A team of Diggers

The main formula to use here is distance = (rate)x(time). But instead of distance traveled, here we are talking about work completed. Let $n$ stand for the number of students, and $r$ stand for the rate at which they work. The first part tells us that 1 job is completed in 24 hours by $n$ students each working at a rate of $r$ . This translates to the equation $1=nr(24)$ . The second part tells us that there was an equal spacing between times that each started working. So the first boy works $t$ hours, then the second boy works $t-a$ hours, and so on, until finally the last boy works $t-(n-1)a$ hours. Then we are told the first boy works 11 times longer than the last. So, let $x$ stand for the amount of time the last boy works. This means the first boy works $11x$ hours. So, $t=11x$ and $t-(n-1)a=x\Rightarrow a=\frac{t-x}{n-1}=\frac{10x}{n-1}$ . Each works at a rate $r$ , so the total work that gets done is $1=r(11x)+r(11x-\frac{10x}{n-1})+r(11x-2\frac{10}{n-1})+...+r(\frac{10x}{n-1}+x)+rx$ . This summation is an arithmetic series with first term $11rx$ and common difference $-r\frac{10x}{n-1}$ . Using the formula for the sum of an arithmetic series we find this summation is equal to $n\frac{11rx+rx}{2}=6nrx=1$ . And since $6rnx=1=24nr$ , we can solve for $x$ to find $x=4$ .