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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 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, \(S = 8\).

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.

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.

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).

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

I think this is a high-quality question and the explanation isn't clear enough, please elaborate. "4 and 2 are the number of legs per C and W accordingly" - wrong

a trully 700-lvl question...I incorrectly answered, but now I see why I did the mistake.

cows have 4 legs chicks have 2 legs. cow and chicks have 1 head each :D suppose cows=K, chicks =C. we thus have: 4K+2C+K+C=100 5K+3C=100 We also know that K+C=3S. we then must find values for K and C so that K+C would be divisible by 3. C=1 - won't work C=2 wont work C=3 wont work C=4 - won't work C=5 -> K=17. K+C=22 - won't work. C=8 - won't work C=9 - won't work C=10 -> K=14 -> C+K=24 - works. C=11 no C=12 no C=15 -> K=11. won't work, since K must be > than C.

as we see, only K=14, and C=10 works. K+C=3S 24=3S S=8

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

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

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:

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