A zoo has w wildebeests, y yaks, z zebras, and no other animals. If an animal is chosen
at random from the zoo, is the probability of choosing a yak greater than the
probability of choosing a zebra?
1) y/w+z>1/2
2) z/w+y<1/2
could you please give me the solution mathematically when we consider both stats.. i need some clarification on inequality problem..thnx
animals in the zoo
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From (1)
y/w+z>1/2
(w+z)/y < 2 (Taking inverse inequality changes)
(w+y+z)/y < 3 (Adding 1 and taking LCM)
y/(w+y+z) > 1/3 (Again taking inverse)
Now similarly from (2) we have;
z/(w+y+z) < 1/3.
Hence P(y) > P(z) C
y/w+z>1/2
(w+z)/y < 2 (Taking inverse inequality changes)
(w+y+z)/y < 3 (Adding 1 and taking LCM)
y/(w+y+z) > 1/3 (Again taking inverse)
Now similarly from (2) we have;
z/(w+y+z) < 1/3.
Hence P(y) > P(z) C
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The question can be rephrased as , is y>z ?
Statement 1
y/w+z>1/2
y>(w+z)/2 Insufficient
Statement 2
z/w+y<1/2
z<(w+y)/2 Insufficient
Statement 1 and 2
As the inequalities are pointing in opposite directions, we can subtract the two, notice that
there are two scenarios, in one case, y>z and in the other y<z
y-z>(w+z)/2-(w+y)/2. simplifying, we get y>z
z-y<(w+y)/2-(w+z)/2. simplifying, we get y>z Sufficient
Statement 1
y/w+z>1/2
y>(w+z)/2 Insufficient
Statement 2
z/w+y<1/2
z<(w+y)/2 Insufficient
Statement 1 and 2
As the inequalities are pointing in opposite directions, we can subtract the two, notice that
there are two scenarios, in one case, y>z and in the other y<z
y-z>(w+z)/2-(w+y)/2. simplifying, we get y>z
z-y<(w+y)/2-(w+z)/2. simplifying, we get y>z Sufficient
I dont know if this question really has solution.. Since we are dealing with animals, here only +ve integers comes into the picture..mehrasa wrote:A zoo has w wildebeests, y yaks, z zebras, and no other animals. If an animal is chosen
at random from the zoo, is the probability of choosing a yak greater than the
probability of choosing a zebra?
1) y/w+z>1/2
2) z/w+y<1/2
could you please give me the solution mathematically when we consider both stats.. i need some clarification on inequality problem..thnx
Statement 1 is insufficient as it can take any values and equation still be fine
Statement 2:- this only works if y=0, and z<w/2 .. Again this statement is insufficient
Going by algebra, we can get proove that y>z, but then z needs to be -ve (as per the statement 2)..
Let me know if i am getting it wrong..
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Almost no math is needed when we combine the two statements.mehrasa wrote:A zoo has w wildebeests, y yaks, z zebras, and no other animals. If an animal is chosen
at random from the zoo, is the probability of choosing a yak greater than the
probability of choosing a zebra?
1) y/w+z>1/2
2) z/w+y<1/2
could you please give me the solution mathematically when we consider both stats.. i need some clarification on inequality problem..thnx
Linking them together, we get:
z/(w+y) < 1/2 < y/(w+z)
z/(w+y) < y/(w+z).
The inequality above compares the following ratios:
zebras : other animals < yaks : other animals.
Clearly, for the relationship above to hold true, there must be fewer zebras than yaks.
SUFFICIENT.
The correct answer is C.
Looking for an algebraic proof will only make the problem take LONGER.
Remember: the GMAT is not a math test; it's a REASONING test.
This problem is similar to the following from GMATPrep:
https://www.beatthegmat.com/probability-t74040.html
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Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
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