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Intern  B
Joined: 27 Mar 2017
Posts: 4
GMAT 1: 720 Q50 V38 Show Tags

It is always so time taking to arrive at the integer values of the equations like this by picking numbers! Is there any material on how we can quicken this step?
Thank you!
Intern  Joined: 30 Oct 2017
Posts: 1

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(5,25) clearly violates the condition that the total number of heads and feet between the cows and chickens equals 100, so C cannot be a possible answer. (5,25) = 5(3) + 25 (5)
= 15 + 125 = 140.

Sachin07 wrote:
A farm has chickens, cows and sheep. The number of chickens and cows combined is 3 times the number of 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?

A. 5
B. 8
C. 10
D. 14
E. 17

Let $$C$$ be a number of chickens, $$W$$ - cows, and $$S$$ - sheep.

Chickens and cows together have 100 feet and heads: $$100=1C+1W+4W+2C$$, assuming that 1 is the number of heads per either unit and 4 and 2 are the number of legs per $$C$$ and $$W$$ accordingly. Therefore: $$100 = 3C + 5W$$.

It is given that $$W \gt C$$, and we also see that $$C$$ should be a multiple of 5 for $$3C + 5W$$ to add up to 100. Picking a few numbers, we get two pairs: (5, 17) and (10, 14).

$$S = \frac{(C+W)}{3}$$; $$5 + 17 = 22$$ which is not divisible by 3, but $$10 + 14 = 24$$ which is divisible by 3, therefore only the second pair will work. So, $$S = 8$$.

another pair of (5,25) is possible and as per this the answer can be C.10
[/quote]
Intern  B
Joined: 16 Jul 2017
Posts: 1

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First post here, so sorry if it is put in the wrong place.

This is a solution to - M06-12

Chicken + Cows = 3 times Sheep
=> Chicken + Cows + Sheep = 3 Sheep + Sheep = 4 Sheep

Therefore, the answer need to be a multiple of 4. Only one choice i.e. 8.

Thanks!
Senior Manager  P
Joined: 27 Dec 2016
Posts: 309

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Hi Bunuel,

I was wondering could you please explain this concept one more time? You mentioned in your OE that, "3C + 5W = 3C + (a multiple of 5) = 100 = (a multiple of 5), thus 3C must also be a multiple of 5, which means that C must be a multiple of 5". I understood that 100 = (a multiple of 5) so we have, 3C + (a multiple of 5) = (a multiple of 5). What I couldn't understand was how did we deduce that 3C is also a multiple of 5 and then concluded that C must be a multiple of 5 as well? Could you please explain this part? Would greatly appreciate it?
Intern  S
Joined: 10 May 2018
Posts: 29

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seems impossible to solve under 2 minutes... It's unlikely to encounter something like this on the GMAT
VP  P
Joined: 14 Feb 2017
Posts: 1200
Location: Australia
Concentration: Technology, Strategy
Schools: LBS '22
GMAT 1: 560 Q41 V26 GMAT 2: 550 Q43 V23 GMAT 3: 650 Q47 V33 GMAT 4: 650 Q44 V36 WE: Management Consulting (Consulting)

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Here's a potential alternative solution I used to solve this correctly.

Equation 1: Chicken + Cow = 3*Sheep

Cow > Chicken

Reverse plug-in the answer choices

Cow + Chicken = 100 feet and heads

Test (b) 8 for sheep in Equation 1

Ch + Cow = 3(8)
Ch + Cow = 24
Cow> Ch
Try Cow = 13, Ch = 11
13*4 legs on a cow + 11*2 feet on chicken + 24 heads = 52 + 22 + 24 = 98

Bit short here, but we can tweak it

Try cow = 14, ch = 10
14*4 + 10*2 + 24 = 56+20+24 = 100

Perfect

I tested AC (D) first and I wasn't able to get it within the desired 100 legs, feet and heads, so I moved straight to (B).
_________________
Goal: Q49, V41

+1 Kudos if I have helped you
Manager  G
Joined: 16 May 2019
Posts: 92

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1
Bunuel wrote:
A farm has chickens, cows and sheep. The number of chickens and cows combined is 3 times the number of 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?

A. 5
B. 8
C. 10
D. 14
E. 17

If there are 14 sheep, then 3 times the number of sheep means that there are 14 * 3, or 42 chickens and cows combined. Knowing that there must be more cows than chickens within this barnyard subgroup, then at a bare minimum, there will be 22 cows. 22 cows = 22 * 4 (legs) + 22 (heads) = 88 + 22 = 110 extremities. Too high. There is no need to even bring in the chickens. Scrap (D) and (E). Importantly, we can now test (B), knowing that the number in the middle of the remaining responses will either be the answer itself or point directly to what the answer needs to be (higher or lower). Assume there are 8 sheep and repeat the process:

If there are 8 sheep, then there will be 8 * 3, or 24 chickens and cows combined. At a bare minimum, there will be 13 cows. 13 * 4 (legs) + 13 (heads) = 52 + 13 = 65 extremities. Time for the chickens: 11 * 2 (legs) + 11 (heads) = 22 + 11 = 33 extremities. 65 (cow extremities) + 33 (chicken extremities) = 98 extremities. Our extremity count is a little low, but 98 is tantalizingly close, and if you can appreciate that the cow-to-chicken extremity ratio (perhaps the first time in the English language these words have been paired together) is 5:3 (from 4 legs + 1 head to 2 legs + 1 head), then you can see that all we need to do is swap out a chicken for a cow. For the must-have-proof nagging voice in your mind...

If there are 14 cows and 10 chickens, then there will be (14 * 4 + 14) + (10 * 2 + 10) extremities, and that we can work quickly: (56 + 14) + (20 + 10) = (70) + (30) = 100. Can we tick all the boxes of the question stem?

Farm Animals:
chickens
cows
sheep

Chickens + cows = 3 times the number of sheep:
10 chickens + 14 cows = 3 * 8 sheep
24 = 24

More cows than chickens or sheep:
14 cows, 10 chickens, 8 sheep

Cows and chickens have a total of 100 feet and heads together:
(See above)

There is no room for doubt. The problem took me a little over a minute, even with the setback in the beginning of misinterpreting the extremity count. I had a lot of fun with this question, even if I felt a bit like a butcher. I hope maybe you enjoyed the problem, too (if you made it this far).

- Andrew Re: M06-12   [#permalink] 06 Oct 2019, 15:56

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