Numerical Applications Test 25

Let the five numbers be $$A, B, C, D, E$$
Given, $$A + B + C + D + E =555$$
$$\displaystyle \frac{A+B}{2} = 75 \Rightarrow A + B = 150 $$ and $$C = 115$$
$$\therefore 150 + 115 + D + E = 555$$
$$\Rightarrow D + E = 555 - 265 = 290$$
$$\therefore$$ Required Average $$= \displaystyle \frac{D + E}{2} = \frac{290}{2} = 145$$

Part of the tank filled by both the taps in 10 minutes $$= 10 \times \left( \displaystyle\frac { 1 }{ 20 } + \displaystyle\frac { 1 }{ 60 }  \right) = 10 \times \left( \displaystyle\frac { 3+1 }{ 60 }  \right) = \displaystyle\frac { 2 }{ 3 }$$
Remaining part $$ = 1 - \displaystyle\frac { 2 }{ 3 } = \displaystyle\frac { 1 }{ 3 }$$
$$\displaystyle\frac { 1 }{ 60 }$$th part of the tank is filled by the second pipe in 1 min.
$$\therefore     \displaystyle\frac { 1 }{ 3 }$$rd of the tank will be filled by the second pipe in $$ 60 \times \displaystyle\frac { 1 }{ 3 } = 20$$  min.

Part of the tank filled by the pump in 1 hour $$= \displaystyle\frac { 1 }{ 3 }$$
Net part of the tank filled by the pump and the leak in 1 hour $$ = \displaystyle\frac { 1 }{ \displaystyle\frac { 7 }{ 2 }  } = \displaystyle\frac { 2 }{ 7 }$$

A's 5 days work$$+$$ B's 7 days work $$+$$ C's 13 days work $$=$$1
$$\therefore$$ $$(A+B)'s$$ 5 days work $$+$$ $$(B+C)'s$$ 2 days work $$+$$ C's 11 days work $$=$$ 1
$$\therefore$$ $$\dfrac{5}{12}+\dfrac{2}{16}$$$$+$$ C's 11 days work$$=1$$
$$\therefore$$ C's 11 days work$$=$$$$1-(\dfrac{5}{12}+\dfrac{2}{16}$$) $$=$$ $$\dfrac{11}{24}$$
$$\therefore$$ C's 1 days work $$=$$$$ (\dfrac{11}{24} \times \dfrac{1}{11}$$) $$=\dfrac{1}{24}$$
$$\therefore$$ C alone can finish the work in 24 days.

A can do a piece of work in 7 days of 9 hr each or in 63 hrs.

Ronald work at a rate of 32 pages per 6 hrs$$=\dfrac{32}{6}=\dfrac{16}{3} pages/hr$$

Elan work at a rate of  40 pages per 5 hr$$=\dfrac{40}{4}=8   Pages/hr$$

 

If they  work together then they work at the rate of$$=\dfrac{16}{3}+8=\dfrac{40}{3}  pages/hr$$

Let t time is required them to type 110 pages then

$$\dfrac{40}{3}t=110$$

$$40t=330$$

$$t=\dfrac{330}{40}=\dfrac{33}{4}=8\dfrac{1}{4}=8  hr  15 min.$$

$${\textbf{Step  - 1: Finding number of work units for the first case}}{\text{.}}$$

                  $${\text{Total 36 men work for a total of 18 days to complete it}}{\text{.}}$$

                  $${\text{So total work units}} = 36 \times 18  = 648$$

                  $${\text{Thus, for the first case ,it's a total of 648 work units}}{\text{.}}$$

$${\textbf{Step  - 2: Finding number of days for the second case}}{\text{.}}$$

                  $${\text{Let it takes }}x{\text{ days for the workers to complete the task,so we can write}}$$

                  $${\text{Number of work units}} = 27  \times x = 27x$$

                  $${\text{Also, the work is same so the work units must be same}}$$

                  $${\text{So, equating the work units for the two cases, we get}}$$

                  $$648 = 27x$$

                  $$x = \dfrac{{648}}{{27}} = 24$$

$${\textbf{Hence option D is correct}}$$

$$Let \, \,the \, \,total \, \,work \, \,be \, \,x.$$
$$\therefore \, \, The \, \, speed \, \, at \, \, which \, \, amount \, \,of \, \,work \, \,A \, \,does \, \,in \, \,20 \, \,days \, \,is=\dfrac{x}{25}$$
$$Let \, \,the \, \,speed \, \,of \, \,B \, \,be \, \,B \, \,work \, \,per \, \,day.$$
$$Since,  \, \, the \, \, work \, \,done \, \,by \, \,A \, \,is \, \,\dfrac{4}{5}x$$
$$\therefore \, \,Remaining \, \,work \, \,is \, \,x-\dfrac{4}{5}x=\dfrac{x}{5}$$
$$If \, \,A \, \,and \, \,B \, \,work \, \,together \, \,then \, \,we \, \,have,$$

$$\dfrac{\dfrac{x}{5}}{\dfrac{x}{25}+B}=3$$
$$\therefore \, \, speed \, \, of \, \, B=\dfrac{2x}{75} \, \,work \, \,per \, \,day.$$
$$If \, \,B \, \, alone \, \,works \, \, then \, \, the \, \,work \, \, will \, \,be \, \, completed \, \,in \, \,\dfrac{x}{\dfrac{2x}{75}}=\, \, 37 \dfrac{1}{2} \, \, days.$$