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 MATH SAMPLE QUESTION PAPER Class X Session 2023-24 MATHEMATICS STANDARD (Code No.041) Question 28 $$Solve:\frac{2}{\sqrt{x}}+\frac{3}{\sqrt{y}}=2;\frac{4}{\sqrt{x}}-\frac{9}{\sqrt{y}}=-1,x,y>0$$ Explanation:   Let's consider : $$a=\frac{1}{\sqrt{x}}andb=\frac{1}{\sqrt{y}}=-1$$  Now equation becomes  2a+3b = 2 -----------------(1) 4a - 9b = -1 -------------------------(2) now solve both equations (1) and (2) First, multiply equation (1) by 2 to align the coefficients of a and subtract from each other  2x(2a+3b) - (4a -9b ) = 4 -(-1) $$2\times (2a+3b)-(4a-9b)=4-(-1)$$ $$or4a+6b-4a+9b=4+1$$ $$or15b=5$$ $$orb=\frac{5}{15}$$ $$orb=\frac{1}{3}$$ Now, substitute b in equation (1) $$\Rightarrow 2a+3\times \frac{1}{3}=2$$ $$or2a+1=2$$ $$or2a=1$$ $$\therefore a=\frac{1}{2}$$ Now we have a and b so we can find the values of x and y  \[\ a = \fr...

MATH SAMPLE QUESTION PAPER Class X Session 2023-24 MATHEMATICS STANDARD (Code No.041) Question 28 The sum of a two digit number and the number obtained by reversing the digits is 66. If the digits of the number differ by 2, find the number. How many such numbers are there? Explanation:   \begin{flalign} & Let's\; consider\; the \; unit\; digit\; be\; a\; and \; ten\; digit\; be \; b\; &\ & so, Original \; number \; is\; a+10b ---(1) \; where\; a\; \neq b &\ & By \;reversing\; the\; digit\,\; we\; get\; 10b+a ----(2) &\ & According \;to\; the\; question\;,\; we \; get & \ & (a+10b) + (10b+a) = 66 &\ &\Rightarrow 11a +11b = 66 & \ & or \; a+b = 6 --(3) & \ & since\; given \; that\; digit \;of \;the\; number \;differ \;2 &\ & \therefore a-b = 2 ---(4) &\ & by \; adding \; equations \;2\; and \;3\; we \;get &\ & (a-b) +(a+b) = 2+6 &\ & 2a = 8 &\ & \therefore a = 4 ...

 MATH SAMPLE QUESTION PAPER Class X Session 2023-24 MATHEMATICS STANDARD (Code No.041) Question 27 $$\begin{array}{rlcr}& if,\alpha \beta areZeroesofquadraticpolynomial5x^2+5x+10findthevalueof(1){\alpha }^2+{\beta }^2,(2){\alpha }^{-}1+{\beta }^{-}1& & \end{array}$$ Explanation:   \begin{flalign} & since\; \alpha\; and \; \beta\; are \;the \;root\; of\; 5x^2 + 5x + 10 as \;we \;know\; that\; if \; quadratic\; equation;&\ & ax^2 + bx +c =0 \;then \; SUM \;of \; zeroes\; \alpha + \beta\ = -\frac{b}{a}; and \;product\; of \;Zeroes\; =\; \frac{c}{a}&\ & since\; \alpha\; and \; \beta\; are \;the \;root\; of\;5x^2 + 5x + 10 \;similarly \;\ from\; the \;given\; equation\; \alpha + \beta = & \ & \frac{-5}{5} and \; \alpha \beta \;= \frac{1}{5} , \; now \; lets \;find\; \alpha^2 + \beta^2 &\ &\Rightarrow (\alpha + \beta)^2 = \a...

 MATH SAMPLE QUESTION PAPER Class X Session 2023-24 MATHEMATICS STANDARD (Code No.041) Question 26 Find the area of the unshaded region shown in the given figure.  Solution: modified view of shared image The total horizontal or vertical extent of the region is 8 cm and extent includes the side length of the square (𝑎)  and the diameters of the semicircles on either side of the square. Given that each semicircle has a radius 𝑟, the side length of the square is 𝑎=8−2𝑟. Given the figure, the radius 𝑟 of each semicircle is 2cm Side Length of the Square 𝑎 = 8-2x2  = 4 cm Area of the Square:  𝑎^2 = 4^2 = 16 Area of One Semicircle = $$\frac{1}{2}\pi r^2=2\pi c{m}^{2}wherer=2$$ Combined Area of Four Semicircles: $$4\times 2\pi c{m}^2$$ Area of the unshaded region=Area of the square + 4 x Area of  one semicircle = $$16c{m}^2+4\times 2\pi c{m}^2$$ $$(16+8\pi )c{m}^2$$ 

