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16 Sep 2014, 00:27
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On a farm, there are chickens, cows, and sheep. The total number of chickens and cows is three times the number of sheep. If there are more cows than either chickens or sheep, and the combined total of heads and feet of chickens and cows is 100, how many sheep are on the farm?
A. 5
B. 8
C. 10
D. 14
E. 17
A. 5
B. 8
C. 10
D. 14
E. 17
16 Sep 2014, 00:27
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Official Solution:
On a farm, there are chickens, cows, and sheep. The total number of chickens and cows is three times the number of sheep. If there are more cows than either chickens or sheep, and the combined total of heads and feet of chickens and cows is 100, how many sheep are on the farm?
A. 5
B. 8
C. 10
D. 14
E. 17
Let's denote the number of chickens as \(c\), cows as \(w\), and sheep as \(s\).
We know that the combined total of heads and feet for chickens and cows is 100, so we can construct the equation: \(100 = (1c + 1w) + (2c + 4w)\). In this equation, 1 represents the number of heads per animal, while 2 and 4 denote the numbers of legs for a chicken (\(c\)) and a cow (\(w\)) respectively. Simplifying, we get: \(100 = 3c + 5w\).
Since both 100 and \(5w\) are multiples of 5, to satisfy \(100 = 3c + 5w\), \(c\) must also be a multiple of 5. If it isn't, then \(3c + 5w\) would become a sum of a non-multiple of 5 and a multiple of 5, which cannot be a multiple of 5. Considering multiples of 5 for \(c\) and keeping in mind the constraint of \(w > c\) (it's specified that there are more cows than either chickens or sheep), we find two fitting pairs: \(c=5\), \(w=17\) and \(c=10\), \(w=14\).
Finally, since the total number of chickens and cows is three times the number of sheep, we have \(c + w = 3s\). The pair \(c=5\), \(w=17\) doesn't satisfy this equation, as \(5 + 17 = 22\) isn't divisible by 3. However, \(10 + 14 = 24\) is divisible by 3, making this pair valid. As a result, for \(10 + 14 = 3s\), we find that \(s = 8\).
Answer: B
On a farm, there are chickens, cows, and sheep. The total number of chickens and cows is three times the number of sheep. If there are more cows than either chickens or sheep, and the combined total of heads and feet of chickens and cows is 100, how many sheep are on the farm?
A. 5
B. 8
C. 10
D. 14
E. 17
Let's denote the number of chickens as \(c\), cows as \(w\), and sheep as \(s\).
We know that the combined total of heads and feet for chickens and cows is 100, so we can construct the equation: \(100 = (1c + 1w) + (2c + 4w)\). In this equation, 1 represents the number of heads per animal, while 2 and 4 denote the numbers of legs for a chicken (\(c\)) and a cow (\(w\)) respectively. Simplifying, we get: \(100 = 3c + 5w\).
Since both 100 and \(5w\) are multiples of 5, to satisfy \(100 = 3c + 5w\), \(c\) must also be a multiple of 5. If it isn't, then \(3c + 5w\) would become a sum of a non-multiple of 5 and a multiple of 5, which cannot be a multiple of 5. Considering multiples of 5 for \(c\) and keeping in mind the constraint of \(w > c\) (it's specified that there are more cows than either chickens or sheep), we find two fitting pairs: \(c=5\), \(w=17\) and \(c=10\), \(w=14\).
Finally, since the total number of chickens and cows is three times the number of sheep, we have \(c + w = 3s\). The pair \(c=5\), \(w=17\) doesn't satisfy this equation, as \(5 + 17 = 22\) isn't divisible by 3. However, \(10 + 14 = 24\) is divisible by 3, making this pair valid. As a result, for \(10 + 14 = 3s\), we find that \(s = 8\).
Answer: B
08 Dec 2014, 05:03
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mat88
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
