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Question: If $A=\left[\begin{array}{lll}0 1 1 \\ 1 0 1 \\ 1 1 0\end{array}\right]$, find $A^{-1}$ and show that $A^{-1}=\frac{1}{2}\left(A^{2}-3 I\right)$ Solution: $A=\left[\begin{array}{lll}0 1 1 \\ 1 0 1 \\ 1 1 0\end{array}\right]$ $A^{-1}=\frac{1}{|A|}$ Adj. $A$ Now, $|A|=\left|\begin{array}{lll}0 1 1 \\ 1 0 1 \\ 1 1 0\end{array}\right|$ $=-1(-1)+1(1)$ $=2$ Now, to find Adj. $A$ $A_{11}=(-1)^{1+1}(-1)=-1 \quad A_{21}=(-1)^{2+1}(-1)=1 \quad A_{31}=(-1)^{3+1}(1)=1$ $A_{12}=(-1)^{1+2}(-1)=1 \qu...
Read More →Metallic spheres of radii 6 cm, 8 cm and 10 cm, respectively are melted to form a single solid sphere.
Question: Metallic spheres of radii 6 cm, 8 cm and 10 cm, respectively are melted to form a single solid sphere. Find the radius of the resulting sphere. Solution: We have, the radii $r_{1}=6 \mathrm{~cm}, r_{2}=8 \mathrm{~cm}$ and $r_{3}=10 \mathrm{~cm}$ Let the radius of the resulting sphere be $R$. As, Volume of resulting sphere $=$ Volume of three metallic spheres $\Rightarrow \frac{4}{3} \pi R^{3}=\frac{4}{3} \pi r_{1}^{3}+\frac{4}{3} \pi r_{2}^{3}+\frac{4}{3} \pi r_{3}{ }^{3}$ $\Rightarrow...
Read More →Balanced diet should contain 12% of proteins, 25% of fats and 63% of carbohydrates.
Question: Balanced diet should contain 12% of proteins, 25% of fats and 63% of carbohydrates. If a child needs 2600 calories in this food daily, find in calories the amount of each of these in his daily food intake. Solution: In a balanced diet of 2600 calories, $12 \%$ is protein. $\therefore$ Amount of protein in food intake $=12 \%$ of 2600 $=\frac{12}{100} \times 2600$ $=312$ Similarly, a balanced diet contain $s 25 \%$ fats. $\therefore$ Amount of fats in food intake $=25 \%$ of 2600 $=\fra...
Read More →A cone of height 20 cm and radius of base 5 cm is made up of modelling clay.
Question: A cone of height 20 cm and radius of base 5 cm is made up of modelling clay. A child reshapes it in the form of a sphere. Find the diameter of the sphere. Solution: We have, the base radius of the cone, $r=5 \mathrm{~cm}$ and the height of the cone, $h=20 \mathrm{~cm}$ Let the radius of the sphere be $R$. As, Volume of sphere $=$ Volume of cone $\Rightarrow \frac{4}{3} \pi R^{3}=\frac{1}{3} \pi r^{2} h$ $\Rightarrow R^{3}=\frac{\pi r^{2} h \times 3}{3 \times 4 \pi}$ $\Rightarrow R^{3}=...
Read More →A solid metallic cuboid of dimensions 9 m × 8 m × 2 m is melted and recast into solid cubes of edge 2 m.
Question: A solid metallic cuboid of dimensions 9 m 8 m 2 m is melted and recast into solid cubes of edge 2 m. Find the number of cubes so formed. Solution: The volume of solid metallic cuboid is $9 \times 8 \times 2=144 \mathrm{~m}^{3}$. This cuboid has been recasted into solid cubes of edge $2 \mathrm{~m}$ whose volume is given by $2^{3}=8 \mathrm{~m}^{3}$. Therefore, the total number of cubes so formed $=\frac{\text { the volume of solid metallic cuboid }}{\text { the volume of solid cubes }}...
Read More →Draw an angle of 110° with the help
Question: Draw an angle of 110 with the help of a protractor and bisect it. Measure , each angle. Solution: Draw BXA = 110 with the help of a protractor. Now, we use the following steps for required construction Taking X as centre and any radius daw an arc to Intersect the rays XA and XB, say at E and D, respectively. Taking D and E as centres and with the radius more than DE, draw arcs to intersecteach other, say at F. Draw the ray XF.Thus, ray XF is the required bisector of the angle B X A. On...
Read More →A cistern can be filled by a tap in 4 hours and emptied by an outlet pipe in 6 hours.
Question: A cistern can be filled by a tap in 4 hours and emptied by an outlet pipe in 6 hours. How long will it take to fill the cistern if both the tap and the pipe are opened together? Solution: Time taken by the tap to fill the cistern $=4$ hours $\therefore$ Tap fills $\frac{1}{4}$ th part of the cistern in 1 hour. Time taken by the pipe to empty the cistern $=6$ hours $\therefore$ Pipe empties out $\frac{1}{6}$ th part of the cistern in 1 hour. Thus, in 1 hour, $\left(\frac{1}{4}-\frac{1}{...
