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.

Answer: 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 Reply

Answer:Step-by-step explanation:75