Question on arithmetic. Time and Work. > Three diggers dug a ditch 324 m deep in six days working simultaneously. During one shift, the third digger digs as many metres more than the second as the second digs more than the first. The third digger's work in 10 days is equal to the first digger's work in 14 days. How many metres does the first digger dig per shift? I don't understand what is the duration of each shift. Is my approach correct? (I haven't arrived at the answer.) My approach: Let the 3 diggers be A,B,C. Let they dig a m/day, b m/day, c m/day respectively. ATQ: 10c=14a or, 5c=7a.------(1) also, (a+b+c)*6=324----(2) and Let duration of each shift be t (c-b)*t =(b-a)*t c-b =b-a-----(3) Solving 3 equations we get b=18 m/day. They have asked to find how many metres does the first digger dig per shift. I can't understand what they have meant, hence not able to find that.

We are told that the third digger digs as many metres more than the second digs than the second digs more than the first. Let $d$ be that difference. Then $d = c - b = b - a$. The amount the first digger digs is $a = b - d = 18~\text{m} - d$ and the amount the third digger digs is $c = b + d = 18~\text{m} + d$. Since we know that $7a = 5c$, we obtain \begin{align*} 5c & = 7a\\\ 5(18~\text{m} + d) & = 7(18~\text{m} - d)\\\ 90~\text{m} + 5d & = 126~\text{m} - 7d\\\ 12d & = 36~\text{m}\\\ d & = 3~\text{m} \end{align*} from which we conclude that $a = 18~\text{m} - 3~\text{m} = 15~\text{m}$.