M06-12

Official Solution:

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, \(S = 8\).


Answer: B
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I did this slightly differently.
We have 100 = 3C + 5W = 3 (C+W) + 2W = 9S + 2W.
Now we know W>S. If S = 14, 17, W will be negative which is not possible. So D, E are eliminated.
If S = 10, W=2 not possible since W>S. Not C.
If S = 5, then W = 55/2 which is not integer. Not possible.
So S has to be 8, which gives W = 14. Looks good.
Answer = B.

Could you please explain why C should be a multiple of 5?

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.

WHEN THE SUM OR THE DIFFERENCE OF NUMBERS IS A MULTIPLE OF AN INTEGER
1. If integers \(a\) and \(b\) are both multiples of some integer \(k>1\) (divisible by \(k\)), then their sum and difference will also be a multiple of \(k\) (divisible by \(k\)):
Example: \(a=6\) and \(b=9\), both divisible by 3 ---> \(a+b=15\) and \(a-b=-3\), again both divisible by 3.

2. If out of integers \(a\) and \(b\) one is a multiple of some integer \(k>1\) and another is not, then their sum and difference will NOT be a multiple of \(k\) (divisible by \(k\)):
Example: \(a=6\), divisible by 3 and \(b=5\), not divisible by 3 ---> \(a+b=11\) and \(a-b=1\), neither is divisible by 3.

3. If integers \(a\) and \(b\) both are NOT multiples of some integer \(k>1\) (divisible by \(k\)), then their sum and difference may or may not be a multiple of \(k\) (divisible by \(k\)):
Example: \(a=5\) and \(b=4\), neither is divisible by 3 ---> \(a+b=9\), is divisible by 3 and \(a-b=1\), is not divisible by 3;
OR: \(a=6\) and \(b=3\), neither is divisible by 5 ---> \(a+b=9\) and \(a-b=3\), neither is divisible by 5;
OR: \(a=2\) and \(b=2\), neither is divisible by 4 ---> \(a+b=4\) and \(a-b=0\), both are divisible by 4.


For more check here: divisibility-multiples-factors-tips-and-hints-174998.html
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first of all, I would like to thank you for your help
Actually, I didn't understand how 100 equal 1C+1W+4W+2C as well as why C should be a multiple of 5 for 3C+5W to add up to 100. and How we get these two pairs: (5, 17) and (10, 14).

Highly appreciate your help

1. We are given that together, cows and chickens have a total of 100 feet and heads. A cow has 1 head and 4 feet and a chicken has 1 head and 2 feet, so 100=1C+1W+4W+2C. This is also answered in the solution. Check highlighted parts below.

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: \(100 = 3C + 5W\).

2. Your second question is answered here: m06-183709.html#p1452616

3. This is also addressed in the solution - it's done by number picking.
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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, \(S = 8\).


Answer: B[/quote]


another pair of (5,25) is possible and as per this the answer can be C.10

Dear Sachin07, please note that all questions here are correct, so if you are getting answer which differs from the OA it means that you've done something wrong.

For example, c = 5 and w = 25 doe not give a sum of 100 for 3c + 5w:

3*10 + 5*25 =155.
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Let \(K\) be a number of chickens, \(C\) - cows, and \(S\) - sheep.

Translation the word question into formulas:

chickens and cows combined is 3 times the number of sheep: \(K\) + \(C\) = 3\(S\).......1

Chickens and cows together have 100 feet and heads: \(3K + 5C=100\). It is known that chicken has 1 head and 2 feet and cow has 1 head and 4 feet...........2

cows than chickens or sheep: C> K & S............3

From 1: K=3S - C

Apply into 2 , we get: 9S+2C=100..... Here I want to make one equation in sheep and cow to make use of the info in (3).

Let's analyze the equation above, 100 is even & 2C is even so 9S MUST be even too. Hence S is even number...........Eliminate choices A & E

Rearrange 2: C= 50- (9S)/2

Plug in choice B, S=8 .......C=14 ......C>S........our target answer.

