Easy Word Problem to Equation Question

[StClone]

Okay, this is an easy math question (to get the answer with simple intuition) but how the heck can it be reduced to a solvable equation?

Here it is:

My car travels 20 miles as another, traveling 20 miles/hr faster, going 30 miles in the same amount of time. How much time does it take?

There seems to be enough information, and I won't give "amount of time" answer. But what would an equation look like to solve for time?

 

[FOREVERTRUE]

Would: 20+20/T=30, be the correct equation?

Never mind that is wrong.

 

[cyclonespiker33]

Okay, this is an easy math question (to get the answer with simple intuition) but how the heck can it be reduced to a solvable equation?

Here it is:

My car travels 20 miles as another, traveling 20 miles faster, going 30 miles in the same amount of time. How much time does it take?

There seems to be enough information, and I won't give "amount of time" answer. But what would an equation look like to solve for time?

"Traveling 20 miles faster"

If this means 20 mph faster, I believe an equation can be made. If it just means the other car travels 20 miles in a shorter amount of time than OP's car drives 20 miles, there are multiple correct answers.

 

[pourcyne]

My car travels 20 miles as another,

What??? Your car has a separate identity?

 

[StClone]

"Traveling 20 miles faster"

If this means 20 mph faster, I believe an equation can be made. If it just means the other car travels 20 miles in a shorter amount of time than OP's car drives 20 miles, there are multiple correct answers.

Both take the same amount of time at the speed they are traveling one to go 20 miles and the other to go 30 miles.

 

[cyfanatic]

"How much time does it take?" Is that a question asking how long it takes your car to make that trip or the other car? Not sure there is enough info to make this a proper question without some stuff being cleared up.

 

[charlie_B]

distance traveled 1 = 20 miles
distance traveled 2 = 30 miles
speed 1 = x
speed 2 = (20+x)
time = y

20 = x*y
30 = (20+x)*y

Solve the system of equations (I think). This could come out to the equation above.

I get speed 1 is 40 mph and time is 1/2 hour. pretty sure only 1 positive combination exists

 

[StClone]

"Traveling 20 miles faster"

If this means 20 mph faster, I believe an equation can be made. If it just means the other car travels 20 miles in a shorter amount of time than OP's car drives 20 miles, there are multiple correct answers.

It's 20 miles/hr faster.

 

[Dante]

start with 2 equations: t = 20/s and t=30/(s+20)

solve for s in one of them: s=20/t

plug it into the other: t=30/(20/t + 20)

t=1/2

 

[GBlade]

A = my car speed, B = another car speed, x = time

A * x = 20
B * x = 30
A + 20 = B

Substituting for B in the second equation:
A * x + 20 * x = 30
Substituting for A * x:
20 + 20 * x = 30
20 * x = 10
x = 1/2

Not sure what the units are though.

 

[cyclonespiker33]

distance traveled 1 = 20 miles
distance traveled 2 = 30 miles
speed 1 = x
speed 2 = (20+x)
time = y

20 = x*y
30 = (20+x)*y

Solve the system of equations (I think). This could come out to the equation above.

I get speed 1 is 40 mph and time is 1/2 hour. pretty sure only 1 positive combination exists

This is correct based on the clarification above

 

[cyclonespiker33]

I solved it this way

v=velocity
d=distance
t=time

v=d/t thus t=d/v

The time traveled is the same for both vehicles, so d1/v1=d2/v2

d1=20 miles
v1 is unknown
d2=30 miles
v2=v1+20

20/v1=30/(v1+20)
Solve that and get v1=40
t=20/40 so t=0.5

 

[StClone]

distance traveled 1 = 20 miles
distance traveled 2 = 30 miles
speed 1 = x
speed 2 = (20+x)
time = y

20 = x*y
30 = (20+x)*y

Solve the system of equations (I think). This could come out to the equation above.

I get speed 1 is 40 mph and time is 1/2 hour. pretty sure only 1 positive combination exists

That seems to work. I did not want to use two variables but the built-in ratio makes it needed. Thanks.

A = my car speed, B = another car speed, x = time

A * x = 20
B * x = 30
A + 20 = B

Substituting for B in the second equation:
A * x + 20 * x = 30
Substituting for A * x:
20 + 20 * x = 30
20 * x = 10
x = 1/2

Not sure what the units are though.

Hours thanks

 

[StClone]

I solved it this way

v=velocity
d=distance
t=time

v=d/t thus t=d/v

The time traveled is the same for both vehicles, so d1/v1=d2/v2

d1=20 miles
v1 is unknown
d2=30 miles
v2=v1+20

20/v1=30/(v1+20)
Solve that and get v1=40
t=20/40 so t=0.5

You must like physics!

 

[cyclonespiker33]

You must like physics!

Haven't used it since the PE exam but sure!

 

[BuffettClone]

This thread "sounds" like the house painting word problem from the movie Little Big League

 

[dahliaclone]

distance traveled 1 = 20 miles
distance traveled 2 = 30 miles
speed 1 = x
speed 2 = (20+x)
time = y

20 = x*y
30 = (20+x)*y

Solve the system of equations (I think). This could come out to the equation above.

I get speed 1 is 40 mph and time is 1/2 hour. pretty sure only 1 positive combination exists

This is why I'm in comms. I see x and y and parenthesis and my brain shuts off.

 

[StClone]

A = my car speed, B = another car speed, x = time

A * x = 20
B * x = 30
A + 20 = B

Substituting for B in the second equation:
A * x + 20 * x = 30
Substituting for A * x:
20 + 20 * x = 30
20 * x = 10
x = 1/2

Not sure what the units are though.

Your explanation comes closest to my solution but got hung in the middle of it.

 

[PSYclone22]

Okay, this is an easy math question (to get the answer with simple intuition) but how the heck can it be reduced to a solvable equation?

Here it is:

My car travels 20 miles as another, traveling 20 miles/hr faster, going 30 miles in the same amount of time. How much time does it take?

There seems to be enough information, and I won't give "amount of time" answer. But what would an equation look like to solve for time?

30/(V+20) = 20/V

Solve for V, convert to hours.

30V = 20V + 400
10V = 400
V = 40

My car is going 40 miles per hour.

20 miles / 40 miles per hour = 0.5 hours

 

[keepngoal]

Goodgawd, we can’t even ‘English’ a math problem correctly from a science and tech university fan site.