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31 May, 10:33

A rectangular piece of metal is 30 in longer than it is wide. Squares with sides 6 in long are cut from the four corners and the flaps are folded upward to form an open box. If the volume of the box is 3354 incubed , what were the original dimensions of the piece of metal?

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  1. 31 May, 10:44
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    Answer = 55,25 inches

    Solution -

    let's take x as length and y as width of the metal piece. As per the question x is 30 more than y,

    ⇒ x = y + 30

    Then four square pieces of side 6 are cut from each corner,

    so the new length and width are

    x-12, y-12

    Then the volume of the new box created will be

    (x-12) (y-12) 6

    in the question the volume of the given figure is given to be 3354

    so (x-12) (y-12) 6 = 3354

    putting the value of x in the the above equation

    ⇒ (y+30 - 12) (x-12) = 3354/6 = 559

    ⇒ (y+18) (y-12) = 559

    ⇒ y² + 6y - 775 = 0

    ⇒ y² + 31y - 25y - 775 = 0

    ⇒ (y+31) (y-25) = 0

    ⇒ y = - 31, 25

    as length can not be - ve, so y = 25

    then x = 25+30 = 55

    Hence the dimensions of the metal piece are 55, 25 inches
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