x$ and $y=\sin x$ when $0 \leq x \leq \frac{\pi}{2}$ A. $2(\sqrt{2}-1)$ B. $\sqrt{2}-1$ C. $\sqrt{2}+1$ D. $\sqrt{2}$ Solution: The given equations are y= cosx (1) And,y= sinx (2) Required area = Area (ABLA) + area (OBLO) $=\int_{\frac{1}{\sqrt{2}}}^{1} x d y+\int_{0}^{\frac{1}{\sqrt{2}}} x d y$ $=\int_{\frac{1}{\sqrt{2}}}^{1} \cos ^{-1} y d y+\int_{0}^{\frac{1}{\sqrt{2}}} \sin ^{-1} x d y$ Integrating by parts, we obtain $=\left[y \cos ^{-1} y...
Read More →An open box is made of wood 3 cm thick. Its external length,
Question: An open box is made of wood 3 cm thick. Its external length, breadth and height are 1.48 m, 1.16 m and 8.3 dm. Find the cost of painting the inner surface of Rs 50 per sq. metre. Solution: Given Data: Outer Dimensions Length = 148 cm Breadth = 116 cm Height = 83 cm Inner Dimensions Length = 148 - (2 * 3) = 142 cm Breadth = 116 - (2 * 3) = 110 cm Height = 83 - 3 = 80 Surface Area of the Inner region = 2h(l + b) + lb = 2 * 80(142 + 110) + 142 * 110 = 2 * 80 * 252 + 142 * 110 $=55940 \mat...
Read More →In the given figure, each of PA, QB, RC and SD is perpendicular to l.
Question: In the given figure, each of PA, QB, RC and SD is perpendicular tol. If AB = 6 cm, BC = 9 cm, CD = 12 cm and PS = 36 cm, then determine PQ, QR and RS. Solution: Given $A B=6 \mathrm{~cm}, B C=9 \mathrm{~cm}, C D=12 \mathrm{~cm}, A D=27 \mathrm{~cm}$ and $P S=36 \mathrm{~cm}$ PA,QB, RCandSDis perpendicular tol, Therefore, by the corollory of basic proportionality theorem, we have $\frac{A B}{A D}=\frac{P Q}{P S}$ $\frac{B C}{A D}=\frac{Q R}{P S}$ $\frac{C D}{A D}=\frac{R S}{P S}$ $\Righ...
Read More →Ravish wanted to make a temporary shelter for his car by making a box-like structure with the tarpaulin that covers all the four sides and the top of the car ( with the front face of a flap which can be rolled up).
Question: Ravish wanted to make a temporary shelter for his car by making a box-like structure with the tarpaulin that covers all the four sides and the top of the car ( with the front face of a flap which can be rolled up). Assuming that the stitching margins are very small, and therefore negligible, how many tarpaulins would be required to make the shelter of height 2.5 m with base dimensions 4 m 3m? Solution: Given That, Shelter length = 4 m Breadth = 3 m Height = 2.5 m The tarpaulin will be ...
Read More →The area of the circle
Question: The area of the circle $x^{2}+y^{2}=16$ exterior to the parabola $y^{2}=6 x$ is A. $\frac{4}{3}(4 \pi-\sqrt{3})$ B. $\frac{4}{3}(4 \pi+\sqrt{3})$ C. $\frac{4}{3}(8 \pi-\sqrt{3})$ D. $\frac{4}{3}(4 \pi+\sqrt{3})$ Solution: The given equations are $x^{2}+y^{2}=16$ ...(1) $y^{2}=6 x$ ...(2) Area bounded by the circle and parabola $=2[\operatorname{area}(\mathrm{OADO})+\operatorname{area}(\mathrm{ADBA})]$ $=2\left[\int_{0}^{2} \sqrt{6 x} d x+\int_{2}^{4} \sqrt{16-x^{2}} d x\right]$ $=2 \in...
Read More →A closed iron tank 12 m long, 9 m wide and 4 m deep is to be made.
