A car travels through a path 1 mile long at a speed of r miles per hou

Question

A car travels through a path 1 mile long at a speed of r miles per hour. If it travels at a speed of r+20 miles per hour, it would cover the path in 4 minutes less. What is the value of r?

A) 30
B) 20
C) 10
D) 8
E) 5

Bunuel · Answer

First of all, since the rate, r is in miles per  hour , then convert 4 minutes into hours: 4 minutes = 4/60 = 1/15 hours.

Next, construct equation 1/r = 1/(r+20) + 1/15. Plug options into the equation. C fits: 1/10 = 1/(10+20) + 1/15.

Answer: C.

beattheheat · Answer

Solution: 
Distance = Speed * Time, Distance is same in both case
r * t = (r + 20) (t - 1/15) -- (4 mins =1/15 hrs)
rt = rt + 20t - r/15 - 20/15 -- (substitute t = 1/r)
r^2 + 20r - 300 = 0
r = {10, -30}
Ignoring negative value, r=10 Answer C

EMPOWERgmatRichC · Answer

Hi pbarrocas,

This question can be quickly solved by TESTing THE ANSWERS. You will have to do a bunch of small calculations though...

Since the question mentions distance, rate and time, we'll need the Distance Formula:

Distance = (Rate)(Time)

We're given the distance (1 mile) and asked to consider 2 rates (R and R+20; both in miles/hour). We're looking for the two resulting "times" to differ by 4 minutes. The question asks us for the value of R.

Let's start with Answer B:
If R = 20 mph, then...
1mi = (20mph)(T)
1/20 hours = T 
T = 3 minutes

R + 20 = 40mph, then...
1mi = (40mph)(T)
1/40 hours = T
T = 1.5 minutes

Difference = 3 - 1.5 = 1.5 minutes, which is too little (the car was driving too fast, so we need to try a slower speed).

Normally, I would TEST Answer D next, but since 8 is an "awkward" value to plug into this equation, I'm going to TEST Answer C.

Answer C
If R = 10 mph, hten
1mi = (10mph)(T)
1/10 hours = T
T = 6 minutes

R + 20 = 30 mph
1mi = (30mph)(T)
1/30 hours = T
T = 2 minutes

Difference = 6 - 2 = 4 minutes, which is a MATCH!

Final Answer:  C

GMAT assassins aren't born, they're made,
Rich

Temurkhon · Answer

1h=60min

60*(1/r) - 60*(1/r+20)=4

60/r-60/(r+20)=4 (we can backsolve here)

60/r=(4r+140)/(r+20)

60r+1200=4r^2+140r

4r^2+80r-1200=0

r^2+20r-300=0

(r+30)*(r-10)=0

r=-30 or 10

it is C

PareshGmat · Answer

.................... Distance ................... Speed ...................... Time

Initial >>>>> 1 ............................. r ...............................  \frac{1}{r}  (Calculated)

After changes .. 1 ............................. r+20 .........................  \frac{1}{r} - \frac{4}{60}  (4 minutes = 4/60 Hours)

Setting up the equation

1 = (r+20)(\frac{1}{r} - \frac{4}{60})

r^2 + 20r - 300 = 0

r = 10

Answer = C

Princ · Answer

OA:C

\frac{1}{r}-\frac{1}{r+20}=\frac{4}{60}

\frac{r+20-r}{r(r+20)}=\frac{4}{60}

\frac{20}{r(r+20)}=\frac{4}{60}

300=r(r+20)

r^2+20r-300=0

r^2+30r-10r-300=0

r(r+30)-10(r+30)=0

(r-10)(r+30)

r=10, -30 (-30 rejected)

r=10

Kritika2008poddar · Answer

I found solving this type of questions using options are more helpful

Posted from my mobile device

fskilnik · Answer

Units control is one of the most powerful tools of the GMATH method:

\frac{{1\,\,mile}}{{r\,\,mph}}\,\, = \,\,\frac{1}{r}\,\,h\,\,\,\left( {\frac{{60\min }}{{1h}}} ight)\,\,\,\, = \,\,\,\frac{{60}}{r}\,\,\,\min \,\,\,\,\,\,\,\left( {{	ext{original}}\,\,{	ext{time}}} ight) 
 \frac{{1\,\,mile}}{{\left( {r + 20} ight)\,\,mph}}\,\, = \,\,\frac{1}{{r + 20}}\,\,h\,\,\,\left( {\frac{{60\min }}{{1h}}} ight)\,\,\,\, = \,\,\,\frac{{60}}{{r + 20}}\,\,\,\min \,\,\,\,\,\,\,\left( {{	ext{less}}\,\,{	ext{time}}} ight) 
 \frac{{60}}{r} - \frac{{60}}{{r + 20}} = 4

Now simply PLUG-IN the alternative choices.

The solution above follows the notations and rationale taught in the GMATH method.

CAMANISHPARMAR · Answer

The key to solving this problem is realizing that the distance is same. We can apply the speed/distance/time formula and solve for r.

Correct answer is C.

ScottTargetTestPrep · Answer

We can let t = the time, in hours, it takes the car to travel 1 mile at a speed of r mph. Thus we have:

rt = 1

t = 1/r

We are given that if the speed is r + 20 mph, the time to travel 1 mile would be 4 minutes, or 4/60 = 1/15 hour, less. Using the formula rate x time = distance, we have:

(r + 20)(1/r - 1/15) = 1

1 - r/15 + 20/r - 4/3 = 1

-r/15 + 20/r - 4/3 = 0

Multiply the equation by -15r, we have:

r^2 - 300 + 20r = 0

r^2 + 20r - 300 = 0

(r - 10)(r + 30) = 0

r = 10 or r = -30

Since r can’t be negative, r = 10.

Alternate Solution:

At r mph, it takes the car 1/r hours to travel 1 mile. At (r + 20) mph, it takes the car 1/(r + 20) hours to travel 1 mile. Since we are given that at (r + 20) mph, it takes 4 minutes, or 1/15 hours, less to travel 1 mile, we can create the equation:

1/r - 1/(r + 20) = 1/15

/r(r + 20) = 1/15

20/r(r+20) = 1/15

r^2 + 20r = 300

r^2 + 20r - 300 = 0

(r - 10)(r + 30) = 0

r = 10 or r = -30

Since r can’t be negative, r = 10.

Answer: C

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A car travels through a path 1 mile long at a speed of r miles per hou

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