Question

A two digit number is 3 more than 4 time the sum of its digit of 18 is added to the number the digit are revese find the number​

Answer 1

$\textbf{\underline{\underline{According\:to\:the\:Question}}}$

Assumption

Tens place be p

Unit place be n

$\textbf{\underline{Two\;digit\;no.\;is\;3\;more\;than\;4 \;times\;the\;sum\;if\;its\;digits}}$

10p + n = 4(p + n) + 3

10p + n = 4p + 4n + 3

10p - 6p + n - 4n = 3

4p - 3n = 3

$\large{\boxed{\sf\:{2p-n=1.....(1)}}}$

$\textbf{\underline{If\;18\;added\;to\;the\;no.\;its\;digits\; are\;reversed}}$

10p + n + 18 = 10n + p

10p - p + n - 10n = -18

9p - 9n = -18

$\large{\boxed{\sf\:{p-n=-2.....(2)}}}$

$\textbf{\underline{Subtracting\;(2)\;from\;(1)}}$

-p = -3

p = 3

$\large{\boxed{\sf\:{Now}}}$

2p - n = 1

2 × 3 - n = 1

6 - n = 1

-n = -5

n = 5

$\large{\boxed{\sf\:{Therefore}}}$

$\large{\boxed{\sf\:{The\;Number}}}$

= 10p + n

= 10 × 3 + 5

= 35

$\Large{\boxed{\sf\:{The\; Number\;is\;35}}}$

Answer 2

ANSWER:-

Given:

A two digit number is 3 more than 4times the sum of its digit of 18 is added to the number the digit are reverse.

To find:

Find the number.

Solution:

Let the unit digit be 'x'.

Let the digit at ten's place be 'y'.

The original number will be 10y+x

Given,

Number is 3 more than 4times the sum of its digits.

=) 10y +x =4(x+y)+3

=) 10y +x = 4x + 4y +3

=) 10y-4y + x-4x =3

=) 6y -3x = 3

=) 2y -x = 1...........(1)

Also,

If the digits are interchanged,

Reversed number will be = 10x +y

As,

reversed number exceeds the original number by 18.

=) (10x+y) - (10y+x)= 18

=) 10x +y -10y -x = 18

=) 9x -9y= 18

=) x - y = 2

=) 2y -1- y=2 [using equation(1)]

=) y= 2+1

=) y= 3

Eliminate the value of y in equation (1), we get;

=) 2y -x =1

=) 2(3) -x = 1

=) 6 - x = 1

=) -x = 1-6

=) -x = -5 [Minus Cancel]

=) x= 5

Hence,

The original number is 10y + x;

=) 10(3) +5

=) 30 +5

=) 35.

Hope it helps ☺️ [rajmaur]