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algebra

Expert replies
by manasi_sh » Sat Oct 06, 2007 8:42 pm
. 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
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Source: — Problem Solving |

by samirpandeyit62 » Sat Oct 06, 2007 11:03 pm
Rate * time = Work

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

t = n-1/x
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Samir
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by apple100 » Mon Mar 23, 2009 10:04 pm
samirpandeyit62 wrote: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?
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by nitya34 » Tue Mar 24, 2009 1:34 am
should be (n-1)*(1/x)
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by sacx » Tue Mar 24, 2009 2:42 am
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
SACX
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