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CAT 2018 Slot 1 — QA Question 12

Working in ShiftsEasy

When they work alone, B needs 25% more time to finish a job than A does. They two finish the job in 13 days in the following manner: A works alone till half the job is done, then A and B work together for four days, and finally B works alone to complete the remaining 5% of the job. In how many days can B alone finish the entire job?

Answer & solution

Correct answer: 20

  • A

    18

  • B

    22

  • C

    16

  • 20

Solution

Easy

B needs 25%25\% more time than A, so the time ratio A::B is 4:54:5 and the efficiency ratio is 5:45:4. A does the first 50%50\% alone, then A and B together for 4 days, then B alone does the last 5%5\%. So A and B together complete 45%45\% in those 4 days. Split that 45%45\% by efficiency to find B's share, hence B's full-job time.

1

Efficiency ratio. "B needs 25%25\% more time" means tB=54tAt_B = \tfrac54 t_A, so times are 4:54:5 and efficiencies invert.

tA:tB=4:5⇒ efficiency A:B=5:4\begin{aligned} &t_A : t_B = 4 : 5\\ &\Rightarrow\ \text{efficiency } A : B = 5 : 4 \end{aligned}
2

Work done together. A alone does 50%50\%, B alone does the last 5%5\%, so the joint 4-day phase covers the rest.

together=100%−50%−5%=45%\begin{aligned} &\text{together} = 100\% - 50\% - 5\% = 45\% \end{aligned}
3

B's share of the joint work. Split 45%45\% in the efficiency ratio 5:45:4; B gets the 49\tfrac{4}{9} part.

B in 4 days=49×45%=20%\begin{aligned} &\text{B in 4 days} = \frac{4}{9}\times 45\% = 20\% \end{aligned}
4

B's full-job time. B does 20%20\% (i.e. 15\tfrac15) in 4 days.

B alone=4×5=20 days\begin{aligned} &\text{B alone} = 4 \times 5 = 20 \text{ days} \end{aligned}
20 days  ⇒  option (d)\boxed{20 \text{ days} \;\Rightarrow\; \text{option (d)}}

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CAT 2018 Slot 1 QA Q12: When they work alone, B needs 25% more time to finish a job than A does. They two finish the job in 13 days in — Solution | TheCATExam