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Question 262003: Find a two-decimal-place number between 0 and 1 such that the sum of its digits is 9 and such that if the digits are reversed the number is increased by .27.
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! Find a two-decimal-place number between 0 and 1 such that the sum of its digits is 9 and such that if the digits are reversed the number is increased by .27.
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Let x = 10th digit
Let y = 100th digit
:
"a two-decimal-place number between 0 and 1 such that the sum of its digits is 9"
x + y = 9
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"if the digits are reversed the number is increased by .27
.1y + .01x = .1x + .01y + .27
:
multiply by 100, get rid of those annoying decimals
10y + x = 10x + y + 27
10y - y = 10x - x - 27
9y = 9x - 27
simplify, divide by 9
y = x + 3
:
In the 1st equation, x + y = 9, replace y with (x+3)
x + x + 3 = 9
2x = 9 - 3
2x = 6
x = 3
then
y = 3 + 3
y = 6
:
The two decimal number: .36
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