The area of the sector of angle θ° of a circle with radius R is
Question: The area of the sector of angleof a circle with radiusRis (a) $\frac{2 \pi R \theta}{180}$ (b) $\frac{\pi R^{2} \theta}{180}$ (c) $\frac{2 \pi R \theta}{360}$ (d) $\frac{\pi R^{2} \theta}{360}$ Solution: (d) $\frac{\pi R^{2} \theta}{360}$...
Read More →In making 1000 revolutions, a wheel covers 88 km.
Question: In making 1000 revolutions, a wheel covers 88 km. The diameter of the wheel is(a) 14 m(b) 24 m(c) 28 m(d) 40 m Solution: (c) 28 m Distance covered by the wheel in 1 revolution $=\left(\frac{88 \times 1000}{1000}\right) \mathrm{m}$ $=88 \mathrm{~m}$ We have:Circumference of the wheel = 88 mNow, let the diameter of the wheel bedm.Thus, we have: $\pi d=88$ $\Rightarrow \frac{22}{7} \times d=88$ $\Rightarrow d=\left(88 \times \frac{7}{22}\right)$ $\Rightarrow d=28 \mathrm{~m}$...
Read More →Solve each of the following equation and also verify your solution:
Question: Solve each of the following equation and also verify your solution: $\frac{x}{2}+\frac{x}{3}+\frac{x}{4}=13$ Solution: $\frac{x}{2}+\frac{x}{3}+\frac{x}{4}=13$ $\Rightarrow \frac{\mathrm{x} \times 6+\mathrm{x} \times 4+\mathrm{x} \times 3}{12}=13$ $\Rightarrow \frac{13 \mathrm{x}}{12}=13$ $\Rightarrow \mathrm{x}=13 \times \frac{12}{13}=12$ Verification : L. H.S. $=\frac{12}{2}+\frac{12}{3}+\frac{12}{4}=6+4+3=13=$ R.H.S....
Read More →In making 1000 revolutions, a wheel covers 88 km.
Question: In making 1000 revolutions, a wheel covers 88 km. The diameter of the wheel is(a) 14 m(b) 24 m(c) 28 m(d) 40 m Solution: (c) 28 m Distance covered by the wheel in 1 revolution $=\left(\frac{88 \times 1000}{1000}\right) \mathrm{m}$ $=88 \mathrm{~m}$ We have:Circumference of the wheel = 88 mNow, let the diameter of the wheel bedm.Thus, we have: $\pi d=88$ $\Rightarrow \frac{22}{7} \times d=88$ $\Rightarrow d=\left(88 \times \frac{7}{22}\right)$ $\Rightarrow d=28 \mathrm{~m}$...
Read More →Solve each of the following equation and also verify your solution:
Question: Solve each of the following equation and also verify your solution: $\frac{5 x}{3}+\frac{2}{5}=1$ Solution: $\frac{5 \mathrm{x}}{3}+\frac{2}{5}=1$ $\Rightarrow \frac{5 \mathrm{x}}{3}=1-\frac{2}{5}$ $\Rightarrow \frac{5 \mathrm{x}}{3}=\frac{3}{5}$ $\Rightarrow \mathrm{x}=\frac{3}{5} \times \frac{3}{5}=\frac{9}{25}$ Verification : L.H.S. $=\frac{5}{3} \times \frac{9}{25}+\frac{2}{5}=\frac{3}{5}+\frac{2}{5}=1$ R.H.S. $=1$ $\therefore \mathrm{L} . \mathrm{H} . \mathrm{S} .=\mathrm{R} . \ma...
Read More →The diameter of a wheel is 40 cm. How many revolutions will it make in covering 176 m?
Question: The diameter of a wheel is 40 cm. How many revolutions will it make in covering 176 m?(a) 140(b) 150(c) 160(d) 166 Solution: (a) 140 Distance covered by the wheel in 1 revolution $=\pi d$ $=\left(\frac{22}{7} \times 40\right) \mathrm{cm}$ $=\frac{880}{7} \mathrm{~cm}$ $=\frac{880}{7 \times 100} \mathrm{~m}$ Number of revolutions required to cover $176 \mathrm{~m}=\left(\frac{176}{\frac{880}{7 \times 100}}\right)$ $=\left(176 \times 100 \times \frac{7}{880}\right)$ $=140$...
