SOLUTION: Given that {{{ a + 1/ (b+2/c) = 17/5 }}}, where a is the integer part of the mixed number, and c is the numerator of the fraction part of the mixed number when the mixed number is

Algebra ->  Percentage-and-ratio-word-problems -> SOLUTION: Given that {{{ a + 1/ (b+2/c) = 17/5 }}}, where a is the integer part of the mixed number, and c is the numerator of the fraction part of the mixed number when the mixed number is       Log On


   



Question 1198342: Given that +a+%2B+1%2F+%28b%2B2%2Fc%29+=+17%2F5+, where a is the integer part of the mixed number, and c is the numerator of the fraction part of the mixed number when the mixed number is written in the least terms, find the product ABC.
A) 9
B) 18
C) 6
D) 30
E) 3
@josgarithmetic(37819), if you don't know how to answer this question, then don't say anything.

Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
+a+%2B+1%2F+%28b%2B2%2Fc%29+=+17%2F5+

Change the 17%2F5 to mixed number 3%262%2F5

+a+%2B+1%2F+%28b%2B2%2Fc%29+=+3%2B2%2F5+

Since a is the whole part, a = 3, so subtract 'a' from the
left side and '3' from the right side:

1%5E%22%22%2F+%28b%2B2%2Fc%29+=+2%2F5+

Since c is the numerator of the fraction part of the mixed number,
c=2

1%5E%22%22%2F+%28b%2B2%2F2%29+=+2%2F5+

1%2F+%28b%2B1%29+=+2%2F5+


Cross-multiply:

5=2%28b%2B1%29

5=2b%2B2

3=2b

3%2F2=b

So a=3, b=3%2F2, c=2, and 

Edwin