Problem Solving

In a group of 68 students, each student is registered for at least one of three classes - History, Math and English. Twenty-five students are registered for History, twenty-five students are registered for Math, and thirty-four students are registered for English. If only three students are registered for all three classes, how many students are registered for exactly two classes?

(A) 13
(B) 10
(C) 9
(D) 8
(E) 7

Is the answer 9 ..that is C..

IMO B

need to draw a venn dig..

let the portion only H as a, only M as b and only E as c

let only H+M as x
only M+E as y
and only E+H as z

now a+x+z+3+b+y+c=68 or
a+b+c+x+y+z=65

and a+b+C+2(x+y+z)+9=84 or
a+b+c+2(x+y+z)=75
using bth the eqn x+y+z=10

EMAN wrote:In a group of 68 students, each student is registered for at least one of three classes - History, Math and English. Twenty-five students are registered for History, twenty-five students are registered for Math, and thirty-four students are registered for English. If only three students are registered for all three classes, how many students are registered for exactly two classes?

(A) 13
(B) 10
(C) 9
(D) 8
(E) 7

Venn diagrams are a great way to solve this type of question.

Another method we can use is formula-based. There are two useful formulas for 3-group overlapping set questions:

True # of objects = (total # in group 1) + (total # in group 2) + (total # in group 3) - (# in only groups 1/2) - (# in only groups 1/3) - (# in only groups 2/3) - 2(# in all 3 groups) + (# in no groups)

or, compacting the middle section:

True # of objects = (total # in group 1) + (total # in group 2) + (total # in group 3) - (# in exactly 2 groups) - 2(# in all 3 groups) + (# in no groups)

The second formula is a bit different:

True # of objects = (# in exactly 1 group) + (# in exactly 2 groups) + (# in all 3 groups) + (# in no groups)

[this is the formula that xcuseme generated from the Venn diagram].

We want to use the first formula to solve this particular question.

We know that:

True # of students = 68
Total # in history = 25
Total # in math = 25
Total # in english = 34
# in all 3 = 3
# in none = 0 (everyone is in at least 1 class)

So:

68 = 25 + 25 + 34 - (# in exactly 2 classes) - 2(3) + 0

68 = 78 - (# in exactly 2 classes)

(# in exactly 2 classes) = 10

Excellent. This is extremely helpful. Can you explain the logic in the formula explanation why you have to put a coefficient of 2 in front of the 3 (number of students in all groups???

EMAN wrote:Excellent. This is extremely helpful. Can you explain the logic in the formula explanation why you have to put a coefficient of 2 in front of the 3 (number of students in all groups???

The students in all 3 groups have been counted 3 times: once as part of group 1, once as part of group 2 and once as part of group 3.

Since we want to count each subsection only once, and since we counted the triple group 3 times, we need to subtract it twice.

Stuart Kovinsky wrote:

EMAN wrote:Excellent. This is extremely helpful. Can you explain the logic in the formula explanation why you have to put a coefficient of 2 in front of the 3 (number of students in all groups???

The students in all 3 groups have been counted 3 times: once as part of group 1, once as part of group 2 and once as part of group 3.

Since we want to count each subsection only once, and since we counted the triple group 3 times, we need to subtract it twice.

Hi Stuart,

https://www.beatthegmat.com/210-bags-con ... 31800.html

Im unable to solve the problem in the link, with mentioned formula, not sure if this would apply here..

Can you please help in solving this?

Tx

kumadil2011 wrote:

Stuart Kovinsky wrote:

EMAN wrote:Excellent. This is extremely helpful. Can you explain the logic in the formula explanation why you have to put a coefficient of 2 in front of the 3 (number of students in all groups???

The students in all 3 groups have been counted 3 times: once as part of group 1, once as part of group 2 and once as part of group 3.

Since we want to count each subsection only once, and since we counted the triple group 3 times, we need to subtract it twice.

Hi Stuart,

https://www.beatthegmat.com/210-bags-con ... 31800.html

Im unable to solve the problem in the link, with mentioned formula, not sure if this would apply here..

Can you please help in solving this?

Tx

I posted a solution to the almonds problem here:

https://www.beatthegmat.com/can-you-plea ... 86595.html

I still don't get the formula. It makes sense for this problem, but I may not be able to relate the formule to a different problem of same type.

Can you please explain some logic, explanation for the Venn Diagram formula? Where can we get more examples to practice such problems?

@Mitch
Why can't this one be solved as follows?
Let h be the number of only history students
Let e be the number of only english students
Let m be the number of only maths students

x be the number of both maths and english
y be the number of both english and history
z be the number of both history and maths

we need to fin the value of x+y+z

68=25+25+34-x-y-z+3
68=84-(x+y+z)+3
x+y+z=19

Why it can't be solved this way?

kakz wrote: x be the number of both maths and english
y be the number of both english and history
z be the number of both history and maths

we need to fin the value of x+y+z

68=25+25+34-x-y-z+3
68=84-(x+y+z)+3
x+y+z=19

Why it can't be solved this way?

The method you've used is only valid if x,y, and z are also counting the people who are taking all three classes. So if you try to use this method, when you do x+y+z, the people who are taking all three classes are counted three times. But we don't want to count these people at all, so we have to subtract them 3 times, which would give us 19-3(3)=10.

Analyzing your method step-by-step: Let's say we define x,y, and z to represent the number of people taking EXACTLY two of the classes, which I assume was your intent if you expected x+y+z to be the right answer. When you add 25+25+34 you count the people who are taking exactly one class once, the people who are taking exactly two classes twice, and the people who are taking exactly three classes three times. When you subtract x,y,z you are now counting the people who are taking exactly one class once, the people who are taking exactly two classes once, but you are still counting the number of people taking exactly three classes three times because when we subtract x,y,z, we're not subtracting ANY of the people taking all three classes, because we defined the variables to only include those taking EXACTLY two classes. When you add the number of people taking all three classes at the end, now you're counting the people taking exactly one class once, the people taking exactly two classes once, and the people taking exactly three classes FOUR times. These people are being vastly overcounted, so you have no reasonable expectation that this should all add up to the total number of people, 68, making your equation invalid. Instead of adding three, you should have subtracted six, because you WERE counting them 3 times and to get everything to add to 68 you should only be counting everybody once, so you need to subtract 3 from this total twice. This would make x+y+z=10

Remember that the point of adding in the number of people who are in all three groups in a problem like this is because they usually get undercounted once you subtract out the people who are in at least two sets. Normally you add everyone in at least one set and the people in exactly one, two, and three sets are counted 1,2 and 3 times respectively. Then you subtract out the people in at least two sets and the people are counted 1, 1, and 0 times. Then you have to add in the people in all the sets to get it back to 1, 1, 1 so that everybody is counted exactly once. But here the people in all three sets are not undercounted because they weren't part of the groups we were subtracting out. And if they're not undercounted, it makes no sense to add them back in.