Aclass of 51 students,25 offer mathematics,21 offer phyiscs,31 offer chemistry,12 offer mathematics and chemistry,15 offer phyiscs and mathematics and 10 offer phyiscs and chemistry.Find the number of students who offers all 3 subjects and who offer mathematics only, chemistry only, phyiscs only and even draw avenn diagram for them.
It would help if you would proofread your work before you post it.
Since when do students "offer" classes?
Also, we cannot post a Venn diagram here.
x do all three
Math alone = 25 - (12-x) - (15-x) - x = x - 2
Chem alone = 31 - (12-x) - (10-x) -x = x + 9
Physics alone = 21 - (15-x) - (10-x) -x = x - 4
sum of all
51 =x-2 +x+9 +x-4 + 15-x +12-x +10-x + x
51 = x + 40
x = 11 took all
check it and go back and get the rest
By the way, I drew a Venn diagram.
I think this usage of "offer" is prevalent in Uganda and maybe some other African nations where English has evolved its own dialects.
To find the number of students who offer all three subjects (mathematics, physics, and chemistry) and who offer mathematics only, chemistry only, and physics only, we can use the principles of set theory and Venn diagrams.
Let's start by creating a Venn diagram. Draw three overlapping circles and label them as mathematics (M), physics (P), and chemistry (C).
Now, let's fill in the given information:
- We are told that 25 students offer mathematics (M), 21 offer physics (P), and 31 offer chemistry (C).
- We know that 12 students offer both mathematics and chemistry (M ∩ C), 15 offer mathematics and physics (M ∩ P), and 10 offer physics and chemistry (P ∩ C).
We can use the information above to fill in the overlapping regions of the Venn diagram.
Let's calculate the number of students who offer all three subjects (M ∩ P ∩ C). To do this, we need to find the intersection of all three circles. We know that:
- M ∩ C = 12
- M ∩ P = 15
- P ∩ C = 10
To find M ∩ P ∩ C, we need to subtract the number of students who offer M ∩ C, M ∩ P, and P ∩ C from the smallest circle, which is physics (P). Therefore:
P - (M ∩ P) - (P ∩ C) = 21 - 15 - 10 = 21 - 25 = -4
We have a negative number, which doesn't make sense in this context. This indicates that there might be an error in the given information. Please double-check the provided numbers or check for any inconsistencies.
However, we can still find the number of students who offer mathematics only, chemistry only, and physics only by using the Venn diagram:
To find the number of students who offer mathematics only (M - (M ∩ C) - (M ∩ P)):
M - (M ∩ C) - (M ∩ P) = 25 - 12 - 15 = 25 - 27 = -2
To find the number of students who offer chemistry only (C - (M ∩ C) - (P ∩ C)):
C - (M ∩ C) - (P ∩ C) = 31 - 12 - 10 = 31 - 22 = 9
To find the number of students who offer physics only (P - (M ∩ P) - (P ∩ C)):
P - (M ∩ P) - (P ∩ C) = 21 - 15 - 10 = 21 - 25 = -4
Again, we have negative values, which indicates an inconsistency in the given information.
Please review the provided numbers or make sure there are no contradictions or errors in the question.