Question #91298

3 Answers
Jun 25, 2017

Answer:

The mass of the #30%# alloy is #=10kg# and the mass of the #70%# alloy is #=40kg#

Explanation:

This is a mass balance with respect to copper.

Let the mass of the #30%# be #=xkg#

The mass of the #70%# is #=(50-x)kg#

So,

#x*30/100+(50-x)*70/100=50*62/100#

#0.3x+35-0.7x=31#

#0.4x=35-31=4#

#x=4/0.4=10kg#

Jun 25, 2017

Answer:

#10# kg of #30%# copper containing metal alloy and

#40# kg of #70%# copper containing metal alloy were combined.

Explanation:

Let #x# kg of #30%# copper containing metal alloy and

#(50-x)# kg of #70%# copper containing metal alloy were combined.

Then , balancing the input and output of copper content , we get ,

#x*0.3 +(50-x)*0.7 = 50*0.62 or 0.3x-0.7x =31-35 # or

# -0.4x = -4.0 or 0.4x = 4.0 or x=10 :. 50-x=50-10=40#

Hence #10# kg of #30%# copper containing metal alloy and

#40# kg of #70%# copper containing metal alloy were combined to

achieve #50 kg# of #62%# copper alloy. [Ans]

Jun 25, 2017

Answer:

Just for the hell of it this is a different approach:

40 kg at 70% content
10 Kg at 30% content

Explanation:

Set material 1 #M_1 -> 30% #copper
Set material 2 #M_2->70% #copper
Set target as #T->62% # copper

Set the amount of 70% alloy be #x#

The mass is fixed at 50Kg thus if you increase the amount of #M_1# then the amount of #M_2# must reduce to maintain this weight. Thus if we consider just one of these the other is directly implied. Consequently we can and may use a straight line graph situation to represent the blend. #color(brown)("This really does work!")#

It models the numbers only approach. It is just that it looks different.

Tony B

The gradient of part is the same as the gradient of the whole:

#("mass")/("copper content")->50/(70-30)-=x/(62-30)#

#x=(50xx32)/(40)=40#

So we have:

40 kg at 70% content
10 Kg at 30% content

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Check:

#10/(40+10)xx30% = color(white)(5)6%#
#40/(40+10)xx70% = ul(56% larr" Add")#
#" "62%#