Question

Mousumi divides a two digit number with the sum of its digits and gets quotient as 6 and
remainder as 6. But if she divides the number interchanging the digits with the sum of its
digits she will get quotient as 4 and remainder as 9. Let us determine the number that
Mousumi has taken by forming simultaneous equations.​

Answer 1

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$\sf Let \: the \: unit \: digit \: be \: x \: and \: tenth \: digit \: be \: y$

$\sf original \: number = 10x+y$

$\sf interchanged \: number = 10y+x$

$\sf in \: case -1$

$\sf Dividend= 10x+y$

$\sf Divisor=x+y$

$\sf Quotient=6$

$\sf Remainder= 6$

$\sf\blue{To \: be \: noted:}$

$\sf {\boxed {\bf Dividend {\bf = {\bf Divisor {\bf × {\bf Quotient {\bf+{\bf remainder}}}}}}}}$

$\sf 10x+y=6(x+y)+6$

$\sf ⟹10x+y= 6x+6y+6$

$\sf ⟹10x-6x+y-6y-6=0$

$\sf ⟹4x-5y-6=0-------(i)$

$\sf in \: case -2$

$\sf Dividend= 10y+x$

$\sf Divisor=x+y$

$\sf Quotient=4$

$\sf Remainder=9$

$\sf ⟹10y+x=4(x+y)+9$

$\sf ⟹10y+x=4x+4y+9$

$\sf ⟹10y-4y+x-4x-9=0$

$\sf ⟹6y-3x-9=0$

$\sf ⟹3(2y-x-3)=0$

$\sf ⟹(2y-x-3)= 0/3 (taking \: common)$

$\sf ⟹(2y-x-3)=0 ---------(ii)$

$\sf from \: both \: equations \: i \: and \: ii$

$\sf multiplying \: equation \: ii \: by \: 4$

$\sf 4x-5y-6=0$

$\sf -4x+8y-12=0 \: \: \: \: \: \: [ by \: method \: of \: elimination].$.

______________ $\sf (by \: adding)$

$\sf 0x+3y-18=0$

$\sf ⟹3y=18$

⟹y=$\dfrac{18}{3}$

$\sf ⟹y=6$

$\sf putting \: the \: value \: of \: y \: in \: equation \: 1, \: we \: get$

$\sf 4x-5×6-6=0$

$\sf ⟹4x-36=0$

$\sf ⟹4x=36$

⟹x=$\dfrac{36}{4}$

$\sf Therefore, \: the \: required \: no. \: is \: 96$