prep question

[jamesk486]

A subway train made eleven stops on its route. If an average(arith mean) of 30 passagenters boarded at each of the first 9 stops and at each stop, begining with the 2nd, 5 fewer passengers boraded than at the previous stop, how many passengers boarded the train at the first stop?

A. 60
B. 50
C. 30
D. 20
E. 10

[jayhawk2001]

A subway train made eleven stops on its route. If an average(arith mean) of 30 passagenters boarded at each of the first 9 stops and at each stop, begining with the 2nd, 5 fewer passengers boraded than at the previous stop, how many passengers boarded the train at the first stop?

A. 60
B. 50
C. 30
D. 20
E. 10

Use arithmetic progression for this. We know sum of people who boarded
on the first 9 stops = 9*30 = 270.

Sum = n/2 ( 2a + (n-1)d) -- a = first term, n = num terms, d = difference

270 = 9/2 * (2a + 8*(-5))
60 = 2a - 40
a = 50

Hence B

[Cybermusings]

Yup thats the best approach

[Scott@TargetTestPrep]

A subway train made eleven stops on its route. If an average(arith mean) of 30 passagenters boarded at each of the first 9 stops and at each stop, begining with the 2nd, 5 fewer passengers boraded than at the previous stop, how many passengers boarded the train at the first stop?

A. 60
B. 50
C. 30
D. 20
E. 10

Solution:

Since at each stop, beginning with the second, 5 fewer passengers boarded the train than at the previous stop, the number of passengers boarding the train at each of the first nine stops forms an evenly-spaced set. Therefore, the 5th stop must have the median, or the average, number of passengers, which is 30. So, the 4th stop has 35, the third has 40, the second has 45, and finally, the first stop has 50 passengers.

Answer: B