If I walk at 5 km/hour; I miss a train by 7 minutes. If however, I walk at 6 km/hour I reach the station 5 minutes before the departure of the train. Then the distance between my house and the station is
5(t+7) = 6(t-5) solve for t
5t+35=6t-30 35+30=t t=65
it just a start after u got the value of t, subtitute the value of t to lefft side or right side the equation above, u will get its distance
5(t+7)= 5(65+7)=360
smart :)
okay , but one thing it is direct or indirect proprrotion
actually, it is an inverse proportion... also it just the formula of distance (d) if given speed(s) and its time (t) : d = s t or s = d/t
k
but time is in minutes and speed is in km/hour
opss... my bad, not read it correctly 7 m = 7/60 h 5 m = 5/60 so, it should be 5(t + 7/60) = 6(t - 5/60) solve fot t again :)
wait i solve it
actually, it just /60 from ur answer before
yup
so, 360/60 = 6 km
t= 13/12
5(t + 7/60) = 6(t - 5/60) 5t + 35/60 = 6t - 30/60 35/60 + 30/60 = 6t - 5t 65/60 = t or t=13/12 u are right :)
now, vat
vat ?
next step
subtitute t to one of side : 5(t + 7/60) = 6(t - 5/60)
according to the statement can we write the equation as 5(t + 7/60) - 6(t - 5/60)=10/60
how come u got 10/60, in right side ?
oops now i did mistke
1 : 1 :)
k
i thing i didn't understand why we have to put the value of t in 5(t + 7/60) or 6(t - 5/60) First of all we didn;t need to assume the distance between my house and the station is
wait can we do in this way the difference in time is 12 minutes since time = distance/speed, d/5 = d/6 + (12/60) d = 6
the distance if u walk late is same is u walk by high speed the difference just to speed and using it's time
d1 = d2
yup
sorry, i didnt conect what relation about d/5 = d/6 + (12/60) maybe, it juts simply form from calculating above ??
yup, d= distance
oh, yes.. manipulating algebra i think
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