Question: A closed iron tank 12 m long, 9 m wide and 4 m deep is to be made. Determine the cost of iron sheet used at the rate of Rs 5 per meter sheet, a sheet being 2 m wide. Solution: Length (l) = 12 m Breadth (b) = 9 m Height (h) = 4 m Total surface area of the tank = 2[lb + bh + hl] = 2[12*9 + 9*4 + 12*4] = 2[108 + 36 + 48] $=384 \mathrm{~m}^{2}$ The Length of the Iron sheet $=\frac{\text { Area of the Iron Sheet }}{\text { Width of the Iron Sheet }}$ =384/2 = 192 m. Cost of the Iron Sheet =...
Read More →Prove that:
Question: Prove that: $\sec \left(\frac{3 \pi}{2}-x\right) \sec \left(x-\frac{5 \pi}{2}\right)+\tan \left(\frac{5 \pi}{2}+x\right) \tan \left(x-\frac{3 \pi}{2}\right)=-1$ Solution: $\mathrm{LHS}=\sec \left(\frac{3 \pi}{2}-x\right) \sec \left(x-\frac{5 \pi}{2}\right)+\tan \left(\frac{5 \pi}{2}+x\right) \tan \left(x-\frac{3 \pi}{2}\right)$ $=\sec \left(\frac{3 \pi}{2}-x\right) \sec \left[-\left(\frac{5 \pi}{2}-x\right)\right]+\tan \left(\frac{5 \pi}{2}+x\right) \tan \left[-\left(\frac{3 \pi}{2}-x\...
Read More →The dimensions of a rectangular box are in the ratio of 2: 3: 4 and the difference between the cost of covering it
Question: The dimensions of a rectangular box are in the ratio of $2: 3: 4$ and the difference between the cost of covering it with a sheet of paper at the rates of Rs 8 and Rs $9.50$ per $m^{2}$ is Rs 1248 . Find the dimensions of the box. Solution: Let the ratio be x Length (l) = 2x Breadth (b) = 3x Height (h) = 4x Therefore, Total Surface area = 2[lb + bh + hl] $=2\left(6 x^{2}+12 x^{2}+8 x^{2}\right)$ $=52 x^{2} \mathrm{~m}^{2}$ When the cost is at Rs. 8 per $\mathrm{m}^{2}$ Therefore, the t...
Read More →The diagonals of quadrilateral ABCD intersect at O. Prove that
Question: The diagonals of quadrilateral $\mathrm{ABCD}$ intersect at $\mathrm{O}$. Prove that $\frac{\operatorname{ar}(\Delta \mathrm{ACB})}{\operatorname{ar}(\Delta \mathrm{ACD})}=\frac{\mathrm{BO}}{\mathrm{DO}}$. Solution: We are given the following quadrilateral withOas the intersection point of diagonals To Prove: $\frac{\operatorname{ar}(\Delta A C B)}{\operatorname{ar}(\triangle A C D)}=\frac{B O}{D O}$ GivenACBandACDare two triangles on the same baseAC Considerhas the distance between tw...
Read More →Prove that:
Question: Prove that: $\sin ^{2} \frac{\pi}{18}+\sin ^{2} \frac{\pi}{9}+\sin ^{2} \frac{7 \pi}{18}+\sin ^{2} \frac{4 \pi}{9}=2$ Solution: $\mathrm{LHS}=\sin ^{2} \frac{\pi}{18}+\sin ^{2} \frac{\pi}{9}+\sin ^{2} \frac{7 \pi}{18}+\sin ^{2} \frac{4 \pi}{9}$ $=\sin ^{2} \frac{\pi}{18}+\sin ^{2} \frac{2 \pi}{18}+\sin ^{2} \frac{7 \pi}{18}+\sin ^{2} \frac{8 \pi}{18}$ $=\sin ^{2} \frac{\pi}{18}+\sin ^{2} \frac{2 \pi}{18}+\sin ^{2}\left(\frac{7 \pi}{18}\right)+\sin ^{2}\left(\frac{8 \pi}{18}\right)$ $=\...
