A woker is paid Rs 200 for 8 days work.
Question: A woker is paid Rs 200 for 8 days work. If he works for 20 days, how much will he get? Solution: Let Rsxbe the income for 20 days of work. Since the income and the number of working days are in direct variation, we have: $\frac{200}{x}=\frac{8}{20}$ $\Rightarrow 200 \times 20=8 x$ $\Rightarrow x=\frac{200 \times 20}{8}$ $=\frac{4000}{8}$ $=500$ Thus, the worker will get Rs 500 for working 20 days....
Read More →Find the matrix X for which
Question: Find the matrixXfor which Solution: Let $A=\left[\begin{array}{ll}3 2\end{array}\right.$ $\left.\begin{array}{ll}7 5\end{array}\right], B=\left[\begin{array}{ll}-1 1\end{array}\right.$ $-2 \quad 1]$ and $C=\left[\begin{array}{ll}2 -1\end{array}\right]$ $\left.\begin{array}{ll}0 4\end{array}\right]$ Now, $|A|=\mid \begin{array}{ll}3 2\end{array}$ $7 \quad 5 \mid=15-14=1$ $|B|=\mid-1 \quad 1$ $\begin{array}{ll}-2 1\end{array}=-1+2=1$ Since, $|A| \neq 0$ and $|B| \neq 0$ Hence, $A \ B$ ar...
Read More →A worker is paid Rs 210 for 6 days work.
Question: A worker is paid Rs 210 for 6 days work. If his total income of the month is Rs 875, for how many days did he work? Solution: Letxbe the number of days for which the worker is paid Rs 875. Since the income of the worker and the number of working days are in direct variation, we have: $\frac{210}{875}=\frac{6}{x}$ $\Rightarrow 210 \times x=875 \times 6$ $\Rightarrow x=\frac{875 \times 6}{210}$ $=\frac{5250}{210}$ $=25$ Thus, the required number of days is 25 ....
Read More →A rocket is in the form of a circular cylinder closed at the lower end and a cone of the same radius is attached to the top.
Question: A rocket is in the form of a circular cylinder closed at the lower end and a cone of the same radius is attached to the top. The radius of the cylinder is 2.5 m, its height is 21 m and the slant height of the cone is 8 m. Calculate the total surface area of the rocket. Solution: Radius of the cylinder = 2.5 mHeight of the cylinder = 21 m Curved surface area of the cylinder $=2 \pi \mathrm{rh}=2 \times \frac{22}{7} \times 2.5 \times 21=330 \mathrm{~m}^{2}$ Radius of the cone = 2.5 mSlan...
Read More →Solve the following
Question: 11 men can dig $6 \frac{3}{4}$ metre long trench in one day. How many men should be employed for digging 27 metre long trench of the same type i one day? Solution: Letxbe the number of men required to dig a trench of 27 metre. Since the length of the trench and the number of men are in direct variation, we have: $\frac{11}{x}=\frac{27 / 4}{27}$ $\Rightarrow 11 \times 27=x \times \frac{27}{4}$ $\Rightarrow x=\frac{11 \times 27 \times 4}{27}$ $=44$ Thus, 44 men will be required to dig a ...
Read More →A circus tent is cylindrical to a height of 3 m and conical above it.
Question: A circus tent is cylindrical to a height of 3 m and conical above it. If its base radius is 52.5 m and the slant height of the conical portion is 53 m, find the area of canvas needed to make the tent. Solution: Radius of the cylinder = 52.5 mRadius of the base of the cone = 52.5 mSlant height (l) of the cone = 53 mHeight of the cylinder (h) = 3 m Curved surface area of the cylindrical portion $=2 \pi r h=\left(2 \times \frac{22}{7} \times 52.5 \times 3\right) \mathrm{m}^{2}=990 \mathrm...
