Knewton Challenge Discussion – GMAT Problem Solving (Cello-Viola Pairs)

Posted on November 17, 2010 by Knewton

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

Posted in Challenge Discussions, GMAT, Quant Guide | 17 comments

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

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Knewton Challenge Discussion – GMAT Problem Solving (Cello-Viola Pairs)

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