Read More →The area bounded by the curve
Question: The area bounded by the curve $y=x|x|, x$-axis and the ordinates $x=-1$ and $x=1$ is given by [Hint: $y=x^{2}$ if $x0$ and $y=-x^{2}$ if $x0$ ] A. 0 B. $\frac{1}{3}$ C. $\frac{2}{3}$ D. $\frac{4}{3}$ Solution: Required area $=\int_{-1}^{1} y d x$ $=\int_{-1}^{1} x|x| d x$ $=\int_{-1}^{0} x^{2} d x+\int_{0}^{1} x^{2} d x$ $=\left[\frac{x^{3}}{3}\right]_{-1}^{0}+\left[\frac{x^{3}}{3}\right]_{0}^{1}$ $=-\left(-\frac{1}{3}\right)+\frac{1}{3}$ $=\frac{2}{3}$ units Thus, the correct answer i...
Read More →Each edge of a cube is increased by 50%.
Question: Each edge of a cube is increased by 50%. Find the percentage increase in the surface area of the cube. Solution: Let 'a' be the edge of the cube Therefore the surface area of the cube $=6 a^{2}$ i.e., $\mathrm{s}_{1}=6 \mathrm{a}^{2}$ We get a new edge after increasing the edge by 50% The new edge =a + 50/100a = 3/2a Considering the new edge, the new surface area is $=6 *(3 / 2 a)^{2}$ i.e., $\mathrm{S}_{2}=6 * \frac{9}{4} \mathrm{a}^{2}$ $\mathrm{S}_{2}=\frac{27}{2} \mathrm{a}^{2}$ Th...
Read More →In ∆ABC, AD and BE are altitude. Prove that
Question: In $\triangle \mathrm{ABC}, \mathrm{AD}$ and $\mathrm{BE}$ are altitude. Prove that $\frac{\operatorname{ar}(\Delta \mathrm{DEC})}{\operatorname{ar}(\Delta \mathrm{ABC})}=\frac{\mathrm{DC}^{2}}{\mathrm{AC}^{2}}$. Solution: Given:ΔABC in which AD and BE are altitudes on sides BC and AC respectively. SinceADB =AEB = 90, there must be a circle passing through point D and E having AB as diameter. We also know that, angle in a semi-circle is a right angle. Now, join DE. So, ABDE is a cyclic...
Read More →Prove that
Question: Prove that (i) $\frac{\cos (2 \pi+x) \operatorname{cosec}(2 \pi+x) \tan (\pi / 2+x)}{\sec (\pi / 2+x) \cos x \cot (\pi+x)}=1$ (ii) $\frac{\operatorname{cosec}\left(90^{\circ}+x\right)+\cot \left(450^{\circ}+x\right)}{\operatorname{cosec}\left(90^{\circ}-x\right)+\tan \left(180^{\circ}-x\right)}+\frac{\tan \left(180^{\circ}+x\right)+\sec \left(180^{\circ}-x\right)}{\tan \left(360^{\circ}+x\right)-\sec (-x)}=2$ (iii) $\frac{\sin \left(180^{\circ}+x\right) \cos \left(90^{\circ}+x\right) \...
Read More →Prove that
Question: Prove that (i) $\frac{\cos (2 \pi+x) \operatorname{cosec}(2 \pi+x) \tan (\pi / 2+x)}{\sec (\pi / 2+x) \cos x \cot (\pi+x)}=1$ (ii) $\frac{\operatorname{cosec}\left(90^{\circ}+x\right)+\cot \left(450^{\circ}+x\right)}{\operatorname{cosec}\left(90^{\circ}-x\right)+\tan \left(180^{\circ}-x\right)}+\frac{\tan \left(180^{\circ}+x\right)+\sec \left(180^{\circ}-x\right)}{\tan \left(360^{\circ}+x\right)-\sec (-x)}=2$ (iii) $\frac{\sin \left(180^{\circ}+x\right) \cos \left(90^{\circ}+x\right) \...
