1200 men can finish a stock of food in 35 days.
Question: 1200 men can finish a stock of food in 35 days. How many more men should join them so that the same stock may last for 25 days? Solution: Letx be the number of additional men required to finish the stock in 25 days. Since the number of men and the time taken to finish a stock are in inverse variation, we have: $1200 \times 35=25 x$ $\Rightarrow x=\frac{1200 \times 35}{25}$ $=1680$ $\therefore$ Required number of men $=1680-1200=480$ Thus, an additional 480 men should join the existing ...
Read More →In the given figure, ABCD is a square each of whose sides measures 28 cm.
Question: In the given figure,ABCDis a square each of whose sides measures 28 cm. Find the area of the shaded region. Solution: Letrbe the radius of the circle.Thus, we have: $r=\frac{28}{2} \mathrm{~cm}$ $=14 \mathrm{~cm}$ Now, Area of the shaded region $=$ (Area of the square $A B C D$ ) $-4$ (Area of the sector where $r=14 \mathrm{~cm}$ and $\theta=90^{\circ}$ ) $=\left|(28 \times 28)-4\left(\frac{22}{7} \times 14 \times 14 \times \frac{90}{360}\right)\right| \mathrm{cm}^{2}$ $=|784-4(154)| \...
Read More →Solve this
Question: If $A=\left[\begin{array}{ccc}2 -1 1 \\ -1 2 -1 \\ 1 -1 2\end{array}\right]$. Verify that $A^{3}-6 A^{2}+9 A-4 I=O$ and hence find $A^{-1}$. Solution: $A=\left[\begin{array}{lll}2 -1 1\end{array}\right.$ $\begin{array}{lll}-1 2 -1\end{array}$ $\left.\begin{array}{lll}1 -1 2\end{array}\right]$ $\Rightarrow|A|=\mid 2 \quad-1 \quad 1$ $\begin{array}{lll}-1 2 -1\end{array}$ $\begin{array}{lll}1 -1 2 \mid=2 \times(4-1)+1(-2+1)+1(1-2)=6-1-1=4\end{array}$ Since, $|A| \neq 0$ Hence, $A^{-1}$ e...
Read More →A work-force of 420 men with a contractor can finish a certain piece of work in 9 months.
Question: A work-force of 420 men with a contractor can finish a certain piece of work in 9 months. How many extra men must he employ to complete the job in 7 months? Solution: Letxbe the extra number of men employed to complete the job in 7 months. Since the number of men hired and the time required to finish the piece of work are in inverse variation, we have : $420 \times 9=7 x$ $\Rightarrow x=\frac{420 \times 9}{7}$ $=540$ Thus, the number of extra men required to complete the job in 7 month...
Read More →In the given figure, ABCD is a square each of whose sides measures 28 cm.
Question: In the given figure,ABCDis a square each of whose sides measures 28 cm. Find the area of the shaded region. Solution: Letrbe the radius of the circle.Thus, we have: $r=\frac{28}{2} \mathrm{~cm}$ $=14 \mathrm{~cm}$ Now, Area of the shaded region $=$ (Area of the square $A B C D$ ) $-4$ (Area of the sector where $r=14 \mathrm{~cm}$ and $\theta=90^{\circ}$ ) $=\left|(28 \times 28)-4\left(\frac{22}{7} \times 14 \times 14 \times \frac{90}{360}\right)\right| \mathrm{cm}^{2}$ $=|784-4(154)| \...
Read More →A work force of 50 men with a contractor can finish a piece of work in 5 months.
Question: A work force of 50 men with a contractor can finish a piece of work in 5 months. In how many months the same work can be completed by 125 men? Solution: Letxbe the number days required to complete a piece of work by 125 men. Since the number of men engaged and the number of days taken to do a piece of work are in inverse variation, we have: $50 \times 5=125 x$ $\Rightarrow x=\frac{50 \times 5}{125}$ $=2$ Thus, the required number of months is 2 ....
Read More →OACB is a quadrant of a circle with centre O and its radius is 3.5 cm.