Read More →A wooden toy is in the shape of a cone mounted on a cylinder, as shown in the figure.
Question: A wooden toy is in the shape of a cone mounted on a cylinder, as shown in the figure. The total height of the toy is 26 cm, while the height of the conical part is 6 cm. The diameter of the base of the conical part is 5 cm and that of the cylindrical part is 4 cm. The conical part and the cylindrical part are respectively painted red and white. Find the area to be painted by each of these colours.$\left[\right.$ Take $\left.\pi=\frac{22}{7}\right]$ Solution: We have, the base radius of...
Read More →Find the adjoint of the matrix
Question: Find the adjoint of the matrix $A=\left[\begin{array}{ccc}-1 -2 -2 \\ 2 1 -2 \\ 2 -2 1\end{array}\right]$ and hence show that $A(\operatorname{adj} A)=|A| I_{3}$. Solution: $A=\left[\begin{array}{ccc}-1 -2 -2 \\ 2 1 -2 \\ 2 -2 1\end{array}\right]$ Now, to find Adj. $A$ $A_{11}=(-1)^{1+1}(-3)=-3 \quad A_{21}=(-1)^{2+1}(-6)=6 \quad A_{31}=(-1)^{3+1}(6)=6$ $A_{12}=(-1)^{1+2}(6)=-6 \quad A_{22}=(-1)^{2+2}(3)=3 \quad A_{32}=(-1)^{3+2}(6)=-6$ $A_{13}=(-1)^{1+3}(-6)=-6 \quad A_{23}=(-1)^{2+3}...
Read More →A ΔABC can be constructed in which ∠B = 60°,
Question: A ΔABC can be constructed in which B = 60, C =45, and AB + BC + CA = 12 cm. Solution: True We know that, sum of angles of a triangle is 180. A + B + C = 180 Here, B + C = 60+ 45 = 105 180, Thus, ΔABC with given conditions can be constructed....
Read More →A ΔABC can be constructed in which ∠B =105°,
Question: A ΔABC can be constructed in which B =105, C = 90 and AB +BC + CA = 10 cm. Solution: False Here, B = 105, C = 90 and AB + BC + CA = 10cm We know that, sum of angles of a triangle is 180. A + B + C = 180 Here, B + C = 105+90 = 195 180 which is not true. Thus, ΔABC with given conditions cannot be constructed....
Read More →A cistern has two inlets A and B which can fill it in 12 hours and 15 hours respectively.
Question: A cistern has two inletsAandBwhich can fill it in 12 hours and 15 hours respectively. An outlet can empty the full cistern in 10 hours. If all the three pipes are opened together in the empty cistern, how much time will they take to fill the cistern completely? Solution: Time taken by tap A to fill the cistern $=12$ hours Time taken by tap B to fill the cistern $=15$ hours Let $C$ be the outlet that can empty the cistern in 10 hours. Time taken by tap $\mathrm{C}$ to empty the cistern ...
Read More →ΔABC can be constructed in which BC = 6 cm,
Question: ΔABC can be constructed in which BC = 6 cm, C = 30 and AC AB =4 cm. Solution: True We know that, a triangle can be constructed if sum of its two sides is greater than third side. i.e., in ΔABC, AB + BC AC = BC AC AB = 6 4, which is true, so ΔABC with given conditions can be constructed....
Read More →ΔABC can be constructed in which AB = 5 cm,
Question: ΔABC can be constructed in which AB = 5 cm, A = 45 and BC + AC = 5 cm. Solution: False We know that, a triangle can be constructed, if sum of its two sides is greater than third side. Here, BC + AC = AB = 5 cm So, ΔABC cannot be constructed....
Read More →An angle of 42.5° can be constructed.
Question: An angle of 42.5 can be constructed. Solution: FalseWe know that, 42.5 = x 85 and an angle of 85 cannot be constructed with the help of ruler and compass....
Read More →The inner diameter of a glass is 7 cm and it has a raised portion in the bottom in the shape of a hemisphere, as shown in the figure.
Question: The inner diameter of a glass is 7 cm and ithas a raised portion in the bottom in theshape of a hemisphere, as shown in thefigure. If the height of the glass is 16 cm, find the apparent capacity and the actual capacity of the glass. Solution: We have, the height of the glass, $h=16 \mathrm{~cm}$ and the base radius of cylinder $=$ the base radius of hemisphere, $r=\frac{7}{2} \mathrm{~cm}$ Now, The apparent capacity of the glass = Volume of the cylinder $=\pi r^{2} h$ $=\frac{22}{7} \t...
Read More →A pipe can fill a cistern in 10 hours.
Question: A pipe can fill a cistern in 10 hours. Due to a leak in the bottom it is filled in 12 hours. When the cistern is full, in how much time will it be emptied by the leak? Solution: When there is no leakage, the pipe can fill the cistern in 10 hours. Thus, the pipe can fill $\frac{1}{10}$ th part of the cistern in 1 hour. When there is leakage, the pipe can fill the cistern in 12 hours. Therefore, in case of leakage, the pipe can fill $\frac{1}{12}$ th part of the cistern in 1 hour. Thus, ...
