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Question: If $A=\left[\begin{array}{l}3 \\ 5 \\ 2\end{array}\right]$ and $B=\left[\begin{array}{lll}1 0 4\end{array}\right]$, verify that $(A B)^{\top}=B^{T} A^{T}$ Solution: Given : $A=\left[\begin{array}{l}3 \\ 5 \\ 2\end{array}\right]$ $A^{T}=\left[\begin{array}{lll}3 5 2\end{array}\right]$ $B=\left[\begin{array}{lll}1 0 4\end{array}\right]$ $B^{T}=\left[\begin{array}{l}1 \\ 0 \\ 4\end{array}\right]$ Now, $A B=\left[\begin{array}{l}3 \\ 5 \\ 2\end{array}\right]\left[\begin{array}{lll}1 0 4\en...

In the given figure, O is the centre of a circle in which ∠OBA = 20° and ∠OCA = 30°.

Question: In the given figure,Ois the centre of a circle in whichOBA= 20 and OCA= 30. Then, BOC= ?(a) 50(b) 90(c) 100(d) 130 Solution: (c) 100In Δ OAB, we have:OA = OB (Radii of a circle)⇒OAB = OBA = 20In ΔOAC, we have:OA = OC (Radii of a circle)⇒OAC = OCA = 30Now,BAC = (20 + 30) = 50BOC = (2BAC) = (2 50) = 100...

In the given figure, O is the centre of a circle.

Question: In the given figure,Ois the centre of a circle. IfOAC= 50, then ODB= ?(a) 40(b) 50(c) 60(d) 75 Solution: (b) 50ODB =OAC= 50 (Angles in the same segment of a circle)...

In a G.P. of positive terms, if any term is equal to the sum of the next two terms.

Question: In a G.P. of positive terms, if any term is equal to the sum of the next two terms. Then the common ratio of the G.P. (a) sin 18 (b) 2 cos 18 (c) cos 18 (d) 2 sin 18 Solution: Let ustn,tn+1andtn+2denotenth, (n+ 1)thand (n+ 2)thterm of geometric progression respectively. According to given condition, tn=tn+1+tn+2 i.earn1=arn+arn+1 where,adenotes first term of g.p andrdenote the common ratio. then $r^{n-1}=r^{n}+r^{n+1}$ i. e $1=r+r^{2}$ i. e $r^{2}+r-1=0$ $\therefore r=\frac{-1 \pm \sqr...

In the given figure, BOC is a diameter of a circle with centre O

Question: In the given figure,BOCis a diameter of a circle with centreO. IfBCA= 30, then CDA= ?(a) 30(b) 45(c) 60(d) 50 Solution: (c) 60Angles in a semi circle measure 90.BAC =90InΔ ABC, we have:BAC + ABC + BCA = 180 (Angle sum property of a triangle) 90 + ABC + 30 = 180⇒ABC = (180 - 120) = 60CDA = ABC = 60 (Angles in the same segment of a circle)...

Let

Question: Let $A=\left[\begin{array}{rr}2 -3 \\ -7 5\end{array}\right]$ and $B=\left[\begin{array}{rr}1 0 \\ 2 -4\end{array}\right]$, verify that (i) $(2 A)^{T}=2 A^{T}$ (ii) $(A+B)^{T}=A^{T}+B^{T}$ (iii) $(A-B)^{\top}=A^{\top}-B^{\top}$ (iv) $(A B)^{T}=B^{T} A^{\top}$ Solution: Given : $A=\left[\begin{array}{cc}2 -3 \\ -7 5\end{array}\right]$ $A^{T}=\left[\begin{array}{cc}2 -7 \\ -3 5\end{array}\right]$ $B=\left[\begin{array}{cc}1 0 \\ 2 -4\end{array}\right]$ $B^{T}=\left[\begin{array}{cc}1 2 \...

In the given figure, ∆ABC and ∆DBC are inscribed in a circle such that

Question: In the given figure,∆ABCand ∆DBCare inscribed in a circle such that BAC= 60 and DBC= 50.(a) 50(b) 60(c) 70(d) 80 Solution: (c) 70BDC = BAC = 60 (Angles in the same segment of a circle)In Δ BDC, we have:DBC +BDC + BCD= 180 (Angle sum property of a triangle) 50 + 60 + BCD = 180⇒BCD = 180- (50 + 60) = (180- 110) = 70...

