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Distance-Time calculation.

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sathishkumarjva9888 Rising GMAT Star Default Avatar
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Distance-Time calculation. Post Thu Apr 26, 2012 6:32 am
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  • Lap #[LAPCOUNT] ([LAPTIME])
    Bill and Sally see each other across a field. They are 500 feet apart. If at the same instant of time they start running toward each other in a direct line how far will Bill have traveled when they meet?

    (1) Bill ran at an average speed that was 50% greater than Sally ’ s average speed.

    (2) Bill ran at an average speed 4 feet per second faster than Sally ’ s average speed.

    A.) Statement (1) BY ITSELF is sufficient to answer the question, but statement (2) by itself is not.
    B.) Statement (2) BY ITSELF is sufficient to answer the question, but statement (1) by itself is not.
    C.) Statements (1) and (2) TAKEN TOGETHER are sufficient to answer the question, even though NEITHER statement BY ITSELF is sufficient.
    D.) EITHER statement BY ITSELF is sufficient to answer the question.
    E.) Statements (1) and (2) TAKEN TOGETHER are NOT sufficient to answer the question, requiring more data pertaining to the problem.


    OA: A

    Can somebody help me how to solve this problem.

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    niketdoshi123 Really wants to Beat The GMAT!
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    Post Thu Apr 26, 2012 7:31 am
    Distance = 500 feet
    let Sally's average speed be = x feet per second
    STATEMENT 1

    Bill's average speed = 1.5x
    While running towards each other their relative speed will be = x+ 1.5x = 2.5x
    Total time taken to travel 500 feet = 500/2.5x= 200/x
    Distance Bill traveled when they meet = 1.5x*200/x = 300 feet

    Hence Sufficient

    Statement 2

    Bill's average speed = x+4 feet per second
    Their relative speed = x+x+4 = 2x+4
    Total time taken to travel 500 feet = 500/(2x+4) = 250/x+2
    distance Bill traveled when they meet = (x+4)*250/x+2

    Hence Insufficient

    so the ans is A

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    sathishkumarjva9888 Rising GMAT Star Default Avatar
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    Post Thu Apr 26, 2012 8:01 am
    Thanks for the explanation doshi.. Smile Smile

    niketdoshi123 Really wants to Beat The GMAT!
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    Post Thu Apr 26, 2012 8:15 am
    In data sufficiency you don't have to solve the problem.
    In this problem you can strait away mark the answer by looking at the statement.

    Consider this example
    A and B are standing d feet apart. Speed of A:Speed of b= x:y
    They start running towards each other. Find the distance traveled by A and B respectively when they meet.
    So
    Distance traveled by A = (x/x+y)*d
    and distance traveled by B = (y/x+y)*d.

    So even if the actual speed is not given, you can still find the distance traveled by an individual using the ratio of their speed.
    Hope this helps.

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    Post Thu Apr 26, 2012 10:57 am
    The total distance covered by the fellows = 500 ft and they take the same time to meet each other.

    S1. Speed of Sally = s
    Speed of Billy = 1.5s
    t = time when they meet
    st = 500/2.5
    t = not solvable
    Not Sufficient

    S2. B = s+4
    S = s
    t = time when they meet
    (s+4)t + st = 500

    Not solvable
    Not Sufficient

    Combination, st = 500/2.5 and t = from second equation

    Solvable
    Sufficient

    (C) is answer

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    Post Thu Apr 26, 2012 7:13 pm
    sathishkumarjva9888 wrote:
    Bill and Sally see each other across a field. They are 500 feet apart. If at the same instant of time they start running toward each other in a direct line how far will Bill have traveled when they meet?

    (1) Bill ran at an average speed that was 50% greater than Sally's average speed.

    (2) Bill ran at an average speed 4 feet per second faster than Sally’s average speed.

    OA: A

    Can somebody help me how to solve this problem.
    Statement 1: Bill ran at an average speed that was 50% greater than Sally's average speed.
    In other words:
    For every 2 feet that Sally travels, Bill travels 3 feet (50% more).
    Thus, Bill travels 3 of every 5 feet that the two travel together.
    Distance for Bill = (3/5)500 = 300 miles.
    SUFFICIENT.

    Statement 2: Bill ran at an average speed 4 feet per second faster than Sally’s average speed.
    Following the same reasoning used to evaluate statement 1:
    If Bill travels 5 feet for every 1 foot that Sally travels, then Bill travels 5 of every 6 feet that the two travel together.
    In this case, the distance that Bill travels = (5/6)500.
    If Bill travels 6 feet for every 2 feet that Sally travels, then Bill travels 6 of every 8 feet that the two travel together.
    In this case, the distance that Bill travels = (6/8)500.
    Since Bill can travel different distances, INSUFFICIENT.

    The correct answer is A.

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