Question: OACBis a quadrant of a circle with centreOand its radius is 3.5 cm. IfOD= 2 cm. find the area of (i) quadrantOACB (ii) the shaded region. Solution: (i) Area of the quadrant $O A C B=\left(\frac{1}{4} \times \frac{22}{7} \times 3.5 \times 3.5\right) \mathrm{cm}^{2}$ $=\left(\frac{1}{4} \times \frac{22}{7} \times \frac{35}{10} \times \frac{35}{10}\right) \mathrm{cm}^{2}$ $=\frac{77}{8} \mathrm{~cm}^{2}$ $=9.625 \mathrm{~cm}^{2}$ (ii) Area of the shaded region = Area of the quadrant $O A ...
Read More →If 36 men can do a piece of work in 25 days,
Question: If 36 men can do a piece of work in 25 days, in how many days will 15 men do it? Solution: Letxbe the number of days in which 15 men can do a piece of work. Since the number of men hired and the number of days taken to do a piece of work are in inverse variation, we have: $36 \times 25=x \times 15$ $\Rightarrow x=\frac{36 \times 25}{15}$ $=\frac{900}{15}$ $=60$ Thus, the required number of days is 60 ....
Read More →The wheel of a cart is making 5 revolutions per second.
Question: The wheel of a cart is making 5 revolutions per second. If the diameter of the wheel is 84 cm, find its speed in km per hour. Solution: Distance covered in 1 revolution $=\pi \times d$ $=\left(\frac{22}{7} \times 84\right) \mathrm{cm}$ $=264 \mathrm{~cm}$ Distance covered in 1 second $=(5 \times 264) \mathrm{cm}$ $=1320 \mathrm{~cm}$ Distance covered in 1 hour $=(60 \times 60 \times 1320) \mathrm{cm}$ $=4752000 \mathrm{~cm}$ $=\left(\frac{4752000}{1000 \times 100}\right) \mathrm{km}$ $...
Read More →A wire when bent in the form of an equilateral triangle encloses an area of
Question: A wire when bent in the form of an equilateral triangle encloses an area of $121 \sqrt{3} \mathrm{~cm}^{2}$. If the same wire is bent into the form of a circle, what will be the area of the circle? Solution: Letacm be the side of the equilateral triangle.Now, Area of the equilateral triangle $=\frac{\sqrt{3}}{4} a^{2}$ We have: $\frac{\sqrt{3}}{4} a^{2}=121 \sqrt{3}$ $\Rightarrow \frac{a^{2}}{4}=121$ $\Rightarrow a^{2}=484$ $\Rightarrow a=22$ Perimeter of the triangle = Circumference o...
Read More →A wire when bent in the form of an equilateral triangle encloses an area of
Question: A wire when bent in the form of an equilateral triangle encloses an area of $121 \sqrt{3} \mathrm{~cm}^{2}$. If the same wire is bent into the form of a circle, what will be the area of the circle? Solution: Letacm be the side of the equilateral triangle.Now, Area of the equilateral triangle $=\frac{\sqrt{3}}{4} a^{2}$ We have: $\frac{\sqrt{3}}{4} a^{2}=121 \sqrt{3}$ $\Rightarrow \frac{a^{2}}{4}=121$ $\Rightarrow a^{2}=484$ $\Rightarrow a=22$ Perimeter of the triangle = Circumference o...
Read More →It is known that for a given mass of gas, the volume v varies inversely as the pressure p.
Question: It is known that for a given mass of gas, the volumevvaries inversely as the pressurep. Fill in the missing entries in the following table: Solution: Since the volume and pressure for the given mass vary inversely, we have: $v p=k$ For $v=60$ and $p=\frac{3}{2}$, we have : $k=60 \times \frac{3}{2}$ $=90$ For $p=2$ and $k=90$, we have : $2 v=90$ $\Rightarrow v=\frac{90}{2}$ $=45$ For $v=48$ and $k=90$, we have : $48 p=90$ $\Rightarrow p=\frac{90}{48}$ $=\frac{15}{8}$ For $p=1$ and $k=90...
Read More →In the given figure, the sectors of two concentric circles of radii 7 cm
Question: In the given figure, the sectors of two concentric circles of radii 7 cm and 3.5 cm are shown. Find the area of the shaded region. Solution: Area of the shaded region $=$ (Area of the sector with $r=7 \mathrm{~cm}$ and $\theta=30^{\circ}$ ) $-$ (Area of the sector with $r=3.5 \mathrm{~cm}$ and $\theta=30^{\circ}$ ) $=\left|\left(\frac{22}{7} \times 7 \times 7 \times \frac{30}{360}\right)-\left(\frac{22}{7} \times 3.5 \times 3.5 \times \frac{30}{360}\right)\right| \mathrm{cm}^{2}$ $=\le...