You may want to examine choice C, S=10..........C=5......C<S... incorrect

Answer: B

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!

I am surprised more people in the comments above have not attempted an answer-first approach, given that this question is asking about a single unknown--i.e. what is the value of such-and-such? Such an approach can lead to a quick and assuredly accurate answer. (An aside: I agree that the phrasing of the question could be clarified a bit: instead of "a total of 100 feet and heads" which I myself interpreted as 100 feet and 100 heads (before realizing that such an interpretation proved untenable), the question could probably use a switch to "a total of 100 feet and heads combined," or the "and together" could be moved and tweaked, as in, "and cows and chickens have a total of 100 feet and heads altogether." No more room for confusion.) Why not start with a (B) or (D) answer to use as a gauge? I started with (D):

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
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I am unable to wrap my head around this equation. 100=1C+1W+4W+2C. I understand that co-efficient 4 & 2 are taken for legs, and co-efficients 1 & 1 are taken for heads. But how can the variables be the same? Number of cows and chickens are assumed W & C, so in the equation it means 4 cows and 2 chickens and not legs. Same for heads. Should we not assign different variables which would complicate the answer but would make sense?

Can you help me fill the gap here? Bunuel bb KarishmaB chetan2u

We are calculating number of heads and legs.
Now, a normal cow will have 4 legs and 1 head, so total 5 items per cow are to be calculated. If each cow has 5 and number of cows are C, then total items are 5C. Similarly, in case of chickens with 2 legs and a head, total number is 3X.
Total 5C+3X.

Rather if you have sheep, cow and horses and number of total heads + legs are 100, then 5A= 100 or A=20, that is total number of animals on the farm are 20.
We don’t require separate variables as they represent same thing, a thing with 4 heads and 1 leg.
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Different ways-

(1) Use options

Start with middle number - 10, so number of cows and chickens =30.
Minimum Number of heads and legs of cows and chickens combined = 16*5+14*3 = 80+42 =122>100
So, it has to be less than 10.

Next number - 8, so number of cows and chickens =24.
Minimum Number of heads and legs of cows and chickens combined = 13*5+11*3 = 65+33 =98<100
Max number = 24*5=120
So, it fits in as we can have more cows- 14, 15 and so on. We will check the option again if A also fits in.

Next number - 5, so number of cows and chickens =15.
Minimum Number of heads and legs of cows and chickens combined = 8*5+7*3 = 40+21 =71<100
Maximum number of heads and legs = 5*15=75, which is still less than 100.


(2) Algebraic

Let the number of cows, sheep and chickens be c, s and x.
Each cow and sheep will have 4 legs and 1 head, so 5, while each chicken will have 3.
Thus, 5c+3x=100. We also know c>x and we can say that x is a multiple of 5 and c will not be a multiple of 3.

So we take x=5 to get 5C+15=100 or c=17.
Next, x=10 to get 5C+30=100 or c=14.
Next, x=15 to get 5C+45=100 or c=11. But here on x>c so discard.
Possible values of (c,x): (17,5) and (14,10)

Now, c+x=3*s, so c+x has to be a multiple of 3.
17+5=22, which is not a multiple of 3. Discard
14+10=24=8*3. Perfect and x=8.


B
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Take some values to understand this. Say I have 1 cow. How many feet and heads do I have? 4 feet and 1 head so total 5 feet and heads.
Say I have 1 cow and 1 chicken. How many feet and heads do I have now? Total = 4 feet (from cow) + 2 feet (from chicken) + 1 head from cow + 1 head from chicken = 8
Now what if I have 3 cows and 2 chickens?
I will have Total = 4*3 feet (from cows) + 2*2 feet (from chickens) + 1*3 heads from cows + 1*2 heads from chickens

So if C were the number of cows and K the number of chickens,
Total = 4*C + 2*K + 1*C + 1*K

This is the equation used.
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