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Math Help - SAT QOTD - July 09

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    Senior Member vaironxxrd's Avatar
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    SAT QOTD - July 09

    Hello guys this is a SAT Question of the day I got today July 09. Some of you know that I'm really bad at math and I always struggle even when given the answer.

    So what I will do is if the forum doesn't mind, post Math questions every time there's an SAT Math question . The reason is because I need to start preparing from this moment for my SAT in about January.

    Here is the QOTD(Question of the day)

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    Member Goku's Avatar
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    Re: SAT QOTD - July 09

    answer is B.
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    MHF Contributor Siron's Avatar
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    Re: SAT QOTD - July 09

    Maybe varionxxrd wants an explanation? ...
    B is correct because, you can write B as:
    -0,2<w-2,95<0,2
    <-> 2,75<w<3,15
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    Re: SAT QOTD - July 09

    Hello, vaironxxrd!

    The weights of 10 bags of apples range from 2.75 lbs to 3.15 lbs.
    If w is the weight, in pounds, of one of these bags,
    which of the following is true?

    . . \begin{array}{c} (A)\;|w\!-\!2.75| < 0.2 \quad (B)\;|w\!-\!2.95| < 0.2 \quad (C)\;|w\!+\!2.95| < 0.2 \\ \\[-2mm] (D)\;|w\!-\!0.2| < 2.75 \quad (E)\;|w\!-\!10| < 2.95 \end{array}

    Graph the information on a number line.

    . . \begin{array}{c} w \\  -- \; \overbrace{\bullet \; ----- \; \bullet} \; -- \\ \underbrace{2.75 \qquad\qquad\quad 3.15}_{\text{average 2.95}} \end{array}

    w\text{ is in the interval }[2.75,\:3.15]
    \text{The midpoint of the interval is: }\:\tfrac{2.75 + 3.15}{2} \:=\:2.95
    \text{Hence, }w\text{ is within 0.2 lbs of the midpoint.}

    \begin{array}{ccccccc}\text{Therefore:} & \text{-}0.2 & \le & w-2.95 & \le & 0.2 \\ \\[-3mm] &&& |w-2.95| &\le & 0.2 \\ \\[-3mm] &&& |w-2.95| & < & 0.2 & \text{ ... answer (B)}\end{array}

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    Re: SAT QOTD - July 09

    Quote Originally Posted by Soroban View Post
    Hello, vaironxxrd!


    Graph the information on a number line.

    . . \begin{array}{c} w \\  -- \; \overbrace{\bullet \; ----- \; \bullet} \; -- \\ \underbrace{2.75 \qquad\qquad\quad 3.15}_{\text{average 2.95}} \end{array}

    w\text{ is in the interval }[2.75,\:3.15]
    \text{The midpoint of the interval is: }\:\tfrac{2.75 + 3.15}{2} \:=\:2.95
    \text{Hence, }w\text{ is within 0.2 lbs of the midpoint.}

    \begin{array}{ccccccc}\text{Therefore:} & \text{-}0.2 & \le & w-2.95 & \le & 0.2 \\ \\[-3mm] &&& |w-2.95| &\le & 0.2 \\ \\[-3mm] &&& |w-2.95| & < & 0.2 & \text{ ... answer (B)}\end{array}

    It looks to me like your last line is uneccesary and incorrect. You want "[tex]\le[itex]", not "<".
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    Senior Member vaironxxrd's Avatar
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    Re: SAT QOTD - July 09

    Quote Originally Posted by Soroban View Post
    Hello, vaironxxrd!


    Graph the information on a number line.

    . . \begin{array}{c} w \\  -- \; \overbrace{\bullet \; ----- \; \bullet} \; -- \\ \underbrace{2.75 \qquad\qquad\quad 3.15}_{\text{average 2.95}} \end{array}

    w\text{ is in the interval }[2.75,\:3.15]
    \text{The midpoint of the interval is: }\:\tfrac{2.75 + 3.15}{2} \:=\:2.95
    \text{Hence, }w\text{ is within 0.2 lbs of the midpoint.}

    \begin{array}{ccccccc}\text{Therefore:} & \text{-}0.2 & \le & w-2.95 & \le & 0.2 \\ \\[-3mm] &&& |w-2.95| &\le & 0.2 \\ \\[-3mm] &&& |w-2.95| & < & 0.2 & \text{ ... answer (B)}\end{array}

    Is it 0.2 because if yu do 3.15 - 2.95 The midpoint the result is 0.2?
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