Multiple choice

$A, B$ and $C$ complete a piece of work in $25, 20$ and $24$ days respectively. All work together for $2$ days and then $A$ and $B$ leave the work . $C$ works for next $\displaystyle8\frac{3}{5}$ days and then $A$ along with $D$ join $C$ and they all finish the work in next three days. In how many days $D$ alone can complete the whole work ?

  1. $22$ days
  2. $\displaystyle21\frac{1}{2}$days
  3. $23$ days
  4. $\displaystyle22\frac{1}{2}$days
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D Correct answer
Explanation

Work rates: A=1/25, B=1/20, C=1/24. Work done in 2 days by A, B, C: 2 * (1/25 + 1/20 + 1/24) = 2 * (24+30+25)/600 = 79/300. Remaining work = 221/300. C works for 43/5 days: 43/5 * 1/24 = 43/120. Remaining = 221/300 - 43/120 = (442-215)/600 = 227/600. A+D+C work for 3 days: 3 * (1/25 + 1/24 + 1/D) = 227/600. Solving for D gives 22.5 days.

AI explanation

Let the total work be the LCM of 25, 20 and 24, which is 600 units, making the daily work of A, B and C equal to 24, 30 and 25 units respectively. In the first 2 days, all three complete 2 times 79, which is 158 units, and C alone completes 43 over 5 times 25, which is 215 units over the next 8.6 days. This leaves 600 minus 158 minus 215, which is 227 units of work. A and C working with D complete 227 units in 3 days, so their combined daily rate is 75.66 units. Since A and C contribute 49 units per day, D's daily rate is 75.66 minus 49, which is 26.66 units. D alone would take 600 divided by 26.66, which is 22.5 days.

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