IMO c
chic=x;cow=y;sheep=z
eqn 1.... x+y=3z
eqn 2.....2x+4y=100 (feet- for chick 2 and cow 4)
eqn3 y> x,z

from eqn 2 x+2y=50( to solve this we need some hit and tral..we have to keep in mind y> x ; x+2y has to be div by 5 or 10)=> y=20,x=10, from eqn 1 z=10

OA pls

xcusemeplz2009 wrote:IMO c
chic=x;cow=y;sheep=z
eqn 1.... x+y=3z
eqn 2.....2x+4y=100 (feet- for chick 2 and cow 4)
eqn3 y> x,z

from eqn 2 x+2y=50( to solve this we need some hit and tral..we have to keep in mind y> x ; x+2y has to be div by 5 or 10)=> y=20,x=10, from eqn 1 z=10

OA pls

All of your chicken and cows are headless! The horror!

(Although I'm sure some evil genius is working on genetically modified headless chickens as we speak.)

100 is the total of heads and feet for chickens and cows, so there are actually 3/chicken and 5/cow (by the way, on the real GMAT you're not required to know any farm animal anatomy).

So:

co + ch = 3sh

5co + 3ch = 100

co > ch; co > sh

At this point we can do some trial and error, knowing that the number of chickens has to be a multiple of 5 to get that "100" total.

5 chickens (15 chicken head/feet) would mean we need 85/5 = 17 cows. 5+17 isn't a multiple of 3, so bzzt.

10 chickens (30 chicken head/feet) would mean we need 70/5 = 14 cows. 10+14 = 24, which is a multple of 3... yay! 24/3 = 8 sheep, choose (B).

Just to play:

15 chickens (45 chicken head/feet) would mean we need 55/5 = 11 cows. We now have more chickens than cows, which violates our rules (and leads to chaos on the farm), so we don't have to worry about any other possibilities.

Thanks Stuart for wonderful explanation

Stuart Kovinsky wrote:

xcusemeplz2009 wrote:IMO c
chic=x;cow=y;sheep=z
eqn 1.... x+y=3z
eqn 2.....2x+4y=100 (feet- for chick 2 and cow 4)
eqn3 y> x,z

from eqn 2 x+2y=50( to solve this we need some hit and tral..we have to keep in mind y> x ; x+2y has to be div by 5 or 10)=> y=20,x=10, from eqn 1 z=10

OA pls

All of your chicken and cows are headless! The horror!

(Although I'm sure some evil genius is working on genetically modified headless chickens as we speak.)

100 is the total of heads and feet for chickens and cows, so there are actually 3/chicken and 5/cow (by the way, on the real GMAT you're not required to know any farm animal anatomy).

So:

co + ch = 3sh

5co + 3ch = 100

co > ch; co > sh

At this point we can do some trial and error, knowing that the number of chickens has to be a multiple of 5 to get that "100" total.

5 chickens (15 chicken head/feet) would mean we need 85/5 = 17 cows. 5+17 isn't a multiple of 3, so bzzt.

10 chickens (30 chicken head/feet) would mean we need 70/5 = 14 cows. 10+14 = 24, which is a multple of 3... yay! 24/3 = 8 sheep, choose (B).

Just to play:

15 chickens (45 chicken head/feet) would mean we need 55/5 = 11 cows. We now have more chickens than cows, which violates our rules (and leads to chaos on the farm), so we don't have to worry about any other possibilities.

oh man what a blunder : .....
thanx stuart for correcting the error.... .

"100 is the total of heads and feet for chickens and cows, so there are actually 3/chicken and 5/cow (by the way, on the real GMAT you're not required to know any farm animal anatomy)."

How do you come up with the above logic under heavy duty stressful exam conditions...?
Can you please explain me how did you come up with the above logic? Both Chic and co can be multiples of 10 instead and still the relation can hold good.

Chicke = C
Sheep = S
Cow = Co

(c + Co) = 3 S ---> (1)

Co > C / S ---> (2)

Co-Head + Co-Feet + C-Head + C-Feet = 100 ---> (3)

S = ??? ---> Question
from equation (3)
(1+4)x No Of Co + (1 + 2)x No Of C = 100
5 Co + 3 C = 100 --->(4)

Combining equation (4) and (1)

2Co + 9S = 100
Now among the ans choice only S = 8 satisfy the conditions

I am finding a hard time comprehending and converting the below statement into an equation.

There are three times the number of chickens and cows than sheep.

Can someone please explain a manner in which I should read this particular sentence or similar sentences in future.

Thank you