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how far has bird flown when two trains cross each other
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Here's the question:
A train leaves point A and heads to point B (which is 100 km away) at 20 km/hr (the track is a straight line between the two points).
Another train leaves point B and heads towards point A at 30 km/hr.
A bird sitting on the firs train takes off as soon as the train starts and flies back and forth between the two trains until the trains pass each other.
If the bird flies 40 km/hr, how far has the bird flown at the time the trains pass each other?
Here's what I've tried:
Distance = 100
Rate of train A = 20
Time it takes train A to travel the track = (100/20) = 5 hours
Time it takes train B to travel the track = (100/30) = 3.33 hours
So is the time that the two trains meet 5 - 3.33 = 1.67 hours
And then I would multiply 40 *1.67 = 66.8 km to get the distance the bird traveled.
Is my logic correct, and did I arrive at the correct answer? If not, please let me know where I have erred.
Thanks!
algebra-precalculus proof-verification word-problem
edited Aug 1 at 17:28
Bram28
54.5k33880
asked Aug 1 at 13:58
rds80
183
add a comment |Â
up vote
1
down vote
favorite
Here's the question:
A train leaves point A and heads to point B (which is 100 km away) at 20 km/hr (the track is a straight line between the two points).
Another train leaves point B and heads towards point A at 30 km/hr.
A bird sitting on the firs train takes off as soon as the train starts and flies back and forth between the two trains until the trains pass each other.
If the bird flies 40 km/hr, how far has the bird flown at the time the trains pass each other?
Here's what I've tried:
Distance = 100
Rate of train A = 20
Time it takes train A to travel the track = (100/20) = 5 hours
Time it takes train B to travel the track = (100/30) = 3.33 hours
So is the time that the two trains meet 5 - 3.33 = 1.67 hours
And then I would multiply 40 *1.67 = 66.8 km to get the distance the bird traveled.
Is my logic correct, and did I arrive at the correct answer? If not, please let me know where I have erred.
Thanks!
algebra-precalculus proof-verification word-problem
edited Aug 1 at 17:28
Bram28
54.5k33880
asked Aug 1 at 13:58
rds80
183
What is your question? Would you like us to check your work?
– amWhy
Aug 1 at 14:01
Yes, if my logic makes sense. And if I have the correct answer.
– rds80
Aug 1 at 14:01
In 1 hour Train A travels 30 km, Train B travels 20 km and the trains have covered half the distance between them.
– Doug M
Aug 1 at 14:05
The trains approach each other at $50$ km/hr, so they meet in $2$ hours.
– saulspatz
Aug 1 at 14:08
"Trick? What trick? All I did was sum the geometric series" - John von Neumann
– NickD
Aug 1 at 17:11
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Here's the question:
A train leaves point A and heads to point B (which is 100 km away) at 20 km/hr (the track is a straight line between the two points).
Another train leaves point B and heads towards point A at 30 km/hr.
A bird sitting on the firs train takes off as soon as the train starts and flies back and forth between the two trains until the trains pass each other.
If the bird flies 40 km/hr, how far has the bird flown at the time the trains pass each other?
Here's what I've tried:
Distance = 100
Rate of train A = 20
Time it takes train A to travel the track = (100/20) = 5 hours
Time it takes train B to travel the track = (100/30) = 3.33 hours
So is the time that the two trains meet 5 - 3.33 = 1.67 hours
And then I would multiply 40 *1.67 = 66.8 km to get the distance the bird traveled.
Is my logic correct, and did I arrive at the correct answer? If not, please let me know where I have erred.
Thanks!
algebra-precalculus proof-verification word-problem
edited Aug 1 at 17:28
Bram28
54.5k33880
asked Aug 1 at 13:58
rds80
183
Here's the question:
A train leaves point A and heads to point B (which is 100 km away) at 20 km/hr (the track is a straight line between the two points).
Another train leaves point B and heads towards point A at 30 km/hr.
A bird sitting on the firs train takes off as soon as the train starts and flies back and forth between the two trains until the trains pass each other.
If the bird flies 40 km/hr, how far has the bird flown at the time the trains pass each other?
Here's what I've tried:
Distance = 100
Rate of train A = 20
Time it takes train A to travel the track = (100/20) = 5 hours
Time it takes train B to travel the track = (100/30) = 3.33 hours
So is the time that the two trains meet 5 - 3.33 = 1.67 hours
And then I would multiply 40 *1.67 = 66.8 km to get the distance the bird traveled.
Is my logic correct, and did I arrive at the correct answer? If not, please let me know where I have erred.