Read More →Solve each of the following equation and also verify your solution:
Question: Solve each of the following equation and also verify your solution: $9 \frac{1}{4}=y-1 \frac{1}{3}$ Solution: $9 \frac{1}{4}=\mathrm{y}-1 \frac{1}{3}$ or $\frac{37}{4}+\frac{4}{3}=\mathrm{y}$ or $\mathrm{y}=\frac{127}{12}$ $\therefore \mathrm{y}=\frac{127}{12}$ for the given equation. Check: L. H.S $=9 \frac{1}{4}$ R. H.S $=\frac{127}{12}-1 \frac{1}{3}=\frac{127}{12}-\frac{4}{3}$...
Read More →Show that
Question: Show that $A=\left[\begin{array}{ll}6 5 \\ 7 6\end{array}\right]$ satisfies the equation $x^{2}-12 x+1=O$. Thus, find $A^{-1}$. Solution: $A=\left[\begin{array}{ll}6 5\end{array}\right.$ $\left.\begin{array}{ll}7 6\end{array}\right]$ $\therefore A^{2}=\left[\begin{array}{ll}71 60\end{array}\right.$ $\left.\begin{array}{ll}84 71\end{array}\right]$ If $I_{2}$ is the identity matrix of order 2, then $A^{2}-12 A+I_{2}=\left[\begin{array}{ll}71 60\end{array}\right.$ $84 \quad 71]-12\left[\b...
Read More →The radius of a wheel is 0.25 m.
Question: The radius of a wheel is 0.25 m. How many revolutions will it make in covering 11 km?(a) 2800(b) 4000(c) 5500(d) 7000 Solution: (d) 7000 Distance covered in 1 revolution $=2 \pi r$ $=\left(2 \times \frac{22}{7} \times 0.25\right) \mathrm{m}$ $=\left(2 \times \frac{22}{7} \times \frac{25}{100}\right) \mathrm{m}$ $=\frac{11}{7} \mathrm{~m}$ Number of revolutions taken to cover $11 \mathrm{~km}=\left(11 \times 1000 \times \frac{7}{11}\right)$ $=7000$...
Read More →The radius of a wheel is 0.25 m.
Question: The radius of a wheel is 0.25 m. How many revolutions will it make in covering 11 km?(a) 2800(b) 4000(c) 5500(d) 7000 Solution: (d) 7000 Distance covered in 1 revolution $=2 \pi r$ $=\left(2 \times \frac{22}{7} \times 0.25\right) \mathrm{m}$ $=\left(2 \times \frac{22}{7} \times \frac{25}{100}\right) \mathrm{m}$ $=\frac{11}{7} \mathrm{~m}$ Number of revolutions taken to cover $11 \mathrm{~km}=\left(11 \times 1000 \times \frac{7}{11}\right)$ $=7000$...
Read More →In figure, AOB is a diameter of the circle
Question: In figure, AOB is a diameter of the circle and C, D, E are any three points on the semi-circle. Find the value of ACD + BED. Solution: Since, $A, C, D$ and $E$ are four point on a circle, then $A C D E$ is a cyclic quadrilateral. $\angle A C D+\angle A E D=180^{\circ} \ldots$ (i) $\quad$ [sum of opposite angles in a cyclic quadrilateral is $180^{\circ}$ ] Now, $\angle A E B=90^{\circ}$ ......(ii) We know that, diameter subtends a right angle to the circle. On adding Eqs. (i) and (ii), ...
Read More →A lady went shopping and spent half of what she had on buying hankies and gave a rupee to a beggar waiting outside the shop.
Question: A lady went shopping and spent half of what she had on buying hankies and gave a rupee to a beggar waiting outside the shop. She spent half of what was left on a lunch and followed that up with a two rupee tip. She spent half of the remaining amount on a book and three rupees on bus fare. When she reached home, she found that she had exactly one rupee left. How much money did she start with? Solution: Suppose, the lady start ed with x rupee $s$. Money spent on shopping $=\frac{x}{2}$ r...
Read More →The areas of two circles are in the ratio 9 : 4.
Question: The areas of two circles are in the ratio 9 : 4. The ratio of their circumferences is(a) 3 : 2(b) 4 : 9(c) 2 : 3(d) 81 : 16 Solution: Let the the radii of the two circles berandR, the circumferences of the circles becandCand the areas of the two circles beaandA.Now, $\frac{a}{A}=\frac{9}{4}$ $\Rightarrow \frac{\pi r^{2}}{\pi R^{2}}=\left(\frac{3}{2}\right)^{2}$ $\Rightarrow \frac{r}{R}=\frac{3}{2}$ Now, the ratio between their circumferences is given by $\frac{c}{C}=\frac{2 \pi r}{2 \p...
