QR Rate problem

[Ocean5]

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Two runners start at the same place and same time and run against eachother around a oval track of 2 miles. One runs 10 mph and the other 15 mph. How long before they meet?

0.48 min
4.8 min
9.6 min
0.16 min
0.8 hur

 

[chessxwizard]

This problem can be solved by setting up one simple equation. The hard part is modeling the situation.

Let's make the track linear by unwinding it and placing the runners at the start position. It says they run against each other, so essentially they'll be running towards each other when making the track linear.

|-------------------------2 mi-------------------------|
x------> 10mph.................................15mph <--------y

At some time t, the runners meet. Thus, the distance not travelled by one runner will be the distance travelled by the other when a critical time is reached.

Hence, we set up an equation.
Distance = Rate x Time

Distance NOT travelled by x at critical time t:
total track length - distance travelled by x
2 - 10t

Distance travelled by y at critical time t:
distance travelled by y
15t

At time t, both distances are equal.

2 - 10t = 15t

t = 2/25hr or 4.8 minutes. Hope this helps! 👍 🙂

 

[dentz]

Suppose the distance at where the runners meet = a
So, if runner 1 travels 'a' distance, runner 2 will travel 2-a
Also, the time (= b) at which both of them meet will be the same for each runner.


Rate x Time = Distance
Runner 1 10m/hr b a
Runner 2 15m/h b 2-a

Thus, 10m/hr x b = a
15m/h x b = 2-a

substituting the a in eq 2,
15m/hr x b = 2 - (10m/h x b)
(15m/60min x b) + (10m/60min x b) = 2 [60 min = 1 hour]
25b/60 = 2
b = 2x60/25
b = 2x60x4/(25x4)
b = 480/100
b = 4.8min

 

[z3u2]

Two runners start at the same place and same time and run against eachother around a oval track of 2 miles. One runs 10 mph and the other 15 mph. How long before they meet?

0.48 min
4.8 min
9.6 min
0.16 min
0.8 hur

here might be a faster way to do it.
sort of like trial and error, without actually calculating it.
suppose both runners run for 1/10 hour (6 min)
Runner A would run 1 mile, runner B would run 1.5 miles.
Which is already greater than 2 miles
meaning the answer is less than 6 min
possible choices are
0.48 min, 4.8 min, 0.16 min
it will definetly take more than one min
so, the only possible answer is 4.8 min

it's really up to you how u solve it.
but for me this is a lot faster than setting up the equation and solving it.

 

[the prodogy]

here might be a faster way to do it.
sort of like trial and error, without actually calculating it.
suppose both runners run for 1/10 hour (6 min)
Runner A would run 1 mile, runner B would run 1.5 miles.
Which is already greater than 2 miles
meaning the answer is less than 6 min
possible choices are
0.48 min, 4.8 min, 0.16 min
it will definetly take more than one min
so, the only possible answer is 4.8 min

it's really up to you how u solve it.
but for me this is a lot faster than setting up the equation and solving it.

Thats how i ended up solving it too... although knowing how to do it with the equation is a lot easier. I think i solved it this way because i didn't know how to set it up, and yet, i solved it faster than most people would using the equation. Funny how knowing less could actually favor you for the exam

 

[dentz]

here might be a faster way to do it.
sort of like trial and error, without actually calculating it.
suppose both runners run for 1/10 hour (6 min)
Runner A would run 1 mile, runner B would run 1.5 miles.
Which is already greater than 2 miles
meaning the answer is less than 6 min
possible choices are
0.48 min, 4.8 min, 0.16 min
it will definetly take more than one min
so, the only possible answer is 4.8 min

it's really up to you how u solve it.
but for me this is a lot faster than setting up the equation and solving it.

Interesting approach!! Never thought of it that way!! Thanks!!🙂

 

[Streetwolf]

Or the easiest way would be that the total speed of both runners combined is 25mph. When they meet again they've covered a total of 2 miles.

So 25*t = 2
t = 2/25 of an hour, or 4.8 minutes.