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bblast: Legendary Member; Posts: 1077; Joined: Mon Dec 13, 2010 1:44 am; Thanked: 118 times; Followed by:33 members; GMAT Score:710

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by bblast » Mon May 30, 2011 8:20 am

A farm has chickens, cows and sheep. There are three times the number of chickens and cows than sheep. If there are more cows than chickens or sheep, and together, cows and chickens have a total of 100 feet and heads, how many sheep live at the farm?

5
8
10
14
17

B

picking numbers looks like the only solution after forming the equation and pulling on answer choices. But too cumbersome.

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cans: Legendary Member; Posts: 1309; Joined: Mon Apr 04, 2011 5:34 am; Location: India; Thanked: 310 times; Followed by:123 members; GMAT Score:750

by cans » Mon May 30, 2011 8:31 am

let no. of cows = c
no. of chickens = k
and no. of sheep = s
given: c+k=3s (Also c,k,s are integers)
Also cows have 4 feet and 1 head and chickens have 2 feet and a head
Thus 5c+3k =100
For k to be a number, k should be a multiple of 5.
If k=5 => c=17 but also s=(c+k)/3 = 22/3 not possible (not integer)
if k=10 => c=14 => s=24/3=8 (possible)
Thus no. of sheep = 8
IMO B

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bblast: Legendary Member; Posts: 1077; Joined: Mon Dec 13, 2010 1:44 am; Thanked: 118 times; Followed by:33 members; GMAT Score:710

by bblast » Mon May 30, 2011 9:08 am

cans wrote:let no. of cows = c
no. of chickens = k
and no. of sheep = s
given: c+k=3s (Also c,k,s are integers)
Also cows have 4 feet and 1 head and chickens have 2 feet and a head
Thus 5c+3k =100
For k to be a number, k should be a multiple of 5.
If k=5 => c=17 but also s=(c+k)/3 = 22/3 not possible (not integer)
if k=10 => c=14 => s=24/3=8 (possible)
Thus no. of sheep = 8
IMO B

Ya, thats actually the official solution and the way I solved it. But took me 3 min nd 21 seconds during the test.

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cans: Legendary Member; Posts: 1309; Joined: Mon Apr 04, 2011 5:34 am; Location: India; Thanked: 310 times; Followed by:123 members; GMAT Score:750

by cans » Mon May 30, 2011 11:39 am

Alternative solution :-
let no. of cows = c
no. of chickens = k
and no. of sheep = s
given: c+k=3s (Also c,k,s are integers)
Also cows have 4 feet and 1 head and chickens have 2 feet and a head
Thus 5c+3k =100
or 2c+3(c+k)=100 => 2c + 9s =100
Now use the options. s=5 =>2c=55 => c is not an integer. Wrong
s=8 =>c=14 correct
s=10 =>c=5 but its given that c>s, and thus not correct
s=14 => c<0 wrong
s=17 => c<0 wrong
Thus B

Thus no. of sheep = 8

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by Anurag@Gurome » Mon May 30, 2011 9:19 pm

bblast wrote:A farm has chickens, cows and sheep. There are three times the number of chickens and cows than sheep. If there are more cows than chickens or sheep, and together, cows and chickens have a total of 100 feet and heads, how many sheep live at the farm?

5
8
10
14
17

B

picking numbers looks like the only solution after forming the equation and pulling on answer choices. But too cumbersome.

Solution:
Let x - chicken, y - cows, z - sheep. So, y > x and y > z.
So, (x+y) = 3z.
Also, 3x+5y = 100.
Or 9z+2y = 100.
Or 2y =100-9z.
Now, 100-9z is a multiple of 2 and positive. So z has to be even and either 8 or 10. (14 will make it negative).
If z is 8, x+y =24 and 3x+5y = 100, x = 10 and y = 14 - possible because y > x and y > z.
If z is 10, x+y = 30 and 3x+5y = 100, x = 25 and y = 5 - not possible because y < x and y < z.
Hence, z = 8 is the correct answer

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