Read More →Corresponding sides of two similar triangles are in the ratio 1 : 3.
Question: Corresponding sides of two similar triangles are in the ratio $1: 3$. If the area of the smaller triangle in $40 \mathrm{~cm}^{2}$, find the area of the larger triangle. Solution: Since the ratio of areas of two similar triangles is equal to the ratio of the squares of any two corresponding sides. $\frac{\text { Area of smaller triangle }}{\text { Area of larger triangle }}=\frac{\text { (Corresponding side of smaller triangle) }^{2}}{\text { (Corresponding side of larger triangle) }^{...
Read More →Area bounded by the curve
Question: Area bounded by the curve $y=x^{3}$, the $x$-axis and the ordinates $x=-2$ and $x=1$ is A. $-9$ B. $-\frac{15}{4}$ C. $\frac{15}{4}$ D. $\frac{17}{4}$ Solution: Required Area $=\left|\int_{-2}^{0} y d x\right|+\int_{0}^{1} y d x$ $=\left|\int_{-2}^{0} x^{3} d x\right|+\int_{0}^{1} x^{3} d x$ $=\left|\left[\frac{x^{4}}{4}\right]_{-2}^{0}\right|+\left[\frac{x^{4}}{4}\right]_{0}^{1}$ $=\left|\left[0-\frac{16}{4}\right]\right|+\left[\frac{1}{4}-0\right]$ $=|-4|+\frac{1}{4}$ $=4+\frac{1}{4}...
Read More →Hameed has built a cubical water tank with lid for his house, with each other edge 1.5 m long.
Question: Hameed has built a cubical water tank with lid for his house, with each other edge 1.5 m long. He gets the outer surface of the tank excluding the base, covered with square tiles of side 25 cm. Find how much he would spend for the tiles if the cost of tiles is Rs 360 per dozen. Solution: Given that Hameed is getting 5 outer faces of the tank covered with tiles, he would need to know the surface area of the tank, to decide on the quantity of tiles required. Edge of the cubical tank (a) ...
Read More →The area of two similar triangles are 36 cm2 and 100 cm2. I
Question: The area of two similar triangles are $36 \mathrm{~cm}^{2}$ and $100 \mathrm{~cm}^{2}$. If the length of a side of the smaller triangle in $3 \mathrm{~cm}$, find the length of the corresponding side of the larger triangle. Solution: Since the ratio of areas of two similar triangles is equal to the ratio of the squares of any two corresponding sides. $\frac{\text { Area of smaller triangle }}{\text { Area of larger triangle }}=\frac{(\text { Corresponding side of smaller triangle })^{2}...
Read More →Find the area of the region
Question: Find the area of the region $\left\{(x, y): y^{2} \leq 4 x, 4 x^{2}+4 y^{2} \leq 9\right\}$ Solution: The area bounded by the curves, $\left\{(x, y): y^{2} \leq 4 x, 4 x^{2}+4 y^{2} \leq 9\right\}$, is represented as The points of intersection of both the curves are $\left(\frac{1}{2}, \sqrt{2}\right)$ and $\left(\frac{1}{2},-\sqrt{2}\right)$. The required area is given by OABCO. It can be observed that area OABCO is symmetrical aboutx-axis. Area OABCO = 2 Area OBC Area OBCO = Area OMC...
Read More →The length of a hall is 18 m and the width 12 m.
Question: The length of a hall is 18 m and the width 12 m. The sum of the areas of the floor and the flat roof is equal to the sum of the areas of the four walls. Find the height of the hall. Solution: Length of the hall = 18 m Width of the hall = 12 m Now given, Area of the floor and the flat roof = sum of the areas of four walls ⇒ 2lb = 2lh + 2bh ⇒ lb = lh + bh $\Rightarrow \mathrm{h}=\frac{\mathrm{lb}}{\mathrm{l}+\mathrm{b}}$ $=\frac{18 * 12}{18+12}$ $=\frac{216}{30}=7.2 \mathrm{~m}$...
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