Question

If 6 painters can draw 9 paintings in 5 hours then find the time taken by a painter to complete 10 drawings.
$\left( a \right){\text{ }}\dfrac{5}{2}hours$
$\left( b \right){\text{ 3}}hours$
$\left( c \right){\text{ }}\dfrac{7}{2}hours$
$\left( d \right){\text{ }}\dfrac{9}{2}hours$

Answer 1

Hint:So for solving this question we will use the concept of efficiency which means that the efficiency of $6$ painters who can draw $9$ paintings in $5$ hours will be equal to the efficiency of a painter drawing $10$ drawings in $t$ hours. So the efficiency is given by $\eta = \dfrac{{t \times n}}{w}$ and on equating it with the other we will get the time required.

Formula used:
Efficiency is given by,
$\eta = \dfrac{{t \times n}}{w}$
$t$, will be the time taken by a person to do a work which is $w$
$n$, will be the number of persons involved in the work
$w$, will be the work itself

Complete step by step solution:
From the hint, we knew that the efficiency of $6$ painters can draw $9$ paintings in $5$ hours will be equal to the efficiency of a painter drawing $10$ drawings in $t$ hours.
Therefore, from the statement, the relation between the efficiency will be
$ \Rightarrow \eta = \dfrac{{{t_1} \times {n_1}}}{{{w_1}}} = \dfrac{{{t_2} \times {n_2}}}{{{w_2}}}$
So from this relation, we can find the time taken, therefore on substituting the values, we get
$ \Rightarrow \eta = \dfrac{{6 \times 5}}{9} = \dfrac{{t \times 1}}{{10}}$
By doing the cross-multiplication, we get
$ \Rightarrow 6 \times 5 \times 10 = t \times 1 \times 9$
Now on solving the multiplication term of both the sides, we get the equation as
\[ \Rightarrow 300 = t \times 9\]
For solving for the value of $t$ the constant term will take to the denominator of the left side, on dividing it we get
$ \Rightarrow t = \dfrac{{100}}{3}hours$
And in the decimal form, it can be written as $33.33hours$ .
As we can see that no options are matching from the answer,

Hence, all options are incorrect.

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 ${t_2}$.