Work Problems - Working Together: Difficult Problems with Solutions

Solution:
Let them work together for [tex]x[/tex] days:
[tex]x\left(\dfrac{1}{8}+\dfrac{1}{12}\right)+\dfrac{3}{8}=1 \Rightarrow \dfrac{5x}{24}=\dfrac{5}{8} \Rightarrow x=3[/tex]
[tex]3+3=6[/tex]

Solution:
Alone, each would need [tex]10[/tex] hours, i.e. do [tex]\frac{1}{10}[/tex] of the order per hour. The new rates are [tex]\frac{1}{10}[/tex] and [tex]1.4\times \frac{1}{10}=\frac{7}{50}[/tex]:
[tex]\dfrac{1}{10}+\dfrac{7}{50}=\dfrac{12}{50}=\dfrac{6}{25}[/tex]
[tex]\frac{25}{6}=4\frac{1}{6}[/tex] h = [tex]4[/tex] h [tex]10[/tex] min

Solution:
Let him work [tex]t[/tex] hours at the normal speed ([tex]\frac{1}{8}[/tex] of the quota per hour) and [tex]t[/tex] hours at double speed:
[tex]\dfrac{t}{8}+\dfrac{2t}{8}=1 \Rightarrow t=\dfrac{8}{3}=2\dfrac{2}{3}[/tex]
He worked [tex]2t=5\frac{1}{3}[/tex] h = [tex]5[/tex] h [tex]20[/tex] min, so he left [tex]2[/tex] h [tex]40[/tex] min early.

Solution:
The job is [tex]6\times 45=270[/tex] worker-hours. If it took [tex]T[/tex] hours:
[tex]6T+3\times \dfrac{T}{2}=270 \Rightarrow 7.5T=270 \Rightarrow T=36[/tex]

Solution:
The job is [tex]30\times 10=300[/tex] worker-days. In the first [tex]3[/tex] days [tex]90[/tex] are done and [tex]210[/tex] remain. The [tex]25[/tex] remaining workers each do [tex]1.4[/tex] units a day, i.e. [tex]35[/tex] units together:
[tex]210\div 35=6, 3+6=9[/tex]
The job was finished on the [tex]9[/tex]th day, one day ahead of schedule.

Solution:
Tom paints [tex]\frac{0.75}{2}=\frac{3}{8}[/tex] of the fence per hour and Huck [tex]\frac{0.6}{2}=\frac{3}{10}[/tex]. Tom's [tex]10\%[/tex] alone took:
[tex]0.1\div \dfrac{3}{8}=\dfrac{4}{15}[/tex]
i.e. [tex]16[/tex] min. Together they paint [tex]\frac{3}{8}+\frac{3}{10}=\frac{27}{40}[/tex] per hour, so the remaining [tex]90\%[/tex] takes [tex]0.9\div \frac{27}{40}=\frac{4}{3}[/tex] h = [tex]1[/tex] h [tex]20[/tex] min. In total: [tex]1[/tex] h [tex]36[/tex] min.

Solution:
Let the pipes fill [tex]a[/tex], [tex]b[/tex], [tex]c[/tex] of the pool per hour. Adding the three conditions:
[tex]2(a+b+c)=\dfrac{1}{20}+\dfrac{1}{28}+\dfrac{1}{35}=\dfrac{7+5+4}{140}=\dfrac{16}{140}[/tex]
[tex]a+b+c=\dfrac{8}{140}=\dfrac{2}{35}[/tex]
[tex]\frac{35}{2}=17\frac{1}{2}[/tex] h = [tex]17[/tex] h [tex]30[/tex] min

Solution:
Let [tex]t[/tex] hours pass after [tex]6{:}00[/tex]. The unploughed parts are [tex]1-\frac{t}{8}[/tex] and [tex]1-\frac{t-1}{10}[/tex]:
[tex]\dfrac{11-t}{10}=2\times \dfrac{8-t}{8} \Rightarrow 4(11-t)=10(8-t) \Rightarrow 6t=36 \Rightarrow t=6[/tex]
At [tex]12{:}00[/tex] the unploughed parts are [tex]\frac{1}{4}[/tex] and [tex]\frac{1}{2}[/tex].