If x, y, z are positive integers then value of the expression (x + y) (y + z) (z + x) is

Question: Ifx,y,zare positive integers then value of the expression (x+y) (y+z) (z+x) is (a) = 8xyz (b) 8xyz (c) 8xyz (d) = 4xyz Solution: Letx,y,zbe positive integers then (x + y) (y + z) (z + x) =?? Since $\frac{x+y}{2}\sqrt{x y}$ (Since arithmetic mean $\geq$ geometric mean) i. e $x+y2 \sqrt{x y}$ Similarly $y+z2 \sqrt{y z}$ and $z+x2 \sqrt{z x}$ $\therefore(x+y)(y+z)(z+x)8 \sqrt{x y} \sqrt{y z} \sqrt{z x}$ $=8 \sqrt{x^{2} y^{2} z^{2}}$ $=8 x y z$ $\therefore(x+y)(y+z)(z+x)8 x y z$ Hence, the...

Angles in the same segment of a circle area are

Question: Angles in the same segment of a circle area are(a) equal(b) complementary(c) supplementary(d) none of these Figure Solution: (a) equalThe angles in the same segment of a circle are equal....

Prove that the points (−4,−1), (−2, 4),

Question: Prove that the points (4,1), (2, 4), (4, 0) and (2, 3) are the vertices of a rectangle. Solution: Let A (4,1); B (2,4); C (4, 0) and D (2, 3) be the vertices of a quadrilateral. We have to prove that the quadrilateral ABCD is a rectangle. So we should find the lengths of opposite sides of quadrilateral ABCD. $\mathrm{AB}=\sqrt{(-2+4)^{2}+(-4+1)^{2}}$ $=\sqrt{4+9}$ $=\sqrt{13}$ $\mathrm{CD}=\sqrt{(4-2)^{2}+(0-3)^{2}}$ $=\sqrt{4+9}$ $=\sqrt{13}$ Opposite sides are equal. So now we will c...

Mark the correct alternative in the following question:

Question: Mark the correct alternative in the following question: Let Sbe the sum,Pbe the product andRbe the sum of the reciprocals of 3 terms of a G.P. Thenp2R3:S3is equal to (a) 1 : 1 (b) (Common ratio)n: 1 (c) (First term)2: (Common ratio)2 (d) None of these Solution: Let the three terms of the G.P. be $\frac{a}{n}, a, a r$. Then $S=\frac{a}{r}+a+a r$ $=a\left(\frac{1}{r}+1+r\right)$ $=a\left(\frac{1+r+r^{2}}{r}\right)$ $=\frac{a\left(r^{2}+r+1\right)}{r}$ Also, $P=\frac{a}{r} \times a \times...

The angle in a semicircle measures

Question: The angle in a semicircle measures(a) 45(b) 60(c) 90(d) 36 Figure Solution: (c) 90The angle in a semicircle measures 90....

Prove that (4, 3), (6, 4) (5, 6) and (3, 5) are the angular points of a square.

Question: Prove that (4, 3), (6, 4) (5, 6) and (3, 5) are the angular points of a square. Solution: Let A (4, 3); B (6, 4); C (5, 6) and D (3, 5) be the vertices of a quadrilateral. We have to prove that the quadrilateral ABCD is a square. So we should find the lengths of sides of quadrilateral ABCD. $\mathrm{AB}=\sqrt{(6-4)^{2}+(4-3)^{2}}$ $=\sqrt{4+1}$ $=\sqrt{5}$ $B C=\sqrt{(6-5)^{2}+(4-6)^{2}}$ $=\sqrt{1+4}$ $=\sqrt{5}$ $\mathrm{CD}=\sqrt{(3-5)^{2}+(5-6)^{2}}$ $=\sqrt{4+1}$ $=\sqrt{5}$ $\mat...

A trust invested some money in two type of bonds.