Read More →In figure, O is the centre of the
Question: In figure, O is the centre of the circle BCO = 30. Find x and y. Solution: Given, $O$ is the centre of the circle and $\angle B C O=30^{\circ}$. In the given figure join $O B$ and $A C$. In $\triangle B O C$, $C O=B O$ [both are the radius of circle] $\therefore \quad \angle O B C=\angle O C B=30^{\circ}$ [angles opposite to equal sides are equal] $\therefore \quad \angle B O C=180^{\circ}-(\angle O B C+\angle O C E)$ [by angle sum property of a triangle] $=180^{\circ}-\left(30^{\circ}...
Read More →A tent is in the shape of a right circular cylinder up to a height of 3 m and conical above it.
Question: A tent is in the shape of a right circular cylinder up to a height of 3 m and conical above it. The total height of the tent is 13.5 m and the radius of its base is 14 m. Find the cost of cloth required to make the tent at the rate of Rs 80 per square metre. Solution: Radius of the cylinder = 14 mRadius of the base of the cone = 14 mHeight of the cylinder (h) = 3 mTotal height of the tent = 13.5 m Surface area of the cylinder $=2 \pi r h=\left(2 \times \frac{22}{7} \times 14 \times 3\r...
Read More →The cost of 97 metre of cloth is Rs 242.50.
Question: The cost of 97 metre of cloth is Rs 242.50. What length of this can be purchased for Rs 302.50? Solution: Letxmetre be the length of the cloth that can be purchased for Rs 302.50. Since the length of the cloth and its cost are in direct variation, we have : $\frac{97}{x}=\frac{242.50}{302.50}$ $\Rightarrow 97 \times 302.50=x \times 242.50$ $\Rightarrow x=\frac{97 \times 302.50}{242.50}$ $=\frac{29342.50}{242.50}$ $=121$ Thus, the required length will be 121 metre....
Read More →A tent is in the shape of a right circular cylinder up to a height of 3 m and conical above it.
Question: A tent is in the shape of a right circular cylinder up to a height of 3 m and conical above it. The total height of the tent is 13.5 m and the radius of its base is 14 m. Find the cost of cloth required to make the tent at the rate of Rs 80 per square metre. Solution: Radius of the cylinder = 14 mRadius of the base of the cone = 14 mHeight of the cylinder (h) = 3 mTotal height of the tent = 13.5 m Surface area of the cylinder $=2 \pi r h=\left(2 \times \frac{22}{7} \times 14 \times 3\r...
Read More →If the thickness of a pile of 12 cardboards is 35 mm,
Question: If the thickness of a pile of 12 cardboards is 35 mm, find the thickness of a pile of 294 cardboards. Solution: Letxcm be the thickness of a pile of 294 cardboards. Since the pile of the cardboards and its thickness are in direct variation, we have: $\frac{3.5}{x}=\frac{12}{294}$ $\Rightarrow 3.5 \times 294=x \times 12$ $\Rightarrow x=\frac{3.5 \times 294}{12}$ $=\frac{1029}{12}$ $=85.75 \mathrm{~cm}$ Thus, the thickness of a pile of 294 cardboards will be $85.75 \mathrm{~cm}($ or $857...
Read More →The second class railway fare for 240 km of Journey is Rs 15.00.
Question: The second class railway fare for 240 km of Journey is Rs 15.00. What would be the fare for a journey of 139.2 km? Solution: Let Rsxbe the fare for a journey of 139.2 km. Since the distance travelled and the fare are in direct variation, we have: $\frac{240}{139.2}=\frac{15}{x}$ $\Rightarrow 240 \times x=15 \times 139.2$ $\Rightarrow x=\frac{15 \times 139.2}{240}$ $=\frac{2088}{240}$ $=8.7$ Thus, the fare for a journey of $139.2 \mathrm{~km}$ will be $\mathrm{Rs} 8.70$....
Read More →A military tent of height 8.25 m is in the form of a right circular cylinder of base diameter 30 m and height 5.5 m
Question: A military tent of height 8.25 m is in the form of a right circular cylinder of base diameter 30 m and height 5.5 m surmounted by a right circular cone of same base radius. Find the length of canvas used in making the tent, if the breadth of the canvas is 1.5 m. Solution: We have,the radii of bases of the cone and cylinder,r= 15m,the height of the cylinder,h= 5.5 m,the height of the tent = 8.25 m Also, the height of the cone, $H=8.25-5.5=2.75 \mathrm{~m}$ The slant height of the cone, ...
