Bunuel wrote:
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) \(\frac{xt}{y}\)
(B) \(\frac{(x+t)}{xy}\)
(C) \(\frac{xyt}{(x+y)}\)
(D) \(\frac{(x+y+t)}{xy}\)
(E) \(\frac{(y+t)}{x}-\frac{t}{y}\)
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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