algebra

[manasi_sh]

. When ticket sales began, Pat was the nth customer in line for a ticket, and customers purchased their tickets at the rate of x customers per minute. Of the following, which best approximates the time, in minutes, that Pat had to wait in line from the moment ticket sales began?
(A) (n - 1) x
(B) n + x –1
(C) n-1/x
(D) x/n-1
(E) n/x-1
ANS : C

[samirpandeyit62]

Rate * time = Work

so x * t =n-1 (i. e n-1 customers processed before Pat)

t = n-1/x

[apple100]

Rate * time = Work

so x * t =n-1 (i. e n-1 customers processed before Pat)

t = n-1/x

how is n-1= work?

[nitya34]

should be (n-1)*(1/x)

[sacx]

x tickets sold in 1 min
therefore, a ticket is sold every 1/x min

since Pat is nth in the queue, there are n-1 people ahead of him in the queue

to sell (n-1) tickets it will take (1/x) * (n-1) mins

hence (n-1)/x

Eg: 10 tickets are sold in 1 min
therefore, a ticket is sold in every 1/10th of a min
now to sell 100 tickets it will take = 100 * (1/10) mins

Hope it helps