anneinco Posted March 2, 2012 Share Posted March 2, 2012 I have a book around that is supposed help me through the Singapore bar model process but we recently moved and I can not locate it... We reached page 166 in cwp2 today and I know there is a way to solve problems 2 and 4 with the bar model but I am not sure I am on the right path or I lucked out. (3 is probably doable with a bar model too but she used fractions). The problem is: a bicycle has 2 wheels and a tricycle has 3 wheels. There are all together 11 bicycles and tricycles with a total of 28 wheels. How many bicycles and how many tricycles are there? Not sure how to transcribe here how I have it on paper, but I have for every bicycle and tricycle there are 5 wheels. In order to have 11, I would have 5 groups (each with 1 bike and 1 trike) with 1 unknown. The five groups equal 25 wheels, so the 11 must be a trike with 3 wheels. So 5 bikes and 6 trikes... Is this correct? Worked through just guessing and then figured this out and it appears to work for the two problems (2 and 4). Thanks for the help! Quote Link to comment Share on other sites More sharing options...
anneinco Posted March 3, 2012 Author Share Posted March 3, 2012 Bump Quote Link to comment Share on other sites More sharing options...
zoo_keeper Posted March 3, 2012 Share Posted March 3, 2012 Honestly, I think we just tried the different combinations of 11 cycles and wrote down the resulting number of wheels until we stumbled across the answer. Quote Link to comment Share on other sites More sharing options...
anneinco Posted March 3, 2012 Author Share Posted March 3, 2012 Honestly, I think we just tried the different combinations of 11 cycles and wrote down the resulting number of wheels until we stumbled across the answer. That is what I did originally and was showing my daughter the process and st happend to pick the correct numbers the first time. I am still hoping that I am on the correct path with figuring the bar model process out. I have not look at book three much am sure it is going to continue to get more difficult. I really need to locate the book. No clue where it could be. Quote Link to comment Share on other sites More sharing options...
jennynd Posted March 3, 2012 Share Posted March 3, 2012 assume all are bikes. 11*2 .. u get 22 wheels. the different between 22 and 28 (6) is the extra wheel for the trike Quote Link to comment Share on other sites More sharing options...
zoo_keeper Posted March 3, 2012 Share Posted March 3, 2012 assume all are bikes. 11*2 .. u get 22 wheels. the different between 22 and 28 (6) is the extra wheel for the trike Well goodness, now I feel silly! That certainly is easier, thanks :) Quote Link to comment Share on other sites More sharing options...
anneinco Posted March 3, 2012 Author Share Posted March 3, 2012 assume all are bikes. 11*2 .. u get 22 wheels. the different between 22 and 28 (6) is the extra wheel for the trike Thank you That makes sense and is something my daughter will understand. I found a similar problem in cwp3 and used the same idea to figure it out. In the answer there, they show the same idea. Quote Link to comment Share on other sites More sharing options...
letsplaymath Posted March 4, 2012 Share Posted March 4, 2012 your original solution works fine when the numbers of each type of thing are close together. the other solution (assume all the things are one type, and then adjust as needed to make the total come out right) works more generally -- even if there were 9 bikes and only 2 trikes, it would find the answer for you. Quote Link to comment Share on other sites More sharing options...
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