In a group of 30 students, 18 are enrolled in an English class and 16 are enrolled in an Algebra class. How many students are enrolled in both an English and an Algebra class?
(1) 20 are enrolled in exactly one of these two classes.
(2) 3 are not enrolled in either of these classes.
Answer: D
Please explain...thank you in advance.
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- DanaJ
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Well, let's see what we've got here:
First off, let's use some notations:
a = people enrolled in Algebra only
e = people enrolled in English only
ae = people enrolled in both classes
n = people that aren't enrolled in anything.
So you get that:
a + e + ae + n = total = 30 students
a + ae = 18 (the students enrolled in Algebra are the students who are enrolled only in Algebra plus the students enrolled in both classes)
e + ae = 16
Now, let's use each stmt:
1. This one tells us that a + e = 20. You've now got a total of three equations with three unknowns, so you can solve the system:
a + ae = 18
e + ae = 16
a + e = 20.
Solve this one and you get a, e, ae.
So 1 is sufficient.
2. This tells us that n = 3, which in turn leads us to a + e + ae = 30 - 3 = 27. Again, here's your third equation for the system:
a + ae = 18
e + ae = 16
a + e + ae = 27.
Solve for a, e and ae.
So 2 is sufficient as well.
This is why the answer is D
(I have been asked to use the spoiler function in my posts so some other people can try and solve the problems on their own).
First off, let's use some notations:
a = people enrolled in Algebra only
e = people enrolled in English only
ae = people enrolled in both classes
n = people that aren't enrolled in anything.
So you get that:
a + e + ae + n = total = 30 students
a + ae = 18 (the students enrolled in Algebra are the students who are enrolled only in Algebra plus the students enrolled in both classes)
e + ae = 16
Now, let's use each stmt:
1. This one tells us that a + e = 20. You've now got a total of three equations with three unknowns, so you can solve the system:
a + ae = 18
e + ae = 16
a + e = 20.
Solve this one and you get a, e, ae.
So 1 is sufficient.
2. This tells us that n = 3, which in turn leads us to a + e + ae = 30 - 3 = 27. Again, here's your third equation for the system:
a + ae = 18
e + ae = 16
a + e + ae = 27.
Solve for a, e and ae.
So 2 is sufficient as well.
This is why the answer is D
(I have been asked to use the spoiler function in my posts so some other people can try and solve the problems on their own).