Question

A two-digit number has two less units than tens. The difference between twice the number and the number reversed is 93. Find the number.

1. thugiang
75
Step-by-step explanation:
Let the unknown number be AB where A is the digit in the tens place and B the digit in the units place
The actual value of the number is computed as 10A + 1B because the tens place has positional value of 10 and units place has positional value of 1
We are given B = A – 2 or alternatively, A = B + 2
The number reversed is BA which has the value 10B + A
The second fact about the number is translated into the equation
2(10A + B) – (10B + A) = 93
Expanding the brackets we get
20A + 2B -10B – A = 93
Simplifying and collecting the unknowns on the LHS gives us
19A – 8B =93
Substitute A in terms of B where A = B + 2
19(B+2) – 8B = 93
19B + 38 – 8B = 93
11B = 93-38=55
So B = 55/11 = 5 and A = 5 + 2 = 7
So the original number is 75
We can verify this using the second fact
Twice the original number = 2(75) = 150
75 reversed is 57
And we have 150-57 = 93