Hi,

Given:
Average Production : 5n+20, n is the number of workers
In K days 500 Units are produced.
In another K days with additional 5 workers they produced = 1250-500 =750.
To Find:
How many workers were part of later production i.e. => n+5 = ?

Solution:
From K days they produced 500 Units,
So, (Average production) * (No of Days) = Total Production for that period
hence, (5n+20) * (k) = 500 ---------Equation 1
From, 'In another K days with additional 5 workers they produced = 1250-500 =750.'
(5(n+5)+20) * (k) = 750 --------------Equation 2
By Solving Equation 1 & 2 we get K=10 and N= 6.

We need to find n+5 So, Ans is 11.

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 1250. How many workers were part of the latter production run?

A. 6
B. 10
C. 11
D. 15
E. 23.5

In the first k days, the number of units produced = 500.
The total number of units produced over the entire run = 1250.
Thus, the number of units produced during the second run of k days = 1250-500 = 750.

We can plug in the answers, which represent the number of workers for the second run.
When the correct answer is plugged in, the number of days for each run will be the same.

Answer choice C: 11 workers.
Average daily output for 11 workers = 5(11)+20 = 75.
Time to produce 750 units at 75 units per day = 750/75 = 10 days.
Since 5 workers are added for the second run, the number of workers for the first run = 11-5 = 6.
Average daily output for 6 workers = 5(6)+20 = 50.
Time to produce 500 units at 50 units per day = 500/50 = 10 days.
Success!
The number of days for each run is the same: k=10.

The correct answer is C.