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Kaplan Horse PS Problem

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Kaplan Horse PS Problem

by xxpatzz » Wed Jul 06, 2011 7:57 am
I know this one might not be a hard one but I am having some problems to solve it.

The smith horse farm has appaloosa, chestnut and palomino horses. The appaloosa to chestnut ratio is 1:4, and the chestnut to palomino ratio is 1:2/ The Windsor horse farm has appaloosa and chestnut horses in the ration of 1:5. Together, the two farms have 45 chestnut horses. How many appaloosa horses does Smith farm have?

A.5
B.7
C.8
D.9
E.11

Thanks,
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by abhisays » Wed Jul 06, 2011 8:14 am
This we can solve just by analyzing the options.

first of all let's say a1 = appaloosa horses in smith horse farm; c1 = chestnut horses in smith horse farm. now we have a1/c1 = 1/4

for windsor farm, we have a2 = appaloosa horses in windsor farm ; c2 = chestnut horses in windsor farm.

a2/c2 = 1/5

also we are given, c1 + c2 = 45 which means 4a1 + 5a2 = 45, if we analyze the equation we can say a1 has be multiple of 5.

and from options we have only option A which is multiple of five. Hence answer is A
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by Frankenstein » Wed Jul 06, 2011 8:15 am
Hi,
Smith horse farm - a:c 1:4 and c:p = 1:2
So, a:c:p = 1:4:8
i.e, x, 4x, 8x
Windsor horse farm - a:c = 1:5
i.e. a,c are y and 5y
Given that 4x + 5y = 45
Possible values of x and y are (10,1),(5,5)
So, x can be either 10 or 5.
Only 5 is present in options.

Hence, A
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by testprepDublin » Wed Jul 06, 2011 8:27 am
Taking the information given, we have:

Smith
A : C : P
1 : 4 : 8

Windsor
A : C
1 : 5

Cs + Cw = 45
Since Cs has to be a multiple of 4 and Cw has to be a multiple of 5, the only options here are Cs=20, Cw=25 and Cs=40, Cw=5.

If Cs=20, As has to be 5 (As:Cs=1:4). If Cs is 40, As has to be 10.

Looking at the options, I would go with 5. Though 10 seems like an equally acceptable answer to me. Should the question read "How many appaloosa horses could the Smith farm have?" or have I made a mistake?
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by Frankenstein » Wed Jul 06, 2011 8:40 am
testprepDublin wrote: Should the question read "How many appaloosa horses could the Smith farm have?" or have I made a mistake?
Hi,
Agreed. Wording of the question posted isn't great.
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by finites » Wed Jul 06, 2011 10:03 am
suffix 1 is smith's farm
suffix 3 is another guys farm.

a1/c1 = 1/4 ---- lets call it eq 1

a2/c2 = 1/5 ---- lets call it eq 2

now we know c1 + c2 45..

since ratios are given.. actual values will be multiple of it.

just trying to make make the denominators c1,c2 such that they sum to 45..

mulitply and divde 5 for eq 1 and eq2..

it will result in
eq1 becomes a1/c1 = 5/20
eq2 becomes a2/c2 = 5/25


Now u get a1 = 5 as c1 + c2 = 45

Another case can be formed with multiply and divide eq1 with 10 and eq2 with 1.
but it will not give the ans needed.
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by GMATGuruNY » Wed Jul 06, 2011 10:58 am
xxpatzz wrote:I know this one might not be a hard one but I am having some problems to solve it.

The smith horse farm has appaloosa, chestnut and palomino horses. The appaloosa to chestnut ratio is 1:4, and the chestnut to palomino ratio is 1:2/ The Windsor horse farm has appaloosa and chestnut horses in the ration of 1:5. Together, the two farms have 45 chestnut horses. How many appaloosa horses does Smith farm have?

A.5
B.7
C.8
D.9
E.11

Thanks,
We can plug in the answer choices, which represent the number of Appaloosa horses on the Smith farm.
Since Appaloosa:Chestnut = 1:4, the number of Chestnut horses on the Smith farm will be 4 times the number of Appaloosa Horses.

A: 5*4 = 20 Chestnut horses on the Smith farm, implying 45-20 = 25 Chestnut horses on the Windsor farm.
B: 7*4 = 28 Chestnut horses on the Smith farm, implying 45-28 = 17 Chestnut horses on the Windsor farm.
C: 8*4 = 32 Chestnut horses on the Smith farm, implying 45-32 = 13 Chestnut horses on the Windsor farm.
D: 9*4 = 36 Chestnut horses on the Smith farm, implying 45-36 = 9 Chestnut horses on the Windsor farm.
E: 11*4 = 44 Chestnut horses on the Smith farm, implying 45-44 = 1 Chestnut horse on the Windsor farm.

Since on the Windsor Farm the ratio of Appaloosa:Chestnut = 1:5, the number of Chestnut horses on the Windsor farm must be a multiple of 5.
In B, C, D and E, the number of Chestnut horses on the Windsor farm is not a multiple of 5. Eliminate B, C, D and E.

The correct answer is A.
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by amit2k9 » Thu Jul 07, 2011 12:54 am
the tip is that 1:5 should be used,where 5 is a factor for the number of C horses in W's farm.

thus for A=5, C=20 in S' farm.
meaning for A=5,C=25 in W's farm.

thus 20+25=45.
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