Mary was making sandwiches for the class picnic. 13 of the sandwiches had a slice of cheese. 14 had a slice of salami. 13 had a slice of tomato. 8 had a slice of cheese & a slice of tomato. Only 3 had a just a slice of salami. 5 had a slice of tomato, a slice of cheese, & a slice of salami. 8 had a slice of tomato & a slice of salami. How many sandwiches did Mary make for the class picnic?
To find out how many sandwiches Mary made for the class picnic, we can start by analyzing the given information.
Let's break down the information:
- 13 sandwiches had a slice of cheese
- 14 sandwiches had a slice of salami
- 13 sandwiches had a slice of tomato
- 8 sandwiches had a slice of cheese and a slice of tomato
- 3 sandwiches had just a slice of salami
- 5 sandwiches had a slice of tomato, a slice of cheese, and a slice of salami
- 8 sandwiches had a slice of tomato and a slice of salami
From the given information, we can deduce the following:
- 13 sandwiches had cheese, 8 of which also had tomato, so there were 13 - 8 = 5 sandwiches with only cheese.
- 14 sandwiches had salami, 5 of which were also with cheese and tomato, and 3 with only salami. So, 14 - 5 - 3 = 6 sandwiches had only salami.
- 13 sandwiches had a tomato, 8 of which had salami, and 5 with cheese and salami. This leaves us with 13 - 8 - 5 = 0 sandwiches with only tomato.
Now, let's calculate the total number of sandwiches:
- Sandwiches with only cheese: 5
- Sandwiches with only salami: 6
- Sandwiches with only tomato: 0
- Sandwiches with cheese and tomato: 8
- Sandwiches with cheese, salami, and tomato: 5
- Sandwiches with tomato and salami: 8
To find the total number of sandwiches, we sum up all the combinations: 5 + 6 + 0 + 8 + 5 + 8 = 32.
Therefore, Mary made a total of 32 sandwiches for the class picnic.
How does this work though?
Using Venn diagram, start filling in the information
Put 5 in the intersection of all three
C+T --> 8, but we have already accounted for 5 of those, so put 3 into the intersection of only C and T
only S --> 3, so put 3 into S which is not part of any intersection
C+S --> 8 , but we have already accounted for 5 of those, so put 3 into the intersection of only C and S
the whole C circle is 13 , but 3+3+5 or 11 are already counted, so the only C part is 2
We don't know what only S and T is, call it x
For S, x+5+3+3=14
x = 3
For T, we already have 11, but the total of T is 13
so we would need 2 for "only T" part.
Complete C the same way, I get 2 for the "only C" part
Now add up all the components,
for a total of 21 sandwiches