Solution:
Let the trench be [tex]V[/tex] cubic yards. [tex]3[/tex] h [tex]20[/tex] min [tex]=\frac{10}{3}[/tex] h, so the hourly outputs are [tex]\frac{V}{2}[/tex] and [tex]V\div \frac{10}{3}=\frac{3V}{10}[/tex]:
[tex]\dfrac{V}{2}-\dfrac{3V}{10}=24 \Rightarrow \dfrac{V}{5}=24 \Rightarrow V=120[/tex]

Solution:
Let the first pipe be open [tex]t[/tex] hours; the second was open [tex]t-1[/tex] hours:
[tex]\dfrac{t}{9}-\dfrac{t-1}{15}=\dfrac{1}{2} \Rightarrow 10t-6(t-1)=45 \Rightarrow 4t=39 \Rightarrow t=9\dfrac{3}{4}[/tex]
The first pipe was open [tex]9[/tex] h [tex]45[/tex] min (and the second [tex]8[/tex] h [tex]45[/tex] min).

Solution:
In [tex]2[/tex] hours together they type [tex]\frac{2}{6}=\frac{1}{3}[/tex] of the text, so in [tex]10[/tex] hours the first typed [tex]\frac{2}{3}[/tex]; alone she needs [tex]10\div \frac{2}{3}=15[/tex] h. The second types per hour:
[tex]\dfrac{1}{6}-\dfrac{1}{15}=\dfrac{5-2}{30}=\dfrac{1}{10}[/tex]
First typist: [tex]15[/tex] h; second: [tex]10[/tex] h.

Solution:
Let there be [tex]x[/tex] workers and [tex]y[/tex] days. The job is [tex]xy[/tex] worker-days:
[tex]xy=(x+3)(y-2)=(x+12)(y-5)[/tex]
[tex]3y-2x-6=0, 12y-5x-60=0[/tex]
[tex]4(2x+6)=5x+60 \Rightarrow x=12, y=10[/tex]
Check: [tex]12\times 10=15\times 8=24\times 5=120[/tex]

Solution:
Let the taps need [tex]a[/tex] and [tex]b[/tex] hours alone:
[tex]\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{1}{12}, \dfrac{a}{2}+\dfrac{b}{2}=25 \Rightarrow a+b=50[/tex]
[tex]\dfrac{a+b}{ab}=\dfrac{1}{12} \Rightarrow ab=600[/tex]
So [tex]a[/tex] and [tex]b[/tex] are the roots of [tex]z^2-50z+600=0[/tex], i.e. [tex]20[/tex] and [tex]30[/tex] hours.

Solution:
Let the first team need [tex]x[/tex] days and the second [tex]x+3[/tex]. The first worked [tex]10[/tex] days and the second [tex]6[/tex]:
[tex]\dfrac{10}{x}+\dfrac{6}{x+3}=1 \Rightarrow x^2-13x-30=0 \Rightarrow (x-15)(x+2)=0 \Rightarrow x=15[/tex]
Check: [tex]\dfrac{10}{15}+\dfrac{6}{18}=\dfrac{2}{3}+\dfrac{1}{3}=1[/tex]

Solution:
[tex]3[/tex] h [tex]44[/tex] min [tex]=3\frac{44}{60}=\frac{56}{15}[/tex] h. Let the times be [tex]x[/tex] and [tex]x+1[/tex] hours:
[tex]\dfrac{1}{x}+\dfrac{1}{x+1}=\dfrac{15}{56} \Rightarrow 15x^2-97x-56=0 \Rightarrow (x-7)(15x+8)=0 \Rightarrow x=7[/tex]
Check: [tex]\dfrac{1}{7}+\dfrac{1}{8}=\dfrac{15}{56}[/tex]