Aaron will jog from home at x miles per hour and then walk back home

Aaron will jog from home at x miles per hour and then walk back home by the same route at y miles per hour. How many miles from home can Aaron jog so that he spends a total of t hours jogging and walking?

(A) xt/y
(B) (x+t)/xy
(C) xyt/(x+y)
(D) (x+y+t)/xy
(E) (y+t)/x-t/y

A note here:

When one travels at speed x for time t and at a speed y for time t (same amount of time), the average speed for the journey is (x+y)/2 i.e. arithmetic mean of x and y.
This is because average speed = Total Distance/Total time = (xt + yt)/2t = (x + y)/2

When one travels at speed x for distance d and at a speed y for distance d (same distance), the average speed for the journey is 2xy/(x+y)
This is so because average speed = Total Distance/Total time = 2d/(d/x + d/y) = 2xy/(x + y)

You don't really need to learn these formulas but knowing that there are these special cases will increase your speed. Once you use them a couple of times, you will automatically remember these.

In this question, if you know that average speed will be 2xy/(x+y), you multiply it by t to get total distance traveled which is 2xyt/(x+y). So distance one way must be xyt/(x+y)

This is a question for which an algebraic approach would be much faster than number picking.
When working with numbers, one easily forgets about units and meanings, and for example, we have no problem in adding speeds to time. Such operations have no meaning in the real, physical world. Once you see x + t or y + t, don't even bother to check that answer.

So, don't be afraid of letters and a little algebra. And first of all, think of the meaning!!!

Hi, I do not understand how you did this step:

\(d(\frac{1}{x}+\frac{1}{y})=t\) --> \(d=\frac{xyt}{x+y}\)

How do you go from the leftward equation to the right? Can you explain in more detail? Thanks in advance.

\(d(\frac{1}{x}+\frac{1}{y})=t\);

\(d(\frac{y+x}{xy})=t\);

\(d=\frac{xyt}{x+y}\).

Hope it's clear.

We can let d = the distance traveled when walking and jogging.

Thus, the jogging time is d/x and the walking time is d/y. Since the total time is t:

d/x + d/y = t

Multiply the entire equation by xy:

dy + dx = txy

d(y + x) = txy

d = txy/(y + x)

Answer: C

Let d = the number of miles (distance) that Aaron JOGS.
This also means that d = the distance that Aaron WALKS.

Let's start with a WORD EQUATION:

total time = (time spent jogging) + (time spent walking)
In other words: t = (time spent jogging) + (time spent walking)
Since time = distance/speed, we can write: t = d/x + d/y [our goal is to solve this equation for d]
The least common multiple of x and y is xy, so we can eliminate the fractions by multiplying both sides by xy. When we do so, we get...
txy = dy + dx
Factor right side to get: txy = d(x + y)
Divide both sides by (x+y) to get: txy/(x+y) = d


So, the correct answer is C

Cheers,
Brent

Hello Bunuel can you please rephrase the question "How many miles from home can Aaron jog so that he spends a total of t hours jogging and walking?"

I simply could not understand what this question want ? Total distance done by walking and jogging ?

thanks

Say the distance in d miles. Aaron jogs d miles at x mile/hour and then walks back the same d miles at y mile/hour. So, Aaron spends on jogging d/x hours and on walking d/y hours, so total time (from home jogging and back walking) takes him t = d/x + d/y hours.

Now, the question asks to find the distance (d) Aaron jogs (How many miles from home can Aaron jog). So, basically the question asks to express d in terms of x, y, and t.

Hope it's clear.

This question is excerpted from Official Guide 2017 edition
Question# PS Diagnostic test 24

When can we ADD or SUBTRACT things in the real world?
------> ONLY when BOTH have EXACTLY THE SAME UNITS.

REMEMBER:
NEVER pick an answer choice that's absurd in the real world!

Here the question is asking about MILES.
From the question,
x,y=SPEEDS in MILES/HOUR
t=TIME in HOUR
Considering only Numerator part:
B) \(x+t\) ----> \(\frac{miles}{hour}+hour\)
We're mistakenly adding TIME with SPEED. So, out.

D) \(x+y+t\) -----> \(\frac{miles}{hour}+\frac{miles}{hour}+hour\)
Again, we can't add TIME with SPEED. So, bye.

E) \(y+t\) ------> \(\frac{miles}{hour}+hour\)
Again, we're mistakenly adding 2 DIFFERENT units! So, out.

We're eliminating B,D, and, E because they don't make sense in real life.

Now, considering both Denominator and Numerator:
A) \(\frac{xt}{y}\) -----> \(\frac{miles}{hour}/\frac{miles}{hour} × hour\)
---> hour (this is NOT our goal; our goal is MILES). So, out.

C) \(\frac{xyt}{x+y}\) ------> Considering "denominator"
------> Considering Denominator: equation (1)
x+y ----> \(\frac{miles}{hour}\) + \(\frac{miles}{hour}\) gives the unit \(\frac{miles}{hour}\)

------> Considering Numerator: equation (2)
xyt ----> \(\frac{miles}{hour}\) × \(\frac{miles}{hour}\) × hour

Dividing equation (2) by equation (1): We get miles
---> This is our goal.

——
You make the algebraic approach look so easy - I can’t believe I didnt get this one 😞

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­Just use those variables as if they were numbers. Trick is to solve for each time and add the times together to get the total time T they are referring to: