Cards with numbers 2 to 101 are placed in a box.
Question: Cards with numbers 2 to 101 are placed in a box. A card is selected at random from the box. Find the probability that the card which is selected has a number which is a perfect square. Solution: It is given that cards with numbers 2 to 101 are placed in a box. A card is picked at random. We have to find the probability that the card selected has a number which is a perfect square. Perfect squares between 2 and 101 are 4, 9, 16, 25, 36, 48, 64, 81, 100 Total number of perfect squares To...
Read More →The points A (2, 9), B (a, 5), C (5, 5) are the vertices
Question: The pointsA(2, 9),B(a, 5),C(5, 5) are the vertices of a triangleABCright angled atB. Find the value of 'a' and hence the area of ΔABC. Solution: It is given that ABC is a right angled triangle..Vertices of triangle are A( 2,9 ),B(a,5 ),C( 5, 5 ). We have to find the value ofaand area of the triangle. is a right angled triangle. We know that length of a line with coordinates of end points $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ $=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\l...
Read More →Prove that
Question: If $\cot \theta=\frac{15}{8}$ then evaluate $\frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)}$ Solution: Given : $\cot \theta=\frac{15}{8}$ Since, $\cot \theta=\frac{B}{P}$ $\Rightarrow P=8$ and $B=15$ Using Pythagoras theorem, $P^{2}+B^{2}=H^{2}$ $\Rightarrow 8^{2}+15^{2}=H^{2}$ $\Rightarrow H^{2}=64+225$ $\Rightarrow H^{2}=289$ $\Rightarrow H=17$ Therefore, $\sin \theta=\frac{P}{H}=\frac{8}{17}$ $\cos \theta=\frac{B}{H}=\frac{15}{17}$ Now, $\frac{(1+\sin \theta)(...
Read More →If A and B are two matrices such that AB = B
Question: If $A$ and $B$ are two matrices such that $A B=B$ and $B A=A, A^{2}+B^{2}$ is equal to (a) $2 A B$ (b) $2 B A$ (c) $A+B$ (d) $A B$ Solution: (c) $A+B$ Given : $A B=B$ and $B A=A$ $A^{2}+B^{2}=A A+B B$ $\Rightarrow A^{2}+B^{2}=B A B A+A B A B \quad[\because A B=B$ and $B A=A]$ $\Rightarrow A^{2}+B^{2}=B B A+A A B \quad[\because A B=B$ and $B A=A]$ $\Rightarrow A^{2}+B^{2}=B A+A B \quad[\because A B=B$ and $B A=A]$ $\Rightarrow A^{2}+B^{2}=A+B \quad[\because A B=B$ and $B A=A]$...
Read More →Prove that the points A (4, 3),
Question: Prove that the pointsA(4, 3),B(6, 4),C(5, 6) andD(3, 7) in that order are the vertices of a parallelogram. Solution: Given pointsA (4, 3),B(6, 4),C(5, 6) andD(3, 7). We have to prove that these points form vertices of a parallelogram. The above four points will form 4 line segments. AB, BC, CD, AD We know that length of a line segment having coordinates $=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$ Therefore $A B=\sqrt{(6-4)^{2}+(4-3)^{2}}$ $=\sqrt{4+1}$ $=\sqrt{5...
Read More →If 3 tan A = 4 then prove that
Question: If 3 tanA= 4 then prove that (i) $\sqrt{\frac{\sec A-\operatorname{cosec} A}{\sec A+\operatorname{cosec} A}}=\frac{1}{\sqrt{7}}$ (ii) $\sqrt{\frac{1-\sin A}{1+\cos A}}=\frac{1}{2 \sqrt{2}}$ Solution: (i) $\mathrm{LHS}=\sqrt{\frac{\sec \theta-\operatorname{cosec} \theta}{\sec \theta+\operatorname{cosec} \theta}}$ $=\sqrt{\frac{\left(\frac{1}{\cos \theta}-\frac{1}{\sin \theta}\right)}{\left(\frac{1}{\cos \theta}+\frac{1}{\sin \theta}\right)}}$ $=\sqrt{\frac{\left(\frac{\sin \theta-\cos \...
Read More →A tower stands vertically on the ground.