Read More →An angle of 52.5° can be constructed.
Question: An angle of 52.5 can be constructed. Solution: True To construct an angle of 52.5 firstly construct an angle of 90, then construct an angle of 120 and then plot an angle bisector of 120 and 90 to get an angle 105 (90 + 15). Now, bisect this angle to get an angle of 52.5....
Read More →Solve this
Question: If $A=\left[\begin{array}{ccc}1 -2 3 \\ 0 -1 4 \\ -2 2 1\end{array}\right]$, find $\left(A^{T}\right)^{-1}$. Solution: We know that $\left(A^{T}\right)^{-1}=\left(A^{-1}\right)^{T}$. $A=\left[\begin{array}{ccc}1 -2 3 \\ 0 -1 4 \\ -2 2 1\end{array}\right]$ $A^{-1}=\frac{1}{|A|}$ Adj. $A$ Now, $|A|=\left|\begin{array}{ccc}1 -2 3 \\ 0 -1 4 \\ -2 2 1\end{array}\right|$ $=1(-1-8)-2(-8+3)$ $=-9+10$ $=1$ Now, to find Adj. $A$ $A_{11}=(-1)^{1+1}(-9)=-9 \quad A_{21}=(-1)^{2+1}(-8)=8 \quad A_{31...
Read More →The construction of a ΔABC,
Question: The construction of a ΔABC, given that BC = 3 cm, C = 60 is possible when difference of AB and AC is equal to (a)3.2 cm (b)3.1 cm (c)3 cm (d)2.8 cm Solution: (d)Given, BC = 3 cm and C=60 We know that, the construction of a triangle is possible, if sum of two sides is greater than the third side of the triangle i.e., AB+ BC AC = BC AC AB = 3 AC AB So, if AC AB = 2.8 cm, then construction of ΔABC with given conditions is possible....
Read More →Two taps A and B can fill an overhead tank in 10 hours and 15 hours respectively.
Question: Two tapsAandBcan fill an overhead tank in 10 hours and 15 hours respectively. Both the taps are opened for 4 hours and theyBis turned off. How much time willAtake to fill the remaining tank? Solution: Pipe A can fill the tank in 10 hours, and pipe B can fill the tank in 15 hours. $\therefore$ In 1 hour, A can fill $\frac{1}{10}$ th part of the tank. In 1 hour, $B$ can fill $\frac{1}{15}$ th part of the tank. $\therefore$ In 1 hour, $\mathrm{A}$ and $\mathrm{B}$ can fill $\left(\frac{1}...
Read More →The construction of ΔABC,
Question: The construction of ΔABC, given that BC = 6 cm, B = 45 is not possible when difference of AB and AC is equal to (a)6.9 cm (b)5.2 cm (c)5.0 cm (d)4.0 cm Solution: (a)Given, BC = 6 cm and B= 45 We know that, the construction of a triangle is not possible, if sum of two sides is less than or equal to the third side of the triangle. i.e., AB + BC AC = BC AC AB = 6 AC-AB So, if AC AB= 6.9 cm, then construction of ΔABC with given conditions is not possible....
Read More →A toy is in the shape of a right circular cylinder with a hemisphere on one end and a cone on the other.
Question: A toy is in the shape of a right circular cylinder with a hemisphere on one end and a cone on the other. The radius and height of the cylindrical part are 5 cm and 13 cm, respectively. The radii of the hemispherical and the conical parts are the same as that of the cylindrical part. Find the surface area of the toy, if the total height of the toy is 30 cm. Solution: We have, the base radius of cone $=$ the base radius of cylinder $=$ the base radius of hemisphere $=r=5 \mathrm{~cm}$, t...
Read More →A cistern can be filled by one tap in 8 hours, and by another in 4 hours.
Question: A cistern can be filled by one tap in 8 hours, and by another in 4 hours. How long will it take to fill the cistern if both taps are opened together? Solution: Time taken by the first tap to fill the cistern $=8$ hours Time taken by the second tap to fill the cistern $=4$ hours $\therefore$ Work done by the first tap in 1 hour $=\frac{1}{8}$ Work done by the second tap in 1 hour $=\frac{1}{4}$ $\therefore$ Work done by both the taps in 1 hour $=\frac{1}{8}+\frac{1}{4}$ $=\frac{1+2}{8}=...
Read More →A hemisphere of maximum possible diameter is placed over a cuboidal block of side 7 cm.
Question: (i) A hemisphere of maximum possible diameter is placed over a cuboidal block of side 7 cm. Find the surface area of the solid so formed. (ii) A cubical block of side 10 cm is surmounted by a hemisphere. What isthe largest diameter that the hemisphere can have? Find the cost ofpainting the total surface area of the solid so formed, at the rate of₹5per 100 sq cm. [Use= 3.14] Solution: (i)Disclaimer:It is written cuboid in the question but it should be cube. From the figure, it can be ob...
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