Jeremy made a Venn diagram showing the number of students in his class who own types of pets. There are 32 students in his class. In addition to the information in the Venn diagram, Jeremy knows half of the students have a dog, $\frac{3}{8}$ have a cat, six have some other pet and five have no pet at all. How many students have all three types of pets (i.e. they have a cat and a dog as well as some other pet)?

The Venn diagram is attached in the figure.

Total number of students = 32
Number of students have a dog = 0.5 *32 = 16
Number of students have a cat = (3/8) * 32 = 12
Number of students have other pets = 6
Number of students have no pets = 5

From the Venn diagram:
Number of students have a dog = x+y+z+10
Number of students have a cat = x+z+w+9
Number of students have other pets = y+z+w+2

Total number of students have no pets = 32 – (x+y+z+w+21)

So, as a conclusion for the above information

x+y+z+10 = 16 (1)

x+z+w+9 = 12 (2)

y+z+w+2 = 6 (3)

32 – (x+y+z+w+21) = 5 (4)

By solving the system of equations to find x , y , z , w

x = 2 , y = 3 , z = 1 , w = 0

By using Venn diagram

Number of students have all three types of pets = z = 1 student

MindTrickery MindTrickery · Answer 2 · 2021-04-11T02:46:32+02:00

Answer:

1

Step-by-step explanation:

Let's first look at the four things the question tells us: First, the number of dogs is x + y + z + 10 = 32/2, so x + y + z = 6. The number of cats is w + x + z + 9 = 32 x 3/8, so w + x + z = 3 Since 6 people have other pets, w + y + z = 4. The total number of people with pets is w + x + y + z + 2 + 9 + 10 = 32 - 5, so w + x + y + z = 6.

This gives us the system

x + y + z = 6

w + x + z = 3

w + y + z = 4

w + x + y + z = 6.

If we subtract x + y + z = 6 from w + x + y + z = 6 we get w = 0. Similarly, we find y = 3 and x = 2. Plugging these values into any of the equations leads to z = 1. Thus, there is 1 student with cats, dogs, and other pets.