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Kona started running towards D from C at a constant speed

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Kona and Muna

by GMATGuruNY » Thu Nov 10, 2016 5:15 am
emdadul28 wrote:Kona started running towards D from C at a constant speed.At the same time Muna started running towards C from D. After they crossed each other it took Kona 40 mins to reach D and it took Muna 90 mins to reach C. What is the ratio of Kona's speed to Muna's speed?
When two people travel at different speeds, their TIME RATIO to travel the same distance will always be the same.
If Kona takes 1/2 as long as Muna to travel 10 miles, then Kona will take 1/2 as long as Muna to travel 1000 miles.
If Kona takes 3 times as long as Muna to travel 500 miles, then Kona will take three times as long as Muna to travel 2 miles.

Let M = the meeting point.
Let t = the time for Kona and Muna each to travel to M.

Kona:
C ----- t -----> M ----> 40 -----> D
Muna:
C <---- 90 ---- M <----- t ------- D

Since Kona takes t minutes to travel the blue portion, while Muna takes 90 minutes, the time ratio for Kona and Muna to travel the blue portion = t/90.
Since Kona takes 40 minutes to travel the red portion, while Muna takes t minutes, the time ratio for Kona and Muna to travel the red portion = 40/t.
Since the time ratio in each case must be the same, we get:
t/90 = 40/t
t² = 3600
t = 60.

Thus:
Kona's time to travel the blue portion = t = 60.
Muna's time to travel the blue portion = 90.
(Kona's time)/(Muna's time) = 60/90 = 2/3.

Time and rate have a RECIPROCAL relationship.
If Kona takes HALF AS LONG as Muna to travel x miles, then Kona travels TWICE AS FAST as Muna.
If Kona takes ONE-THIRD AS LONG as Muna to travel x miles, then Kona travels THREE TIMES AS FAST as Muna.

Thus, the rate ratio for Kona and Muna must be the RECIPROCAL of their time ratio:
(Kona's speed)/(Muna's speed) = [spoiler]3/2[/spoiler].
Last edited by GMATGuruNY on Sun Feb 17, 2019 4:38 am, edited 1 time in total.
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by Brent@GMATPrepNow » Fri Nov 11, 2016 6:50 am
emdadul28 wrote:Kona started running towards D from C at a constant speed.At the same time Muna started running towards C from D. After they crossed each other it took Kona 40 mins to reach D and it took Muna 90 mins to reach C. What is the ratio of Kona's speed to Muna's speed?
Another approach.

Let T = the time (in minutes for simplicity) it took until Kona and Muni to meet.
This means that Kona's total travel time = T + 40
And Muna's total travel time = T + 90

Also, let Kona's speed = K
And let Muna's speed = M

We'll start with the "word" equation: (Kona's total distance traveled) = (Muna's total distance traveled)
Distance = (rate)(time)
So, we get: (K)(T + 40) = (M)(T + 90)
Divide both sides by M to get: (K)(T + 40)/M = (T + 90)
Divide both sides by (T + 40) to get: K/M = (T + 90)/(T + 40)

We're almost there, now we need only determine the value of T

To determine this, I'm going to borrow a piece of Mitch's solution:
Kona:
C ----- T -----> M ----> 40 ----> D
Muna:
C <---- 90 ----M <----- T ------- D

So, it took Muna 90 minutes to travel the SAME distance that Kona traveled in T minutes
So, we'll start with: (Munu's distance) = (Kona's distance)
Now we can write: (M)(90) = (K)(T)
Rearrange to get: K/M = 90/T

Also, it took Kona 40 minutes to travel the SAME distance that Muna traveled in T minutes
So, we'll start with: (Kona's distance) = (Munu's distance)
Now we can write: (K)(40) = (M)(T)
Rearrange to get: K/M = T/40

Since we solved for K/M both times, we can write: 90/T = T/40
So, T² = 3600
So, T = 60

Now that we know the value of T, we'll plug it into our equation K/M = (T + 90)/(T + 40)
We get: K/M = (60 + 90)/(60 + 40) = 150/100 = 3/2

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by Matt@VeritasPrep » Fri Nov 11, 2016 2:34 pm
This one is simpler than it looks.

Let's say that Kona is running from Columbus to Dayton, that Muna is running from Dayton to Columbus, and that they cross paths in Springfield.

Once they meet, Kona is going to run the distance that Muna has already run: the distance from Springfield to Dayton. Likewise, Muna is going to run the distance that Kona has already run: the distance from Springfield to Columbus.

Let's say that they meet after running for h hours. What took Kona h hours will take Muna (3/2) of an hour, and what took Muna h hours will take Kona (2/3) of an hour.

If Kona's rate is k, and Muna's rate is m, then we've got

k * h = m * (3/2)

and

m * h = k * (2/3)

This gives us h = (3/2)m/k and h = (2/3)k/m,

or

(3/2m)/k = (2/3)k/m,

or

k/m = 3/2.
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by Matt@VeritasPrep » Fri Nov 11, 2016 2:47 pm
You can also use a general equation here. Provided that Kona and Muna leave at the same time, run at constant rates, and don't take any breaks at any point in their runs, we'll always have

Kona's D Before They Meet = Muna's D After They Meet

and

Kona's D After They Meet = Muna's D Before They Meet

If their rates are k and m, the time until the meeting is h hours, and the times after the meeting are x hours (for Kona) and y hours (for Muna), then we'll have

k * h = m * y

and

k * x = m * h

From here:

h = m*y/k = k*x/m

m²*y = k²*x

y/x = (k/m)²

Since k, m, x, and y are all positive, we've then got

√(y/x) = k/m
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by Matt@VeritasPrep » Fri Nov 11, 2016 2:49 pm
Follow up: a general equation for a problem like this isn't something I would bother deriving on the GMAT! The scenario isn't common enough to bother finding and memorizing one either.
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