Question 128157This question is from textbook Structure and Method Book 1
: When the digits of a two-digit number are reversed, the new number is 9 more than the original number, and the sum of the digits of the original number is 11. What is the original number?
This question is from textbook Structure and Method Book 1
Answer by ozymandias(9)   (Show Source):
You can put this solution on YOUR website! Let the two digit number be written in the form yz.
Or - to put it another way, since the "y" is in the "tens" column and the "z" is in the "units" column, this number can be written as :
10y + z
The number reversed is therefore :
10z + y (ie putting the "tens" in the "units" column and vice versa)
The difference (ie subtract the first from the second) is 9z - 9y = 9(z-y)
Now, we are told that this equals 9 , therefore z-y equals 1. ie z is one more than y.
We also are told that z + y = 11
You can now solve this by using simultaneous equations, or alternatively think of two consecutive numbers which add up to 11. (ie 5 and 6)
So the numbers are 56 (which becomes 65)
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