Question: A tower stands vertically on the ground. From a point on the ground which is 20 m away from the foot of the tower, the angle of elevation of the top of the tower is found to be 60. Find the height of the tower. Solution: Given a tower at the ground such that at a distance of 20 m away from foot of tower, the top of tower makes an angle ofwith the ground. We have to find the height of the tower. Let the tower beAB Height of the tower =h Distance of the tower from pointCwhere top of the ...
Read More →If AB = A and BA = B, where A and B are square matrices, then
Question: If $A B=A$ and $B A=B$, where $A$ and $B$ are square matrices, then (a) $B^{2}=B$ and $A^{2}=A$ (b) $B^{2} \neq B$ and $A^{2}=A$ (c) $A^{2} \neq A, B^{2}=B$ (d) $A^{2} \neq A, B^{2} \neq B$ Solution: (a) $B^{2}=B$ and $A^{2}=A$ Here, $A B=A \quad \ldots(1)$ $B A=B \quad \ldots(2)$ $\Rightarrow A B A=A A \quad$ [Multiplying both sides by $A$ ] $B A B=B B$ [Multiplying both sides by $A$ ] $\Rightarrow A B=A^{2} \quad[$ From eq. (2) $]$ $B A=B^{2} \quad$ [From eq. (1) $]$ $\Rightarrow A=A...
Read More →A hemispherical depression is cut out from one face of a cubical
Question: A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter 'l' of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid. Solution: It is given that a hemisphere is cut from a cubical box with edgelsuch that diameter of hemisphere is alsol. We have to find the surface area of the remaining solid. Surface area of the cubical box with side $l=6 l^{2}$ Let $r$ be the radius of hemisphere Surface area of ...
Read More →If A and B are two matrices such that
Question: If $A$ and $B$ are two matrices such that $A B=A$ and $B A=B$, then $B^{2}$ is equal to (a) $B$ (b) $A$ (c) 1 (d) 0 Solution: (a) $B$ Here, $A B=A \quad \ldots(1)$ $B A=B \quad \ldots(2)$ $\Rightarrow B A B=B B \quad[$ Multiplying both sides by $B]$ $\Rightarrow B A=B^{2} \quad$ [From eq. (1)] $\Rightarrow B=B^{2} \quad$ [From eq. (2)]...
Read More →In Fig. 4, OABC is a square inscribed in a quadrant OPBQ.
Question: In Fig. 4,OABCis a square inscribed in a quadrantOPBQ. IfOA = 20 cm, find the area of shaded region [Use = 3.14] Solution: It is given that OABC is a square in the quadrant OPBQ.OA is 20 cm. We have to find the area of the shaded region. OABCis a square , therefore sides of theOABCmust be equal HenceOA, AB, BC, OC= 20 cm. Join the pointsOandBto form a line segmentOB. SinceOBis the diagonal ofOABC,is a right angled triangle. Applying Pythagoras Theorem in It can be seen from the figure ...
Read More →Solve this
Question: If $\sin \theta=\frac{3}{4}$, show that $\sqrt{\frac{\operatorname{cosec}^{2} \theta-\cot ^{2} \theta}{\sec ^{2} \theta-1}}=\frac{\sqrt{7}}{3}$. Solution: $\mathrm{LHS}=\sqrt{\frac{\operatorname{cosec}^{2} \theta-\cot ^{2} \theta}{\sec ^{2} \theta-1}}$ $=\sqrt{\frac{1}{\tan ^{2} \theta}}$ $=\sqrt{\cot ^{2} \theta}$ $=\cot \theta$ $=\sqrt{\operatorname{cosec}^{2} \theta-1}$ $=\sqrt{\left(\frac{1}{\sin \theta}\right)^{2}-1}$ $=\sqrt{\left(\frac{1}{\left(\frac{3}{4}\right)}\right)^{2}-1}$...
Read More →Draw a triangle ABC with side BC = 6 cm,
Question: Draw a triangle $A B C$ with side $B C=6 \mathrm{~cm}, A B=5 \mathrm{~cm}$ and $\angle A B C=60^{\circ}$. Then construct a triangle whose sides are $\frac{3}{4}$ time the corresponding sides of $\triangle A B C$. Solution: We have to draw a triangle $A B C$ with $B C=6 \mathrm{~cm}, A B=5 \mathrm{~cm}, \angle A B C=60^{\circ}$ Then we have to construct a similar triangle with side $\frac{3}{4}$ of $\triangle A B C$ Steps of construction 1. Draw a line segment 2. AtBdrawso that a rayBXi...
