car travels from Town A to Town B at a constant speed. If it increases its speeed by 20%,...

[hello100]

I have this question in a textbook and I have no idea if it has anything to do with algebra, but ill post it anyways.
So, a car travels from Town A to Town B at a constant speed. If it increases its speeed by 20%, it can arrive at its destination 1 hour ahead of schedule. If it increases its speed by 25% after travelling 120km at the usual speed, it can arrive 36min ahead of schedule. How far is Town A to Town B ??

 

[Otis]

Hello. It's hard to know where to begin helping, when students don't show any initial efforts (or at least explain what they're confused about). There's a guidelines notice that tells how to ask for help here. It's located at the top of the page after logging in (unless you closed it).

Start by assigning symbols to represent some unknown values.

Let d = distance from A to B
Let v = scheduled speed (what they've called "a constant speed")
Let t = scheduled time

You ought to have learned the relationship 'distance travelled equals constant speed times elapsed time'. Using our symbols, we can write:

d = v∙t

We're told that increasing the scheduled speed by 20% reduces the scheduled time by 1 hour. Hopefully, you already know how to increase a number by 20% (multiply it by 1.2). Therefore:

d = (1.2∙v)(t - 1)

Now substitute v∙t for d:

v∙t = (1.2∙v)(t - 1)

Solve this equation, to find t=6, by first dividing out v.

In other words, it takes 6 hr to drive from A to B, at constant speed v.

Under a different scenario, we're told that 36 min would be saved, if v were increased by 25% for the remainder of the trip after having driven 120km at speed v.

If a six-hour trip is completed 36 min sooner, how many hours did it take? In other words:

6 hr - 36 min = ?? hr

Determine that number of hours.

Next, we don't know how much time it took to drive 120km at speed v, so let's introduce another symbol for that:

Let T = time required to drive 120km at speed v

v∙T = 120

Putting it all together:

6∙v = v∙T + 1.25∙v∙(?? - T)

Solve for T.

Now that you know T, calculate v using v∙T=120

Then remember: The distance you're looking for is v∙t and you now have values for v and t.

There are other approaches, but you didn't show any work, so I went with my first idea. How did I come up with my approach? By experimenting! That is, I made use of the basic relationship 'distance=velocity×time', I picked symbols for unknown quantities, I used the given scenarios relating these quantities and I tried writing equations and making substitutions, solving for what I could. After a little experimenting, I realized how to determine values for t and v. (Don't be afraid to try stuff; experimenting is good practice, and you can't really break anything by trying stuff.)

If you would like more help, please post your work or ask about the part(s) you don't understand. Cheers

?

 

[Denis]

[1]: A car travels from Town A to Town B at a constant speed.
[2]: If it increases its speed by 20%, it can arrive at its destination 1 hour ahead of schedule.
[3]: If it increases its speed by 25% after travelling 120km at the usual speed,
it can arrive 36min ahead of schedule.

A "li'l picture" helps:

[1]@v.................................................................> t hours

[2]@1.2v...........................................................> t-1 hours

[3]@v...........120............>@1.25v................> t-36/60 hours

 

[hello100]

Hello. It's hard to know where to begin helping, when students don't show any initial efforts (or at least explain what they're confused about). There's a guidelines notice that tells how to ask for help here. It's located at the top of the page after logging in (unless you closed it).

Start by assigning symbols to represent some unknown values.

Let d = distance from A to B
Let v = scheduled speed (what they've called "a constant speed")
Let t = scheduled time

You ought to have learned the relationship 'distance travelled equals constant speed times elapsed time'. Using our symbols, we can write:

d = v∙t

We're told that increasing the scheduled speed by 20% reduces the scheduled time by 1 hour. Hopefully, you already know how to increase a number by 20% (multiply it by 1.2). Therefore:

d = (1.2∙v)(t - 1)

Now substitute v∙t for d:

v∙t = (1.2∙v)(t - 1)

Solve this equation, to find t=6, by first dividing out v.

In other words, it takes 6 hr to drive from A to B, at constant speed v.

Under a different scenario, we're told that 36 min would be saved, if v were increased by 25% for the remainder of the trip after having driven 120km at speed v.

If a six-hour trip is completed 36 min sooner, how many hours did it take? In other words:

6 hr - 36 min = ?? hr

Determine that number of hours.

Next, we don't know how much time it took to drive 120km at speed v, so let's introduce another symbol for that:

Let T = time required to drive 120km at speed v

v∙T = 120

Putting it all together:

6∙v = v∙T + 1.25∙v∙(?? - T)

Solve for T.

Now that you know T, calculate v using v∙T=120

Then remember: The distance you're looking for is v∙t and you now have values for v and t.

There are other approaches, but you didn't show any work, so I went with my first idea. How did I come up with my approach? By experimenting! That is, I made use of the basic relationship 'distance=velocity×time', I picked symbols for unknown quantities, I used the given scenarios relating these quantities and I tried writing equations and making substitutions, solving for what I could. After a little experimenting, I realized how to determine values for t and v. (Don't be afraid to try stuff; experimenting is good practice, and you can't really break anything by trying stuff.)

If you would like more help, please post your work or ask about the part(s) you don't understand. Cheers

?

Thank you for answering, but I have a few questions
6v=vT+1.25v*(??-T)
Here, how do I calculate that last part: 1.25v*(??-T)?
I've tried: T=120/v, but that somehow doesn't work out.

In addition, I'm supposed to explain to the whole class what method I used, so maybe there is a more simple method?
I've heard there's one involving ratios.
I'm only in grade 7 and I didn't know ( or couldn't remember) that D=V*T.
My classmates probably don't either, so I was just wondering.

 

[HallsofIvy]

"So, a car travels from Town A to Town B at a constant speed."
Let the initial speed, in km per hour, be "v", the distance between the towns, in km, be "d". The time require to travel from A to B is d/v.

"If it increases its speed by 20%, it can arrive at its destination 1 hour ahead of schedule."
Now its speed is v+ 0.20v= 1.2v. The distance is still d so the time required is d/1.2v= (1/1.2)(d/v)= (5/6)(d/v)= d/v- 1. We can solve that for d/v: (1/6)(d/v)= 1 so d/v= 6 hours

"If it increases its speed by 25% after travelling 120km at the usual speed, it can arrive 36min ahead of schedule."
120 km at the usual speed takes 120/v hours. The remaining distance is d- 120 km and at v+ .25v= 1.25 v that requires (d- 120)/1.25v. The total time is 120/v+ (d- 120)/1.25v= d/v- 36/60= 6- 0.6= 5.4.

120/v+ (1/1.25)(d/v)- (120/1.25)v= 6/1.25+ (120- 96)v= 4.8+ 24v= 5.4.
Solve 4.8+ 24v= 5.4 for v and then solve d/v= 6 for d.

24v= 0.6 so v= 0.6/24=

 

[HallsofIvy]

"I'm only in grade 7 and I didn't know ( or couldn't remember) that D=V*T."
Well, you know that "speed" is measured in "km per hour", don't you? So that V is "distance divided by time", V=D/T. And multiplying both sides by T, D= VT.