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Difficulty: MediumRatios, Rates, and Percentages

A telecommunications company deploys two installation teams, Team Alpha and Team Beta, to lay fiber-optic cable across a rural region containing both rocky terrain and flat terrain.

- Team Alpha lays cable at a constant rate of 400400 meters per day in rocky terrain and 600600 meters per day in flat terrain. Over a project lasting exactly 2020 working days, Team Alpha laid a total of 10,40010,400 meters of cable.
- Team Beta lays cable at a constant rate of 300300 meters per day in rocky terrain and 700700 meters per day in flat terrain. Over a project also lasting exactly 2020 working days, Team Beta laid a total of 11,60011,600 meters of cable.

Based on the information provided, which of the following correctly pairs the number of days Team Alpha worked in rocky terrain with the number of days Team Beta worked in rocky terrain?

  1. Team Alpha: 8 days; Team Beta: 6 daysAnswer
  2. B
    Team Alpha: 6 days; Team Beta: 8 days
  3. C
    Team Alpha: 12 days; Team Beta: 14 days
  4. D
    Team Alpha: 8 days; Team Beta: 14 days
  5. E
    Team Alpha: 12 days; Team Beta: 6 days

Answer

Team Alpha worked 8 days in rocky terrain and Team Beta worked 6 days in rocky terrain.
For Team Alpha, setting 400dA+600(20−dA)=10,400400 d_A + 600(20 - d_A) = 10,400 simplifies to 200dA=1,600200 d_A = 1,600, giving dA=8d_A = 8 days in rocky terrain. For Team Beta, setting 300dB+700(20−dB)=11,600300 d_B + 700(20 - d_B) = 11,600 simplifies to 400dB=2,400400 d_B = 2,400, giving dB=6d_B = 6 days in rocky terrain. Thus, the pairing 'Team Alpha: 8 days; Team Beta: 6 days' is correct.

Step-by-Step Solution

1
Set up the work-rate equation for Team Alpha.
Let dAd_A be the number of days Team Alpha worked in rocky terrain. The days worked in flat terrain is (20−dA)(20 - d_A). The equation is 400dA+600(20−dA)=10,400400 d_A + 600(20 - d_A) = 10,400.
Total distance laid equals rate in rocky terrain times rocky days plus rate in flat terrain times flat days.
2
Solve for dAd_A (Team Alpha's rocky terrain days).
400dA+12,000−600dA=10,400  ⟹  12,000−200dA=10,400  ⟹  200dA=1,600  ⟹  dA=8400 d_A + 12,000 - 600 d_A = 10,400 \implies 12,000 - 200 d_A = 10,400 \implies 200 d_A = 1,600 \implies d_A = 8 days.
Simplifying the algebraic equation yields the exact number of rocky terrain days for Team Alpha.
3
Set up the work-rate equation for Team Beta.
Let dBd_B be the number of days Team Beta worked in rocky terrain. The days worked in flat terrain is (20−dB)(20 - d_B). The equation is 300dB+700(20−dB)=11,600300 d_B + 700(20 - d_B) = 11,600.
Total distance laid equals rate in rocky terrain times rocky days plus rate in flat terrain times flat days for Team Beta.
4
Solve for dBd_B (Team Beta's rocky terrain days).
300dB+14,000−700dB=11,600  ⟹  14,000−400dB=11,600  ⟹  400dB=2,400  ⟹  dB=6300 d_B + 14,000 - 700 d_B = 11,600 \implies 14,000 - 400 d_B = 11,600 \implies 400 d_B = 2,400 \implies d_B = 6 days.
Simplifying the algebraic equation yields the exact number of rocky terrain days for Team Beta.

Key Concept

Solving simultaneous linear work-rate problems with two terrain types and time constraints.
Estimated Time:2m 0s
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Topics

Ratios, Rates, and Percentages

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