Read More →A military tent of height 8.25 m is in the form of a right circular cylinder of base diameter 30 m and height 5.5 m
Question: A military tent of height 8.25 m is in the form of a right circular cylinder of base diameter 30 m and height 5.5 m surmounted by a right circular cone of same base radius. Find the length of canvas used in making the tent, if the breadth of the canvas is 1.5 m. Solution: We have,the radii of bases of the cone and cylinder,r= 15m,the height of the cylinder,h= 5.5 m,the height of the tent = 8.25 m Also, the height of the cone, $H=8.25-5.5=2.75 \mathrm{~m}$ The slant height of the cone, ...
Read More →In a library 136 copies of a certain book require a shelf-length of 3.4 metre.
Question: In a library 136 copies of a certain book require a shelf-length of 3.4 metre. How many copies of the same book would occupy a shelf-length of 5.1 metres? Solution: Let $x$ be the number of copies that would occupy a shelf-length of $5.1 \mathrm{~m}$. Since the number of copies and the length of the shelf are in direct variation, we have: $\frac{136}{x}=\frac{3.4}{5.1}$ $\Rightarrow 136 \times 5.1=x \times 3.4$ $\Rightarrow x=\frac{136 \times 5.1}{3.4}$ $=204$ Thus, 204 copies will occ...
Read More →Find the matrix X satisfying the matrix equation
Question: Find the matrixXsatisfying the matrix equation Solution: Let: $A=\left[\begin{array}{ll}5 3\end{array}\right.$ $-1 \quad-2]$ $\Rightarrow|\mathrm{A}|=\mid 5 \quad 3$ $-1-2 \mid=-10+3=-7 \neq 0$ Hence, $A$ is invertible. If $C_{i j}$ is a cofactor of $a_{i j}$ in $A$, then $\mathrm{C}_{11}=-2, \mathrm{C}_{12}=1, \mathrm{C}_{21}=-3$ and $\mathrm{C}_{22}=5$. Now, adj $A=\left[\begin{array}{ll}-2 1\end{array}\right.$ $\left.\begin{array}{ll}-3 5\end{array}\right]^{\mathrm{T}}=\left[\begin{...
Read More →AB and AC are two chords of a circle of radius
Question: AB and AC are two chords of a circle of radius r such that AB = 2AC. If p and q are the distances of AB and AC from the centre, prove that 4q2= p2+ 3r2. Thinking Process Firstly, use the Pythagoras theorem in ΔOAM and ΔOAN and further adjust them to prove the required result. Solution: Given In a circle of radius $r$, there are two chords $A B$ and $A C$ such that $A B=2 A C$. Also, the distance of $A B$ and $A C$ from the centre are $p$ and $q$, respectively. To prove $4 q^{2}=p^{2}+3...
Read More →68 boxes of a certain commodity require a shelf-length of 13.6 m.
Question: 68 boxes of a certain commodity require a shelf-length of 13.6 m. How many boxes of the same commodity would occupy a shelf length of 20.4 m? Solution: Let $x$ be the number of boxes that occupy a shelf-length of $20.4 \mathrm{~m}$. If the length of the shelf increases, the number of boxes will also increase. Therefore, it is a case of direct variation. $\frac{68}{x}=\frac{13.6}{20.4}$ $\Rightarrow 68 \times 20.4=x \times 13.6$ $\Rightarrow x=\frac{68 \times 20.4}{13.6}$ $=\frac{1387.2...
Read More →The surface area of a sphere is 2464 cm2.
Question: The surface area of a sphere is 2464 cm2. If its radius be doubled, then what will be the surface area of the new sphere? Solution: Let the radii of the given sphere and the new sphere berandR, respectively. We have, $R=2 r$ and the surface area of the given sphere $=2464 \mathrm{~cm}^{2}$ i.e. $4 \pi r^{2}=2464 \mathrm{~cm}^{2}$ The surface area of the new sphere $=4 \pi R^{2}$ $=4 \pi(2 r)^{2}$ $=4 \pi\left(4 r^{2}\right)$ $=4\left(4 \pi r^{2}\right)$ $=4 \times 2464$ $=9856 \mathrm{...
