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Difficult Math Problem #100 - Sets

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800guy Master | Next Rank: 500 Posts Default Avatar
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Difficult Math Problem #100 - Sets

Post Wed Feb 21, 2007 9:15 am
Elapsed Time: 00:00
  • Lap #[LAPCOUNT] ([LAPTIME])
    There are 70 students in Math or English or German. Exactly 40 are in Math, 30 in German, 35 in English and 15 in all three courses. How many students are enrolled in exactly two of the courses? Math, English and German.

    from diff math doc; oa coming after some people answer/explain

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    gmat_enthus Senior | Next Rank: 100 Posts Default Avatar
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    Post Wed Feb 21, 2007 12:44 pm
    Exactly 2 = 75.

    Not posting the solution so others can try

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    Neo2000 Legendary Member
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    Post Wed Feb 21, 2007 5:59 pm
    I am getting 5

    aim-wsc Legendary Member
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    Post Wed Feb 21, 2007 10:03 pm
    gmat_enthus wrote:
    Exactly 2 = 75.

    Not posting the solution so others can try
    how come exactly 2=75

    Total number of students are 70 so you cant exceed the figure, right?
    I think you forgot to subtract 70 out of it.

    40 Math,
    30 German
    35 English

    makes it 40+30+35=105 total choices out of 70 students.

    A student can choose only one subject, two subjects or all three subjects.
    the students belong to third category are 15.

    70 + 2(15) + two subjects students + 0(single subject studnet) = total choices


    now somebody should come up with better explanation.
    I too think asnwer is 5.

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    jayhawk2001 Community Manager
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    Post Wed Feb 21, 2007 10:34 pm
    Math = 25 + 15 (all-3)
    German = 15 + 15 (all-3)
    English = 20 + 15 (all-3)

    i.e. we have 15 students taking all-3 + 60 students
    not taking all-3 for a total of 75.

    Since there are 70 students in all, 5 of them should be
    taking 2 courses at the same time.

    Is the OA 5 ?

    banona Senior | Next Rank: 100 Posts Default Avatar
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    Post Fri Feb 23, 2007 6:30 am
    let :
    M : students in maths, G : all students in geman ; E : all students in English
    DM : are students either with nath and english or with maths and german; so are DG and DE
    T : students in 3 class

    ( M - DM -T) +( G - DG- T) +( E-DE - T) = 70


    summimg the following expessions : M+G+E - ( DM+DE+DG) - 3T= 70

    Given M+E+G = 70
    T = 15
    BUT, (carefully) DM+DE+DG =2 sum of students with all double subjects , because DM has double components

    LEt the sum of all students with double subjects be x;

    70 =105 - 2X- 45

    Here I can not go farther, because x would be negative,

    Am I wrong.
    Any comment,

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    banona Senior | Next Rank: 100 Posts Default Avatar
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    Post Fri Feb 23, 2007 10:39 am
    Oooops, I was wrong, Embarassed but I can not give up
    in fact, the equation USING Venn Diagrams is, (SEE ATTACHED FIGURE)

    LEt the sum of all students with double subjects be x;

    M+G+E - x - 2T= 70

    Given M+E+G = 40+35+30 = 105
    T = 15

    then x = 5

    I hope it's clear
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    800guy Master | Next Rank: 500 Posts Default Avatar
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    Post Fri Feb 23, 2007 3:52 pm
    here's the oa:

    MuEuG = M + E + G - MnE - MnG - EnG - 2(MnEnG)
    MnE + MnG + EnG = M + E + G - 2(MnEnG) - MuEuG
    MnE + MnG+ EnG = 40 + 30 + 35 - 2(15) - 70 = 105 - 30 - 70 = 5

    Whenever an intersection occurs between 2 sets, (MnEnG) is counted twice, therefore you deduct one of it. If the intersection occurs between 3 sets, it is counted thrice; therefore you deduct two of it. And so forth.
    If there are four sets, then the formula is A + B + C + D -(two) -(three)*2 -(four)*3 = total

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