Question: A trust invested some money in two type of bonds. The first bond pays 10% interest and second bond pays 12% interest. The trust received ₹ 2800 as interest. However, if trust had interchanged money in bonds, they would have got ₹ 100 less as interes. Using matrix method, find the amount invested by the trust. Solution: Let Rsxbe invested in the first bond and Rsybe invested in the second bond.Let A be the investment matrix and B be the interest per rupee matrix. Then, $\mathrm{A}=\left...

An equilateral triangle of side 9 cm is inscribed in a circle.

Question: An equilateral triangle of side 9 cm is inscribed in a circle. The radius of the circle is (a) 3 cm (b) $3 \sqrt{2} \mathrm{~cm}$ (c) $3 \sqrt{3} \mathrm{~cm}$ (d) 6 cm Figure Solution: (c) $3 \sqrt{3} \mathrm{~cm}$ Let ΔABC be an equilateral triangle of side 9 cm.Let AD be one of its medians. Then AD BC and BD = 4.5 cm $\therefore \mathrm{AD}=\sqrt{\mathrm{AB}^{2}-\mathrm{BD}^{2}}=\sqrt{(9)^{2}-\left(\frac{9}{2}\right)^{2}}=\sqrt{81-\frac{81}{4}}=\sqrt{\frac{324-81}{4}}=\sqrt{\frac{24...

In a G.P. if the

Question: In a G.P. if the (m+n)thterm ispand (mn)thterm isq, then itsmthterm is (a) 0 (b)pq (c) $\sqrt{p q}$ (d) $\frac{1}{2}(p+q)$ Solution: (c) $\sqrt{p q}$ Here, $a_{(m+n)}=p$ $\Rightarrow a r^{(m+n-1)}=p \quad \ldots \ldots(\mathrm{i})$ Also, $\mathbf{a}_{(m-n)}=q$ $\Rightarrow a r^{(m-n-1)}=q \quad \ldots \ldots$ (ii) Mutliplying (i) and (ii): $\Rightarrow a r^{(m+n-1)} a r^{(m-n-1)}=p q$ $\Rightarrow a^{2} r^{(2 m-2)}=p q$ $\Rightarrow\left(a r^{(m-1)}\right)^{2}=p q$ $\Rightarrow a r^{(m...

Prove that the points (4, 5) (7, 6), (6, 3) (3, 2)

Question: Prove that the points (4, 5) (7, 6), (6, 3) (3, 2) are the vertices of a parallelogram. Is it a rectangle. Solution: Let A (4, 5); B (7, 6); C (6, 3) and D (3, 2) be the vertices of a quadrilateral. We have to prove that the quadrilateral ABCD is a parallelogram. We should proceed with the fact that if the diagonals of a quadrilateral bisect each other than the quadrilateral is a parallelogram. Now to find the mid-pointof two pointsandwe use section formula as, $\mathrm{P}(x, y)=\left(...

The two geometric means between the numbers 1 and 64 are

Question: The two geometric means between the numbers 1 and 64 are (a) 1 and 64 (b) 4 and 16 (c) 2 and 16 (d) 8 and 16 (e) 3 and 16 Solution: (b) 4 and 16 Let the two G.M.s between 1 and 64 be $G_{1}$ and $G_{2}$. Thus, $1, G_{1}, G_{2}$ and 64 are in G.P. $64=1 \times r^{3}$ $\Rightarrow r=\sqrt[3]{64}$ $\Rightarrow r=4$ $\Rightarrow G_{1}=a r=1 \times 4=4$ And, $G_{2}=a r^{2}=1 \times 4^{2}=16$ Thus, 4 and 16 are the required G.M.s....

In the given figure, AB is a chord of a circle with centre O and BOC is a diameter.

Question: In the given figure,ABis a chord of a circle with centreOandBOCis a diameter. IfODABsuch thatOD= 6 cm, thenAC= ?(a) 9 cm(b) 12 cm(c) 15 cm(d) 7.5 cm Solution: (b) 12 cmOD ABi.e., D is the mid point of AB.Also, O is the mid point of BC.Now, in Δ BAC, D is the mid point of AB and O is the mid point of BC. $\therefore \mathrm{OD}=\frac{1}{2} \mathrm{AC}$ (By mid point theorem) ⇒ AC = 2OD = (2 6) cm = 12 cm...

