Beth R. answered 03/18/16
Tutor
2
(1)
Biology & Chemistry Degrees w/ Teaching & Tutoring Experience
Hi Jose,
The trial and error method will take forever to answer this question; your best bet is to use simple algebra.
We know that we have 124 total pounds of coffee and some cost $4.35, some cost $5.45 and the total cost was $602.10.
So as an equation, it looks like this:
(Pounds of coffee purchased for $4.35)*$4.35 + (Pounds of coffee purchased for $5.45)*$5.45 = $602.10
What we need to do now is find a way to express the pounds of coffee in terms of x to get a solvable equation.
We know that the total number of pounds purchased was 124, so if we define x as the number of pounds purchased for $4.35, we know that the number purchased for $5.45 will be 124-x.
If we substitute these values into the equation we wrote above, we get:
(x)*$4.35 + (124-x)*5.45=602.10
Now we use the distributive property to simply the equation:
4.35x + 675.8-5.45x=602.10
Now Combine like terms:
(4.35x-5.45x) + 675.8 = 602.10
-1.1x+675.8=602.10
subtract 675.8 from both sides:
-1.1x=-73.7
Divide both sides by -1.1, and you get x=67.
Remember our definitions from above:
Pounds purchased at $4.35=x
Pounds purchased at $5.45=124-x
So, pounds purchased at $4.35 =x =67
Pounds Purchased at $5.45 = 124-x = 57.
When solving problems like this, it does not matter which term you define as x; if we had defined pounds purchased at $5.45 as x and pounds purchased at $4.35 as 124-x, we would have had the same answer:
(x)*$5.45 + (124-x)*$4.35 = 602.10
5.45x+539.4-4.35x=602.10
1.1x+539.4=602.10
1.1x=62.7
x=57
Pounds at $4.35 = 124-x = 67
Pounds purchased at $5.45 =x=57.
Using algebra to solve these types of problems is quicker than guessing and checking, and if your teacher expects you to show your work you will not get credit if you use guess and check. Also, the answers are not always whole numbers, so you could be doing a whole lot of checking if you don't use the algebraic method to solve these.
Please feel free to message me if you have any questions.
Thanks
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