 MATH SAMPLE QUESTION PAPER Class X Session 2023-24 MATHEMATICS STANDARD (Code No.041) Question 25 Find the value of x if $$\begin{array}{rlcr}& 2cose{c}^2(30)+xsi{n}^2(60)-\frac{3}{4}ta{n}^2(30)=10& & \end{array}$$ Explanation:    Now simplify the equation below  $$\begin{array}{rlcr}& 2cose{c}^2(30)+xsi{n}^2(60)-\frac{3}{4}ta{n}^2(30)=10& & \end{array}$$ $$\Rightarrow 2\times (2{)}^2+x{\left(\frac{\sqrt{3}}{2}\right)}^2-\frac{3}{4}{\left(\frac{1}{\sqrt{3}}\right)}^2=10$$ we know that value of cosec(30) , sin(60)and tan(30) putting their values $$\Rightarrow 2\times 4+x\frac{3}{4}-(\frac{3}{4}\times \frac{1}{3})=10$$ $$OR8+\frac{3x}{4}-\frac{1}{4}=10$$ $$OR\frac{3x}{4}=\frac{1}{4}+(10-8)$$ $$OR\frac{3x}{4}=\frac{1}{4}+2\Rightarrow \frac{9}{4}$$ $$OR\frac{3x}{4}=\frac{9}{4}$$ $$\Rightarrow 3x=9$$ $$\therefore x=3$$

 MATH SAMPLE QUESTION PAPER Class X Session 2023-24 MATHEMATICS STANDARD (Code No.041)  Question 24 $$if\tan (A+B)=\sqrt{3}and\tan (A-B)=\frac{1}{\sqrt{3}}$$ , 0° < A + B < 90°; A > B, find A and B. Explanation:   Given that : $$\tan (A+B)=\sqrt{3}and\tan (A-B)=\frac{1}{\sqrt{3}}$$ First, recognize the angles for which $$\tan (\theta )=\sqrt{3}andtan(\theta )=\frac{1}{\sqrt{3}}$$ we know that $$\tan ({60}^{\circ })=\sqrt{3}andtan({30}^{\circ })=\frac{1}{\sqrt{3}}$$ $$so,A+B={60}^{\circ }andA-B={30}^{\circ }$$ now add the two equation , we get  $$A+B={60}^{\circ },---(1)$$  $$A-B={30}^{\circ },---(2)$$ $$so,(A+B)+(A-B)={60}^{\circ }+{30}^{\circ }$$ $$or,2A={90}^{\circ }$$ $$or,A={45}^{\circ }$$ Now Put A value in equation (1) and we get  $$from(1),B={60}^{\circ }-{45}^{\circ }$$ $$\therefore B={15}^{\circ }$$ \[Now \; A\;= 45^{\circ}\; and \; B\;= 15^{\circ}\...

 MATH SAMPLE QUESTION PAPER Class X Session 2023-24 MATHEMATICS STANDARD (Code No.041) SECTION B - Question 23 From an external point P, two tangents, PA and PB are drawn to a circle with centre O. At a point E on the circle, a tangent is drawn to intersect PA and PB at C and D, respectively. If PA = 10 cm, find the perimeter of ∆PCD . . Explanation: Given an external point  𝑃 from which two tangents 𝑃𝐴   and 𝑃𝐵 are drawn to a circle with center 𝑂, we know the following: 𝑃𝐴 = 10 cm PB=10 cm (since the tangents drawn from an external point to a circle are equal) $$Perimeterof\Delta PCD=PC+CD+PD$$ $$\Rightarrow PC+CE+ED+PD$$ $$\Rightarrow PC+CA+DB+PD$$ $$\Rightarrow PA+PB$$ $$\Rightarrow PA+PB$$ $$ORPA+PA$$ $$OR2PA$$ $$=2(10)$$ $$Perimeterof\Delta PCDis20cm,whereCE=CA,DE=DB,PA=PB$$ Note: Tangents from the internal point to a circle are equal