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

SOLUTION

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

Algebraic approach:

Say the distance Aaron jogs is \(d\) miles, notice that the distance Aaron walks back will also be \(d\) miles (since he walks back home on the same route).

Next, total time \(t\) would be equal to the time he spends on jogging plus the time he spends on walking: \(\frac{d}{x}+\frac{d}{y}=t\) --> \(d(\frac{1}{x}+\frac{1}{y})=t\) --> \(d=\frac{xyt}{x+y}\).

Answer: C.

Number picking approach:

Say the distance in 10 miles, \(x=10\) mile/hour and \(y=5\) mile/hour (pick x and y so that they will be factors of 10).

So, Aaron spends on jogging 10/10=1 hour and on walking 10/5=2 hours, so total time \(t=1+2=3\) hours.

Now, we have that \(x=10\), \(y=5\) and \(t=3\). Plug these values into the answer choices to see which gives 10 miles. Only answer choice C fits: \(\frac{xyt}{x+y}=\frac{10*5*3}{10+5}=10\).

Answer: C.

Note that for plug-in method it might happen that for some particular number(s) more than one option may give "correct" answer. In this case just pick some other numbers and check again these "correct" options only.

Hope it helps.
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The correct answer should have units of length (distance).
Also, we cannot add/subtract velocities and time, it has no meaning. Therefore, we can immediately eliminate answers B, D and E.
The expression in A has units of time, so it cannot be the correct answer.
Only C is left, and it is the correct answer. It also has the correct units of length.

Formally, if we denote by D the distance we are looking for, we can write the equation:
D/x + D/y = t, from which, solving for D we get D =xyt/(x+y).
(time spent jogging + time spent walking = total time)

Answer: C

Note: The answer should not depend on whether Aaron first jogs and then walks, or the other way around. This means that the expression for the distance should be symmetrical in x and y (meaning, it doesn't change if we switch between x and y). The only expressions on the list of answers which fulfill this condition are C and D. But D has velocities added to time, therefore it cannot be correct.
This is just another way to easily pinpoint the correct answer.
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[quote="Bunuel"]The Official Guide for GMAT® Review, 13th Edition - Quantitative Questions Project

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!!!
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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.
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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
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Always Average is equal to total distance by total time.
In this case
(D/x+D/Y) = t
Hence
D = xyt/(x+y)
Hence 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.
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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.
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lets consider the distance is d.

so time taken (t1) for jogging is d/x
time taken (t2) for walking is d/y

Given that the t1+t2=t

t=(d/x)+(d/y)
d=t* (xy)/(x+y)

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

Posted from my mobile device

Picking Numbers is the way to go on the test. Get to the answer quickly. But Algebra method I used:


Set up R-T-D Chart for both Going and Coming Back.

Both legs of the round trip have EQUAL Distance = d


Distance = Speed * Time
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Distance Going = d = X (mph) * (Total Time T - Time spent traveling BACK)


Time = Distance / Speed


Thus the Time spent traveling BACK = Tb = (d miles) / (Y mph)

———substitute this equation into the 1st equation ———-



d = x * [ T - (d/y) ]

d = x * [ (Ty - d) / (y) ]

——multiply both sides by the +pos. DEN of y AND Multiply x through the parenthesis ——

yd = xTy - xd


—- since the Question is asking for the Distance he can travel for 1 leg of the round trip —> ISOLATE d——

yd + xd = xTy

d * (y + x) = xTy

d = xTy / (y + x)

Answer -C-

Posted from my mobile device

Hello Guys,

Why can't we start directly by making

D total = V total x T total .ie

D total = (xt + yt)*(D/x + D/y)... Is it possible ?

How can we know that we need and must to get through the total time to find the answer ?

Thank you for your help !

Video solution from Quant Reasoning starts at 15:45
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The question is asking us to find a value of distance Aaron can jog if he spends T hours jogging and walking. Since we know that the distance traveled of jogging and walking are equal to each other we can create the below table.

Using the table below can say that the sum of two individual travel times are equal to the total travel time T. So D/X + D/Y = T, and with this equation we can solve for D which is equal to D = XYT/X+y and C is our answer.