Correct Answer - Option 3 : 17
Let the efficiency of one girl from school P is x and the efficiency of one girl from school Q is y.
We know that
\(\frac{{{m_1} \times {h_1} \times {d_1}}}{{{w_1}}} = \frac{{{m_2} \times {h_2} \times {d_2}}}{{{w_2}}}\;\)
Where m, h, d and w respectively are number of person, number of hour, number of days and work done
So, 10 × [6x + 5y] = 9 × [5x + 7y]
⇒ 60x + 50y = 45x + 63y
⇒ 60x – 45x = 63y – 50y
⇒ 15x = 13y
⇒ x/y = 13/15
⇒ x : y = 13 : 15
One day work by one girl from school P = 13 units
One day work by one girl from school Q = 15 units
So, Total work = 10 × [6(13) + 5(15)] = 10 × [78 + 75] = 1530 units
According to question,
Let ‘n’ girls from school Q can complete the exhibition in 6 days.
Total work = n × y × 6 = 1530
⇒ n × 15 × 6 = 1530
⇒ n × 90 = 1530
\( \Rightarrow {\rm{n}} = \frac{{1530}}{{90}}\)
⇒ n = 17
∴ 17 girls from school Q can set up the exhibition in 6 days.
