A can complete a work in 12 days. B is 60% more efficient than A. The number of days taken by B to finish the same work is
A can complete the work in 12 days.
A’s efficiency = Work done per day = W / 12 units/day.
B is 60% more efficient than A.
B’s efficiency = A’s efficiency + 60% of A’s efficiency
B’s efficiency = A’s efficiency * (1 + 60/100) = A’s efficiency * (1 + 0.60) = 1.60 * A’s efficiency.
B’s efficiency = 1.60 * (W / 12) units/day.
B’s efficiency = (1.6 / 12) * W units/day
B’s efficiency = (16 / 120) * W units/day
B’s efficiency = (2 / 15) * W units/day.
Let the number of days taken by B to finish the same work be D days.
Work done by B = B’s efficiency * Number of days
W = (2W / 15) * D
To find D, we can divide both sides by W (assuming W > 0, which is true for work):
1 = (2 / 15) * D
D = 15 / 2
D = 7.5 days.
7.5 days is equal to 7½ days.
– If A takes $T_A$ days, A’s efficiency is proportional to $1/T_A$.
– If B is X% more efficient than A, B’s efficiency = $(1 + X/100) \times$ A’s efficiency.
Ratio of efficiencies (A:B) = 100 : 160 = 10 : 16 = 5 : 8.
Ratio of time taken (A:B) = 1/Efficiency ratio (B:A) = 1/(160:100) = 160:100 reciprocal = 100:160 = 5:8.
So, Time taken by A / Time taken by B = 8 / 5.
12 / D = 8 / 5
8 * D = 12 * 5
D = (12 * 5) / 8 = 60 / 8 = 15 / 2 = 7.5 days.
This approach using ratios confirms the result.