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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