Read More →Two circles with centres O and O’ intersect
Question: Two circles with centres O and O intersect at two points A and B. A line PQ is drawn parallel to OO through A (or B) intersecting the circles at P and Q. Prove that PQ =2 OO. Solution: Given, draw two circles having centres O and O intersect at points A and 8. Also, draw line $P Q$ parallel to $O O^{\prime}$. Construction Join $O O^{\prime}, O P, O^{\prime} Q, O M$ and $O^{\prime} N$. To prove$P Q=200^{\prime}$ Proof in $\Delta O P B$, $B M=M P$ [OM is the perpendicular bisector of $P ...
Read More →The sum of the ages of Anup and his father is 100.
Question: The sum of the ages of Anup and his father is 100. When Anup is as old as his father now, he will be five times as old as his son Anuj is now. Anuj will be eight years older than Anup is now, when Anup is as old as his father. What are their ages now? Solution: Let Anup's age be $x$ years. Therefore, his father's age will be $(100-\mathrm{x})$ years. When Anup is as old as his father after $(100-2 \mathrm{x})$ years, Anuj's age $=\left(\frac{100-\mathrm{x}}{5}+100-2 \mathrm{x}\right)$ ...
Read More →The circumferences of two circles are in the ratio 3 : 4.
Question: The circumferences of two circles are in the ratio 3 : 4. The ratio of their areas is(a) 3: 4(b) 4 : 3(c) 9 : 16(d) 16: 9 Solution: Let the the radii of the two circles berandR, the circumferences of the circles becandCand the areas of the two circles beaandA.Now, $\frac{c}{C}=\frac{3}{4}$ $\Rightarrow \frac{2 \pi r}{2 \pi R}=\frac{3}{4}$ $\Rightarrow \frac{r}{R}=\frac{3}{4}$ Now, the ratio between their areas is given by $\frac{a}{A}=\frac{\pi r^{2}}{\pi R^{2}}$ $=\left(\frac{r}{R}\ri...
Read More →Show that
Question: Show that $A=\left[\begin{array}{cc}5 3 \\ -1 -2\end{array}\right]$ satisfies the equation $x^{2}-3 x-7=0$. Thus, find $A^{-1}$. Solution: $A=\left[\begin{array}{ll}5 3\end{array}\right.$ $-1-2]$ $A^{2}=\left[\begin{array}{ll}22 9\end{array}\right.$ $\left.\begin{array}{ll}-3 1\end{array}\right]$ If $I_{2}$ is the identity matrix of order 2, then $A^{2}-3 A-7 I_{2}=\left[\begin{array}{ll}22 9\end{array}\right.$ $-3 \quad 1]-3\left[\begin{array}{ll}5 3\end{array}\right.$ $-1-2]-7\left[\...
Read More →The length of a rectangle exceeds its breadth by 9 cm.
Question: The length of a rectangle exceeds its breadth by 9 cm. If length and breadth are each increased by 3 cm, the area of the new rectangle will be 84 cm2more than that of the given rectangle. Find the length and breath of the given rectangle. Solution: Let the breadth of the rectangle be $x \mathrm{~cm}$. Therefore, the length of the rectangle will be $(\mathrm{x}+9) \mathrm{cm}$. $\therefore$ Area of the rectangle $=\mathrm{x}(\mathrm{x}+9) \mathrm{cm}^{2}$. If the length and breadth are ...
Read More →The areas of two concentric circles are 1386 cm2 and 962.5 cm2.
Question: The areas of two concentric circles are 1386 cm2and 962.5 cm2. The width of the ring is(a) 2.8 cm(b) 3.5 cm(c) 4.2 cm(d) 3.8 cm Solution: (b) 3.5 cmLetrcm andRcm be the radii of two concentric circles.Thus, we have: $\pi \mathrm{R}^{2}=1386$ $\Rightarrow \frac{22}{7} \times R^{2}=1386$ $\Rightarrow R^{2}=\left(1386 \times \frac{7}{22}\right) \mathrm{cm}^{2}$ $\Rightarrow R^{2}=441 \mathrm{~cm}^{2}$ $\Rightarrow R=21 \mathrm{~cm}$ Also, $\pi \mathrm{r}^{2}=962.2$ $\Rightarrow \frac{22}{...