Read More →A chord of a circle of radius 14 cm a makes a right angle at the centre.
Question: A chord of a circle of radius 14 cm a makes a right angle at the centre. Find the area of the sector. Solution: Letrcm be the radius of the circle andbe the angle.We have: $r=14 \mathrm{~cm}$ and $\theta=90^{\circ}$ Area of the sector $=\frac{\pi r^{2} \theta}{360}$ $=\left(\frac{22}{7} \times 14 \times 14 \times \frac{90}{360}\right) \mathrm{cm}^{2}$ $=154 \mathrm{~cm}^{2}$...
Read More →Which of the following quantities vary inversely as each other?
Question: Which of the following quantities vary inversely as each other? (i) The number ofxmen hired to construct a wall and the timeytaken to finish the job. (ii) The lengthxof a journey by bus and priceyof the ticket. (iii) Journey (xkm) undertaken by a car and the petrol (ylitres) consumed by it. Solution: (i) If the number of men is more, the time taken to construct a wallwill be less. Therefore, it is in inverse variation. (ii) If the length of a journey is more, the price of the ticket wi...
Read More →The perimeter of a sector of a circle of radius 5.6 cm is 27.2 cm.
Question: The perimeter of a sector of a circle of radius 5.6 cm is 27.2 cm. Find the area of the sector. Solution: LetObe the centre of the circle with radius 5.6 cm andOACBbe its sector with perimeter 27.2 cm.Thus, we have: $O A+O B+\operatorname{arc} A B=27.2$ $\Rightarrow 5.6+5.6+\operatorname{arc} A B=27.2$ $\Rightarrow \operatorname{arc} A B=16 \mathrm{~cm}$ Now, Area of the sector $O A C B O=\left(\frac{1}{2} \times\right.$ Radius $\left.\times l\right)$ square units $=\left(\frac{1}{2} 5...
Read More →The perimeter of a sector of a circle of radius 5.6 cm is 27.2 cm.
Question: The perimeter of a sector of a circle of radius 5.6 cm is 27.2 cm. Find the area of the sector. Solution: LetObe the centre of the circle with radius 5.6 cm andOACBbe its sector with perimeter 27.2 cm.Thus, we have: $O A+O B+\operatorname{arc} A B=27.2$ $\Rightarrow 5.6+5.6+\operatorname{arc} A B=27.2$ $\Rightarrow \operatorname{arc} A B=16 \mathrm{~cm}$ Now, Area of the sector $O A C B O=\left(\frac{1}{2} \times\right.$ Radius $\left.\times l\right)$ square units $=\left(\frac{1}{2} 5...
Read More →It x and y vary inversely, fill in the following blanks:
Question: Itxandyvary inversely, fill in the following blanks: (i) (ii) (iii) Solution: (i) Since $x$ and $y$ vary inversely, we have : $x y=k$ For $x=16$ and $y=6$, we have : $16 \times 6=k$ $\Rightarrow k=96$ For $x=12$ and $k=96$, we have : $x y=k$ $\Rightarrow 12 y=96$ $\Rightarrow y=\frac{96}{12}$ $=8$ For $y=4$ and $k=96$, we have : $x y=k$ $\Rightarrow 4 x=96$ $\Rightarrow x=\frac{96}{4}$ $=24$ For $x=8$ and $k=96$, we have : $x y=k$ $\Rightarrow 8 y=96$ $\Rightarrow y=\frac{96}{8}$ $=12$...
Read More →Show that the matrix,
Question: Show that the matrix, $A=\left[\begin{array}{ccc}1 0 -2 \\ -2 -1 2 \\ 3 4 1\end{array}\right]$ satisfies the equation, $A^{3}-A^{2}-3 A-I_{3}=O$. Hence, find $A^{-1}$. Solution: We have, $A=\left[\begin{array}{lll}1 0 -2\end{array}\right.$ $\begin{array}{lll}-2 -1 2\end{array}$ $\begin{array}{lll}3 4 1]\end{array}$ $\Rightarrow|A|=\mid \begin{array}{lll}1 0 -2\end{array}$ $\begin{array}{lll}-2 -1 2\end{array}$ $3 \quad 4 \quad 1 \mid=1(-9)+0-2(-8)=-9+16=7$ Since, $|A| \neq 0$ Hence, $A...
