Math

Two boys decided to buy a computer. The second boy had 5/6 of the money the first had. The first boy had 7/8 of the price of the computer. Together they had $696.00 more than they need to pay. What was the price for the computer?

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  1. b1 = 7/8 C
    b2 = 5/6 b1 = 35/ 48 C

    7/8 C + 35/48 C = C + 696
    42/48 C + 35/48 C = 48/48 C + 696 * 48/48
    (42 + 35 - 48 ) C = 696 * 48
    29 C = 33408
    C =1152.

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    Damon
  2. Computer: $X.
    5x/6 + 7x/8 = x + 690.
    20x/24 + 21x/24 = x + 690
    41x/24 = x + 690
    41x = 24x + 24*690
    17x = 24*690
    X = $974.12.

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  3. Correction: 17x = 24*696
    X = $982.60.

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  4. Let the price of the computer = x
    This means:
    1st boy = (7/8)x {because the 1st boy has 7/8 of the price of the computer}

    2nd boy = (5/6*(7/8)x) {because the 2nd boy has 5/6 of the first boy who has 7/8 of the price of the computer}

    2nd boy = (35/48)x {after multiplying (5/6)*(7/8)}

    Now let's set up the equation:

    (7/8)x + (35/48)x = x +696 {The total of the money between both boys is the price of the computer and an extra 696 dollars}

    Next, we subtract (7/8)x from both sides to get rid of the first x on the left side of the equation

    (35/48)x = (1/8)x + 696

    Now, we need to subtract (1/8)x from both sides. This fraction is also (6/48)x

    (29/48)x = 696

    Now we need to multiply the reciprocal of 29/48 which is 48/29.

    x = 1152

    The computer is $1152

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  5. rsm

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