Solve this
Question: If $A^{-1}=\left[\begin{array}{ccc}3 -1 1 \\ -15 6 -5 \\ 5 -2 2\end{array}\right]$ and $B=\left[\begin{array}{ccc}1 2 -2 \\ -1 3 0 \\ 0 -2 1\end{array}\right]$, find $(A B)^{-1}$. Solution: We know that $(A B)^{-1}=B^{-1} A^{-1}$. $B=\left[\begin{array}{ccc}1 2 -2 \\ -1 3 0 \\ 0 -2 1\end{array}\right]$ $B^{-1}=\frac{1}{|B|}$ Adj. $B$ Now, $|B|=\left|\begin{array}{ccc}1 2 -2 \\ -1 3 0 \\ 0 -2 1\end{array}\right|$ $=1(3+0)+1(2-4)$ $=1$ Now, to find Adj. $B$ $|B|=\left|\begin{array}{ccc}1...
Read More →If 9 girls can prepare 135 garlands in 3 hours,
Question: If 9 girls can prepare 135 garlands in 3 hours, how many girls are needed to prepare 270 garlands in 1 hour? Solution: Number of garlands made by 9 girls in 1 hour $=\frac{135}{3}=45$ Number of garlands made by 1 girl in 1 hour $=\frac{45}{9}=5$ $\therefore$ Number of girls required to make 270 garlands in 1 hour $=\frac{270}{5}=54$....
Read More →Working 8 hours a day, Ashu can copy a book in 18 days.
Question: Working 8 hours a day, Ashu can copy a book in 18 days. How many hours a day should he work so as to finish the work in 12 days? Solution: If Ashu works for 8 hours daily, he can finish the work in 18 days. If he works for 1 hour daily, he can finish the work in $(18 \times 8)$ days or 144 days. $\therefore$ Number of hours he should work daily to finish the work in 12 days $=\frac{144}{12}=12$...
Read More →If 25 men earn Rs 1000 in 10 days,
Question: If 25 men earn Rs 1000 in 10 days, how much will 15 men earn in 15 days? Solution: Earnings of 25 men in 10 days $=$ Rs. 1000 $\therefore$ Earnings of 25 men in 1 day $=$ Rs. $\frac{1000}{10}=$ Rs. 100 $\therefore$ Earning of 1 man in 1 day $=$ Rs. $\frac{100}{25}=$ Rs. 4 $\therefore$ Earnings of 1 men in 15 days $=$ Rs. $(15 \times 4)=$ Rs. 60 $\therefore$ Earnings of 15 men in 15 days $=$ Rs. $(60 \times 15)$ $=$ Rs. 900...
Read More →With the help of a ruler and a compass
Question: With the help of a ruler and a compass it is not possible to construct an angle of (a)37.5 (b)40 (c)22.5 (d)67.5 Solution: (b)With the help of a ruler and a compass, we can construct the angels, 90, 60, 45, 22.5, 30 etc., and its bisector of an angle. So, it is not possible to construct an angle of 40....
Read More →Prove that:
Question: Prove that $\mid A$ adj $\left.A|=| A\right|^{n}$. Solution: Let $A=\left[a_{i j}\right]$ be a square matrix of order $n \times n$. If $C_{i j}$ is a cofactor of $a_{i j}$ in $A$, then adj $A=\left[C_{\mathrm{i} j}\right]^{T}=\left[C_{\mathrm{ji}}\right] .$ Also, it is a matrix of order $n \times n$. Because $A$ and adj $A$ are matrices of order $n \times n, A \times(\operatorname{adj} A)$ exists and is of order $n \times n$. $\Rightarrow\{A \times(\operatorname{adj} A)\}_{\mathrm{ij}}...
Read More →If 12 boys earn Rs 840 in 7 days,
Question: If 12 boys earn Rs 840 in 7 days, what will 15 boys earn in 6 days? Solution: Earnings of 12 boys in 7 days $=$ Rs. 840 $\therefore$ Earnings of 12 boys in 1 day $=$ Rs. $\frac{840}{7}=$ Rs. 120 $\therefore$ Earnings of 1 boy in 1 day $=$ Rs. $\frac{120}{12}=$ Rs. 10 $\therefore$ Earnings of 1 boy in 6 days $=$ Rs. $(10 \times 6)=$ Rs. 60 $\therefore$ Earnings of 15 boys in 6 days $=$ Rs. $(60 \times 15)$ $=$ Rs. 900...
