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A drives at a rate of m miles per hour from his home to the  [#permalink]

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8 00:00

Difficulty:   25% (medium)

Question Stats: 79% (02:17) correct 21% (02:37) wrong based on 212 sessions

### HideShow timer Statistics 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. $$\frac{mz}{n}$$

_________________ Support GMAT Club by putting a GMAT Club badge on your blog Originally posted by ezhilkumarank on 26 Aug 2010, 19:33.
Last edited by Bunuel on 19 Apr 2012, 20:58, edited 1 time in total.
Edited the question
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Re: Rate --- GMAT Practice Question.  [#permalink]

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ezhilkumarank wrote:
Question: 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) $$((n+z)/m) - (z/n)$$
B) $$(m+n+z)/mn$$
C) $$(mnz)/(m+n)$$
D) $$(m+z)/mn$$
E) $$mz/n$$

Let the distance between home and park be $$d$$.

From home to park A would need $$\frac{d}{m}$$ hours and from park to home A would need $$\frac{d}{n}$$ hours as A spent total of z hours for round trip then $$\frac{d}{m}+\frac{d}{n}=z$$ --> $$d(\frac{m+n}{mn})=z$$ --> $$d=\frac{mnz}{m+n}$$.

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Senior Manager  Status: Time to step up the tempo
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Re: Rate --- GMAT Practice Question.  [#permalink]

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Bunuel wrote:
ezhilkumarank wrote:
Question: 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) $$((n+z)/m) - (z/n)$$
B) $$(m+n+z)/mn$$
C) $$(mnz)/(m+n)$$
D) $$(m+z)/mn$$
E) $$mz/n$$

Let the distance between home and park be $$d$$.

From home to park A would need $$\frac{d}{m}$$ hours and from park to home A would need $$\frac{d}{n}$$ hours as A spent total of z hours for round trip then $$\frac{d}{m}+\frac{d}{n}=z$$ --> $$d(\frac{m+n}{mn})=z$$ --> $$d=\frac{mnz}{m+n}$$.

Bunuel -- you made it look so simple. I could not get my mind cranked up so quickly. Guess I have to practice more. _________________ Support GMAT Club by putting a GMAT Club badge on your blog EMPOWERgmat Instructor V
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GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: A drives at a rate of m miles per hour from his home to the  [#permalink]

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1
Hi All,

While this is an older post, the prompt is still representative of the types of 'variable-heavy' prompts that you can see on the Official GMAT. This question can be solved by TESTing VALUES.

Let's TEST...
Distance to park = 6 miles
M = 2 miles per hour (driving to the park)
N = 3 miles per hour (driving from the park)
Z = 2+3 = 5 total hours driving

We're asked for the distance from the home to the park. Thus, we're looking for an answer that equals 6, when we plug in M=2, N=3 and Z=5 into the answer choices.

Answer A: 8/2 - 5/3 = 2 1/3 NOT a match
Answer B: 10/6 NOT a match
Answer C: 30/5 = 6 This IS a match
Answer D: 7/6 NOT a match
Answer E: 10/3 NOT a match

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Re: A drives at a rate of m miles per hour from his home to the  [#permalink]

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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. $$\frac{mz}{n}$$

SHORTCUT

if m=n, time taken for 1 way will be z/2
therefore distance will be mz/2

replace n =m in the options

only option C will give the value mz/2

option C
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Re: A drives at a rate of m miles per hour from his home to the  [#permalink]

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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. $$\frac{mz}{n}$$

Let the distance between home and park be d. Driving from home to the park, the driver spends d/m hours, and, driving from the park to home, the driver spends d/n hours. Since we are given that the total time spent is z, we can create the following equation:

d/m + d/n = z

Let’s multiply each side by mn:

dn + dm = mnz

d(n + m) = mnz

d = mnz/(n + m)

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Re: A drives at a rate of m miles per hour from his home to the  [#permalink]

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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!

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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