Carcass wrote:
In a certain box of cookies, \(\frac{3}{4}\) of all the cookies have nuts and \(\frac{1}{3}\) of all the cookies have both nuts and fruit. What fraction of all the cookies in the box have nuts but no fruit?
(A) \(\frac{1}{4}\)
(B) \(\frac{5}{12}\)
(C) \(\frac{1}{2}\)
(D) \(\frac{7}{12}\)
(E) \(\frac{5}{6}\)
This is an excellent question for the double matrix method!
Draw a 3x3 box.
Down the column have Nuts, No Nuts, and Total Nuts.
Across the row have Fruits, No Fruits, and Total Fruits.
The box where Total Fruits and Total Nuts meet is the Total number of Cookies.
Pick a multiple of 3 and 4 since we need to divide by both of those numbers. 60 or 120 work fine, I chose 60.
In the Total Number of Cookies box, fill in 60.
We know that \(\frac{3}{4}\) of all the cookies have nuts, so Total Nuts Box = 45.
We also know that \(\frac{1}{3}\) of all the cookies have nuts and fruit, so the Nuts and Fruits box has 20.
We're asked to find the fraction of all the cookies that have Nuts but No Fruit.
The Nuts and No Fruit Box is in between 20 and 45. You can fill in 25 into that box.
Mathematically speaking (because I don't have a sketch):
[Total Cookies with Nuts] = [Cookies with Nuts and Fruit] + [Cookies with Nuts and No Fruit]
45 = 20 + [Cookies with Nuts and No Fruit]
25 = [Cookies with Nuts and No Fruit]
We're asked to find the fraction of
all the cookies that have
Nuts but No Fruit.
All the cookies in the box = 60
Nuts but no Fruit = 25
\(\frac{25}{60}\) = \(\frac{5}{12}\)
Therefore B is our answer.
15 Jul 2019, 02:00 
83% (01:45) correct