Find the ratio in which the line segment joining (−2, −3)

Question: Find the ratio in which the line segment joining (2, 3) and (5, 6) is divided by (i)x-axis (ii)y-axis. Also, find the coordinates of the point of division in each case. Solution: The ratio in which the $x$-axis divides two points $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ is $\lambda: 1$ The ratio in which the $y$-axis divides two points $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ is $\mu: 1$ The co-ordinates of the point dividing two points $\left(x_{1...

The monthly incomes of Aryan and Babban are in the ratio 3 : 4 and their monthly expenditures are in the ratio 5 : 7.

Question: The monthly incomes of Aryan and Babban are in the ratio 3 : 4 and their monthly expenditures are in the ratio 5 : 7. If each saves ₹ 15,000 per month, find their monthly incomes using matrix method. This problem reflects which value? Solution: Let the monthly incomes of Aryan and Babban be 3xand 4x, respectively.Suppose their monthly expenditures are 5yand 7y, respectively.Since each saves Rs 15,000 per month, Monthly saving of Aryan : $3 x-5 y=15,000$ Monthly saving of Babban : $4 x-...

In the given figure, AB is a chord of a circle with centre O and AB is produced to C such that BC = OB.

Question: In the given figure,ABis a chord of a circle with centreOandABis produced toCsuch thatBC=OB. Also,COis joined and produced to meet the circle inD. IfACD= 25, then AOD= ?(a) 50(b) 75(c) 90(d) 100 Solution: (b) 75OB = BC (Given)⇒ BOC = BCO = 25Exterior OBA = BOC + BCO = (25 + 25) = 50OA = OB (Radius of a circle)⇒OAB = OBA = 50In Δ AOC, side CO has been produced to D. Exterior AOD = OAC + ACO = OAB + BCO = (50 + 25) = 75...

The product (32),

Question: The product (32), (32)1/6(32)1/36... to is equal to (a) 64 (b) 16 (c) 32 (d) 0 Solution: (a) 64 $32 \times 32^{\frac{1}{6}} \times 32^{\frac{1}{36}} \times \ldots \infty$ $=32\left(1+\frac{1}{6}+\frac{1}{36}+\ldots \infty\right)$ $=32\left(\frac{1}{1-\frac{1}{6}}\right)[\because$ it is a G.P. $]$ $=32^{\left(\frac{6}{5}\right)}$ $=\left(2^{5}\right)^{\left(\frac{6}{5}\right)}$ $=2^{6}$ $=64$...

Determine the ratio in which the straight line

Question: Determine the ratio in which the straight line x y 2 = 0 divides the line segment joining (3, 1) and (8, 9). Solution: Let the linedivide the line segment joining the points A (3,1) and B (8, 9) in the ratioat any point Now according to the section formula if point a point P divides a line segment joiningandin the ratio m: n internally than, $\mathrm{P}(x, y)=\left(\frac{n x_{1}+m x_{2}}{m+n}, \frac{n y_{1}+m y_{2}}{m+n}\right)$ So, $\mathrm{P}(x, y)=\left(\frac{8 \lambda+3}{\lambda+1}...

Let x be the A.M. and y, z be two G.M.s between two positive numbers.

Question: Let $x$ be the A.M. and $y, z$ be two G.M.s between two positive numbers. Then, $\frac{y^{3}+z^{3}}{x y z}$ is equal to (a) 1 (b) 2 (c) $\frac{1}{2}$ (d) none of these Solution: (b) 2 Let the two numbers be $a$ and $b$. $a, x$ and $b$ are in A.P. $\therefore 2 x=a+b$ ...(i) Also, $a, y, z$ and $b$ are in G.P. $\therefore \frac{y}{a}=\frac{z}{y}=\frac{b}{z}$ $\Rightarrow y^{2}=a z, y z=a b, z^{2}=b y$ ...(ii) Now, $\frac{y^{3}+z^{3}}{x y z}$ $=\frac{y^{2}}{x z}+\frac{z^{2}}{x y}$ $=\fra...

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