Thanks!
algebra-precalculus proof-verification word-problem
edited Aug 1 at 17:28
Bram28
54.5k33880
asked Aug 1 at 13:58
rds80
183
edited Aug 1 at 17:28
Bram28
54.5k33880
edited Aug 1 at 17:28
Bram28
54.5k33880
edited Aug 1 at 17:28
Bram28
54.5k33880
54.5k33880
asked Aug 1 at 13:58
rds80
183
asked Aug 1 at 13:58
rds80
183
asked Aug 1 at 13:58
rds80
183
183
What is your question? Would you like us to check your work?
– amWhy
Aug 1 at 14:01
Yes, if my logic makes sense. And if I have the correct answer.
– rds80
Aug 1 at 14:01
In 1 hour Train A travels 30 km, Train B travels 20 km and the trains have covered half the distance between them.
– Doug M
Aug 1 at 14:05
The trains approach each other at $50$ km/hr, so they meet in $2$ hours.
– saulspatz
Aug 1 at 14:08
"Trick? What trick? All I did was sum the geometric series" - John von Neumann
– NickD
Aug 1 at 17:11
add a comment |Â
What is your question? Would you like us to check your work?
– amWhy
Aug 1 at 14:01
Yes, if my logic makes sense. And if I have the correct answer.
– rds80
Aug 1 at 14:01
In 1 hour Train A travels 30 km, Train B travels 20 km and the trains have covered half the distance between them.
– Doug M
Aug 1 at 14:05
The trains approach each other at $50$ km/hr, so they meet in $2$ hours.
– saulspatz
Aug 1 at 14:08
"Trick? What trick? All I did was sum the geometric series" - John von Neumann
– NickD
Aug 1 at 17:11
What is your question? Would you like us to check your work?
– amWhy
Aug 1 at 14:01
What is your question? Would you like us to check your work?
– amWhy
Aug 1 at 14:01
Yes, if my logic makes sense. And if I have the correct answer.
– rds80
Aug 1 at 14:01
Yes, if my logic makes sense. And if I have the correct answer.
– rds80
Aug 1 at 14:01
In 1 hour Train A travels 30 km, Train B travels 20 km and the trains have covered half the distance between them.
– Doug M
Aug 1 at 14:05
In 1 hour Train A travels 30 km, Train B travels 20 km and the trains have covered half the distance between them.
– Doug M
Aug 1 at 14:05
The trains approach each other at $50$ km/hr, so they meet in $2$ hours.
– saulspatz
Aug 1 at 14:08
The trains approach each other at $50$ km/hr, so they meet in $2$ hours.
– saulspatz
Aug 1 at 14:08
"Trick? What trick? All I did was sum the geometric series" - John von Neumann
– NickD
Aug 1 at 17:11
"Trick? What trick? All I did was sum the geometric series" - John von Neumann
– NickD
Aug 1 at 17:11
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
0
down vote
accepted
Well, your last idea to find the distance the bird flew is nice. However, the first part kind of isn't. For the first part, what you need to do is to find $x$ such that train $A$ travels $x$ in the same amount of time $B$ needs to go $100-x$ kilometers. This leads to solving:
$$ fracx20 = frac100 - x30 $$
This is a linear equation and it should not be too hard for you to solve by yourself.
To illustrate the next part, let me give away that the answer is given by $x=40$. This means that $A$ travels $40$ km in the same time as $B$ $60$ km. The time it took the two trains to meet is thus given by $frac4020 left(=frac100-4030right) = 2$ hours. This then implies that the bird flew in total $80 ( = 2cdot 40)$ km.
answered Aug 1 at 14:09
Stan Tendijck
1,277110
add a comment |Â
up vote
0
down vote
Not quite correct:
The trains do not pass each other after $1.67$ hours
Try to check you answer by seeing where they each are after that time, and where they are after $2$ hours
You should consider how fast the distance between them reduces
answered Aug 1 at 14:05
Henry
92.7k469147
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
Well, your last idea to find the distance the bird flew is nice. However, the first part kind of isn't. For the first part, what you need to do is to find $x$ such that train $A$ travels $x$ in the same amount of time $B$ needs to go $100-x$ kilometers. This leads to solving:
$$ fracx20 = frac100 - x30 $$
This is a linear equation and it should not be too hard for you to solve by yourself.
