ezhilkumarank wrote:
A drives at a rate of m miles per hour from his home to the park. On his return trip he drives at a rate of n miles per hour. How far away from his home is the part if he spends a total of z hours in the car, making no stops along the way?
A. \(\frac{n+z}{m}-\frac{z}{n}\)
B. \(\frac{m+n+z}{mn}\)
C. \(\frac{mnz}{m+n}\)
D. \(\frac{m+z}{mn}\)
E. \(\frac{mz}{n}\)
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 (distance).
From the question,
m,n=SPEEDS in MILES/HOUR
z=TIME in HOUR
Considering only Numerator part:A) \(n+z\) ------> \(\frac{miles}{hour}+hour\)
Again, we're mistakenly adding 2 DIFFERENT units! So, out.
B) \(m+n+z\) -----> \(\frac{miles}{hour}+\frac{miles}{hour}+hour\)
Again, we can't add TIME with SPEED. So, bye.
D) \(m+z\) ----> \(\frac{miles}{hour}+hour\)
We're mistakenly adding TIME with SPEED. So, out.
We're eliminating A,B, and D, because they don't make sense in real life.
Now, considering both Denominator and Numerator:E) \(\frac{mz}{n}\) -----> \(\frac{miles}{hour}/\frac{miles}{hour} × hour\)
--->
hour (this is NOT our goal; our goal is MILES). So, out.
C) \(\frac{mnz}{m+n}\) ------> Considering "denominator"
------>
Considering Denominator: equation (1)
\(m+n\) ----> \(\frac{miles}{hour}\) + \(\frac{miles}{hour}\) gives the unit \(\frac{miles}{hour}\)
------>
Considering Numerator: equation (2)
\(mnz\) ----> \(\frac{miles}{hour}\) × \(\frac{miles}{hour}\) × hour
Dividing equation (2) by equation (1): We get
miles---> This is our goal.
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