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TriC-MathMOOC 17 Online
OpenStudy (anonymous):

A machine makes big widgets.Another machine makes small widgets.Three small widgets have the same mass as two big widgets.It takes the same amount of time to make 3 big widgets as it does to make 5 small widgets.The machines start at the same time and make widgets until the total mass of all of the widgets made is equal to the mass of 380 small widgets.What is the total no. of widgets made?

OpenStudy (kropot72):

Let x be the mass of a small widget and let y be the mass of a big widget. 3x = 2y y = (3/2)x Let n be the total number of widgets made. The number of small widgets made = (5/8)n The number of big widgets made = (3/8)n Do you follow the above working?

OpenStudy (kropot72):

@tanusreebag Are you there?

OpenStudy (kropot72):

The total mass is the mass of 380 small widgets which equals 380x. So now we can write the following equation: \[(\frac{5}{8}n \times x)+(\frac{3}{8}n \times \frac{3}{2}x)=380x\ .........(1)\] Equation (1) can be simplified to: \[n(\frac{5}{8}x+\frac{9}{16}x)=380x\ .........(2)\] which further simplifies to: \[n(\frac{19}{16})x=380x\ ...........(3)\] So finally we get: \[n=\frac{380\times16}{19}=you\ can\ calculate\]

OpenStudy (anonymous):

Thank u very much for ur reply.

OpenStudy (kropot72):

You're welcome :)

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