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let fare be (f), constant be (k) and inverse proportion be (c)
f=k+c/n
when, n=40, then f=240; when n=50, f= 200
i) 240=k+c/40
ii) 200= k+c/50we remove the denominator by multiplying through by the denominator,
i.e. (240*40)= 40k+ (c/40)*40
9600= 40k+ c ..............................................(i)
& (200*50)= 50k+ (c/50)*50
10000= 50k+c .............................................(ii)
solving the two equations (i) and (ii),
we subtract through directly since (c) is constant in both the equations
10000 = 50k + c
- 9600 = 40k + c
= 400 = 10k
hence, k = 400/10
= 40
we input the value of k in one of the equations (i) or (ii)
(i) 9600 = 40*40 + c
9600 = 1600 + c
c = 9600 - 1600
c = 8000
now we form our equation from including only values for (c) and (k)
from; f = k + c/n
f = 40 + 8000/n
lets go to the question; we are given (n) as 100 to find (f)
feeding (n) in our equation; f = 40 + 8000/100
f = 40 + 80
hence, f = 120
therefore, the fare per passenger when there are 100 passengers, is 120.
Eng melau answered the question on January 31, 2018 at 05:49
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