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
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
Log On
Question 1198341: Given that , 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 Found 5 solutions by josgarithmetic, MarkSingh-, math_tutor2020, MathTherapy, greenestamps:Answer by josgarithmetic(39617) (Show Source):
You can put this solution on YOUR website! The way the equation appears and the description of it doe not match.
THE DESCRIPTION DOES NOT CORRESPOND TO THE EQUATION.
You can put this solution on YOUR website!
Rewrite the improper fraction 17/5 into its mixed number format.
17/5 = (15+2)/5
17/5 = (15/5)+(2/5)
17/5 = 3+(2/5)
17/5 = 3 & 2/5
'a' is the integer part, or whole part, of the mixed number.
Therefore a = 3.
c is the numerator of the fractional part when the fraction is fully reduced.
We have c = 2.
Let's simplify the left hand side of the given equation.
Then plug in a = 3 and c = 2.
Also, let's replace the 17/5 with 3 + 2/5 we found earlier.
It's clear that the fractional part must be equal to 2/5.
Furthermore, the denominator must be 5.
You can put this solution on YOUR website! Given that , 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
We see clearly that a = 3
In addition, since c is the numerator of the reduced fraction in the mixed number, then c = 2
Now, we need to change the 2 in the numerator in the mixed number on the right () to MATCH
the 1 in the numerator in the mixed number on the left .
How do we achieve that?
We simply divide the 2 in the numerator in the mixed number on the right by 2 to get 1. We then do the same to the
denominator (divide it by 2), as follows:
We now see that: ---- Substituting 2 for c
5 = 2b + 2 --- Denominators are equal and so are the numerators
With , we finally get:
=======
CHECK
=======