Multiple choice

$3$ men and $4$ boys can do a piece of work in $14$ days, while $4$ men and $6$ boys do the same work in $10$ days. How long would it take one man to finish the same work ?

  1. $40$ days
  2. $20$ days
  3. $35$ days
  4. $70$ days
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let M be the work of a man and B be the work of a boy. 14(3M + 4B) = 10(4M + 6B) => 42M + 56B = 40M + 60B => 2M = 4B => M = 2B. Total work = 14(3(2B) + 4B) = 14(10B) = 140B. Time for one man = 140B / M = 140B / 2B = 70 days.

AI explanation

Let the total work be the LCM of 14 and 10, which is 70 units, so 3 men and 4 boys do 5 units per day and 4 men and 6 boys do 7 units per day. Setting up the equations 3m plus 4b equals 5 and 4m plus 6b equals 7, we multiply the first by 3 and the second by 2 to get 9m plus 12b equals 15 and 8m plus 12b equals 14. Subtracting them gives 1 man equals 1 unit per day. Therefore, one man alone will take 70 divided by 1, which is 70 days to finish the work.

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