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M06 #12

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M06 #12 [#permalink] New post 17 Sep 2009, 04:55
A farm has chickens, cows and sheep with three times the number of chickens and cows, than there are sheep. If there are more cows present than chicken or sheep and together, cows and chickens have a total of 100 feet and heads, how many sheep are at the farm?

5
8
10
14
17
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Re: M06 #12 [#permalink] New post 17 Sep 2009, 21:58
tejal777 wrote:
A farm has chickens, cows and sheep with three times the number of chickens and cows, than there are sheep. If there are more cows present than chicken or sheep and together, cows and chickens have a total of 100 feet and heads, how many sheep are at the farm?

5
8
10
14
17


(ck+cw) = 3s
cw > ck
cw > s
cw > (ck+s)

3ck + 5cw = 100

1. if cw = 17, ck = 5
total = 22 that is not exactly divisible by 3 to get s.

2. if cw = 14, ck = 10
total = 24 that is exactly divisible by 3 to get s = 8. That works...

B.
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Re: M06 #12 [#permalink] New post 18 Sep 2009, 05:19
chicken=x,
cow=y,
sheep=z

3x+5y=100
x+y=3z
y>x, y>z, y>x+z we have all these in hand,
3x+3y=9z so I replace this in the first equation,
9z+2y=100,
z=100-2y/9 so this number should be divisible by 9,
z=10, z=8 both satisfies this but when z=10, y should be 5, because y>z, this does not satisfy the equation and thus we are left with z=8, y=14, which satisfy the inequality
so 8 is the answ.
Re: M06 #12   [#permalink] 18 Sep 2009, 05:19
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