Read More →From a cubical piece of wood of side 21 cm,
Question: From a cubical piece of wood of side 21 cm, a hemisphere is carved out in such a way that the diameter of the hemisphere is equal to the side of the cubical piece. Find the surface area and volume of the remaining piece. Solution: We have, the edge of the cubical piece, $a=21 \mathrm{~cm}$ and the radius of the hemisphere, $r=\frac{a}{2}=\frac{21}{2} \mathrm{~cm}$ The surface area of the remaining piece $=$ TSA of cube + CSA of hemisphere - Area of circle $=6 a^{2}+2 \pi r^{2}-\pi r^{2...
Read More →Neha types 75 pages in 14 hours.
Question: Neha types 75 pages in 14 hours. How many pages will she type in 20 hours? Solution: It is given that Neha types 75 pages in 14 hours. $\therefore$ Number of pages typed by her in 1 hour $=\frac{75}{14}$ $\therefore$ Number of pages typed by her in 20 hours $=\left(\frac{75}{14} \times 20\right)$ pages $=\frac{1500}{14}$ pages $=\frac{750}{7}$ pages $=107 \frac{1}{7}$ pages Thus, Neha types $107 \frac{1}{7}$ pages in 20 hours....
Read More →A rhombus whose diagonals are 4 cm and 6 cm in lengths.
Question: A rhombus whose diagonals are 4 cm and 6 cm in lengths. Solution: We know that, all sides of a rhombus are equal and the diagonals of a rhombus are perpendicular bisectors of one another. So, to construct a rhombus whose diagonals are 4 cm and 6 cm use the following steps. Draw the diagonal say AC = 4 cm Taking A and C as centres and radius more than AC draw arcs on both sides of the line segment AC to intersect each other. Cut both arcs intersect each other at P and Q, then join PQ. L...
Read More →The adjoining figure represents a solid consisting of a cylinder surmounted by a cone at one end and a hemisphere at the other.
Question: The adjoining figure represents a solid consisting of a cylinder surmounted by a cone at one end and a hemisphere at the other. Find the volume of the solid. Solution: Diameter of the cylindrical part =7 cmTherefore,radius of the cylindrical part = 3.5 cm Volume of hemisphere $=\frac{2}{3} \pi \mathrm{r}^{3}=\frac{2}{3} \times \frac{22}{7} \times 3.5 \times 3.5 \times 3.5=89.83 \mathrm{~cm}^{3}$ Volume of the cylinder $=\pi r^{2} \mathrm{~h}=\frac{22}{7} \times 3.5 \times 3.5 \times 6....
Read More →Solve this
Question: If $A=\left[\begin{array}{lll}1 2 2 \\ 2 1 2 \\ 2 2 1\end{array}\right]$, find $A^{-1}$ and prove that $A^{2}-4 A-5 I=O$ Solution: $A=\left[\begin{array}{lll}1 2 2\end{array}\right.$ $\begin{array}{lll}2 1 2\end{array}$ $\left.\begin{array}{lll}2 2 1\end{array}\right]$ $\Rightarrow|A|=\mid \begin{array}{lll}1 2 2\end{array}$ $\begin{array}{lll}2 1 2\end{array}$ $2 \quad 2 \quad 1 \mid=1(1-4)-2(2-4)+2(4-2)=-3+4+4=5$ Since, $|A| \neq 0$ Hence, $A$ is invertible. Now, $A^{2}=\left[\begin{...
Read More →The adjoining figure represents a solid consisting of a cylinder surmounted by a cone at one end and a hemisphere at the other.
