Question

The digit in in tens place of two digit number is more than its digit in the the units place by 5 the sum of of its digit is 1/8 of the number. Find the number???

Answer 1

Answer :

72

Solution :

Let the tens digit and the unit digit of the required two digits number be x and y respectively .

Thus ,

The required number = 10x + y

Now ,

According to the question ,

The tens digit of the required number is 5 more than its unit digit .

Thus ,

x = y + 5 ----------(1)

Also ,

It is given that ,

The sum of both the digits of the number is ⅛ of the number .

Thus ,

=> x + y = ⅛ of (10x + y)

=> x + y = ⅛•(10x + y)

=> 8(x + y) = 10x + y

=> 8x + 8y = 10x + y

=> 8x + 8y - 10x - y = 0

=> 7y - 2x = 0 ----------(2)

Now ,

Putting x = y + 5 in eq-(2) , we get ;

=> 7y - 2x = 0

=> 7y - 2(y + 5) = 0

=> 7y - 2y - 10 = 0

=> 5y - 10 = 0

=> 5y = 10

=> y = 10/5

=> y = 2

Now ,

Putting y = 2 in eq-(1) , we get ;

=> x = y + 5

=> x = 2 + 5

=> x = 7

Hence ,

The required number is 72 .