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A train covered the distance of 400 km between A and B at a certain speed. On the way back it covered 2/5 of the distance at th...Asked by Anish Panda
A train covered the distance of 400 km between A and B at a certain speed. On the way back it covered 2/5 of the distance at that same speed and then it decreased its speed by 20 km/hour. Find the speed of the train at the end of its journey from B back to A, if the entire trip took 11 hours.
Answers
Answered by
Reiny
I did this for you here
http://www.jiskha.com/display.cgi?id=1458920959
hoping you would be able to find the equation yourself
then Steve finished it for you here:
http://www.jiskha.com/display.cgi?id=1458939085
http://www.jiskha.com/display.cgi?id=1458920959
hoping you would be able to find the equation yourself
then Steve finished it for you here:
http://www.jiskha.com/display.cgi?id=1458939085
Answered by
Anish Panda
I cant solve it though
Answered by
Reiny
400/s + 160/s + 240/(s-20) = 11
ok, hint:
multiply each term by s(s-20) to get a quadratic.
Hint: it comes out to be a nice whole number
ok, hint:
multiply each term by s(s-20) to get a quadratic.
Hint: it comes out to be a nice whole number
Answered by
Edy
400(s-20)+160(s-20)+240(s)=11(s)(s-20)
(400+160)(s-20)+240s=11(s^2-20s)
560(s-20)+240s=11s^2-220s
560s-11200+240s=11s^2-220
800s-11200=11s^2-220
ā11s2+1020sā11200=0
(ā11s+140)(sā80)=0
s=80? or 140/11?
It still does not work
(400+160)(s-20)+240s=11(s^2-20s)
560(s-20)+240s=11s^2-220s
560s-11200+240s=11s^2-220
800s-11200=11s^2-220
ā11s2+1020sā11200=0
(ā11s+140)(sā80)=0
s=80? or 140/11?
It still does not work
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