To illustrate the next part, let me give away that the answer is given by $x=40$. This means that $A$ travels $40$ km in the same time as $B$ $60$ km. The time it took the two trains to meet is thus given by $frac4020 left(=frac100-4030right) = 2$ hours. This then implies that the bird flew in total $80 ( = 2cdot 40)$ km.
answered Aug 1 at 14:09
Stan Tendijck
1,277110
add a comment |Â
up vote
0
down vote
accepted
Well, your last idea to find the distance the bird flew is nice. However, the first part kind of isn't. For the first part, what you need to do is to find $x$ such that train $A$ travels $x$ in the same amount of time $B$ needs to go $100-x$ kilometers. This leads to solving:
$$ fracx20 = frac100 - x30 $$
This is a linear equation and it should not be too hard for you to solve by yourself.
To illustrate the next part, let me give away that the answer is given by $x=40$. This means that $A$ travels $40$ km in the same time as $B$ $60$ km. The time it took the two trains to meet is thus given by $frac4020 left(=frac100-4030right) = 2$ hours. This then implies that the bird flew in total $80 ( = 2cdot 40)$ km.
answered Aug 1 at 14:09
Stan Tendijck
1,277110
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Well, your last idea to find the distance the bird flew is nice. However, the first part kind of isn't. For the first part, what you need to do is to find $x$ such that train $A$ travels $x$ in the same amount of time $B$ needs to go $100-x$ kilometers. This leads to solving:
$$ fracx20 = frac100 - x30 $$
This is a linear equation and it should not be too hard for you to solve by yourself.
To illustrate the next part, let me give away that the answer is given by $x=40$. This means that $A$ travels $40$ km in the same time as $B$ $60$ km. The time it took the two trains to meet is thus given by $frac4020 left(=frac100-4030right) = 2$ hours. This then implies that the bird flew in total $80 ( = 2cdot 40)$ km.
answered Aug 1 at 14:09
Stan Tendijck
1,277110
Well, your last idea to find the distance the bird flew is nice. However, the first part kind of isn't. For the first part, what you need to do is to find $x$ such that train $A$ travels $x$ in the same amount of time $B$ needs to go $100-x$ kilometers. This leads to solving:
$$ fracx20 = frac100 - x30 $$
This is a linear equation and it should not be too hard for you to solve by yourself.
To illustrate the next part, let me give away that the answer is given by $x=40$. This means that $A$ travels $40$ km in the same time as $B$ $60$ km. The time it took the two trains to meet is thus given by $frac4020 left(=frac100-4030right) = 2$ hours. This then implies that the bird flew in total $80 ( = 2cdot 40)$ km.
answered Aug 1 at 14:09
Stan Tendijck
1,277110
answered Aug 1 at 14:09
Stan Tendijck
1,277110
answered Aug 1 at 14:09
Stan Tendijck
1,277110
answered Aug 1 at 14:09
Stan Tendijck
1,277110
1,277110
add a comment |Â
add a comment |Â
up vote
0
down vote
Not quite correct:
The trains do not pass each other after $1.67$ hours
Try to check you answer by seeing where they each are after that time, and where they are after $2$ hours
You should consider how fast the distance between them reduces
answered Aug 1 at 14:05
Henry
92.7k469147
add a comment |Â
up vote
0
down vote
Not quite correct:
The trains do not pass each other after $1.67$ hours
Try to check you answer by seeing where they each are after that time, and where they are after $2$ hours
You should consider how fast the distance between them reduces
answered Aug 1 at 14:05
Henry
92.7k469147
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Not quite correct:
The trains do not pass each other after $1.67$ hours
Try to check you answer by seeing where they each are after that time, and where they are after $2$ hours
You should consider how fast the distance between them reduces
answered Aug 1 at 14:05
Henry
92.7k469147
Not quite correct:
The trains do not pass each other after $1.67$ hours
Try to check you answer by seeing where they each are after that time, and where they are after $2$ hours
You should consider how fast the distance between them reduces
answered Aug 1 at 14:05
Henry
92.7k469147
answered Aug 1 at 14:05
Henry
92.7k469147
answered Aug 1 at 14:05
Henry
92.7k469147
answered Aug 1 at 14:05
Henry
92.7k469147
92.7k469147
add a comment |Â
add a comment |Â
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What is your question? Would you like us to check your work?
– amWhy
Aug 1 at 14:01
Yes, if my logic makes sense. And if I have the correct answer.
– rds80
Aug 1 at 14:01
In 1 hour Train A travels 30 km, Train B travels 20 km and the trains have covered half the distance between them.
– Doug M
Aug 1 at 14:05
The trains approach each other at $50$ km/hr, so they meet in $2$ hours.
– saulspatz
Aug 1 at 14:08
"Trick? What trick? All I did was sum the geometric series" - John von Neumann
– NickD
Aug 1 at 17:11