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Comment on Cole’s Travel Time
Hey, IF we can take 2-W for
That's correct
That's correct
Hey ... If I use the first
Sure thing.
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
You bet.
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
You bet.
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
Thanks for the helpful video,
If we take the fraction 87.5
If we take the fraction 87.5/75 and multiply numerator and denominator by 2, we get the equivalent fraction 175/150
Sir i solved this Question
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
Another valid approach. Great
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.
Great question!
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
I'd say the most relevant
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
That's 100% correct. Since we
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.