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ICSE Class X Math 2014 Paper

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Question : 44 of 52
Marks: +1, -0
A two digit positive number is such that the product of its digits is 6 . If 9 is added to the number, the digits interchange their places. Find the number.
Solution:  
Let the required two digit number be 10x+10 x+ yy
Given : xy=6x y=6 and 10x+y+9=10y+x10 x+y+9=10 y+x
  10x−x+y−10y+9  =0\;10 x-x+y-10 y+9\;=0
⇒  9x−9y+9  =0\Rightarrow\;9 x-9 y+9\;=0
⇒  x−y+1  =0\Rightarrow\;x-y+1\;=0
⇒  y  =x+1\Rightarrow\;y\;=x+1
  xy  =6   (given)   \;x y\;=6 \;\text{ (given) }\;
⇒  x(x+1)  =6\Rightarrow\;x(x+1)\;=6
⇒x2+x−6  =0\Rightarrow x^2+x-6\;=0
⇒  x2+3x−2x−6  =0\Rightarrow\;x^2+3 x-2 x-6\;=0
⇒  (x+3)(x−2)  =0\Rightarrow\;(x+3)(x-2)\;=0
⇒  x  =−3,2\Rightarrow\;x\;=-3,2
     Rejecting     x  =−3\;\;\text{ Rejecting }\;\;x\;=-3
   When   x=2,y=x+1=2+1=3\;\text{ When }\; x=2, y=x+1=2+1=3
∴\therefore The required two digit number
  =10x+y\;=10 x+y
  =10×2+3=23\;=10 \times 2+3=23
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