A student took five papers in an examination, where the maximum marks were same for each paper. His marks in these papers were in the proportion of 3: 4: 9: 11: 13. In all papers together, the candidate obtained 60% of the total marks. Then, the number of papers in which he got less than 70% marks is .

To solve this problem, we need to follow these steps:

1. First, we need to understand that the student's marks in the five papers are in the proportion of 3:4:9:11:13. This means that if we add these proportions together, we get the total proportion of marks the student got in all papers. So, 3+4+9+11+13 = 40.

2. The problem states that the student obtained 60% of the total marks in all papers. This means that the total proportion of marks the student got (40) is equal to 60% of the total marks.

3. To find out the proportion that represents 100% of the marks, we can set up the equation 40/60 = x/100, where x is the total marks. Solving for x, we get x = (40*100)/60 = 66.67. This means that the total marks for each paper is 66.67.

4. Now, we need to find out in how many papers the student got less than 70% marks. To do this, we need to calculate 70% of the total marks for each paper, which is 0.7*66.67 = 46.67.

5. Comparing this with the student's marks in each paper (which are 3, 4, 9, 11, and 13 times the common proportion), we find that the student got less than 46.67 marks in the first two papers (3 and 4 times the common proportion).

So, the student got less than 70% marks in 2 papers.