Question

If 18 pumps can raise 2170 tonnes water in $10$ days, working ${\text{7 hours}}$ a day then in how many days will $16$ pump raise ${\text{1736 tonnes}}$ of water, working ${\text{9 hours}}$ a day?
$\left( a \right){\text{ 6}}$
$\left( b \right){\text{ 7}}$
$\left( c \right){\text{ 8}}$
$\left( d \right){\text{ 9}}$

Answer 1

Hint: In this problem first we will find out the number of days a pump can raise for ${\text{1 tonnes}}$ of water in $1hour$ by using the unitary method and then we will make the efficiency equal and by using its formula which is given by $\eta = \dfrac{{d \times h \times n}}{w}$ and comparing both the efficiencies we will find the number of days.

Formula used:
Efficiency is given by
$\eta = \dfrac{{d \times h \times n}}{w}$
Here, $d \times h$, will be the time taken by the pump to do a work
$n$, will be the number of pumps involved in the work
$d$, will be the days
$h$, will be the hours
$w$, will be the work

Complete step by step solution:
 So it is given that the efficiency of $18$ pumps raising $2170{\text{ tonnes}}$ of water in $10$ days and working ${\text{7 hours}}$ a day is equal to $16$ pumps raising ${\text{1736 tonnes}}$ of water, working ${\text{9 hours}}$ a day in $t$ days.
So we have to find the $t$ days.
If $18$ pumps can raise $2170{\text{ tonnes}}$ of water in $10$ days, working ${\text{7 hours}}$.
So, $18$ pumps can raise $2170{\text{ tonnes}}$ of water in $70$ days, working ${\text{1 hour}}$.
Therefore, $1$ pump can raise $2170{\text{ tonnes}}$ water in $70 \times 18$ days, working ${\text{1 hour}}$.
Hence, $1$ pump can raise $2170{\text{ tonnes}}$ water in $\dfrac{{7 \times 10 \times 18}}{{2170}}$ days, working ${\text{1 hour}}$.
So, we have made efficiency equal, which may not affect any parameter.
Therefore, the formula $\eta = \dfrac{{d \times h \times n}}{w}$ will be termed as equation $1$
Therefore, from the relation which we had discussed in starting the efficiency can be written as
$ \Rightarrow \eta = \dfrac{{{d_1} \times {h_1} \times {n_1}}}{{{w_1}}} = \dfrac{{{d_2} \times {h_2} \times {n_2}}}{{{w_2}}}$
Now on substituting the known values, we get
$ \Rightarrow \dfrac{{18 \times 10 \times 7}}{{2170}} = \dfrac{{16 \times 9 \times t}}{{1736}}$
Now on solving the above equation, we get
$ \Rightarrow t = 7$
So the number of days required by $16$ pumps raising ${\text{1736 tonnes}}$ of water, working ${\text{9 hours}}$ a day is $7$ days.

Hence, the option $\left( b \right)$ is correct.

Note:
In this type of question, we should always remember that the efficiency will be equal that is the efficiency of ${n_1}$ the number of persons doing work ${w_1}$ in time ${t_1}$ is always equal to the ${n_2}$ number of persons doing work ${w_2}$ in time ${t_2}$.