Thursday

December 18, 2014

December 18, 2014

Posted by **angleica** on Thursday, February 14, 2008 at 9:40pm.

- algebra 1 -
**Reiny**, Thursday, February 14, 2008 at 10:45pmlet the number of reds be r, let the number of whites be w

r + w = 12 and r<w and both r and w must be whole numbers

let the price of a white be x cents, then the price of a red is x+3 cents.

clearly x also must be a whole number.

then r(x+3) + wx = 129

rx + 3r + wx = 129

x(r+w) = 129-3r, but r = 12-w

x(12-w+w) = 129-3r

x = (129-3r)/12

we stated that x had to be a whole number, and the only choices for r are:

0,1,2,3,4,5

the only value which gives a whole number for x is when r = 3

then w = 9, and x = 10

so a white costs 10 cents, and a red costs 13 cents.

**but 3(10) + 9(13) is not equal to 129**

so your data is inconsistent and there is no solution to your question.

- algebra 1 -
**Reiny**, Thursday, February 14, 2008 at 10:49pmArhgghhh!!!!

Ignore the last two lines of my previous posts, I subbed in the price for the wrong colours.

Obviously 3(13) + 9(10) = 129

So there were 3 reds and 9 whites.

- algebra 1 -

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