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At their usual efficiency levels, AA and BB together finish a task in 1212 days. If AA had worked half as efficiently as she usually does, and BB had worked thrice as efficiently as he usually does, the task would have been completed in 99 days. How many days would AA take to finish the task if she works alone at her usual efficiency?

Solution

✅ Correct Option: 3

We have two workers, A and B, with different efficiency levels. We need to find how long A takes to complete the task alone.

Given information:

A and B together finish the task in 12 days (at usual efficiency)

If A works at half efficiency and B works at triple efficiency, they finish in 9 days

Find: Days for A to complete the task alone


We think about work rates in terms of "units of work per day":

We let A normally do 2x2x units per day

We let B normally do yy units per day

We choose 2x2x for A because this makes the math cleaner when A works at "half efficiency" - it becomes just xx units per day.


Scenario 1: Normal efficiency

A does 2x2x units/day, B does yy units/day

Together they complete the task in 12 days

Total work = 12(2x+y)12(2x + y) units

Scenario 2: Changed efficiency

A works at half efficiency = xx units/day

B works at triple efficiency = 3y3y units/day

Together they complete the task in 9 days

Total work = 9(x+3y)9(x + 3y) units

The total work is the same in both scenarios, so:

12(2x+y)=9(x+3y)12(2x + y) = 9(x + 3y)


Expanding both sides:

24x+12y=9x+27y24x + 12y = 9x + 27y

24x−9x=27y−12y24x - 9x = 27y - 12y

15x=15y15x = 15y

Therefore: x=yx = y


Since x=yx = y, we can choose any convenient value. We use x=y=1x = y = 1.

This gives us:

A's normal rate = 2x=2×1=22x = 2 \times 1 = 2 units per day

B's normal rate = y=1y = 1 unit per day

Total work = 12(2+1)=12×3=3612(2 + 1) = 12 \times 3 = 36 units


If A works alone at her usual efficiency:

Time = Total workA’s rate=362=18\frac{\text{Total work}}{\text{A's rate}} = \frac{36}{2} = 18 days

A would take 18 days to finish the task alone.

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