Alice Posted March 23, 2012 Share Posted March 23, 2012 How would you solve this...they haven't yet done Algebra or dividing by fractions. I can't figure out how to solve this without either Algebra or with the bar diagrams we keep ending up needing to divide by a fraction. There has to be a way but I just can't see it.... Thanks in advance... There were three times as many jelly beans in Jar A as in Jar B. After 2685 beans in Jar A were sold, Jar B had twice as many jelly beans as Jar A. How many more jelly beans were there in Jar A than in Jar B at first? I'm not good at drawing the bar diagram on the computer but we did that and we've gotten to the point of seeing that 2685 is equal to 2 1/2 B but he hasn't yet done dividing by fractions and I can't figure out how to do it without teaching that. I'm wondering if there is a way to set up the problem so we don't end up dividing by a fraction. Quote Link to comment Share on other sites More sharing options...
jennynd Posted March 23, 2012 Share Posted March 23, 2012 before A |--|--|--| B |--| After A |-| B |--| so 5 units get sold the child need to know break the A unit in half Quote Link to comment Share on other sites More sharing options...
nmoira Posted March 23, 2012 Share Posted March 23, 2012 There were three times as many jelly beans in Jar A as in Jar B. After 2685 beans in Jar A were sold, Jar B had twice as many jelly beans as Jar A. How many more jelly beans were there in Jar A than in Jar B at first? Before: Jar A (I've given six rather than three divisions to avoid dealing with a half in the after diagram) |-----|-----|-----|-----|-----|-----| Jar B |-----|-----| After: Jar A |----|<------------2658-----------> Jar B |-----|-----| The difference between Jar A before and after is 2685, so 5 of it's 6 divisions represent 2684. So one part is 537 and the original six parts are 3222. Quote Link to comment Share on other sites More sharing options...
Alice Posted March 23, 2012 Author Share Posted March 23, 2012 before A |--|--|--| B |--| After A |-| B |--| so 5 units get sold the child need to know break the A unit in half THANK YOU! That makes perfect sense. I just had one of those blocks before and couldn't see it. Quote Link to comment Share on other sites More sharing options...
Ali in OR Posted March 23, 2012 Share Posted March 23, 2012 I'm guilty of wanting to do Singapore problems with algebra. Instead of a pure bar diagram approach, we often used little "boxes" to represent problems, and my boxes could be thought of as representing a variable in a pure algebra approach. For this problem, I would have drawn two big jars. In jar B, I would have drawn two small little boxes. Why two? Because at the end of the problem, B ends up with *twice* as many as A. The amount of beans in B never changes, so those two boxes are the amount at the beginning and the amount at the end. The amount in A does change, but at the beginning of the problem, it has 3 times as many beans as B. I would have my dd figure out that if B has 2 boxes, how many should A have (6--3 times as many). How many are left in A at the end? Half as many as B, so only 1 left. So the difference, 5 boxes, has to account for all of the jelly beans that left the jar, 2685 beans. If 5 boxes have 2685 beans, how many are in one box? (divide by 5--537). Now, we found the amount in one of the little boxes. What did the book ask us to find? How many more in A than B. A was 6 boxes=3222. B was 2 boxes=1074. The difference is 2148. Is that right? Quote Link to comment Share on other sites More sharing options...
SKL Posted March 23, 2012 Share Posted March 23, 2012 My brain would have done this: 3B = A at first B = 2A at end C = A at end = B/2 A = 6C at first C = 2685/5 = 537 A at first = 537*6 or 537+2685 = 3222 B = 537*2 = 1074 Quote Link to comment Share on other sites More sharing options...
Recommended Posts
Join the conversation
You can post now and register later. If you have an account, sign in now to post with your account.