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
OA is C
Source is MGMAT Word Translations.
Hope someone can explain. Thanks.
5-Day Free Trial
5-day free, full-access trial TTP
Available with Beat the GMAT members only code
MORE DETAILS
Probability
This topic has expert replies
-
liferocks
- Legendary Member
- Posts: 576
- Joined: Sat Mar 13, 2010 8:31 pm
- Thanked: 97 times
- Followed by:1 members
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
probability of choosing a yak =y/(W+y+z)
probability of choosing a zebra =z/(W+y+z)
from 1 we get y/(w+z)>1/2 or y/(w+y+z)>1/3-->not sufficient
from 2 we get z/(w+y)<1/2 or z/(w+y+z)<1/3-->not sufficient
together we can say z/(w+y+z)<y/(w+y+z) -->sufficient ans C
note: I have assumed that condition is like y/(w+z)>1/2 and z/(w+y)<1/2 if this is not correct than above logic does not hold good.
(1) y/w+z > 1/2
(2) z/w+y < 1/2
probability of choosing a yak =y/(W+y+z)
probability of choosing a zebra =z/(W+y+z)
from 1 we get y/(w+z)>1/2 or y/(w+y+z)>1/3-->not sufficient
from 2 we get z/(w+y)<1/2 or z/(w+y+z)<1/3-->not sufficient
together we can say z/(w+y+z)<y/(w+y+z) -->sufficient ans C
note: I have assumed that condition is like y/(w+z)>1/2 and z/(w+y)<1/2 if this is not correct than above logic does not hold good.
-
sanju09
- GMAT Instructor
- Posts: 3650
- Joined: Wed Jan 21, 2009 4:27 am
- Location: India
- Thanked: 267 times
- Followed by:80 members
- GMAT Score:760
great explanationliferocks 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
probability of choosing a yak =y/(W+y+z)
probability of choosing a zebra =z/(W+y+z)
from 1 we get y/(w+z)>1/2 or y/(w+y+z)>1/3-->not sufficient
from 2 we get z/(w+y)<1/2 or z/(w+y+z)<1/3-->not sufficient
together we can say z/(w+y+z)<y/(w+y+z) -->sufficient ans C
note: I have assumed that condition is like y/(w+z)>1/2 and z/(w+y)<1/2 if this is not correct than above logic does not hold good.
For those who can be misled by the bold part in the above post must know that w, y, and z are positive integers and:
when y/ (w + z) > 1/2; (w + z)/y < 2,
or [(w + z)/y] + 1 < 2 + 1,
or (w + y + z)/y < 3,
or y/(w + y + z) > 1/3; and
when z/ (w + y) < 1/2; (w + y)/z > 2,
or [(w + y)/z] + 1 > 2 + 1,
or (w + y + z)/z > 3,
or z/(w + y + z) < 1/3.
Therefore comparison is possible.
[spoiler]C[/spoiler]
The mind is everything. What you think you become. -Lord Buddha
Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001
www.manyagroup.com
Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001
www.manyagroup.com
-
digvijay.rajput
- Newbie | Next Rank: 10 Posts
- Posts: 4
- Joined: Fri Aug 02, 2013 12:34 am
- Location: New Delhi, India
In the above example, can you please explain these steps....
when y/ (w + z) > 1/2; (w + z)/y < 2,
or [(w + z)/y] + 1 < 2 + 1,
or (w + y + z)/y < 3,
Thanks in advance
when y/ (w + z) > 1/2; (w + z)/y < 2,
or [(w + z)/y] + 1 < 2 + 1,
or (w + y + z)/y < 3,
Thanks in advance
-
sanju09
- GMAT Instructor
- Posts: 3650
- Joined: Wed Jan 21, 2009 4:27 am
- Location: India
- Thanked: 267 times
- Followed by:80 members
- GMAT Score:760
Take suitable numbers for the variables when you are in fix.digvijay.rajput wrote:In the above example, can you please explain these steps....
when y/ (w + z) > 1/2; (w + z)/y < 2,
or [(w + z)/y] + 1 < 2 + 1,
or (w + y + z)/y < 3,
Thanks in advance
E.g. if 3/4 > 1/2, then 4/3 < 2 RIGHT!
