Caitilin Posted October 20, 2011 Share Posted October 20, 2011 Can someone help me understand how to solve this problem? I don't understand how to solve it the SM way, and I can't come up with the answer key's answer doing it algebraically. Help, please? :D Thanks! Here is the problem: Larry spent 1/2 of his money on a camera, and another 1/8 on a radio. The camera cost $120 more than the radio. How much money did he have at first? Quote Link to comment Share on other sites More sharing options...
cbollin Posted October 20, 2011 Share Posted October 20, 2011 (edited) I found it helps to think equivalent fractions and draw the bar diagrams for it. 1/2 is 4/8.. so the camera is 4 out of 8 bars. camera ___ ____ ___ ___ radio ___ other money is ___ ___ ___ the camera is 120 more than radio. That means 120 dollars is divided over those 3 more bars. each bar is 40 there are 8 bars total. 40*8=320 sound right? -crystal Edited October 20, 2011 by cbollin Quote Link to comment Share on other sites More sharing options...
Snickerdoodle Posted October 20, 2011 Share Posted October 20, 2011 Here courtesy of my son: Quote Link to comment Share on other sites More sharing options...
Caitilin Posted October 20, 2011 Author Share Posted October 20, 2011 Yes, thanks! It is so simple once you see it, but before you see it...:tongue_smilie: Thank-you! Quote Link to comment Share on other sites More sharing options...
Guest homework_help Posted April 24, 2013 Share Posted April 24, 2013 :confused1: :closedeyes: :confused1: Not getting thisugh why cant this be easier Quote Link to comment Share on other sites More sharing options...
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