A group of 12 persons containing men and women can do a piece of work in 13 days, other group of 10 persons containing men and women can do the same piece of work in 16 days. Number of men in first group is 1 more than the number of men in second group. If there is another group of 2 men and 6 women which can do the same piece of work in 26 days. What is the ratio of efficiency of 1 man to that of 1 woman?

This problem can be solved using the concept of work and efficiency.

Step 1: Let's denote the efficiency of one man as 'm' and the efficiency of one woman as 'w'.

Step 2: From the problem, we know that the first group of 12 people (men and women) can complete the work in 13 days. So, the total work can be represented as 12 * 13 = 156 units (considering 1 day work of one person as one unit).

Step 3: Similarly, the second group of 10 people can complete the same work in 16 days. So, the total work can also be represented as 10 * 16 = 160 units.

Step 4: Since the total work should be the same, we equate 156 units and 160 units to get the equation: 156m + 156w = 160m + 160w.

Step 5: Simplifying the equation, we get 4m = 4w, or m = w. This means that the efficiency of one man is equal to the efficiency of one woman.

Step 6: The third group of 2 men and 6 women can complete the work in 26 days. So, the total work can also be represented as (2m + 6w) * 26.

Step 7: Since we know that m = w, we can substitute m in place of w in the equation to get: (2m + 6m) * 26 = 156m.

Step 8: Comparing this with the total work, we get 156m = 156, or m = 1.

Step 9: Since m = w, we also get w = 1.

Step 10: Therefore, the ratio of the efficiency of one man to that of one woman is 1:1.