Question

In a two-digit number, the digit at the units place is thrice the digit in the tens place. The number exceeds the sum of its digits by 27. Find the number.

Answer 1

Hint:
Here, assume two digits numbers as 10x + y, and form the equations based on given statements in question. Then solve the two equations to get the values of x and y. Also verify the answer by putting values of x and y, in equations.

Complete step-by-step answer:
Let the number be 10x + y, where x is the tens place digit and y be the units place digit.
Given, the digit at the units place is thrice the digit in the tens place.
 y = 3x …(i)
Also given, the number exceeds the sum of its digits by 27.
10x + y = (x + y) + 27 …(ii)
Rearranging equation (iii), we get
10x + y = x + y + 27
⇒ 10x – x = 27
⇒ 9x = 27
⇒ x = $\dfrac{{27}}{9} = 3$
Putting value of x in equation (i), we have
y = 3 × 3 = 9
Therefore, the required number is 10 × 3 + 9 = 30 + 9 = 39.
Verification:
Unit digit is three times its tens digit.
i.e., 9 = 3 × 3
Also, the number exceeds the sum of its digits by 27.
Sum of digits = 3 + 9 = 12
i.e., 39 = 12 + 27
Both the statements in question are satisfied.

Note:
In these types of questions, two equations are formed using statements. Solve the equations using substitution method or elimination method. Always keep in mind to assume 2-digits numbers and 3-digits numbers. Two-digits number i.e. 10x + y and three-digits number i.e. 100x + 10y + z. Also, xy represents the product of x and y. Do not confuse between xy and 10x + y.