Question

A two digit number is less by 8 than 5 times the sum of its digit and a number obtained interchanging the digit is greater by 27 than the original number. Find the original number?​

Answer 1

Answer:

let the digit at tens place be X and digit at ones place be Y,Then:

the number = 10X + Y

According To the question ,

( 10X + Y )  = 5 *( X + Y ) - 8

or, 10X + Y  = 5X + 5Y - 8

or, 10X - 5X + Y - 5Y = -8

or, 5X - 4Y = -8  equn (1)

According To the Second condition,

10Y + X = 27 + ( 10X + Y )

or, 10Y - Y + X - 10X = 27

or, 9Y -9X = 27

or, 9( Y -X ) = 27

or, X  = Y -3  equn(2)

Again, Putting The Value of X in equn (1) We get,

or, 5*(Y -3) - 4Y = -8

or, 5Y - 15 - 4Y + 8 = 0

Therefore, Y  = 7 and X = 7 - 3 = 4

Therefore , The required Number is 10*4 + 7 = 40 + 7 = 47 Answer

Step-by-step explanation: