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Option 2 : 9 hours

As we know, if M1 men can do W1 work in D1 days working for H1 hours per day and M2 men can do W2 work in D2 days working for H2 hours per day at the same efficiency, then,

⇒ (M1 × D1 × H1)/W1 = (M2 × D2 × H2)/W2

Let extra hours of work per day be ‘x’ hours

Let work done on normal days be ‘100 units’

Work done in festive season = (100 + 60)% of 100 = 160 units

∵ Men and children are equally efficient and there number remains the same, hence we can write, for 1 day work,

⇒ /100 = /160

⇒ 15/100 = (15 + x)/160

⇒ 15 + x = 24

⇒ x = 24 - 15 = 9 hours

∴ Extra work of 9 hours is to be done
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