A and B completed work together in 5 days. Had A worked at twice his own speed…

Concept

In a work-rate problem, if a person completes a job alone in a certain number of days, their one-day work is 1 divided by that number of days. When two people work together, their combined one-day work is the sum of their individual one-day works, equal to 1 divided by the number of days taken together. Changing a person's speed scales their one-day work by that factor - doubling speed doubles the one-day work, halving speed halves it. Two such combined-rate conditions give a pair of simultaneous equations in the two individual work-rates, which can be solved algebraically.

Solving for A and B's Individual Times

1. Let A's one-day work be 1/x and B's one-day work be 1/y, where x and y are the number of days A and B would individually take to complete the job alone.

2. Working together at normal speed for 5 days: 1/x + 1/y = 1/5, which gives 5(x + y) = xy ... equation (1)

3. Working together with A at twice speed and B at half speed for 4 days: 2/x + 1/(2y) = 1/4, which gives 8y + 2x = xy ... equation (2)

4. Equate the two expressions for xy from equations (1) and (2): 8y + 2x = 5x + 5y, which simplifies to 3y = 3x, so y = x.

5. Substitute y = x into equation (1): 1/x + 1/x = 1/5, i.e., 2/x = 1/5, giving x = 10.

Cross-check

Check equation (1): 1/10 + 1/10 = 2/10 = 1/5 - satisfied. Check equation (2): 2/10 + 1/(2 x 10) = 0.2 + 0.05 = 0.25 = 1/4 - satisfied. Both conditions hold, confirming the individual times.

Hence, A alone would take 10 days to complete the job.