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

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

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

.................... 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

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)

\(r=10\)

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

Posted from my mobile device

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

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

Now simply PLUG-IN the alternative choices.

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

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.

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 + 20) - r]/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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