Question

A number consisting of a two digit number is 7 times the sum of its digit. when 27 is subtracted from the number the digits are reversed .find the number

Answer 1

Given : a number consisting of two digits is 7 times the sum of its digits. when 27 is subtracted from the number the digits are reversed. 

 

To find : the number.

 

As the number is a two digit , then let x be tens's place and y be one's place.

Therefore the two digit number would form as 10 x +y 

 

according to the question,

a number consisting of two digits is 7 times the sum of its digits.

=> 10 x + y = 7( x+y) 

=> 3x-6y=0

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

 

when 27 is subtracted from the number the digits are reversed.

After reversing the digits, the number = 10 y +x 

=> (10 x+ y )- 27 = 10y + x

=> 9 x- 9y = 27

=>x -y = 3 ...............(2)

 

Substituting the value of x from eq (1) to eq(2) , we get

=> 2y-y = 3

=> y = 3

 

x= 2y

=> x = 2(3) = 6

Therefore the number is =  10 (6) + 3 = 63


Answer : the required two digit number is 63

Answer 2

Let the unit place digit be x .
and tens place digit be y.
then,the no. formed = 10y+x
=(10y+x)= 7(x+y)
= 10y+x= 7x+7y
= 3y-6x= 0
= 3y= 6x
= y= 2x.........(i)
No. formed after reversed= 10x+y
10y+x-27= 10x+y
9y-9x= 27
y-x= 3
putting value of x from equation (i) in (ii)
2x-x= 3
x= 3
so ,y= 6
hence ,the no. obtained = 63