Question: The adjoining figure represents a solid consisting of a cylinder surmounted by a cone at one end and a hemisphere at the other. Find the volume of the solid. Solution: Diameter of the cylindrical part =7 cmTherefore,radius of the cylindrical part = 3.5 cm Volume of hemisphere $=\frac{2}{3} \pi \mathrm{r}^{3}=\frac{2}{3} \times \frac{22}{7} \times 3.5 \times 3.5 \times 3.5=89.83 \mathrm{~cm}^{3}$ Volume of the cylinder $=\pi r^{2} \mathrm{~h}=\frac{22}{7} \times 3.5 \times 3.5 \times 6....
Read More →Reema weaves 35 baskets in 25 days.
Question: Reema weaves 35 baskets in 25 days. In how many days will she weave 55 baskets? Solution: It is given that Reema weaves 35 baskets in 25 days. $\therefore$ Time taken by her to weave 1 basket $=\frac{25}{35}$ day $\therefore$ Time taken by her to weave 55 baskets $=\left(\frac{25}{35} \times 55\right)$ days $=\frac{1375}{35}$ days $=\frac{275}{7}$ days $=39 \frac{2}{7}$ days...
Read More →A spherical glass vessel has a cylindrical neck that is 7 cm long and 4 cm in diameter.
Question: A spherical glass vessel has a cylindrical neck that is 7 cm long and 4 cm in diameter. The diameter of the spherical part is 21 cm. Find the quantity of water the vessel can hold. Solution: Diameter of spherical part = 21 cmRadius of the spherical part = 10.5 cm Volume of spherical part of the vessel $=\frac{4}{3} \pi r^{3}=\frac{4}{3} \times \frac{22}{7} \times 10.5 \times 10.5 \times 10.5=4851 \mathrm{~cm}^{3}$ Diameter of cylinder $=4 \mathrm{~cm}$ Radius of cylinder $=2 \mathrm{~c...
Read More →A spherical glass vessel has a cylindrical neck that is 7 cm long and 4 cm in diameter.
Question: A spherical glass vessel has a cylindrical neck that is 7 cm long and 4 cm in diameter. The diameter of the spherical part is 21 cm. Find the quantity of water the vessel can hold. Solution: Diameter of spherical part = 21 cmRadius of the spherical part = 10.5 cm Volume of spherical part of the vessel $=\frac{4}{3} \pi r^{3}=\frac{4}{3} \times \frac{22}{7} \times 10.5 \times 10.5 \times 10.5=4851 \mathrm{~cm}^{3}$ Diameter of cylinder $=4 \mathrm{~cm}$ Radius of cylinder $=2 \mathrm{~c...
Read More →An equilateral triangle,
Question: An equilateral triangle, if its altitude is 3.2 cm. Solution: We know that, in an equilateral triangle all sides are equal and all angles are equal i.e., each angle is of 60. Given, altitude of an equilateral triangle say ABC is 3.2 cm. To construct the ΔABC use the following steps. 1. Draw a line PQ. 2. Take a point $D$ on $P Q$ and draw a ray $D E \perp P Q$. 3. Cut the line segment $A D$ of length $3.2 \mathrm{~cm}$ from $D E$. 4. Make angles equal to $30^{\circ}$ at $A$ on both sid...
Read More →A metallic cylinder has radius 3 cm and height 5 cm.
Question: A metallic cylinder has radius $3 \mathrm{~cm}$ and height $5 \mathrm{~cm}$. To reduce its weight, a conical hole is drilled in the cylinder. The conical hole has a radius of $\frac{3}{2} \mathrm{~cm}$ and its depth is $\frac{8}{9} \mathrm{~cm}$. Calculate the ratio of the volume of metal left in the cylinder to the volume of metal taken out in conical shape. Solution: We have, the base radius of the cylinder, $R=3 \mathrm{~cm}$, the height of the cylinder, $H=5 \mathrm{~cm}$, the base...
Read More →A right triangle when one side is 3.5 cm
Question: A right triangle when one side is 3.5 cm and sum of other sides and the hypotenuse is 5.5 cm. Solution: Let given right triangle be $\mathrm{ABC}$. Then, given $B C=3.5 \mathrm{~cm}, \angle B=90^{\circ}$ and sum of other side and hypotenuse i.e., $A B+A C=5.5 \mathrm{~cm}$ To construct $\triangle \mathrm{ABC}$ use the following steps 1. Draw the base $B C=3.5 \mathrm{~cm}$ 2. Make an angle $X B C=90^{\circ}$ at the point $B$ of base $B C$. 3. Cut the line segment BD equal to AB + AC i....
