Question 1035728: The sum of the digits of a two-digit number is 11. When the numbers are reversed, the new number is 20 less than twice the original. What is the original number?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! let the 2 digits of the number be a and b.
a is the tens digit.
b is the units digit.
the value of the 2 digit number is 10a + b.
the sum of the digits of the 2 digit number is 11.
this means that a + b = 11.
when the numbers are reversed, the new number is 20 less than twice the original number.
this means that 10b + a = 2 * (10a + b) - 20.
simplify this to get 10b + a = 20a + 2b - 20.
subtract 20a and 2b from both sides of this equation to get 10b - 2b + a - 20a = -20.
simplify this to get 8b - 19a = -20.
from a + b = 11, solve for a to get a = 11 - b.
replace a with 11 - b in the equation of 8b - 19a = -20 to get 8b - 19 * (11 - b) = -10.
simplify this to get 8b - 209 + 19b = -20
combine like terms to get 27b - 209 = -20
add 209 to both sides of this equation to get 27b = 189.
divide both sides of this equation by 27 to get b = 189/27 = 7
you have b = 7.
since a + b = 11, this means that a = 4.
your original number is ab = 47.
when you reverse the digits, you get ba = 74.
your original statement says that when the numbers are reversed, the new number is 20 less than twice the original.
this means that 74 is equal to 20 less than 2 * 47.
2 * 47 = 94.
94 - 20 = 74.
74 is equal to 74, confirming that the solution is correct.
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