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?
No replies. Was it that tough? Let me tell you this should not have taken more than a min. Forget about normal set approaches. There are more ways of doing this, which will infact help you save time in quants. Here is the approach.
Let number of students registered for 1 class only be I, 2 classes only be II and all 3 classes only be III than
I+II+III= 68---(1)
Now if you add 25 from History, 25 from Math and 34 from English, you actually add I, you add two twice (say if 2 people are in History and Math than you account 2 while adding history and while adding math, hence twice) and you add III thrice (same reason as above)
Hence I+2II+3III = 84 ----(2)
(2)-(1) II+2III=16
we know that III =3 hence
II+6=16 II=10
hence B
Note: Eqn 1 and Eqn 2 formed in this question is a standard one and will apply to any kind of set problems in the above format. When you see this question you can straight away put these two eqns and finding value after that should be 20 secs job.
Before proceeding Have a look at this diagram http://rapidshare.com/files/270564539/untitled.bmp.html
If you add maths and English you add "d" twice, Adding english and history you add "e" twice and adding maths and history you add "f" twice. Hence number of students who have taken atleast two subjects, which is II, can be represented as 2II. If you add maths, english and history you are adding g thrice. Hence number of students studying 3 subjects, which is III, can be represented as 3III. Let me know if you have further questions.
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No replies. Was it that tough? Let me tell you this should not have taken more than a min. Forget about normal set approaches. There are more ways of doing this, which will infact help you save time in quants. Here is the approach.
ReplyDeleteLet number of students registered for 1 class only be I, 2 classes only be II and all 3 classes only be III than
I+II+III= 68---(1)
Now if you add 25 from History, 25 from Math and 34 from English, you actually add I, you add two twice (say if 2 people are in History and Math than you account 2 while adding history and while adding math, hence twice) and you add III thrice (same reason as above)
Hence I+2II+3III = 84 ----(2)
(2)-(1)
II+2III=16
we know that III =3 hence
II+6=16
II=10
hence B
Note: Eqn 1 and Eqn 2 formed in this question is a standard one and will apply to any kind of set problems in the above format. When you see this question you can straight away put these two eqns and finding value after that should be 20 secs job.
Let me know if you have any queries.
Thanks,
Quant-Master
can u kindly elaborate more that how did u form the 2nd equation?
ReplyDeleteWe have 3 subjects - Maths, English and History.
ReplyDeleteBefore proceeding Have a look at this diagram
http://rapidshare.com/files/270564539/untitled.bmp.html
If you add maths and English you add "d" twice, Adding english and history you add "e" twice and adding maths and history you add "f" twice. Hence number of students who have taken atleast two subjects, which is II, can be represented as 2II. If you add maths, english and history you are adding g thrice. Hence number of students studying 3 subjects, which is III, can be represented as 3III. Let me know if you have further questions.
Thanks,
Quant-Master