8/12/2023 0 Comments Cow pig chicken math problem![]() ![]() Mr Daley bought 10 pigs, 24 goats and 66 sheep. Going back to the first equation we get s = 66. So 1 ≤ 36 m ≤ 59 (depending on the value of g). And since the original question implies that there is at least one of each animal, g is at least 1, and so 60 – g can’t equal 60. What values can m actually take? Since 36 m is positive (the original question implies that there is at least one of each animal), 60 – g has to be strictly positive. We can now substitute this back into the last equation to giveĪt the moment this doesn’t appear to be any better than the equation 18 p = 300 – 5 g, after all, it still has two unknowns, m and g. On the other hand, we know from the question that p has to be even. Since 18 and 5 have no factors in common (they are relatively prime) then 5 is a factor of p. It's interesting (and crucial) to note that 5 is a factor of the right hand side of this last equation. Can we solve this in some way? At this point we could continue with a systematic list using guess and check.Īnother approach, but one that will probably be too complex for most students involves rearranging the equation to give: If we eliminate s from these two equations we get 18p + 5 g = 300. The problem now is to get the equations into a form that will allow us to solve them. However, here we have the added fact that the solutions are whole numbers – we can’t have a fraction of an animal. In general there are no unique solutions in such situations. The difficulty that we are now faced with is that we have two equations and three unknowns. Let p, g, s be the numbers of pigs, goats, sheep, respectively. Solutionįrom the information in the question we can make up two equations. Swap your problem with someone else in your class. Reword this problem to put it into another setting, and write your own similar problem. What does this mean about possible answers for p? Questions that focus on the parameters for the variables can help the students to use number strategies instead of the systematic list approach:.A systematic list can be used once the s has been eliminated.Support in working with three variables may be needed but it is a good opportunity to apply elimination skills to the problem.Encourage the students to think about applying algebra to the problem. Have students try a systematic list, but there are a number of combinations to try due to the three variables.If we just use guess and check how will we know if we have all the possible answers? What strategy can we use to start this problem? How many of each animal did Mr Daley buy? Teaching Sequence It gives a really nice basic overview of what is required to raise chickens, pigs, sheep, goats, cows, horses and rabbits and might also help with your planning. Mr Daley paid $21 for each pig, $8 for each goat and $3 for each sheep. Suppose there were exactly twice as many chickens as. Then each group of 2 chickens and 1 cow would have a total of 11 legs and heads.6 groups of 2 chickens and 1 cow would have a total of 16 × 11 176 legs and heads. ![]() Altogether he bought 100 animals and spent $600. How many chickens and cows have 8 legs and heads altogetherlearly, 1 chicken and 1 cow have 8 legs and heads altogether. Mr Daley bought some pigs, goats and sheep. ![]()
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