Question

A number of two digits exceeds four times the sum of its digits by 6 and it is increased by 9 on reversing the digits. find the number​

Answer 1

$\huge\red{\mathbb{Answer}}$

See in attachment

Answer 2

$\sf\large\underline{Let:}$

• $\rm{The\: ones\: digit=x}$
• $\rm{The\: tens\: digit=y}$
• $\rm{The\: Number=10y+x}$

$\sf\large\underline{To\: Find:}$

• $\rm{The\:Number=?}$

$\sf\large\underline{Solution:}$

$\sf\underline{Given\:in\:1st\:case:}$

• A number of two digits exceeds four times the sum of its digits by 6]

$\rm{\implies 10y+x=4(x+y)+6}$

$\rm{\implies 10y+x=4x+4y+6}$

$\rm{\implies 4x-x+4y-10y=-6}$

$\rm{\implies 3x-6y=-6}$

• Dividing by 3 on both sides]

$\rm{\implies x-2y=-2-------(1)}$

$\sf\underline{Given\:in\:2nd\:case:}$

• When it is increased by 9 then the digit is reversed]

$\rm{\implies 10y+x+9=10x+y}$

$\rm{\implies 10x-x+y-10y=9}$

$\rm{\implies 9x-9y=9}$

• Dividing by 9 on both sides]

$\rm{\implies x-y=1-----(2)}$

• Now, solving equation I and 2]

$\rm{\implies x-2y=-2}$

$\rm{\implies x-y=1}$

• By solving we get here]

$\rm{\implies -y=-3}$

$\rm{\implies y=3}$

• Putting the value of y=3 in eq 1]

$\rm{\implies x-2y=-2}$

$\rm{\implies x-2\times\:3=-2}$

$\rm{\implies x-6=-2}$

$\rm{\implies x=-2+6}$

$\rm{\implies x=4}$

$\sf\large{Hence,}$

$\rm{\implies The\:Number=10y+x}$

$\rm{\implies The\: Number=10\times\:3+4}$

$\rm{\implies The\: Number=30+4}$

$\rm\purple{\implies The\: Number=34}$