Read More →Solve this
Question: If $A=\left[\begin{array}{ll}i 0 \\ 0 i\end{array}\right], n \in N$, then $A^{4 n}$ equals (a) $\left[\begin{array}{ll}0 i \\ i 0\end{array}\right]$ (b) $\left[\begin{array}{ll}0 0 \\ 0 0\end{array}\right]$ (c) $\left[\begin{array}{ll}1 0 \\ 0 1\end{array}\right]$ (d) $\left[\begin{array}{ll}0 i \\ i 0\end{array}\right]$ Solution: (c) $\left[\begin{array}{ll}1 0 \\ 0 1\end{array}\right]$ Here, $A=\left[\begin{array}{ll}i 0 \\ 0 i\end{array}\right]$ $\Rightarrow A^{2}=\left[\begin{array...
Read More →In Fig. 3, two tangents PQ are PR are drawn to a
Question: In Fig. 3, two tangentsPQarePRare drawn to a circle with centreOfrom an external pointP. Prove that QPR= 2 OQR. Solution: Given a figure as shown. We have to prove that JoinOR We know that sum of opposite angles of a cyclic quadrilateral Therefore in quadrilateralPQOR, $\angle Q O R=180^{\circ}-\angle Q P R$ In $\triangle O Q R$ $O Q=O R$ [Since $O Q, O R$ are radii of the circle] Therefore $\triangle O Q R$ is an isosceles triangle. Hence[Angles opposite to equal side of isosceles tri...
Read More →Solve this
Question: If $\tan \theta=\frac{1}{\sqrt{7}}$ then prove that $\left(\frac{\operatorname{cosec}^{2} \theta+\sec ^{2} \theta}{\operatorname{cosec}^{2} \theta-\sec ^{2} \theta}\right)=\frac{4}{3}$. Solution: Let us consider a right $\triangle \mathrm{ABC}$, right angled at $\mathrm{B}$ and $\angle C=\theta$. Now it is given that $\tan \theta=\frac{A B}{B C}=\frac{1}{\sqrt{7}}$. So, if $\mathrm{AB}=k$, then $\mathrm{BC}=\sqrt{7} k$, where $k$ is a positive number. Using Pythagoras theorem, we have:...
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Question: If $A=\left[\begin{array}{rrr}1 0 0 \\ 0 1 0 \\ a b -1\end{array}\right]$, then $A^{2}$ is equal to (a) a : matrix (b) a unit matrix (c) $-A$ (d) $A$ Solution: (b) a unit matrix $A^{2}=A A$ $\Rightarrow A^{2}=\left[\begin{array}{ccc}1 0 0 \\ 0 1 0 \\ a b -1\end{array}\right]\left[\begin{array}{ccc}1 0 0 \\ 0 1 0 \\ a b -1\end{array}\right]$ $\Rightarrow A^{2}=\left[\begin{array}{lll}1+0+0 0+0+0 0+0-0 \\ 0+0+0 0+1+0 0+0-0 \\ a+0-a 0+b-b 0+0+1\end{array}\right]$ $\Rightarrow A^{2}=\left[...
Read More →Find the sum of the integers between 100
Question: Find the sum of the integers between 100 $=\frac{11}{2} \times 306$d 200 that are divisible by 9. Solution: We have to find the sum of integers between 100 and 200 that are divisible by 9. Integers divisible by 9 between 100 and 200 are 108, 117, 126, 198 The above equation forms an Arithmetic Progression (A.P) Let there be n such integers in the above A.P We know that the nth term of an A.P $a_{n}=a+(n-1) d$ Where= First term of A.P = Common difference of successive members Therefore ...
Read More →Solve this
Question: For the matrix $A=\left[\begin{array}{ll}2 3 \\ 5 7\end{array}\right]$, find $A+A^{T}$ and verify that it is a symmetric matrix. Solution: The given matrix is $A=\left[\begin{array}{ll}2 3 \\ 5 7\end{array}\right]$ ...(1) $\therefore A^{T}=\left[\begin{array}{ll}2 5 \\ 3 7\end{array}\right]$ ....(2) Adding $(1)$ and $(2)$, we get $A+A^{T}=\left[\begin{array}{ll}2 3 \\ 5 7\end{array}\right]+\left[\begin{array}{ll}2 5 \\ 3 7\end{array}\right]$ $\Rightarrow A+A^{T}=\left[\begin{array}{ll}...
