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Question 1176416: The local takeaway shop sells hamburgers with the choice of tomato sauce, cheese and pickles. On a certain day they sold 768 hamburgers; 420 had cheese, 429 had tomato sauce, 252 had tomato sauce and pickles, 186 had cheese and pickles and 60 had none.
a) Draw a neat Venn diagram to illustrate this data.
b) How many hamburgers were sold with pickles only?
c) How many hamburgers were sold with no pickles?
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
Unfortunately, there isn't enough info. Here's why:
Start by drawing out the Venn Diagram like so
We have
U = universal set = set of all hamburgers sold
C = cheese
T = tomato sauce
P = pickles
The letters in blue (a through h) represent the 8 various regions of which we can place a number.
More specifically we have these region definitions- a = number of hamburgers sold that had cheese only (no tomato sauce, no pickles)
- b = number of hamburgers sold that had cheese and tomato sauce (but no pickles)
- c = number of hamburgers sold that had tomato sauce only (no cheese, no pickles)
- d = number of hamburgers sold that had cheese and pickles (but tomato sauce)
- e = number of hamburgers sold that had all three ingredients (cheese, tomato sauce, pickles).
- f = number of hamburgers sold that had tomato sauce and pickles (but no cheese)
- g = number of hamburgers sold that had pickles only (no cheese, no tomato sauce)
- h = number of hambugers that had none of the three ingredients (no cheese, no tomato sauce, no pickles)
Of course, it is assumed that every burger sold has two buns and meat of some kind.
Your teacher then provides you with these six facts- There are 768 hamburgers sold in total
- There are 420 that had cheese
- There are 429 that had tomato sauce
- There are 252 that had tomato sauce and pickles
- There are 186 that had cheese and pickles
- There are 60 that had none (no cheese, no tomato sauce, no pickles)
We can then translate each of those six sentences into six mathematical equations like so
Fact 1) a+b+c+d+e+f+g+h = 768
Fact 2) a+b+d+e = 420
Fact 3) b+c+e+f = 429
Fact 4) e+f = 252
Fact 5) d+e = 186
Fact 6) h = 60
So we have this system of equations
We have 6 equations and 8 variables. If we ignore h = 60, then we have 5 equations and 7 variables.
We would need to have the same number of equations as variables in order to solve for each value. Otherwise, we'll get more than one possible solution.
For instance, consider this smaller system of equations
x+y+z = 10
x+y-z = 4
There are 2 equations and 3 variables. This system has more than one solution.
Adding those two equations leads to
2x+2y = 14
which reduces to
x+y = 7
and solves to
y = -x+7
You would also find that z = 3 through substitution
So we have infinitely many solutions of the form (x,y,z) = (k,-k+7,3)
k is the free variable and could be any real number, so that's why there are infinitely many solutions.
This smaller example can be used to see why we don't have enough information here. The current set of information leads to more than one possible answer.
Here is a small subset of possible solutions for the values a,b,c,e,f,g,h
a = 125, b = 109, c = 68, d = 14, e = 172, f = 80, g = 140, h = 60
a = 124, b = 110, c = 67, d = 14, e = 172, f = 80, g = 141, h = 60
a = 123, b = 111, c = 66, d = 14, e = 172, f = 80, g = 142, h = 60
a = 122, b = 112, c = 65, d = 14, e = 172, f = 80, g = 143, h = 60
I'll let you check that each row satisfies the 6 equations mentioned.
Many other solutions are possible.
Since there isn't enough information given, you'll have to contact your teacher to ask for clarification on this problem.
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