A shop produces sarongs. The daily average production is given by 5n + 20, where n is the number of workers aside from the owner. In the first k days, 500 units are produced, and then 5 workers are added to the team. After another k days, the cumulative total is 1100. How many workers were part of the latter production run?	Answers.
A) 26	B) 10	C) 11	D) 15	E) 23.5

Rationale: the daily average production is given by 5 n + 20 - given in the first k days , 500 units are produced = ( 5 n + 20 ) k = 500 k = 500 / 5 n + 20 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 5 workers were added = 5 ( n + 5 ) + 20 = 5 n + 45 cumulative is 1100 . . thus for the current period = 1100 - 500 = 600 ( 5 n + 45 ) k = 600 k = 600 / 5 n + 45 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 equate 1 and 2 500 / 5 n + 20 = 600 / 5 n + 45 500 ( 5 n + 45 ) = 600 ( 5 n + 20 ) 2500 n + 22500 = 3000 n + 12000 - 500 n = - 10500 n = 21 thus n + 5 = 26 hence a
Answer: A