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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 : \[a =  \; \frac{1}{\sqrt{x}} \; and \; b = \frac{1}{\sqrt{y}} = -1\]  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) \[2\times (2a+3b) - (4a-9b) = 4-(-1)\] \[ or \; 4a+6b -4a + 9b = 4+1\] \[ or \; 15b = 5\] \[ or \; b = \frac{5}{15}\] \[ or \; b = \frac{1}{3}\] Now, substitute b in equation (1) \[\Rightarrow 2a + 3\times \frac{1}{3} = 2\] \[\ or\; 2a + 1 = 2\] \[\ or\; 2a = 1\] \[\therefore a = \frac{1}{2}\] Now we have a and b so we can find the values of x and y  \[\ a = \frac{1}{\sqrt{x}} \Rightarrow \frac{1}{2} = \frac{1}{\sqr

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 &\\ &am

 MATH SAMPLE QUESTION PAPER Class X Session 2023-24 MATHEMATICS STANDARD (Code No.041) Question 27 \begin{flalign} & if\;,\;\alpha\; \beta \; are \; Zeroes \; of \; quadratic\;polynomial\; 5x^2 + 5x + 10 \;find \; the\; value\; of\; (1) \;\alpha^2\;+ \beta^2 \; ,\;(2)\;\alpha^-{1} + \beta^-{1} &\ \end{flalign} 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 = \alpha^2+\beta^2