Question

a two-digit number is 3 more than 4 times the sum of its digits.if 18 is added to the number,its digits are reversed.find the number

Answer 1

$let \: the \: digit \: in \: ones \: place \: be \: x \: and \: tens \: place \: be \: y \\ original \: number \: = 10y + x \\ given \\ 10y + x = 4(x + y) + 3 \\ 10y + x - 4x - 4y = 3 \\ 6y - 3x = 3 \\ 2y - x = 1 \: \: \: \: \: \: \: ....(1) \\ given \\ 10y + x + 18 = 10x + y \\ 9x - 9y = 18 \\ x - y = 2 \: \: \: \: \: \: ...(2) \\ add \: (1) \: and \: (2) \: we \: get \\ 2y - x + x - y = 1 + 2 \\ y = 3 \\ put \: y = 3 \: in \: equ.(1) \\ 2(3) - x = 1 \\ 6 - x = 1 \\ - x = 1 - 6 \\ - x = - 5 \\ x = 5 \\ hence \: the \: required \: number = 35$

Answer 2

Answer:

Step-by-step explanation:

let units digit be y adn tens be 10x

atq,

number=10x+y

∵

(10x+y) - 4(x+y)= 3.....(i)

and

(10x+y)+ 18=10y+x.....(ii)

Solving (i)

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

or,6x-3y=3

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

Solving (ii)

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

or, 9y-9x=18

or,y-x=2...(2)

(1)+(2)

2x-y+y-x=1+2

or,x=3

Thus, y=x+2 (from (2))

y=5

∴ the number= 10x+y=10x3+5=35