What will be the time taken by 30 managers and 40 clerks in doing a work, if 6 managers and 8 clerks can do the piece of work in 10 days while 26 managers and 48 clerks can do the same piece of work in 2 days?

Let the one-day work of one manager be M and one clerk be C. From the first condition, 6M plus 8C equals 1/10. From the second condition, 26M plus 48C equals 1/2. If you multiply the first equation by 6, you get 36M plus 48C equals 6/10, or 3/5. Subtracting the second equation from this result yields 10M equals 3/5 minus 1/2, which simplifies to 1/10. Therefore, the one-day work of 10 managers is 1/10, which is identical to the combined rate of 6 managers and 8 clerks. Because 30 managers and 40 clerks is exactly 5 times the original team of 6 managers and 8 clerks, their combined daily work is 5 times 1/10, or 1/2. Thus, they will complete the work in 2 days.

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