Question 1085902: Dear Tutor,
please help me.
20 students take a test with three questions A,B and C. 16 students answer at least one question correctly. 10 answer question A correctly, 8 answer question B correctly, 6 answer question C correctly. 3 students answer both A and B correctly, 4 answer A and C correctly and 1 student answers all three questions correctly. How many students answer questions B and C correctly?
Could you please explain steps taken while solving this problem?
Thank you.
Found 2 solutions by Theo, ikleyn:
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! the basic formula says:
A or B or C = A + B + C - AB - AC - BC + ABC
in your problem, this becomes:
A or B or C = 10 + 8 + 6 - 3 - 4 - 0 + 1 = 18.
the problem states 20 students, 16 of which answered at least 1 of the problem correctly.
these numbers indicate that 18 students answered 1 of the problems correctly.
it does appear that you are missing a category, namely BC.
i put a 2 in the BC category and then the numbers came out correctly.
A or B or C = A + B + C - AB - AC - BC + ABC
with 2 in BC category, formula becomes:
A or B or C = 10 + 8 + 6 - 3 - 4 - 2 + 1 = 16
add the 4 that didn't answer any correctly and you get a total of 20.
bottom line is your problem doesn't come out correctly because there is something wrong with the numbers used.
i used the following venn diagram generator to confirm that the number don't come out correct as shown, but do come out correct when i put a 2 in the BC category.
first 2 displays are with numbers as given.
next 2 displays are with numbers as given plus BC category has 2 in it.
to show you how it works logically, i'll use the numbers given, but will add a category of BC = 2.
you are given, plus what i added:
total students = 20
total students who got at least 1 of the problems correct = 16
total who got none correct = 4 (i created this category).
A = 10
B = 8
C = 6
AB = 3
AC = 4
BC = 2 (this is what i added to make the problem come out corect).
ABC = 1
ABC is pure since there are no other categories included in it.
from AB and AC and BC, subtract ABC pure to get:
AB pure = 2
AC pure = 3
BC pure = 1
ABC pure = 1
From A, subtract AB pure, AC pure, ABC pure, to get A pure = 10-2-3-1 = 4
from B, subtract AB pure, BC pure, ABC pure, to get B pure = 8-2-1-1 = 4
from C, subtract AC pure, BC pure, ABC pure, to get C pure = 6-3-1-1 = 1
your pure categories are:
none pure = 4
A pure = 4
B pure = 4
C pure = 1
AB pure = 2
AC pure = 3
BC pure = 1
ABC pure = 1
add them up and you get a total of 20, as you should.
you can do it by formula or you can do it by logic.
the total should be 20 students.
it is, but only after i added the category of BC with 2 in it.
bottom line:
you won't get a good answer with the numbers as given, but you will get a good answer with the number as given plus the numbers i added into the BC category.
some reference that you might find helpful.
https://www.easycalculation.com/algebra/venn-diagram-3sets.php
https://www.thoughtco.com/probability-union-of-three-sets-more-3126263
http://www.probabilityformula.org/union-of-events.html
Answer by ikleyn(52803)  (Show Source):
You can put this solution on YOUR website! .
Let "a" be the set of students that answered the question A correctly.
Let "b" be the set of students that answered the question B correctly.
Let "c" be the set of students that answered the question C correctly.
Let ab be the set of students that answered both question A and B correctly.
Let bc be the set of students that answered both question B and C correctly.
Let ac be the set of students that answered both question A and C correctly.
Let abc be the set of students that answered all three question A, B and C correctly.
Next, for any finite set "S" in this problem i will denote by |S| the number of its elements.
From the elementary set theory, this formula is well known
|a U b U c| = |a| + |b| + |c| - |ab| - |ac| - |ab| + |abc| (1) (see the reference at the end of my post)
In our case |a U b U c| = 16: it is the set of students who correctly answered at least one question.
Further, from the condition
|a| = 10, |b| = 8, |c| = 6, |ab| = 3, |ac| = 4, |abc| = 1.
The value |bc| is unknown, and we EASILY will find it from the equation (1) after substituting all other known values to the equation:
16 = 10 + 8 + 6 - 3 - 4 - |bc| + 1.
It gives |bc| = 10 + 8 + 6 - 3 - 4 + 1 - 16 = 2.
Solved.
Answer. 2 students answered correctly questions B and C.
Regarding the formula (1), read these two lessons in this site
- Counting elements in sub-sets of a given finite set
- Advanced problems on counting elements in sub-sets of a given finite set
Also, you have this free of charge online textbook in ALGEBRA-I in this site
- ALGEBRA-I - YOUR ONLINE TEXTBOOK.
The referred lessons are the part of this online textbook under the topic
"Miscellaneous word problems".
Ignore everything the tutor "Theo" wrote in his post: it is not relevant . . . Unfortunately.
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Usually, I read attentively the students' posts, as well as the tutors' posts.
(Except statistic problems, where I am not an expert).
But I never read the tutor's Theo posts: they always are so long that are simply unreadable.
I do not believe that the right Math problem may have long formulation.
I also do not believe that the right solution to elementary Math problem should be long.
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