tag:blogger.com,1999:blog-7142973019586201955.post9156529020274571061..comments2015-03-10T16:10:32.865-05:00Comments on Math Rescue: Math Brain Teasers: The SolutionsVictoria Tnoreply@blogger.comBlogger4125tag:blogger.com,1999:blog-7142973019586201955.post-20652435824868116392015-03-10T16:10:32.865-05:002015-03-10T16:10:32.865-05:00Richard, That is a very good way to approach this ...Richard, That is a very good way to approach this problem. I like how you've laid things out into columns, with a row for each operation/step in the word problem. It provides a helpful visual. :-)Victoriahttp://www.blogger.com/profile/03915348031548489924noreply@blogger.comtag:blogger.com,1999:blog-7142973019586201955.post-80034490689900015262015-03-10T15:33:17.959-05:002015-03-10T15:33:17.959-05:00My solution is essentially the same as Victoria...My solution is essentially the same as Victoria's, but I broke it up into little pieces.<br /><br />Let j and s be the number of coins that Jones and Smith start out with, respectively. Then the story goes like this:<br /><br />......................................... Jones.......................Smith........<br /><br />Starting out.......................... j...............................s<br /><br />Jones gives Smith<br />what Smith has................. j - s..........................2s<br /><br />Smith gives Jones<br />what Jones has..............2(j - s)..................2s - (j - s)<br /><br />or......................................2j - 2s.....................3s - j<br /><br />which is..............................36..........................42<br /><br /><br />Now we have a system of two simultaneous<br />equations, namely<br /><br />...................... 2j - 2s = 36<br />........................ 3s - j = 42<br /><br />which gives us<br /><br />..................... j = 48......and......s = 30<br /><br />So Jones started out with 48 coins, and Smith started out with 30.<br /><br />--Richard<br />Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-7142973019586201955.post-83617834602128651492014-12-03T10:15:33.628-06:002014-12-03T10:15:33.628-06:00That is an interesting way of approaching these pr...That is an interesting way of approaching these problems. However, this can only be done if the final values are given. I can see how it is much faster, though!Victoriahttp://www.blogger.com/profile/03915348031548489924noreply@blogger.comtag:blogger.com,1999:blog-7142973019586201955.post-62153021025964481802014-12-03T10:10:09.600-06:002014-12-03T10:10:09.600-06:00Without equations, I tend to solve problems like c...Without equations, I tend to solve problems like coin counting by working backwards<br />Jones ended with 36 and Smith with 42<br />The previous action doubled Jones coins (meaning he had 18 and got 18 more from Smith), which meant that Smith had 42 + 18 = 60 before that.<br /><br />Now this 60 that Smith had was double his original number of 30 (after he go 30 from Jones), which means that Jones started with 18 + 30 = 48<br /><br />To confirm: <br /> Smith Jones<br />Start 30 48<br />Jones gives 30 to Smith<br /> 60 18<br />Smith gives 18 to Jones<br /> 42 36Anonymousnoreply@blogger.com