Question

if a two digit number is 4 times of sum of its digit and number obtained by interchanging the place is 9 less than twice of original number then find original number

Answer 1

hope this answer will help you

Answer 2

let the number be xy

tens place = 10x 

ones plce = y

 two digit number is 4 times of sum of its digit

s0,

10x +y = 4 ( x +y)

10x +y = 4x + 4y

-4y - y = 4x - 10x

-3y = -6x


remove - sign

y = 2x

now

number obtained by interchanging the place is 9 less than twice of original number

so

10y +x = 2(10x +y ) - 9

10y + x = 20x + 2y -9

now since y = 2x

10y + x = 20x + 2y - 9

=> 10(2x) + x = 20x + 2(2x) - 9

=>20x +x = 20x + 4x - 9

=>21x = 24x - 9

=>-21x - 24x = -9

=>  -3x = -9 

x = -9/-3

x = 3


now

y = 2x

=>2*3

=>6

so original no. is 36

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