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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.

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Answer:

75

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

User Gabe Hackebeil
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