OR
If 5 > 1/2, then 1/5 < 2; the statement would consistently remain true. Do the remaining math likewise.
The mind is everything. What you think you become. -Lord Buddha
Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001
www.manyagroup.com
Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001
www.manyagroup.com
GMAT/MBA Expert
-
[email protected]
- Elite Legendary Member
- Posts: 10392
- Joined: Sun Jun 23, 2013 6:38 pm
- Location: Palo Alto, CA
- Thanked: 2867 times
- Followed by:511 members
- GMAT Score:800
Hi C,
This DS question can be beaten by TESTing values and a bit of logic:
The question asks if the probability of picking a yak is greater than the probability of picking a zebra.
Since we're dealing with a certain number of animals (and positive integers only), this question can be rewritten as: "Is Y > Z"?
Fact 1 tells us that Y/(W+Z) > 1/2
We could have
Y = 4
W = 1
Z = 1
The answer to the question is YES
We could also have
Y = 4
W = 1
Z = 5
The answer to the question is NO
Inconsistent = INSUFFICIENT
Fact 2 tells us that Z/(W+Y) < 1/2
We could have
Y = 4
W =1
Z = 1
The answer to the question is YES
We could also have
Y = 4
W = 10
Z = 5
The answer to the question is NO
Inconsistent = INSUFFICIENT
Now, we have to COMBINE Facts (and we get a number property shortcut). Since the W is a "fixed" value in both inequalities, it's the Y and Z that have to change to fit the > or <. In this scenario, by definition, the Y would have to be greater than the Z.
Final Answer: C
GMAT assassins aren't born, they're made,
Rich
This DS question can be beaten by TESTing values and a bit of logic:
The question asks if the probability of picking a yak is greater than the probability of picking a zebra.
Since we're dealing with a certain number of animals (and positive integers only), this question can be rewritten as: "Is Y > Z"?
Fact 1 tells us that Y/(W+Z) > 1/2
We could have
Y = 4
W = 1
Z = 1
The answer to the question is YES
We could also have
Y = 4
W = 1
Z = 5
The answer to the question is NO
Inconsistent = INSUFFICIENT
Fact 2 tells us that Z/(W+Y) < 1/2
We could have
Y = 4
W =1
Z = 1
The answer to the question is YES
We could also have
Y = 4
W = 10
Z = 5
The answer to the question is NO
Inconsistent = INSUFFICIENT
Now, we have to COMBINE Facts (and we get a number property shortcut). Since the W is a "fixed" value in both inequalities, it's the Y and Z that have to change to fit the > or <. In this scenario, by definition, the Y would have to be greater than the Z.
Final Answer: C
GMAT assassins aren't born, they're made,
Rich
-
Matt@VeritasPrep
- GMAT Instructor
- Posts: 2630
- Joined: Wed Sep 12, 2012 3:32 pm
- Location: East Bay all the way
- Thanked: 625 times
- Followed by:119 members
- GMAT Score:780
Maybe an easier way:
(If I'm reading this right, the original statements are y/(w+z) > 1/2 ... not (y/w) + z > 1/2. If I'm mistaken, ignore this post.)
The question is really "Are there more yaks than zebras?" or "is y > z?"
S1::
y/(w+z) > 1/2
2y > w + z
y + y > w + z
So y > w or y > z, but we don't know which. (Both could be true, for that matter.)
S2::
1/2 > z/(w+y)
w + y > 2z
w + y > z + z
Much like the above: w > z or y > z, but we don't know which.
Combining the two statements, we have two options:
Either y > z ... or y > w and w > z, implying y > z. So y > z.
(If I'm reading this right, the original statements are y/(w+z) > 1/2 ... not (y/w) + z > 1/2. If I'm mistaken, ignore this post.)
The question is really "Are there more yaks than zebras?" or "is y > z?"
S1::
y/(w+z) > 1/2
2y > w + z
y + y > w + z
So y > w or y > z, but we don't know which. (Both could be true, for that matter.)
S2::
1/2 > z/(w+y)
w + y > 2z
w + y > z + z
Much like the above: w > z or y > z, but we don't know which.
Combining the two statements, we have two options:
Either y > z ... or y > w and w > z, implying y > z. So y > z.
