Question

The “Steal a Deal” gift shop specializes in heavily discounted merchandise for the holiday season. The store prices are so appealing that lines of people queue up in the early AM hours in front of the store doors in order to secure a place in the line.

On the morning after Thanksgiving, one of the busiest days of the year, the store opened its doors at 8:00AM. At that time there were 122 customers already waiting in line. From 8:00AM to 10:00AM, new customers arrive at the rate of 5 per minute. After 10:00AM, the rate reduces to 2 per minute. The store admits customers at the rate of 4 per minute.

a) Jacob plans to arrive to the store at 9:00AM. How long should he expect to wait?

b) Rachel does not want to wait more than 15 minutes. When should she show up?

Answer 1

Answer -1) At 8.00 am , their were already 122 customers. Jacob plans to arrive at store at 9.00 am. The customers arrive at store at 5/min . So at 9.00 am , the number of people standing in front of Jacob are- 122 + (60*5) = 422 . The store attends customer at 4/min

The store opens at 8.00 am and till 9.00am the store attends (60*4) = 240 customers . The customers who are left are 182 (422-240) . The time which will be taken to attend jacob is = 182/4 = 45.5 min. So the time jacob will wait is 60+45.5= 105.5 minutes.

Answer-2) So till 10 am the customers came in total is (122+ 600) = 722. The customers who have entered the store -( 120min *4) =480 . The Customers still in line after 10 am is 242 and then customers arrive bat rate of 2/min. So the amount of time by which the left customers can be treated 242/4 = 60 min. the new customers which came = 60*2 =120. They will be treated in 120/4 = 30 min. The new customers came in 30*2= 60 customers. 60 customers will be treated in 60/4 = 15 min. Hence Rachel shall shows up - at 10.am + 60 min +30 min = 11.30 am.