Work Problems - Working Together: Problems with Solutions

Solution:
In one hour the pipes fill [tex]\frac{1}{12}[/tex] and [tex]\frac{1}{8}[/tex] of the pool. Together:
[tex]\dfrac{1}{12}+\dfrac{1}{8}=\dfrac{2+3}{24}=\dfrac{5}{24}[/tex]
[tex]1\div \dfrac{5}{24}=\dfrac{24}{5}=4\dfrac{4}{5}[/tex]
[tex]4\frac{4}{5}[/tex] h = [tex]4[/tex] h [tex]48[/tex] min

Solution:
Together they eat [tex]\frac{1}{6}[/tex] of the plate per minute and Karlsson alone [tex]\frac{1}{8}[/tex]. The Kid eats:
[tex]\dfrac{1}{6}-\dfrac{1}{8}=\dfrac{4-3}{24}=\dfrac{1}{24}[/tex]
So the Kid needs [tex]24[/tex] minutes.

Solution:
Together, in one minute they do:
[tex]\dfrac{1}{30}+\dfrac{1}{40}+\dfrac{1}{24}=\dfrac{4+3+5}{120}=\dfrac{12}{120}=\dfrac{1}{10}[/tex]
So they need [tex]10[/tex] minutes.

Solution:
Let the slower pipe fill [tex]r[/tex] of the pool per hour; the faster one fills [tex]2r[/tex]:
[tex]r+2r=\dfrac{1}{2} \Rightarrow r=\dfrac{1}{6}[/tex]
The slower pipe needs [tex]6[/tex] hours and the faster one [tex]3[/tex] hours.

Solution:
In [tex]6[/tex] minutes the first machine makes [tex]15\times 3=45[/tex] cups and the second [tex]20\times 2=40[/tex] cups, together [tex]85[/tex] cups:
[tex]255\div 85=3 \Rightarrow 3\times 6=18[/tex]

Solution:
In [tex]2[/tex] hours the two tractors plough:
[tex]2\left(\dfrac{1}{8}+\dfrac{1}{6}\right)=2\times \dfrac{7}{24}=\dfrac{7}{12}[/tex]
[tex]\frac{5}{12}[/tex] of the field is left, and the third tractor ploughs [tex]\frac{1}{12}[/tex] per hour, so it needs [tex]5[/tex] hours.

Solution:
The second tap needs [tex]12-3=9[/tex] minutes. Together the taps fill per minute:
[tex]\dfrac{1}{12}+\dfrac{1}{9}=\dfrac{3+4}{36}=\dfrac{7}{36}[/tex]
[tex]\dfrac{7}{9}\div \dfrac{7}{36}=\dfrac{7}{9}\times \dfrac{36}{7}=4[/tex]

Solution:
Each minute [tex]\frac{1}{8}[/tex] of the tub flows in and [tex]\frac{1}{10}[/tex] flows out. Since [tex]\frac{1}{8}\gt \frac{1}{10}[/tex], the tub does fill:
[tex]\dfrac{1}{8}-\dfrac{1}{10}=\dfrac{5-4}{40}=\dfrac{1}{40}[/tex]
It fills in [tex]40[/tex] minutes; meanwhile [tex]40\times \frac{1}{10}=4[/tex] bathtubs of water drain away for nothing.

Solution:
The first worker works all [tex]6[/tex] days and does [tex]\frac{6}{10}=\frac{3}{5}[/tex] of the job. The second does the remaining [tex]\frac{2}{5}[/tex] in [tex]2[/tex] days, i.e. [tex]\frac{1}{5}[/tex] per day, so alone he needs [tex]5[/tex] days.

Solution:
Let Greg work [tex]t[/tex] minutes; Mike worked [tex]t+25[/tex]:
[tex]2t+3(t+25)=400 \Rightarrow 5t=325 \Rightarrow t=65[/tex]
[tex]65+25=90[/tex]

Solution:
In minutes the rates are [tex]\frac{1}{120}[/tex] and [tex]\frac{1}{90}[/tex] of the road. In [tex]20[/tex] minutes together they clear:
[tex]20\left(\dfrac{1}{120}+\dfrac{1}{90}\right)=20\times \dfrac{3+4}{360}=\dfrac{7}{18}[/tex]
[tex]\dfrac{11}{18}\div \dfrac{1}{90}=\dfrac{11}{18}\times 90=55[/tex]