Question 817273
The key to this problem is to focus on the fraction of freshmen who were not cut, not on the fraction that were cut.<br>
Let x = the initial number of freshmen who tried out.
After the first round 1/2 were cut. If 1/2 were cut then 1/2 were not cut. So
{{{(1/2)x}}}
would be the number who were not cut during the first round. In the next round 1/3 were cut. So 2/3 were not cut making:
{{{(2/3)(1/2)x}}}
the number of freshmen not cut after two rounds. Next 1/4 were cut and 3/4 were not. So:
{{{(3/4)(2/3)(1/2)x}}}
were still in it after 3 rounds. Then 1/5 were cut and 4/5 were not making
{{{(4/5)(3/4)(2/3)(1/2)x}}}
the number of freshmen who were not cut after 4 rounds. Next 1/6 were cut and 5/6 were not. So:
{{{(5/6)(4/5)(3/4)(2/3)(1/2)x}}}
After all this there were 10 left. So:
{{{(5/6)(4/5)(3/4)(2/3)(1/2)x = 10}}}<br>
Now we solve for x. A lot of canceling can be done on the left side:
{{{(cross(5)/6)(cross(4)/cross(5))(cross(3)/cross(4))(cross(2)/cross(3))(1/cross(2))x = 10}}}
leaving:
{{{(1/6)(1/1)(1/1)(1/1)(1/1)x = 10}}}
which simplifies to:
{{{(1/6)x = 10}}}
Multiplying by 6:
{{{x = 60}}}<br>
Check:
60
Keep 1/2 => 30
Keep 2/3 => 20
Keep 3/4 => 15
Keep 4/5 => 12
Keep 5/6 => 10 Check!