Question

erica hiked up a mountain trial at 3 km/h and returned at 4 km/h. the entire trip took 5 hours and 10 minuites, including the half-hour she spent at the top. how long was the trail
how does this work

Answer 1

If I may help?

Answer 2

?

Answer 3

can u? please...

Answer 4

Thank you.

Answer 5

mp

Answer 6

entire trip took 5 hours and 10 minuites, what can you conclude from that?

Answer 7

that it was 310 min?

Answer 8

Great!

Answer 9

yay!

Answer 10

"including the half-hour she spent at the top",
what is the net time for just go and back?

Answer 11

280 min

Answer 12

oh

Answer 13

ye

Answer 14

Excellent!

Answer 15

what is the speed for going?

Answer 16

3 km/pm

Answer 17

3 km/hr you mean, ok?

what in km/min?

Answer 18

180km/pm?

Answer 19

ye i meant hours but thought it would be easier in minutes...

Answer 20

\[3km/hr=\frac{ 3 }{ 60 }km/\min\]

Answer 21

ok.

Answer 22

what for back?

Answer 23

4/60 km/min

Answer 24

and we know that
\[speed=\frac{ distance }{ time }\]
so that
\[diatance=speed×time\]
ok?

Answer 25

ok

Answer 26

and she go and come back the same distance twice ok?

Answer 27

ok

Answer 28

so now we can equate the (speed×time) for the go one to the back one
can you share your work?

Answer 29

so it's \[3=x/280\]

Answer 30

or 3+x/280=4+x/280

Answer 31

No man
280 is for both together not for one trip

Answer 32

oh
ok...

Answer 33

I mean\[v_1*t_1=v_2*t_2\]
and we have v_1 and v_2
can you show me?

Answer 34

whats v and t

Answer 35

v stands for speed and t for time and x for distance

Answer 36

Can you figure out the relation between t_1 and t_2?

Answer 37

for km/min:
\[\frac{ 3 }{ 60 }*t_1=\frac{ 4 }{ 60 }*t_2\]

ok?

Answer 38

ok, great

Answer 39

so now do we need to find out the time?

Answer 40

1 min plz

Answer 41

ok

Answer 42

Sorry for late.

Answer 43

So as we get the final equation, we can substitute with it into the first one!

Answer 44

\[\frac{ 3 }{ 60 }*t_1=\frac{ 4 }{ 60 }*t_2........3t_1=4t_2..............\huge t_1=\frac{ 4 }{ 3 }t_2\]

Answer 45

@raphibomb
Can you proceed?

Answer 46

\[\huge t_1+t_2=280\min\]

Answer 47

\[\huge t_1=\frac{ 4 }{ 3 }t_2\]
\[\huge t_1+t_2=280\min\]
\[\frac{ 4 }{ 3 }t_2+t_2=280\]
\[\frac{ 7 }{ 3 }t_2=280\]
\[t_2=120\min\]
subsitute in any equation, we get t_1:
\[t_1=\frac{ 4 }{ 3 }t_2=\frac{ 4 }{ 3 }*120=160\min\]


For check:
to go =120 min
to come back = 160 min
she stayed here for= 30 min

add theses periods up, you got 310 as we calculated firstly

Answer 48

I am happy to hear that.

Answer 49

@raphibomb
If you have any difficulties, kindly tell me!

Answer 50

Salam!