There are 627 students taking college algebra or calculus. 417 are taking college algebra, 261 are taking calculus, 51 are taking both algebra and calculus. Ho many are taking algebra but not calculus. Please solve and show work. When I solved my answer was 105, but I am not certain. Thank you.
[I'm wondering about the 51 who are taking calculus and algebra, because you
must know algebra before taking calculus. But that is of no concern here.]
The red circle A includes all the people who are taking algebra, regardless of
whether they are taking calculus. The blue circle C includes all who are taking
calculus regardless of whether they are taking algebra.
a = the number of students who are taking algebra BUT NOT calculus
b = the number of students who are taking both algebra AND calculus
c = the number of students who are taking calculus BUT NOT algebra
d = the number of students who are NOT taking algebra AND NOT taking calculus.
(There will be none here, but in other problems we must include the "neithers".)
51 are taking both algebra and calculus.
So b = 51. We fill that in:
417 are taking college algebra,
We have already accounted for 51 students who are taking both maths, so we
subtract 417-51 and get that 366 are taking algebra but not calculus.
So a = 366.
THAT'S THE ANSWER> BUT LET'S FINISH ANYWAY.
We fill that in:
261 are taking calculus,
We have already accounted for 51 students who are taking both maths, so we
subtract 261-51 and get that 210 are taking calculus but not algebra.
So c = 210. We fill that in:
There are 627 students taking college algebra or calculus.
We have accounted for the 366+51+210 and that adds to 627, so we have accounted
for all the stdents. So there are no students that are not taking either. So
d=0. We did not have to consider this for this problem, but as I said above,
in other problems you have to consider the "none"'s.
Answer: 366
Edwin