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    Question - No Options this time

    Of the 200 candidates who were interviewed for a position at a call center, 100 had a two-wheeler, 70 had a credit card and 140 had a mobile phone. 40 of them had both, a two-wheeler and a credit card, 30 had both, a credit card and a mobile phone and 60 had both, a two wheeler and mobile phone and 10 had all three. How many candidates had none of the three?
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    Thanks! I had made a textbook error in the decimal to time conversion and confused myself.
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    (Original post by jjkkll)
    Question - No Options this time

    Of the 200 candidates who were interviewed for a position at a call center, 100 had a two-wheeler, 70 had a credit card and 140 had a mobile phone. 40 of them had both, a two-wheeler and a credit card, 30 had both, a credit card and a mobile phone and 60 had both, a two wheeler and mobile phone and 10 had all three. How many candidates had none of the three?
    10?
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    10 had none?
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    is it 10 ?
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    I get 10.
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    I love How everyone has put a question mark after their answers.

    Yes its 10!

    Would anyone like the working?

    or Next Question?
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    i got 20..:confused:
    how did you guys get 10?
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    (Original post by jjkkll)
    I love How everyone has put a question mark after their answers.

    Yes its 10!

    Would anyone like the working?

    or Next Question?
    please
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    (Original post by jjkkll)
    I love How everyone has put a question mark after their answers.

    Yes its 10!

    Would anyone like the working?

    or Next Question?
    Both? :p:
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    (Original post by jjkkll)
    I love How everyone has put a question mark after their answers.

    Yes its 10!

    Would anyone like the working?

    or Next Question?
    yes!!!!
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    (Original post by Medicine Man)
    i got 20..:confused:
    how did you guys get 10?

    you forgot those with (tw + CC + MP)

    Number of candidates who had none of the three = Total number of candidates - number of candidates who had at least one of three devices.

    Total number of candidates = 200.

    Number of candidates who had at least one of the three = A U B U C, where A is the set of those who have a two wheeler, B the set of those who have a credit card and C the set of those who have a mobile phone.

    We know that AUBUC = A + B + C - {A n B + B n C + C n A} + A n B n C
    Therefore, AUBUC = 100 + 70 + 140 - {40 + 30 + 60} + 10
    Or AUBUC = 190.

    As 190 candidates who attended the interview had at least one of the three gadgets, 200 - 190 = 10 candidates had none of three.
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    Question 3 - No Options
    In a class of 40 students, 12 enrolled for both English and German. 22 enrolled for German. If the students of the class enrolled for at least one of the two subjects, then how many students enrolled for only English and not German?
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    (Original post by jjkkll)
    you forgot those with (tw + CC + MP)

    Number of candidates who had none of the three = Total number of candidates - number of candidates who had at least one of three devices.

    Total number of candidates = 200.

    Number of candidates who had at least one of the three = A U B U C, where A is the set of those who have a two wheeler, B the set of those who have a credit card and C the set of those who have a mobile phone.

    We know that AUBUC = A + B + C - {A n B + B n C + C n A} + A n B n C
    Therefore, AUBUC = 100 + 70 + 140 - {40 + 30 + 60} + 10
    Or AUBUC = 190.

    As 190 candidates who attended the interview had at least one of the three gadgets, 200 - 190 = 10 candidates had none of three.
    anyone got a simpler method?
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    (Original post by jjkkll)
    you forgot those with (tw + CC + MP)

    Number of candidates who had none of the three = Total number of candidates - number of candidates who had at least one of three devices.

    Total number of candidates = 200.

    Number of candidates who had at least one of the three = A U B U C, where A is the set of those who have a two wheeler, B the set of those who have a credit card and C the set of those who have a mobile phone.

    We know that AUBUC = A + B + C - {A n B + B n C + C n A} + A n B n C
    Therefore, AUBUC = 100 + 70 + 140 - {40 + 30 + 60} + 10
    Or AUBUC = 190.

    As 190 candidates who attended the interview had at least one of the three gadgets, 200 - 190 = 10 candidates had none of three.
    ahh
    i used a venn diagram
    mine added up 180
    forgot to add the 10 on that had all 3 to make 190
    silly mistake really!!:rolleyes:
    thanks though!!
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    (Original post by Isometrix)
    anyone got a simpler method?
    100 w (2 wheeler)
    70 c (credit card)
    140 m (mobile)

    People with w, c, or both w+c = 100 + 70 - 40 (40 had both)
    = 130

    People with c, m, or both c+m = 70 + 140 - 30 (30 had both)
    = 180

    People with w, m, or both w+m = 100 + 140 - 60 (60 had both)
    = 180

    As 180 was the highest, I added that to 10 (who had all three)
    = 190

    200 - 190 = 10

    (Probably not a very mathematical way of doing it!)


    Edit: As above, Venn diagram to represent this would probably be simpler!
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    18?
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    (Original post by jjkkll)
    Question 3 - No Options
    In a class of 40 students, 12 enrolled for both English and German. 22 enrolled for German. If the students of the class enrolled for at least one of the two subjects, then how many students enrolled for only English and not German?
    18

    EDIT:
    wait, i've made a mistake..
    just a sec

    EDIT EDIT:
    actually i think 18 is fine
    its either that or 6
    the question is ambiguous
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    (Original post by Myoclonic Jerk)
    100 w (2 wheeler)
    70 c (credit card)
    140 m (mobile)

    People with w, c, or both w+c = 100 + 70 - 40 (40 had both)
    = 130

    People with c, m, or both c+m = 70 + 140 - 30 (30 had both)
    = 180

    People with w, m, or both w+m = 100 + 140 - 60 (60 had both)
    = 180

    As 180 was the highest, I added that to 10 (who had all three)
    = 190

    200 - 190 = 10

    (Probably not a very mathematical way of doing it!)


    Edit: As above, Venn diagram to represent this would probably be simpler!

    That's exactly what the method shows, the only difference is the method iv posted is on less lines.
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    (Original post by Isometrix)
    anyone got a simpler method?
    I don't think the question makes sense, but I'm trying to rationalise it.
 
 
 
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