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We can calculate the number of days in the given week in which one of the students spent more time than the other.

If $X$ is more than $10\%$ of $Y$ then $Y$ should be minimum and vice-versa.

- On Monday, the required percentage $ = \left(\dfrac{70-45}{45}\right) \times 100\% = 55.55\%$
- On Tuesday, the required percentage $ = \left(\dfrac{65-55}{55}\right) \times 100\% = 18.18\%$
- On Wednesday, the required percentage $ = \left(\dfrac{60-50}{50}\right) \times 100\% = 20\%$
- On Thursday, the required percentage $ = \left(\dfrac{60-55}{55}\right) \times 100\% = 9.09%$
- On Friday, the required percentage $ = \left(\dfrac{35-20}{20}\right) \times 100\% = 75\%$
- On Saturday, the required percentage $ = \left(\dfrac{60-50}{50}\right) \times 100\% = 20\%$
- On Sunday, the required percentage $ = \left(\dfrac{65-55}{55}\right) \times 100\% = 18.18\%$

$\therefore$ Total $6$ days are there, when one of the students spent a minimum of $10\%$ more than the other student.

So, the correct answer is $(C).$