Question: Cole’s Travel Time

Comment on Cole’s Travel Time

Hey, IF we can take 2-W for time driving to home. We should be able to use the same equation for 2-W for time driving to work and W for time driving to home right. Cause D.W = D.H.
greenlight-admin's picture

That's correct

Hey ... If I use the first equation with the distance variable and add the two values to equal 2hrs I get a different answer. Could u pls explain how to solve it using the first equation with T1+T2=2
greenlight-admin's picture

Sure thing.
time = distance/speed
Let d = distance EACH way.
Speed going TO work is 75 kmh.
So, time to get to work = d/75

Speed going FROM work is 105 kmh.
So, time to get home = d/105

(time to get to work) + (time to get home) = 2 hours
d/75 + d/105 = 2
Solve for d to get d = 87.5 km

We already know that time to get to work = d/75
So, time = 87.5/75
= 175/150
= 1 25/150
= 1 1/6 hours
= 70 minutes
= B

Could you please show how you arrived at 87.5? thanks.
greenlight-admin's picture

You bet.
We have: d/75 + d/105 = 2
To eliminate the fractions, multiply both sides of the equation by the least common multiple of the denominators 75 and 105
So, multiply both sides of the equation by 525 to get: 7d + 5d = 1050
Simplify: 12d = 1050
Solve: d = 87.5

Could you please show how calculate 175/150 - i don't get the 1 25/150 part.
greenlight-admin's picture

You bet.
How's this: 175/150 = (150 + 25)/150
= 150/150 + 25/150
= 1 + 25/150
= 1 + 5/30
= 1 + 1/6
= 1 1/6 hours
= 1 hour + (1/6 of an hour)
= 60 minutes + 10 minutes
= 70 minutes

Sir i solved this Question using Average speed concept.

we have the formula for Avg speed = (d+d)/(t1+t2).

t1 = time taken to travel from home to office
t2 = time taken to travel from office to work.

for t1= d/75
for t2= d/105

as we know t1+t2 = 2
therefore
d/75 + d/105 = 2
after taking lcm we get
105d+75d =15750
180d = 15750
d=87.5km

therefore t1=d/75 = 87.5/75 = 1.167hours =>1.167x60min we get 70min
therefore 70min is the answer
greenlight-admin's picture

Another valid approach. Great work!

Hi Brent,

Awesome videos. As explained, these problems can be solved in multiple ways. What I am having trouble with is deciding on which method will produce the answer quickest. Would appreciate your thoughts. Cheers.
greenlight-admin's picture

Great question!

In order to determine which approach will yield the quickest answer, you need to roughly predict the number of steps each approach will take.

For example, if a certain word problem can be solved algebraically or by testing the answer choices, you need to predict how long it will take you to create and solve the equations in an algebraic solution and compare that to how long it will take you to test each answer choice.

Cheers,
Brent

hey Brent! I got tripped up by the word "average speed".And tried to use that formula instead... It took me a longer time to get to the answer. Was that piece relevant in this question? If not, how can one tell the difference?
greenlight-admin's picture

I'd say the most relevant piece of information is that the question doesn't ask us to find the average speed; it asks us to find the time it takes Cole to get to work.

Aside: When we're given "average speeds" in a question like this, you can pretty much ignore the word "average" and just assume that the given people/vehicles are travelling at constant speeds.

Is it generally better to use the formula that has the variable in it that one wants to solve, e.g. here we ant to know time so it's better to not use the t = D/S formula if one only knows the speed? Thanks!
greenlight-admin's picture

That's 100% correct. Since we want to find TIME, we should NOT use the equation where we compare TIMES.

That said, if we DO use the equation that compares times, we'll just have one small extra step to answer the question.

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