Here’s a tough Problem Solving question from Session 11′s QC homework. So far, 69.8% of students have missed it. How would you solve it?

Give it a shot, then share your answers, questions, and thought processes in the comments below. Remember, if you’re in our GMAT class now, add your teacher name and session to your comment (e.g., Zwelling, MW 1:30).

A certain music store stocks 800 cellos and 600 violas. Of these instruments, there are 90 cello-viola pairs, such that a cello and a viola were both made with wood from the same tree (each tree can make at most one viola and one cello, so there are no pairs other than these 90). If one viola and one cello are chosen at random, what is the probability that the two instruments are made with wood from the same tree?

[A] 3/16,000
[B] 1/8,100
[C] 3/1,600
[D] 1/90
[E] 2/45

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  • JSS

    Prob (choosing a C-V pair) = Prob (choosing a cello from wood of particular tree)*Prob(choosing viola from the wood from same tree) or vice versa

    Prob (choosing a cello from wood of particular tree) = 90/800
    Prob (choosing viola from the wood from same tree) = 1/600 (note it has to be the same tree, so there is only one chance out of 600)

    Therefore, Prob (choosing a C-V pair) = 90*1/(800*600)=3/16000

    Answer is A.

    Jaspal, MW7:00, Chris W

  • JSS

    Prob (choosing a C-V pair) = Prob (choosing a cello from wood of particular tree)*Prob(choosing viola from the wood from same tree) or vice versa

    Prob (choosing a cello from wood of particular tree) = 90/800
    Prob (choosing viola from the wood from same tree) = 1/600 (note it has to be the same tree, so there is only one chance out of 600)

    Therefore, Prob (choosing a C-V pair) = 90*1/(800*600)=3/16000

    Answer is A.

    Jaspal, MW7:00, Chris W

  • Gtestprep

    Agree with JSS, the answer is A, 3/16000.

  • Gtestprep

    Agree with JSS, the answer is A, 3/16000.

  • keith h

    Don’t you need to do P(Choosing a cello from a paired-tree)*P(Choosing a viola from the same tree) + P(Choosing a viola from a paired-tree)*P(Choosing a cello from the same tree)?

    Then it’d be (90/800)(1/600) + (90/600)(1/800). But that is 3/8,000, which isn’t even an option…

  • keith h

    Don’t you need to do P(Choosing a cello from a paired-tree)*P(Choosing a viola from the same tree) + P(Choosing a viola from a paired-tree)*P(Choosing a cello from the same tree)?

    Then it’d be (90/800)(1/600) + (90/600)(1/800). But that is 3/8,000, which isn’t even an option…

  • Sid

    90c1/600c1 *800c1 =3/1600

  • Sid

    90c1/600c1 *800c1 =3/1600

  • Siddhartha

    this is easy

    select the pair /total option

    90c1/800c1*600c1= 3/16000

    answer A

  • Siddhartha

    this is easy

    select the pair /total option

    90c1/800c1*600c1= 3/16000

    answer A

  • TNO

    N° of options : 600 x 800
    N° of pair = 90
    Prob of choosing a C-V pair = 90/(600*800) = 3/16000

  • TNO

    N° of options : 600 x 800
    N° of pair = 90
    Prob of choosing a C-V pair = 90/(600*800) = 3/16000

  • EvaJager

    Use the basic definition of probability – number of desired outcomes over the total number of outcomes.
    Desired outcomes – 90 (there are exactly 90 pairs of cello & viola, each from the same one tree).
    Total number of cello & viola pairs – 800×600.
    Therefore, the probability is 90/800×600=3/16000 (after reducing by a factor of 3).
    The same simple way as TNO suggested.

  • EvaJager

    Use the basic definition of probability – number of desired outcomes over the total number of outcomes.
    Desired outcomes – 90 (there are exactly 90 pairs of cello & viola, each from the same one tree).
    Total number of cello & viola pairs – 800×600.
    Therefore, the probability is 90/800×600=3/16000 (after reducing by a factor of 3).
    The same simple way as TNO suggested.

  • Guest

    It’s A

  • Favour Elekwachi

    the answer is c

  • Favour Elekwachi

    the answer is a