Read More →The areas of two concentric circles are 1386 cm2 and 962.5 cm2.
Question: The areas of two concentric circles are 1386 cm2and 962.5 cm2. The width of the ring is(a) 2.8 cm(b) 3.5 cm(c) 4.2 cm(d) 3.8 cm Solution: (b) 3.5 cmLetrcm andRcm be the radii of two concentric circles.Thus, we have: $\pi \mathrm{R}^{2}=1386$ $\Rightarrow \frac{22}{7} \times R^{2}=1386$ $\Rightarrow R^{2}=\left(1386 \times \frac{7}{22}\right) \mathrm{cm}^{2}$ $\Rightarrow R^{2}=441 \mathrm{~cm}^{2}$ $\Rightarrow R=21 \mathrm{~cm}$ Also, $\pi \mathrm{r}^{2}=962.2$ $\Rightarrow \frac{22}{...
Read More →A quadrilateral ABCD is inscribed in a circle
Question: A quadrilateral ABCD is inscribed in a circlesuch that AB is a diameter and ADC = 130. Find BAC. Solution: Draw a quadrilateral $A B C D$ inscribed in a circle having centre $O$. Given, $\angle A D C=130^{\circ}$ Since, $A B C D$ is a quadrilateral inscribed in a circle, therefore $A B C D$ becomes a cyclic quadrilateral. $\because$ Since, the sum of opposite angles of a cyclic quadrilateral is $180^{\circ}$, $\therefore$ $\angle A D C+\angle A B C=180^{\circ}$ $\Rightarrow$ $130^{\cir...
Read More →Bhagwanti inherited Rs 12000.00. She invested part of it as 10% and the rest at 12%.
Question: Bhagwanti inherited Rs 12000.00. She invested part of it as 10% and the rest at 12%. Her annual income from these investments is Rs 1280.00. How much did she invest at each rate? Solution: At the rate of $10 \%$, let the investment by Bhagwanti be Rs. $x$. Therefore, at the rate of $12 \%$, the investment will be Rs. $(12000-\mathrm{x})$. At the rate of $10 \%$, her annual income $=\mathrm{x} \times 10 \%$ At the rate of $12 \%$, her annual income $=(12000-\mathrm{x}) \times 12 \%$ So,...
Read More →The radii of two concentric circles are 19 cm and 16 cm respectively.
Question: The radii of two concentric circles are 19 cm and 16 cm respectively. The area of the ring enclosed by these circles is(a) 320 cm2(b) 330 cm2(c) 332 cm2(d) 340 cm2 Solution: (b) 330 cm2Let:R= 19 cm andr= 16 cmThus, we have: Area of the ring $=\pi\left(R^{2}-r^{2}\right)$ $=\pi(R+r)(\mathrm{R}-\mathrm{r})$ $=\left|\frac{22}{7} \times(19+16) \times(19-16)\right| \mathrm{cm}^{2}$ $=\left(\frac{22}{7} \times 35 \times 3\right) \mathrm{cm}^{2}$ $=330 \mathrm{~cm}^{2}$...
Read More →The radii of two concentric circles are 19 cm and 16 cm respectively.
Question: The radii of two concentric circles are 19 cm and 16 cm respectively. The area of the ring enclosed by these circles is(a) 320 cm2(b) 330 cm2(c) 332 cm2(d) 340 cm2 Solution: (b) 330 cm2Let:R= 19 cm andr= 16 cmThus, we have: Area of the ring $=\pi\left(R^{2}-r^{2}\right)$ $=\pi(R+r)(\mathrm{R}-\mathrm{r})$ $=\left|\frac{22}{7} \times(19+16) \times(19-16)\right| \mathrm{cm}^{2}$ $=\left(\frac{22}{7} \times 35 \times 3\right) \mathrm{cm}^{2}$ $=330 \mathrm{~cm}^{2}$...
Read More →A steamer goes downstream from one point to another in 9 hours.
Question: A steamer goes downstream from one point to another in 9 hours. It covers the same distance upstream in 10 hours. If the speed of the stream be 1 km/hr, find the speed of the steamer in still water and the distance between the ports. Solution: It is given that the speed of the stream is $1 \mathrm{~km} / \mathrm{h}$. Let the speed of the steamer in still water be $x \mathrm{~km} / \mathrm{h}$. $\therefore$ Downstream speed $=(\mathrm{x}+1) \mathrm{km} / \mathrm{h}$ Upstream speed $=(\m...
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