Guest Posted November 8, 2008 Report Share Posted November 8, 2008 In the park one day, I saw a great assortment of children and dogs. Counting heads, I saw 22. Counting legs, I saw 68. How many children and dogs were in the park? Quote Link to post Share on other sites

0 Guest Posted November 8, 2008 Report Share Posted November 8, 2008 12 dogs 10 people WAY TOO EASY Quote Link to post Share on other sites

0 unreality 1 Posted November 8, 2008 Report Share Posted November 8, 2008 In the park one day, I saw a great assortment of children and dogs. Counting heads, I saw 22. Counting legs, I saw 68. How many children and dogs were in the park? c = children, d = dogs c+d = 22 2c+4d = 68 subtract the first from the second: (2c+4d)-(c+d) = 68-22 c+3d = 46 subtract the first from this (c+3d)-(c+d) = 46-22 2d = 24 d = 12 which means c has to be 22-12 = 10 10 children, 12 dogs checking: 20 legs + 48 legs = 68 legs ~~~ alternatively, you could do it this way: c+d = 22 2c+4d = 68 divide the second equation by 2 c+d = 22 c+2d = 34 subtract the first from the second, and you get d = 12, and it follows that c = 10 Quote Link to post Share on other sites

0 Guest Posted November 8, 2008 Report Share Posted November 8, 2008 12 dogs= 48 heads +10 people=20 heads -------------------- 22 heads =68 heads Quote Link to post Share on other sites

0 Guest Posted November 10, 2008 Report Share Posted November 10, 2008 Correct; there were 10 children and 12 dogs. Quote Link to post Share on other sites

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In the park one day, I saw a great assortment of children and dogs. Counting heads, I saw 22. Counting legs, I saw 68.

How many children and dogs were in the park?

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