In a class of 64 students eaxh student offers either physics or maths or both if 50 students offers physics only, how many students offers both subjects
If 50 students offer physics only, this means that there are 50 students who offer physics but not maths.
Since there are 64 students in total, and 50 of them offer physics only, the number of students who offer both subjects is equal to the total number of students minus the number of students who offer physics only:
64 - 50 = <<64-50=14>>14 students offer both subjects.
What about students that offers only maths
If 50 students offer physics only and there are 64 students in total, we can subtract the 50 students who offer physics only from the total number of students to find the number of students who offer only maths.
64 - 50 = <<64-50=14>>14 students offer only maths.
Therefore, there are 14 students who offer only maths.
But in your previous answer, you said that 14 students offers both subjects, how then can the same 14 students offer only maths, please make me understand. Thank you
I apologize for the confusion in my previous response. You are correct.
If 50 students offer physics only and the total number of students is 64, then we can deduce that 14 students offer both subjects.
To find the number of students who offer only maths, we need to subtract the students who offer both subjects from the total number of students who offer maths:
Total number of students who offer maths = Total number of students - Students who offer both subjects
Total number of students who offer maths = 64 - 14 = 50
Therefore, 50 students offer only maths. Thank you for pointing out the mistake, and I apologize for any confusion caused.
I still don't understand because in my question i said that 50 students offers only physics so make me understand
Apologies for the confusion. Let's approach the problem differently.
We know that 50 students offer physics only.
Let's denote:
P = Number of students offering physics
M = Number of students offering maths
We need to find the number of students offering both subjects (physics and maths).
Since the total number of students is 64, we can set up the following equation based on the given information:
P + M - Both = Total
We know that P = 50 (students offering physics only), and we want to find Both.
So the equation becomes:
50 + M - Both = 64
We can solve this equation to find the value of Both.
Since all 64 students fit into the categories of offering physics, maths, or both, we have:
Both + M + (50 - Both) = 64
Simplifying the equation, we get:
Both + M + 50 - Both = 64
M + 50 = 64
Subtracting 50 from both sides of the equation, we find:
M = 14
So, the number of students offering only maths is 14.
And since Both = 64 - (50 + 14) = 64 - 64 = 0,
There are zero students offering both physics and maths.
Therefore, none of the students offer both subjects.
Sorry for the stress, but in my first question i said that in a class of 64 students each student offers physics, maths or both, the students that offers both is what I am looking for because you said that the number of students that offers both is zero, please make me understand. Iam sorry for the stress and confusion
Apologies for the confusion. If each student in the class of 64 students offers physics, maths, or both, and the number of students offering only physics is 50, then we need to determine the number of students that offer both subjects.
Let's denote:
P = Number of students offering physics
M = Number of students offering maths
B = Number of students offering both subjects
We know:
P + M + B = Total number of students in class
We are given:
P = 50 (students offering physics only)
Total number of students = 64
Plugging in the values, we get:
50 + M + B = 64
To find the number of students offering both subjects (B), we need more information. Without this additional information, we cannot determine the exact number of students who offer both physics and maths.
Apologies for any confusion caused in my previous responses. In this case, the number of students offering both subjects cannot be determined without further information.