Time relations relative to the team: A alone takes 25 days more than (A + B) together, and B alone takes 49 days more than (A + B) together. How many days do A and B together take to finish the work?

Difficulty: Medium

Correct Answer: 35 days

Explanation:


Introduction / Context:
This problem defines solo times in terms of the team time. Let t be the time taken by A + B together; then A alone is t + 25 and B alone is t + 49. Setting the sum of solo rates equal to the team rate yields a solvable equation for t.


Given Data / Assumptions:

  • (A + B) time = t days.
  • A alone time = t + 25 days.
  • B alone time = t + 49 days.
  • Work size = 1 unit.


Concept / Approach:
Rates add: 1/(t + 25) + 1/(t + 49) = 1/t. Clear denominators to form a quadratic in t, then solve for the positive root to find the team time directly.


Step-by-Step Solution:

1/(t+25) + 1/(t+49) = 1/t((t+49) + (t+25))/((t+25)(t+49)) = 1/t(2t + 74) / ((t+25)(t+49)) = 1/tt(2t + 74) = (t+25)(t+49)2t^2 + 74t = t^2 + 74t + 1225 ⇒ t^2 = 1225 ⇒ t = 35.


Verification / Alternative check:
A alone = 60 days, B alone = 84 days. Check: 1/60 + 1/84 = (7 + 5)/420 = 12/420 = 1/35, matching the team time of 35 days.



Why Other Options Are Wrong:
25, 15, or 45 days do not satisfy the relation between solo and team times defined in the equation; only 35 days does.



Common Pitfalls:
Adding times instead of rates or making algebraic errors while cross-multiplying. Keep denominators intact until the final quadratic emerges.



Final Answer:
35 days

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