Doubt: AR-TWK-012-20260621-0615-q5

Anonymous

26d ago

🔢 QA Doubts

isnt the answer 4 1/3

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Linked Question

A, B, and C can complete a piece of work in 12, 8, and 6 days respectively when working alone. They work in rotating pairs: AB on day 1, BC on day 2, CA on day 3, repeating. In how many days will the work be completed?

A2⅔ days

B5 days

C4 1/7 days✓ Correct

D4⅙ days

▶ Show solution

Step 1 — Individual and pair rates. $A=\frac{1}{12}$ , $B=\frac{1}{8}$ , $C=\frac{1}{6}$ (common denominator 24).

Day 1 AB: $\frac{2}{24}+\frac{3}{24} = \frac{5}{24}$
Day 2 BC: $\frac{3}{24}+\frac{4}{24} = \frac{7}{24}$
Day 3 CA: $\frac{4}{24}+\frac{2}{24} = \frac{6}{24}$

Step 2 — One 3-day cycle. $\frac{5+7+6}{24} = \frac{18}{24} = \frac{3}{4}$ .

Step 3 — Track the remainder.

After cycle (3 days): $\frac{3}{4}$ done, remaining $\frac{1}{4} = \frac{6}{24}$ .
Day 4 (AB, $\frac{5}{24}$ ): since $\frac{5}{24} < \frac{6}{24}$ , continue; remaining $\frac{1}{24}$ .
Day 5 (BC, rate $\frac{7}{24}$ ): leftover $\frac{1}{24}$ takes $\frac{1/24}{7/24} = \frac{1}{7}$ of the day.

Total $= 4 + \frac{1}{7} = 4\tfrac{1}{7}$ days.

Answer: C — 4 1/7 days.

Takeaway: Day 5 is BC's turn (the cycle repeats AB, BC, CA), so divide the leftover by BC's rate $\frac{7}{24}$ . Using CA's rate gives the wrong 4⅙; rounding up gives 5.

2 Answers

Accepted answer

CATLab AI✦ CATLab AI26d ago

This one hinges on tracking the work left after each full 3-day cycle — and on using the correct pair's rate for the final partial day.

Step 1 — Set up rates. Take total work as the LCM of 12, 8, 6 = 24 units. Then per day: A does $\frac{24}{12}=2$ units, B does $\frac{24}{8}=3$ units, C does $\frac{24}{6}=4$ units.

So each pair's daily output:

AB: $2+3=5$ units
BC: $3+4=7$ units
CA: $4+2=6$ units

Step 2 — One full cycle (3 days). The order is AB, BC, CA, so in 3 days they finish $5+7+6=18$ units out of 24, i.e. $\frac{3}{4}$ of the work. Remaining: $24-18=6$ units ( $\frac{1}{4}$ of the work).

Step 3 — Day 4. The cycle repeats, so AB works: 5 units done. Remaining: $6-5=1$ unit.

Step 4 — Day 5. It's BC's turn, and BC finishes 7 units per day. The last 1 unit therefore takes $\frac{1}{7} \text{ of a day.}$

Total time $= 4 + \frac{1}{7} = 4\tfrac{1}{7}$ days.

About the $4\frac{1}{3}$ guess: that comes from dividing the leftover 1 unit by B's individual rate of 3 units/day — but B isn't working alone on day 5. BC together do 7 units/day, giving $\frac{1}{7}$ , not $\frac{1}{3}$ . (Similarly, using C's rate would give $\frac{1}{6}$ , which is the trap in option D.)

Final answer: C

Anonymous24d ago

No the answer is actually 4 complete day and on day 5 the work done was 1/7. So yeah its the right answer

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