Read More →Solve this
Question: If $\tan \theta=\frac{20}{21}$, show that $\frac{(1-\sin \theta+\cos \theta)}{(1+\sin \theta+\cos \theta)}=\frac{3}{7}$ Solution: Let us consider a right $\triangle \mathrm{ABC}$ right angled at $\mathrm{B}$ and $\angle C=\theta$. Now, we know that $\tan \theta=\frac{A B}{B C}=\frac{20}{21}$ So, if AB = 20k, then BC = 21k, wherekis a positive number.Using Pythagoras theorem, we get:AC2= AB2+ BC2⇒ AC2= (20k)2+ (21k)2⇒ AC2= 841k2⇒ AC = 29k Now, $\sin \theta=\frac{A B}{A C}=\frac{20}{29}$...
Read More →A natural number, when increased by 12,
Question: A natural number, when increased by 12, becomes equal to 160 times its reciprocal. Find the number. Solution: It is given that a number when increased by 12 becomes 160 times its reciprocal. We have to find the number. Let the number bex Reciprocal of $x=\frac{1}{x}$ According to the question $x+12=160 \times \frac{1}{x}$ $x^{2}+12 x=160$ $x^{2}+12 x-160=0$ $x^{2}+20 x-8 x-160=0$ $x(x+20)-8(x+20)=0$ $(x+20)(x-8)=0$ $x=-20,8$ since $-20+8 \neq \frac{160}{-20}$ Therefore $x=8$...
Read More →Two dice are thrown at the same time.
Question: Two dice are thrown at the same time. Find the probability of getting different numbers on both dice. Solution: It is given that two dice are thrown at the same time. We have to find the probability of getting different numbers on both dice. Total number of possible choices in rolling a dice = 6 Total number of possible choices in rolling two dice[Using multiplication rule] Probability of getting same number if two dice are thrown $\mathrm{P}($ Same Number $)=\frac{6}{36}$ Probability ...
Read More →Solve this
Question: If $\sec \theta=\frac{17}{8}$ then prove that $\frac{3-4 \sin ^{2} \theta}{4 \cos ^{2} \theta-3}=\frac{3-\tan ^{2} \theta}{1-3 \tan ^{2} \theta}$. Solution: It is given that $\sec \theta=\frac{17}{8}$. Let us consider a right $\triangle \mathrm{ABC}$ right angled at $\mathrm{B}$ and $\angle C=\theta$. We know that $\cos \theta=\frac{1}{\sec \theta}=\frac{8}{17}=\frac{B C}{A C}$ So, if BC = 8k, then AC = 17k, wherekis a positive number.Using Pythagoras theorem, we have:AC2= AB2+ BC2⇒ AB...
Read More →Show that the point P (−4, 2) lies on the line segment
Question: Show that the point P (4, 2) lies on the line segment joining the points A (4, 6) and B (4, 6). Solution: We have to show that pointP(4, 2) lies on line segmentABwith pointsA(4, 6) and B(4, 6) IfP(4, 2) lies on the line segment joiningA(4, 6) andB(4, 6), then the three points must be collinear. Let the three points be not collinear and form a trianglePAB We know that area of a triangle with coordinates of vertices $\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right),\left(x_{3}, y_{3}\...
Read More →Express the following matrix as the sum of a symmetric
Question: Express the following matrix as the sum of a symmetric and skew-symmetric matrix and verify your result: $\left[\begin{array}{rrr}3 -2 -4 \\ 3 -2 -5 \\ -1 1 2\end{array}\right]$ Solution: Here, $A=\left[\begin{array}{ccc}3 -2 -4 \\ 3 -2 -5 \\ -1 1 2\end{array}\right]$ $\Rightarrow A^{T}=\left|\begin{array}{ccc}3 3 -1 \\ -2 -2 1 \\ -4 -5 2\end{array}\right|$ Let $X=\frac{1}{2}\left(A+A^{T}\right)=\frac{1}{2}\left(\left[\begin{array}{ccc}3 -2 -4 \\ 3 -2 -5 \\ -1 1 2\end{array}\right]+\le...
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