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Mathematics 19 Online
OpenStudy (butterflyhope):

Urgent (Medal): An alloy is composed of two parts copper (Cu) to one part aluminum (Al). How much of each metal is needed to make 105 lb. of the alloy? Thanks c:

OpenStudy (anonymous):

The ratio is given to you, for 2 parts of Cu, one part of Al is used.. So: \[\frac{Cu}{Al} = \frac2{1}{}\]

OpenStudy (anonymous):

And \(Cu : Al = 2 : 1\) Let the amount of \(Cu\) used is : \(2x\) So, the amount of \(Al\) used is : \(x\)..

OpenStudy (anonymous):

When ratio is given, let the quantity given to you according to ratios only..

OpenStudy (anonymous):

So, \(2x + x = 105\) \(3x = 105\) Find \(x\) from here.. \(x\) is your \(Al\), for \(Cu\), find \(2x\)..

OpenStudy (anonymous):

And I am not typing.. :P

OpenStudy (butterflyhope):

So: Aluminum = 3x = 105 Divide 3 on both sides, so you're left with x = 35. Copper: 2x = 105 Divide 2 on both sides, so you're left with x = 52.5?

OpenStudy (butterflyhope):

@waterineyes

OpenStudy (anonymous):

No..

OpenStudy (anonymous):

you have found \(x= 35\) which is right.. Find \(2x\) for \(Cu\) means : Multiply \(x\) with 2 after finding \(x\).. :P

OpenStudy (anonymous):

Getting?

OpenStudy (anonymous):

If \(x = 35\), then what is \(2x\) ??

OpenStudy (butterflyhope):

70?

OpenStudy (butterflyhope):

So you need 70% copper and 35% aluminum?

OpenStudy (anonymous):

yep

OpenStudy (butterflyhope):

Thx ^-^

OpenStudy (anonymous):

So, amount of Copper : 70 lb amount of Aluminium : 35 lb.

OpenStudy (anonymous):

\(\text{This is it...}\)

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