In a group of 30 students, 8 are enrolled in an English

Question

In a group of 30 students, 8 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.

ggarr · Answer

1) insuff. how many are enrolled in none?

2) 3 enrolled in none. how many are enrolled in only 1? again, insufficient

choice C tells us that 20 are enrolled in exactly 1 class and that 3 are enrolled in none, 30 (total students) - 20 (students in exactly 1 class) - 3 (students enrolled in neither class) = 7 enrolled in both classes

somethingbetter · Answer

That's what I figured out. But OA says
--- d. EACH statement ALONE is sufficient to answer the question asked.

Any more takers for this puzzle?

It's from Peterson's online test. There is one more question where I do not agree with OA, Please look at the today's post with subject "Guess OA is wrong"

somethingbetter · Answer

ggarr, you said 7 are enrolled in both the classes, means these 7 students should belong to the numbers given for both classes i.e. 16+8=24. If yes, then what justifies 20 students taking only one class which also come from 16+8= 24 ?

ggarr · Answer

I think I was careless the first time around. This is my second go.
Stem: 24 are enrolled in E, A or E and A. 6 are in neither.
1) if 20 are in A or E and 6 are in neither then 4 are in both
2) 30 students in total, 24 are enrolled in E, A or E and A and 3 are in neither. 30 - 3 = 27 students in E, A or E and A. insufficient and inconsistent

now I'm getting A. hmmmm

ggarr · Answer

I completely overlooked the 16+8=24 part. please ignore that post.

somethingbetter · Answer

Still, OA is D , my friend.

IrinaOK · Answer

To find how many students are enrolled in both we can use any of the following two formulas:

Both=ENG+ALG+none-Total
or
Both=Total-(ENG(only)+ALG(only))-none

Let`s plug in what is given in the question, we get;

Both=8+16+none-30
or
Both=30-(ENG(only)+ALG(only))-none, these are the formulas we have;

1st- gives us (ENG(only)+ALG(only)), insuf, still need the # of none
2nd- gives #none- suff

So,  my answer is B

bkk145 · Answer

Set x=students who take both Algebra and English, Venn Diagram formula gives:
16 + 8 - x + neither = 30
-x + neither = 6
We are trying to find x

(1) Knowing that students taking exactly one class = 20, you can obtain:
20 + x + neither = 30
x + neither = 10
Using equation above, you can solve for "neither":
2*neither = 16
neither = 8
Plug this in, you should get x = 2
SUFFICIENT

(2) Neither = 3
Plug in, you get x = -9!!!
I would say it is sufficient, but neither = 3 doesn't make any sense. Think about it. If you have total of 30 students, 3 doesn't take any class, you have 27 students total taking English, Algebra, or both. This number should be no more than (16+8 =) 24. This is impossible.
I would say SUFFICIENT, but bad given number.

somethingbetter · Answer

Explain please. I guess, it's E

IrinaOK · Answer

Yes, agree A.

I forgot about second equation. Given two equation we can find two unknow variables.

Thank you!

Though probably OA is D, assuming we do not calculate...

Ferihere · Answer

I do agree with bkk145, 
1) is sufficient and it is proven
2) is sufficient to find an ANSWER to the problem even if it is -3 persons who take both subjects

So the answer is D

In a group of 30 students, 8 are enrolled in an English

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