Mixture Problems: Very Difficult Problems with Solutions

Solution:
Let [tex]x[/tex] gallons of the [tex]40\%[/tex] and [tex]y[/tex] gallons of the [tex]60\%[/tex] solution be mixed:
[tex]0.4x+0.6y=0.2(x+y+5) \Rightarrow x+2y=5[/tex]
[tex]0.4x+0.6y+0.8\times 5=0.7(x+y+5) \Rightarrow 3x+y=5[/tex]
[tex]x=1, y=2[/tex]

Solution:
Let the first mixture contain [tex]x[/tex] lb of the [tex]50\%[/tex] ore and [tex]y[/tex] lb of the [tex]60\%[/tex] ore: [tex]0.5x+0.6y=2800[/tex]. The second mixture also weighs [tex]x+y[/tex] and contains [tex]2800+500-600=2700[/tex] lb of iron:
[tex]0.54(x+y)=2700 \Rightarrow x+y=5000[/tex]
[tex]0.5x+0.6(5000-x)=2800 \Rightarrow 0.1x=200 \Rightarrow x=2000, y=3000[/tex]

Solution:
The second container needs [tex]5[/tex] quarts of the [tex]4\%[/tex] solution. Its acid becomes [tex]0.08\times 15+0.04\times 5=1.4[/tex] in [tex]20[/tex] quarts, i.e. [tex]7\%[/tex].
The first container now holds [tex]5[/tex] quarts of the [tex]4\%[/tex] solution and needs [tex]15[/tex] quarts of the [tex]7\%[/tex] one: [tex]0.04\times 5+0.07\times 15=1.25[/tex] in [tex]20[/tex] quarts, i.e. [tex]6.25\%[/tex]. The second container keeps [tex]7\%[/tex].

Solution:
In water zinc loses [tex]\frac{5}{36}[/tex] of its weight and lead [tex]\frac{2}{23}[/tex]; the alloy loses [tex]292-261=31[/tex] oz. With [tex]z[/tex] oz of zinc and [tex]l[/tex] oz of lead:
[tex]z+l=292, \dfrac{5z}{36}+\dfrac{2l}{23}=31[/tex]
[tex]115z+72l=25\,668, 72z+72l=21\,024[/tex]
[tex]43z=4644 \Rightarrow z=108, l=184[/tex]
Check: [tex]\dfrac{5\times 108}{36}+\dfrac{2\times 184}{23}=15+16=31[/tex]

Solution:
Let [tex]x[/tex] gallons be poured out each time. After the first time the barrel holds [tex]27-x[/tex] gallons of alcohol, i.e. the fraction [tex]\frac{27-x}{27}[/tex] of the liquid. The second pour keeps the same fraction of the alcohol again:
[tex]27\left(\dfrac{27-x}{27}\right)^2=12 \Rightarrow (27-x)^2=324 \Rightarrow 27-x=18 \Rightarrow x=9[/tex]
Check: [tex]27-9=18, 18\times \dfrac{18}{27}=12[/tex]

Solution:
Let the solution weigh [tex]m[/tex] oz:
[tex]\dfrac{40}{m}-\dfrac{40}{m+200}=\dfrac{1}{10}[/tex]
[tex]400(m+200)-400m=m(m+200) \Rightarrow m^2+200m-80\,000=0[/tex]
[tex](m-200)(m+400)=0 \Rightarrow m=200[/tex]
Water: [tex]200-40=160[/tex] oz; concentration: [tex]\frac{40}{200}=20\%[/tex] (afterwards [tex]\frac{40}{400}=10\%[/tex]).

Solution:
Let each bottle hold [tex]V[/tex] of solution at first, with concentrations [tex]c_1[/tex], [tex]c_2[/tex], [tex]c_3[/tex]. The first two bottles each receive [tex]\frac{V}{2}[/tex] from the third and then hold [tex]\frac{3V}{2}[/tex]:
[tex]\dfrac{c_1+\frac{c_3}{2}}{\frac{3}{2}}=0.8c_1 \Rightarrow c_3=0.4c_1[/tex]
[tex]\dfrac{c_2+\frac{c_3}{2}}{\frac{3}{2}}=1.1c_2 \Rightarrow c_3=1.3c_2[/tex]
[tex]0.4c_1=1.3c_2 \Rightarrow c_1:c_2=13:4[/tex]

Solution:
Let the concentrations (in %) be [tex]a[/tex], [tex]b[/tex], [tex]c[/tex], [tex]d[/tex]:
[tex]3a+2b+c=90, b+c+d=72, a+c=20[/tex]
Subtracting three times the third equation from the first: [tex]2b-2c=30[/tex], i.e. [tex]b-c=15[/tex]. Then:
[tex]2b+d=(b+c+d)+(b-c)=72+15=87 \Rightarrow \dfrac{2b+d}{3}=29[/tex]
The four concentrations cannot be found separately, but the requested mixture is [tex]29\%[/tex].