AEC Posted October 3, 2012 Share Posted October 3, 2012 [note: I checked singaporemath.com forum but all the threads are missing??] I've either missed something or SM has. pg 60, #10 in CWP 4. "2/5 of Ian's marbles is equal to 3/7 of Roy's marbles. Roy has 55 more marbles and Ian. How many marbles to they have together." I keep coming up with Ian and Roy each having a negative number of marbles. this is, of course, not the method they'd suggest, but still... r = i+55 2/5 i = 3/7 r 2/5 i = 3/7 (i+55) 14/35 i = 15/35 i + 165/7 -i/35 = 165/7 i = -825 Is this problem misstated, or am I just lame tonight? Quote Link to comment Share on other sites More sharing options...
Arcadia Posted October 3, 2012 Share Posted October 3, 2012 2/5 i = 3/7 r 14/35 i = 15/35 r i = 15/14 r Ian would have more marbles than Roy Quote Link to comment Share on other sites More sharing options...
SunnyDays Posted October 3, 2012 Share Posted October 3, 2012 I spent some time trying to figure this out late last night. Apparently I need more math education myself, as I'm still not getting it, LOL!! Quote Link to comment Share on other sites More sharing options...
AEC Posted October 3, 2012 Author Share Posted October 3, 2012 2/5 i = 3/7 r 14/35 i = 15/35 r i = 15/14 r Ian would have more marbles than Roy precisely. And yet, the problem then states that Roy has 55 more marbles than Ian. (which is the second equation you need to be able to solve w/ two unknowns). I think the problem statement is broken. I suspect they meant to switch the two names in the last sentence. Or, perhaps it just neglects to mention that both Roy and Ian are running a huge marble debt. :001_smile: Quote Link to comment Share on other sites More sharing options...
Arcadia Posted October 3, 2012 Share Posted October 3, 2012 I think the problem statement is broken. I suspect they meant to switch the two names in the last sentence. Or, perhaps it just neglects to mention that both Roy and Ian are running a huge marble debt. :001_smile: Or just change this math problem into a logic problem of "prove that there is no possible solution assuming answers have to be positive" I did not buy the CWP but I have encountered typing mistakes in other books. Quote Link to comment Share on other sites More sharing options...
JadeOrchidSong Posted October 3, 2012 Share Posted October 3, 2012 (edited) I calculated it out and my answer is that they have 1595 altogether. Is that correct? 4/5 of one boy equal 6/7 of the other. One part of each boy is left. The bigger part is 1.5 the smaller part. So half of the smaller part is 55. Then the whole one part of the smaller one is 110. The total marbles of the small boy has 110x7 and the bigger boy has a total of 1.5x110x5. So the total is 770+825=1595. Hope it helps! I wish I could draw a good graph to show you how I got it. - - - - - - - - - - - - Edited October 3, 2012 by aomom Quote Link to comment Share on other sites More sharing options...
Snickerdoodle Posted October 3, 2012 Share Posted October 3, 2012 I calculated it out and my answer is that they have 1595 altogether. Is that correct?4/5 of one boy equal 6/7 of the other. One part of each boy is left. The bigger part is 1.5 the smaller part. So half of the smaller part is 55. Then the whole one part of the smaller one is 110. The total marbles of the small boy has 110x7 and the bigger boy has a total of 1.5x110x5. So the total is 770+825=1595. Hope it helps! I wish I could draw a good graph to show you how I got it. - - - - - - - - - - - - The way that would work is if Ian had 55 more marbles than Roy. OP: What does the answer say? This is in the new version of CWP? Quote Link to comment Share on other sites More sharing options...
JadeOrchidSong Posted October 3, 2012 Share Posted October 3, 2012 The way that would work is if Ian had 55 more marbles than Roy. OP: What does the answer say? This is in the new version of CWP? I totally forgot the names of the boys. Ian has to have more marbles than Roy because as I stated Ian's one part equals Roy's x1.5, so Ian's 1 part - Roys's 1 part equals 1/2 of Roy's part, which is 55. So Roy's one part is 110 and Ian's one part is 165. I believe there is a mistake for the order of the boys' names. I checked my US edition CWP 4 and didn't find this math problem anywhere in the whole book. It must be the new CWP 4 in print now. Quote Link to comment Share on other sites More sharing options...
Snickerdoodle Posted October 3, 2012 Share Posted October 3, 2012 I totally forgot the names of the boys. Ian has to have more marbles than Roy because as I stated Ian's one part equals Roy's x1.5, so Ian's 1 part - Roys's 1 part equals 1/2 of Roy's part, which is 55. So Roy's one part is 110 and Ian's one part is 165. I believe there is a mistake for the order of the boys' names. I checked my US edition CWP 4 and didn't find this math problem anywhere in the whole book. It must be the new CWP 4 in print now.FWIW, I agree with you. The names in the problem must be reversed. Quote Link to comment Share on other sites More sharing options...
AEC Posted October 4, 2012 Author Share Posted October 4, 2012 replies to several previous posts... I calculated it out and my answer is that they have 1595 altogether. Is that correct? 4/5 of one boy equal 6/7 of the other. that would be correct, except that the problem states 4/5 of one boy = 3/7 of the other. I think we've all agreed - the problem is stated incorrectly. Either they've swapped the names of the boys or there is a typo in one of the fractions. Upon reflection, the answer in the back of the book is consistent with the names of the boy's having been swapped in one of the stmts (or the 3/7th fraction being actually 6/7th) For the person who asked... This is CWP 4, copyright 2010...so yes, I'd guess that's the 'new' one. I didn't see it on the CWP errata list, so I've sent it to them. Quote Link to comment Share on other sites More sharing options...
JadeOrchidSong Posted October 4, 2012 Share Posted October 4, 2012 replies to several previous posts... that would be correct, except that the problem states 4/5 of one boy = 3/7 of the other. I think we've all agreed - the problem is stated incorrectly. Either they've swapped the names of the boys or there is a typo in one of the fractions. Upon reflection, the answer in the back of the book is consistent with the names of the boy's having been swapped in one of the stmts (or the 3/7th fraction being actually 6/7th) For the person who asked... This is CWP 4, copyright 2010...so yes, I'd guess that's the 'new' one. I didn't see it on the CWP errata list, so I've sent it to them. 2/5 x 2 one boy= 3/7 x 2 the other boy What does the answer key say? Is 1595 correct for their total marble #? Quote Link to comment Share on other sites More sharing options...
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