Read More →8 men can do a piece of work in 9 days.
Question: 8 men can do a piece of work in 9 days. In how many days will 6 men do it? Solution: It is given that 8 men can complete a job in 9 days. $\therefore$ Time taken by 1 man to complete the same job $=(8 \times 9)=72$ days $\therefore$ Time taken by 6 men to complete the same job $=\frac{72}{6}=12$ days...
Read More →A ΔPQR, given that QR = 3 cm,
Question: A ΔPQR, given that QR = 3 cm, PQR = 45 and QP PR =2 cm. Solution: Given, inv $\triangle P Q R, Q R=3 \mathrm{~cm}, \angle P Q R=45^{\circ}$ and $Q P-P R=2 \mathrm{~cm}$ since, $C$ lies on the perpendicular bisector $R S$ of $A Y$. To construct $\triangle P Q R$, use the following steps. 1. Draw the base $Q R$ of length $3 \mathrm{~cm}$. 2. Make an angle $X Q R=45^{\circ}$ at point $Q$ of base $Q R$. 3. Cut the line segment $Q S=Q P-P R=2 \mathrm{~cm}$ from the ray $Q X$. 4. Join $S R$ ...
Read More →6 men can complete the electric fitting in a building in 7 days.
Question: 6 men can complete the electric fitting in a building in 7 days. How many days will it take if 21 men do the job? Solution: It is given that 6 men can complete a job in 7 days. $\therefore$ Time taken by 1 man to complete the same job $=(6 \times 7)=42$ days $\therefore$ Time taken by 21 men to complete the same job $=\frac{42}{21}=2$ days...
Read More →From a solid cylinder of height 14 cm and base diameter 7 cm, two equal conical holes each of radius 2.1 cm and height 4 cm are cut off.
Question: From a solid cylinder of height 14 cm and base diameter 7 cm, two equal conical holes each of radius 2.1 cm and height 4 cm are cut off. Find the volume of the remaining solid. Solution: We have, the height of the cylinder, $H=14 \mathrm{~cm}$, the base radius of cylinder, $R=\frac{7}{2} \mathrm{~cm}$, the base radius of each conical holes, $r=2.1 \mathrm{~cm}$ and the height of each conical holes, $h=4 \mathrm{~cm}$ Volume of the remaining solid = Volume of the cylinder - Volume of 2 ...
Read More →A and B can do a piece of work in 6 days and 4 days respectively.
Question: AandBcan do a piece of work in 6 days and 4 days respectively. A started the work; worked at it for 2 days and then was joined byB. Find the total time taken to complete the work. Solution: A can do a work in 6 days, and B can do the same work in 4 days. $\therefore$ Work done by A in 2 days $=\frac{2}{6}=\frac{1}{3}$ Remaining work $=1-\frac{1}{3}=\frac{2}{3}$ $\therefore$ Work done by $(\mathrm{A}+\mathrm{B})$ in 1 day $=\left(\frac{1}{6}+\frac{1}{4}\right)$ $=\frac{2+3}{12}=\frac{5}...
Read More →A triangle if its perimeter is 10.4 cm
Question: A triangle if its perimeter is 10.4 cm and two angles are 45 and 120. Solution: Let ABC be a triangle. Then, given perimeter = 10.4 cm i.e., AB+ BC + CA = 10.4 cm and two angles are 45 and 120. say B = 45 and C = 120 Now, to construct the ΔABC use the following steps. 1. Draw a line segment say $X Y$ and equal to perimeter i.e., $A B+B C+C A=10.4 \mathrm{~cm}$ 2. Make angle $\angle L X Y=\angle B=45^{\circ}$ and $\angle M Y X=\angle C=120^{\circ}$. 3. Bisect $\angle L X Y$ and $\angle ...
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