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Math Help - 2 easy questions: Probability/Percentages and Digits

  1. #1
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    2 easy questions: Probability/Percentages and Digits

    In the life sciences department, 65% of the total number of students watch the news on
    television, 40% read a newspaper and 25% both read a newspaper and watch the news on
    television. What percentage of the department's students do not watch the news on
    television and do not read a newspaper?
    (1) 0%
    (2) 10%
    (3) 20%
    (4) 25%


    My solution:
    100 students.
    40-25 = 15 only newspaper
    65- 25 = 40 only TV


    answer is 2.. 10%
    not sure..
    guess:
    65+25 = 90?
    100-90 is 10?

    not sure...


    DIGITS question:


    25. A, B, C and D are letters representing different digits, so that ABCD is a four-digit
    number.
    Given: AAB = CD
    If we divide ABCD by the number AB, what three-digit number will be obtained?
    (1) 101
    (2) 1AB
    (3) 11A
    (4) 10A


    answer 4
    AB into ABCD..
    start with a 1

    CD is A^2B
    and AB into A^2B is...? 0A
    AB/AAB leave you with.. 1/A

    not sure ..

    THANKS IN ADVANCE
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  2. #2
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    Re: 2 easy questions: Probability/Percentages and Digits

    For the first question, there are just four groups. Those that just watch TV, those that just read, those that do both and those that do neither. You've worked out numbers for the first two and you are given the number for the third group. If all four total 100 you should be able to deduce the fourth. (It's easy to see what's going on if you draw a Venn diagram).

    For the second question, get what's going on straight in your mind by working through a long division using actual numbers. Try for example 1236/12, 12's into 1 don't go, 12's into 12 go once etc.
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  3. #3
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    Re: 2 easy questions: Probability/Percentages and Digits

    Hello, jsvlad!

    25. A, B, C, D are letters representing different digits,
    . . .so that ABCD is a four-digit number.

    Given: . A\cdot AB \,=\, CD
    If we divide ABCD by the number AB, what three-digit number will be obtained?

    . . (1)\;101 \qquad (2)\;1AB \qquad (3)\;11A \qquad (4)\;10A

    Simply "do" the long division . . .


    . . \begin{array}{cccccc}&&&1 & 0 & A \\ && -- & -- & -- & -- \\ A\:\;\;B & | & A & B & C & D \\ && A & B \\ && -- & -- \\ &&&& C \\ &&&& 0 \\ &&&& -- \\ &&&& C & D \\ &&&& C & D \\ &&&& -- & -- \end{array}

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  4. #4
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    Re: 2 easy questions: Probability/Percentages and Digits

    hello..

    how did you pull that off with those awesome graphics/image? im impressed, i would like to knw..



    q1
    15+40+25 = 80
    100-80 = 20


    Quote Originally Posted by Soroban View Post
    Hello, jsvlad!


    Simply "do" the long division . . .


    . . \begin{array}{cccccc}&&&1 & 0 & A \\ && -- & -- & -- & -- \\ A\:\;\;B & | & A & B & C & D \\ && A & B \\ && -- & -- \\ &&&& C \\ &&&& 0 \\ &&&& -- \\ &&&& C & D \\ &&&& C & D \\ &&&& -- & -- \end{array}

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