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Difficulty: Very hardAlgebraic Word Problems and Modeling

Working independently at their respective constant rates, Alex and Blair can complete a certain job together. Alex works alone for 44 hours, after which Blair joins Alex, and together they work for an additional 66 hours to finish the entire job. If Alex takes strictly less time to complete the job working alone than Blair takes working alone, which of the following statements must be true? Select all that apply.

  1. Alex working alone would take less than 1616 hours to complete the job.Answer
  2. The time required for Alex and Blair to complete the job working together from start to finish is strictly between 88 hours and 1010 hours.Answer
  3. Alex completes more than 60%60\% of the entire job.Answer
  4. D
    Blair working alone would take more than 2424 hours to complete the job.
  5. E
    Blair completes more than 35%35\% of the entire job.

Answer

The true statements are that Alex working alone would take less than 16 hours, the combined time to finish the job working together is between 8 and 10 hours, and Alex completes more than 60 percent of the entire job.
By setting up the total work equation 10rA+6rB=110 r_A + 6 r_B = 1 and using the inequality rA>rB>0r_A > r_B > 0, we find that 1/16<rA<1/101/16 < r_A < 1/10. This implies Alex's solo time is strictly less than 16 hours. The combined time TtogetherT_{together} is constrained between 8 and 10 hours because 6/Ttogether=1−4rA6/T_{together} = 1 - 4 r_A. Additionally, Alex's total work contribution is 10rA>10/16=62.5%10 r_A > 10/16 = 62.5\%, which is strictly greater than 60%.

Step-by-Step Solution

1
Set up the work equation using individual rates rAr_A and rBr_B.
4rA+6(rA+rB)=1  ⟹  10rA+6rB=14 r_A + 6(r_A + r_B) = 1 \implies 10 r_A + 6 r_B = 1.
Alex works alone for 4 hours and then both Alex and Blair work together for 6 hours to complete 1 unit of work.
2
Apply the condition that Alex takes strictly less time alone than Blair (rA>rB>0r_A > r_B > 0).
10rA+6rA>10rA+6rB=1  ⟹  16rA>1  ⟹  rA>11610 r_A + 6 r_A > 10 r_A + 6 r_B = 1 \implies 16 r_A > 1 \implies r_A > \frac{1}{16}.
Since Alex's rate rAr_A is strictly greater than Blair's rate rBr_B, replacing rBr_B with rAr_A gives an upper bound on Alex's solo time TA=1/rA<16T_A = 1/r_A < 16 hours.
3
Determine the feasible range for Alex's rate rAr_A.
116<rA<110\frac{1}{16} < r_A < \frac{1}{10}.
From 6rB=1−10rA>06 r_B = 1 - 10 r_A > 0, we get rA<1/10r_A < 1/10. Combined with rA>1/16r_A > 1/16, we have 1/16<rA<1/101/16 < r_A < 1/10.
4
Calculate the combined time Ttogether=1rA+rBT_{together} = \frac{1}{r_A + r_B}.
8<Ttogether<108 < T_{together} < 10.
Since 6(rA+rB)=1−4rA6(r_A + r_B) = 1 - 4 r_A, substituting 1/16<rA<1/101/16 < r_A < 1/10 gives 3/5<6(rA+rB)<3/43/5 < 6(r_A + r_B) < 3/4, which simplifies to 8<Ttogether<108 < T_{together} < 10.
5
Calculate the fraction of total work performed by Alex.
Alex performs 10rA>10×116=0.625=62.5%10 r_A > 10 \times \frac{1}{16} = 0.625 = 62.5\% of the total work.
Alex works for a total of 10 hours (4+64 + 6). Since rA>1/16r_A > 1/16, Alex completes over 62.5%62.5\% of the job, which is greater than 60%60\%.

Key Concept

Formulating algebraic inequalities for work rate problems when relative individual speeds are given.
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Topics

Algebraic Word Problems and Modeling

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