Q2 of 227 Page 79

Formulate the following problems as a pair of equations, and hence find their solutions:

A two digit number is seven times the sum of its digits. The number formed by reversing the digits is 18 less than the given number. Find the given number.

Let x denote the digit in the tenth place and y denote the digit in the unit place. So the number may be written as 10x + y.


When digits are reversed, the number becomes 10y + x.


First condition is


10x + y = 7 (x + y)


10x + y – 7x – 7y = 0


3x – 6y = 0


x – 2y = 0


Second condition is


10y + x = (10x + y) – 18


10y + x – 10x – y + 18 = 0


– 9x + 9y + 18 = 0


– x + y + 2 = 0


For cross multiplication method, we write the coefficients as



Hence, we get = =


= =


= =


x = = 4


y = = 2


The number is 10x + y = 10 (4) + 2 = 40 + 2 = 42.


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