a bus is travelling with 52 passenger. when it arrives at a stop, "y" passengers get off & 4 get on. at the next stop, one-third of the passengers get off & 4 get on. there are now 25 passengers. find "y"

after first stop the number of passengers is :52 -y +4second stop: (52 -y +4) - (52 -y +4) * (1/3) + 4 = 25
so we have to solve : (52 -y +4) - (52 -y +4) * (1/3) + 4 = 25 at the second stop there was (52 -y +4) passengersand you have been told that 1/3 of them get offso we have to subtract (52 -y +4) * (1/3)

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jscuervo1992 jscuervo1992 · Answer 2 · 2019-10-03T02:19:07+02:00

Answer:

The exact answer is y = 24.5 passengers, Between 24 and 25 passengers,

Step-by-step explanation:

The problem is stated so we cant arrive to a whole number of passengers but Ill explain it.

In t= 0 , the initial time we have

Passengers = 52

In t= 1 , first stop

Passengers = 52 - y + 4

In t= 2 , second stop passengers are

Passengers in bus = 52 - y + 4 - 1/3(52 - y + 4) +4

This is the tricky part. Ill explain it slower. In time t = 2 passengers are

Passengers in t2 = Passengers in t1 - 1/3 * Passengers in t1 +4

We subtract a portion of total passengers in t1 , so its 1/3 of the whole expression 52 - y + 4.

In t= 2 the second and final stop, there are 25 passengers so

52 - y + 4 - 1/3(52 - y + 4) +4 = 25 (eq 1)

Using common factors we have :

[52 - y + 4] (1 - 1/3) + 4 = 25

[52 - y + 4] 2/3 = 21

[52 - y + 4] = 31.5

y = 52+4-31.5 = 24.5

If we replace in equation 1

52 - y + 4 - 1/3(52 - y + 4) +4 = 25 (eq 1)

52 - 24.5 + 4 - 1/3(52 - 24.5 + 4) +4 = 25 (eq 1)

31.5 -1/3(31,5) + 4= 25

31,5 - 10.5 + 4= 25

21 +4 = 25

25 = 25