SOLUTION: The sum of the digits of a two-digit counting number was 11. When the digits were reversed, the new number was 10 more than one-half the original number. What was the original nu

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Question 82606: The sum of the digits of a two-digit counting number was 11. When the digits were reversed, the new number was 10 more than one-half the original number. What was the original number?
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
Let x = the 10's digit; let y = the units digit
:
"The sum of the digits of a two-digit counting number was 11."
x + y = 11
or x = (11 - y)
:
"When the digits were reversed, the new number was 10 more than one-half the original number."
10y + x = .5(10x + y) + 10
:
What was the original number?
:
10y + x = .5(10x + y) + 10
10y + x = 5x + .5y + 10
10y -.5y = 5x - x + 10
9.5y = 4x + 10
replace x with (11-y)
9.5y = 4(11-y) + 10
9.5y = 44 - 4y + 10
9.5y + 4y = 54
13.5y = 54
y = 54/13.5
y = 4
:
find x:
x = 11-y
x = 11-4
x = 7
:
The number: 74
:
:
Check using:
"When the digits were reversed, the new number was 10 more than one-half the original number."
47 = .5(74) + 10
47 = 37 + 10