Read More →The surface area of a sphere is 2464 cm2.
Question: The surface area of a sphere is 2464 cm2. If its radius be doubled, then what will be the surface area of the new sphere? Solution: Let the radii of the given sphere and the new sphere berandR, respectively. We have, $R=2 r$ and the surface area of the given sphere $=2464 \mathrm{~cm}^{2}$ i.e. $4 \pi r^{2}=2464 \mathrm{~cm}^{2}$ The surface area of the new sphere $=4 \pi R^{2}$ $=4 \pi(2 r)^{2}$ $=4 \pi\left(4 r^{2}\right)$ $=4\left(4 \pi r^{2}\right)$ $=4 \times 2464$ $=9856 \mathrm{...
Read More →A car is travelling at the average speed of 50 km/hr.
Question: A car is travelling at the average speed of 50 km/hr. How much distance would it travel in 12 minutes? Solution: Let the distance be $x \mathrm{~km}$. If the time taken is less, the distance covered will also be less. Therefore, it is a direct variation. $50: x=60: 12$ $\Rightarrow \frac{50}{x}=\frac{60}{12}$ Applying cross muliplication, we get: $x=\frac{50 \times 12}{60}$ $=10$ Thus, the required distance will be $10 \mathrm{~km}$....
Read More →The surface area of a sphere is 2464 cm2.
Question: The surface area of a sphere is 2464 cm2. If its radius be doubled, then what will be the surface area of the new sphere? Solution: Let the radii of the given sphere and the new sphere berandR, respectively. We have, $R=2 r$ and the surface area of the given sphere $=2464 \mathrm{~cm}^{2}$ i.e. $4 \pi r^{2}=2464 \mathrm{~cm}^{2}$ The surface area of the new sphere $=4 \pi R^{2}$ $=4 \pi(2 r)^{2}$ $=4 \pi\left(4 r^{2}\right)$ $=4\left(4 \pi r^{2}\right)$ $=4 \times 2464$ $=9856 \mathrm{...
Read More →The sum of the radius of the base and the height of a solid cylinder is 37 metres.
Question: The sum of the radius of the base and the height of a solid cylinder is 37 metres. If the total surface area of the cylinder be 1628 sq metres, then find its volume. Solution: Letrandhbe the base radius and the height of the solid cylinder, respectively. We have, $(r+h)=37 \mathrm{~m}$ As, the total surface area of the cylinder $=1628 \mathrm{~m}^{2}$ $\Rightarrow 2 \pi r(r+h)=1628$ $\Rightarrow 2 \times \frac{22}{7} \times r \times 37=1628$ $\Rightarrow r=\frac{1628 \times 7}{2 \times...
Read More →Suneeta types 1080 words in one hour.
Question: Suneeta types 1080 words in one hour. What is her GWAM (gross words a minute rate)? Solution: Let $x$ be her GWAM. If the time taken is less, GWAM will also be less. Therefore, it is a direct variation. $1080: x=60: 1$ $\Rightarrow \frac{1080}{x}=\frac{60}{1}$ Applying cross muliplication, we get: $x=\frac{1080 \times 1}{60}$ $=18$ Thus, her GWAM will be 18 ....
Read More →The sum of the radius of the base and the height of a solid cylinder is 37 metres.
Question: The sum of the radius of the base and the height of a solid cylinder is 37 metres. If the total surface area of the cylinder be 1628 sq metres, then find its volume. Solution: Letrandhbe the base radius and the height of the solid cylinder, respectively. We have, $(r+h)=37 \mathrm{~m}$ As, the total surface area of the cylinder $=1628 \mathrm{~m}^{2}$ $\Rightarrow 2 \pi r(r+h)=1628$ $\Rightarrow 2 \times \frac{22}{7} \times r \times 37=1628$ $\Rightarrow r=\frac{1628 \times 7}{2 \times...
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