A mixture of 30 pounds of candy sells for $1.10 a pound. The mixture consists of chocolates worth $1.50 a pound and chocolates worth 90 cents a pound. How many pounds of the $1.50 chocolate were used to make the mixture?

2 Answers
Jan 7, 2018

Answer:

10 lbs of the more expensive chocolate was used.

Explanation:

Let #x# be the expensive chocolate ($1.50 per pound) and #y# be the cheap chocolate ($0.90 per pound).

Based on total pounds we know: #x + y = 30#
Based on cost we know #(1.50*x+0.90*y)/(x+y) = 1.10#

We can simplify the cost equation:

#(1.50*x+0.90*y)/(x+y) = 1.10#

cross multiply:

#1.50*x+0.90*y = 1.10(x+y)#

distribute:

#1.50*x+0.90*y = 1.10x+1.10y#

move everything to one side:

#0.4x -0.20y=0#

Now we can solve the system:
#x + y = 30# and #0.4x -0.20y=0#

From the first equation we know that #x=30-y#

Substituting into #0.4x -0.20y=0#, we have:

#0.4(30-y) -0.20y=0#

expanding:

#12-.4y-.2y=0#

collecting:

#12-0.60y=0#

subtract 12 from both sides:

#-0.60y=-12#

divide through by #-0.60#:

#y=20#

Taking #y=20# and substituting into #x+y=30# gives #x=10#.

Since #x# is the more expensive chocolate, we used 10 lbs of the more expensive.

Jan 7, 2018

Answer:

This is a simultaneous equation problem.

Chocolate used at $1.50 per pound is 10 pounds

Chocolate used at $0.90 per pound is 20 pounds

Explanation:

Let the weight of chocolates worth $0.90 a pound be #c_(0.90)#
Let the weight of chocolates worth $1.50 a pound be #c_(1.50)#

Initial conditions:

#c_(0.90)+c_(1.50)=30^("lb")" ".....................Equation(1)#

The value of the blend is #($1.10)/("pound")# so the total value of the final blend is #($1.10)/cancel("pound")xx30 color(white)("d")cancel("pound") = $33.00#

Thus we have:

#$0.90c_(0.90)+$1.5c_(1.5)=$33.00" "................Equation(2)#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
You may choose which one you so wish to substitute for. I choose #c_(0.90)#

Consider #Eqn(1) #

Write as #c_(0.90)=30-c_(1.50)" "......Equation(1_a)#

Using #Eqn(1_a)# substitute for #color(red)(c_(0.90))# in #Eqn(2)# giving:
Dropping the units of measerment for now.

#color(green)( 0.90color(red)(c_(0.90))color(white)("dddd.d")+1.5c_(1.50)=33.00) #

#color(green)(0.90( color(red)(30-c_(1.50)))+1.5c_(1.50)=33.00)#

#color(green)(color(white)("d")27-0.9c_(1.50)color(white)("d")+1.5c_(1.50)=33.00)#

#color(green)(color(white)("d")27color(white)("ddddd")+0.6c_(1.50)color(white)("dddd")=33.00)#

Subtract 26 from both sides

#color(green)(0.6c_(1.50)=6#

Divide both sides by #0.6#

#c_(1.50)=10 larr " pounds"#

#c_(0.90)= 30-10=20" pounds"#