Work Problems - Working Together: Very Difficult Problems with Solutions

Solution:
Let the pipes need [tex]x[/tex] and [tex]x+4[/tex] hours. The first fills half the pool in [tex]\frac{x}{2}[/tex] hours:
[tex]\dfrac{x}{2}\left(\dfrac{1}{x}+\dfrac{1}{x+4}\right)=\dfrac{10}{x+4}[/tex]
[tex](x+4)+x=20 \Rightarrow x=8[/tex]
Check: [tex]4\left(\dfrac{1}{8}+\dfrac{1}{12}\right)=\dfrac{5}{6}=\dfrac{10}{12}[/tex]

Solution:
[tex]2[/tex] h [tex]24[/tex] min [tex]=\frac{12}{5}[/tex] h, so together the machines pave [tex]\frac{2}{3}\div \frac{12}{5}=\frac{5}{18}[/tex] of the street per hour. With [tex]b=\frac{2}{3}a[/tex]:
[tex]a+\dfrac{2}{3}a=\dfrac{5}{18} \Rightarrow a=\dfrac{1}{6}, b=\dfrac{1}{9}[/tex]
The first machine paves half the street in [tex]3[/tex] hours, i.e. by [tex]11{:}00[/tex]. [tex]t[/tex] hours after that:
[tex]\dfrac{1}{2}+\dfrac{t}{6}=6\times \dfrac{t}{9} \Rightarrow \dfrac{1}{2}=\dfrac{t}{2} \Rightarrow t=1[/tex]
At [tex]12{:}00[/tex] (the street is not finished yet: [tex]\frac{2}{3}+\frac{1}{9}=\frac{7}{9}[/tex]).

Solution:
Per hour the pipes fill [tex]\frac{1}{5}=0.2[/tex] and [tex]\frac{1}{12.5}=0.08[/tex] of the pool, and the third empties [tex]1\div \frac{20}{3}=0.15[/tex]. If the first two worked together [tex]t[/tex] hours:
[tex]0.28t+5\times 0.08=0.68 \Rightarrow 0.28t=0.28 \Rightarrow t=1[/tex]
Draining from [tex]68\%[/tex] to [tex]50\%[/tex] takes [tex]0.18\div 0.15=1.2[/tex] h. In total: [tex]1+5+1.2=7.2[/tex] h = [tex]432[/tex] min.

Solution:
Let one sailor pump [tex]p[/tex] per hour, let [tex]w[/tex] flow in per hour, and let [tex]V[/tex] be the water at the start:
[tex]V+6w=60p, V+4w=56p \Rightarrow w=2p, V=48p[/tex]
[tex]48p+2pt=18pt \Rightarrow 16t=48 \Rightarrow t=3[/tex]

Solution:
Let the slower team need [tex]a[/tex] days and the faster [tex]b[/tex] days. Since [tex]\frac{75}{7}\%=\frac{3}{28}[/tex]:
[tex]\dfrac{3}{28}a=\dfrac{3}{7}b-6 \Rightarrow a=4b-56[/tex]
[tex]\dfrac{1}{4b-56}+\dfrac{1}{b}=\dfrac{1}{12} \Rightarrow b^2-29b+168=0 \Rightarrow (b-21)(b-8)=0[/tex]
[tex]b=8[/tex] gives [tex]a\lt 0[/tex], so [tex]b=21[/tex] and [tex]a=28[/tex]: the slower team needs [tex]28[/tex] days and the faster [tex]21[/tex] days.

Solution:
Let [tex]T[/tex] be the actual time in minutes and [tex]p[/tex], [tex]q[/tex] the workers' rates (parts per minute):
[tex]2p(T-12)=12, 2q(T+20)=12, (p+q)T=12[/tex]
[tex]\dfrac{6T}{T-12}+\dfrac{6T}{T+20}=12 \Rightarrow 2T^2+8T=2T^2+16T-480 \Rightarrow T=60[/tex]
[tex]p=\frac{6}{48}=\frac{1}{8}[/tex] and [tex]q=\frac{6}{80}=\frac{3}{40}[/tex] parts per minute. In [tex]480[/tex] minutes: [tex]60[/tex] and [tex]36[/tex] parts.

Solution:
Let there be [tex]n[/tex] mowers, each mowing one unit of area per day. The large meadow got:
[tex]n\times \dfrac{1}{2}+\dfrac{n}{2}\times \dfrac{1}{2}=\dfrac{3n}{4}[/tex]
So the small meadow is [tex]\frac{3n}{8}[/tex], and [tex]\frac{n}{2}\times \frac{1}{2}=\frac{n}{4}[/tex] of it was mowed:
[tex]\dfrac{3n}{8}-\dfrac{n}{4}=\dfrac{n}{8}=1 \Rightarrow n=8[/tex]

Solution:
Let there be [tex]n[/tex] workers, each doing [tex]p[/tex] of the job per day, so [tex]24np=1[/tex]. The working times form an arithmetic progression from [tex]T[/tex] (the first) down to [tex]\frac{T}{5}[/tex] (the last); their average is [tex]\frac{1}{2}\left(T+\frac{T}{5}\right)=\frac{3T}{5}[/tex]:
[tex]np\times \dfrac{3T}{5}=1=24np \Rightarrow \dfrac{3T}{5}=24 \Rightarrow T=40[/tex]
The job lasted as long as the first worker worked: [tex]40[/tex] days.