Read More →The minute hand of a clock is 12 cm long.
Question: The minute hand of a clock is 12 cm long. Find the area swept by in it 35 minutes. Solution: Angle described by the minute hand in 60 minutes $=360^{\circ}$ Angle described by the minute hand in 35 minutes $=\left(\frac{360}{60} \times 35\right)^{\circ}$ $=210^{\circ}$ Now, $r=12 \mathrm{~cm}$ and $\theta=210^{\circ}$ $\therefore$ Required area swept by the minute hand in 35 minutes = Area of the sector with $r=12 \mathrm{~cm}$ and $\theta=210^{\circ}$ $=\frac{\pi r^{2} \theta}{360}$ $...
Read More →In a circle of radius 21 cm, an arc subtends an angle of 60°
Question: In a circle of radius 21 cm, an arc subtends an angle of 60at the centre. Find the length of the arc. Solution: Let ACB be the given arc subtending at an angle of60at the centre.Now, we have: $r=21 \mathrm{~cm}$ and $\theta=60^{\circ}$ $\therefore$ Length of the $\operatorname{arc} \mathrm{ACB}=\frac{2 \pi r}{360}$ $=\left(2 \times \frac{22}{7} \times 21 \times \frac{60}{360}\right)$ $=22 \mathrm{~cm}$...
Read More →In a circle of radius 21 cm, an arc subtends an angle of 60°
Question: In a circle of radius 21 cm, an arc subtends an angle of 60at the centre. Find the length of the arc. Solution: Let ACB be the given arc subtending at an angle of60at the centre.Now, we have: $r=21 \mathrm{~cm}$ and $\theta=60^{\circ}$ $\therefore$ Length of the $\operatorname{arc} \mathrm{ACB}=\frac{2 \pi r}{360}$ $=\left(2 \times \frac{22}{7} \times 21 \times \frac{60}{360}\right)$ $=22 \mathrm{~cm}$...
Read More →If non-parallel sides of a trapezium are equal,
Question: If non-parallel sides of a trapezium are equal, prove that it is cyclic. Solution: Given ABCD is a trapezium whose non-parallel sides AD and BC are equal. To prove Trapezium $A B C D$ is a cyclic: Join $B E$, where $B E \| A D$. Proof Since, $A B \| D E$ and $A D \| B E$ Since, the quadrilateral $A B E D$ is a parallelogram. $\therefore$ $\angle B A D=\angle B E D$ $\ldots($ (i) [opposite angles of a paralleiogram are equal] and $A D=B E$ [opposite sides of a parallelogram are equal] B...
Read More →The circumference of a circle is 22 cm. Find its area.
Question: The circumference of a circle is 22 cm. Find its area. Solution: Letrcm be the radius of the circle.Now,Circumference of the circle: $2 \pi r=22$ $\Rightarrow 2 \times \frac{22}{7} \times r=22$ $\Rightarrow r=\left(22 \times \frac{7}{44}\right) \mathrm{cm}$ $\Rightarrow r=\frac{7}{2}$ Also, Area of the circle $=\pi r^{2}$ $=\left(\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}\right) \mathrm{cm}^{2}$ $=\frac{77}{2} \mathrm{~cm}^{2}$ $=38.5 \mathrm{~cm}^{2}$...
Read More →The circumference of a circle is 22 cm. Find its area.
Question: The circumference of a circle is 22 cm. Find its area. Solution: Letrcm be the radius of the circle.Now,Circumference of the circle: $2 \pi r=22$ $\Rightarrow 2 \times \frac{22}{7} \times r=22$ $\Rightarrow r=\left(22 \times \frac{7}{44}\right) \mathrm{cm}$ $\Rightarrow r=\frac{7}{2}$ Also, Area of the circle $=\pi r^{2}$ $=\left(\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}\right) \mathrm{cm}^{2}$ $=\frac{77}{2} \mathrm{~cm}^{2}$ $=38.5 \mathrm{~cm}^{2}$...
Read More →