Class X Session 2023-24 Question 28

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 &\\ & now \; put\; a \; value\; in \; eq\; (3) we\; get\; b = 2 &\\ & the \;number\; is \;given\; is\; by\; putting\; values \;(a) \;and \;(b) in \; eq\; (3) \; (4+10 \times 2) = 24 &\\ & This \; is \; case \;1\; or\; other\; case \;would\; be \; 42 &\\ & Hence \; there\; are\; 2 \; such\; number\; \end{flalign}