Sarah is driving to the airport. After driving at 35 miles

This is pretty simple. He travels the first 35 kms of his journey in 1 hr. He has to travel the remaining distance say x kms in x/35 hrs. With the increased speed, he can reach the venue in x/35-1 hrs. But he reaches the destination in x/3.5-1.5. Therefore 50=x/(x/3.5-1.5)=>x=175.Total distance to be travelled is 210 kms.

My 2 cents:
Sarah drives 1st 35 miles at 35 mph speed in 1 hr.
1) If she drives at same speed (35 mph) she would reach airport 1 hr late.
NOTE: BECAUSE SHE REACHES AIRPORT 1 HR LATE WE HAVE TO ADD 1 (t + 1)
(t+1) = d/35 = d/(5*7)
2) If she drives at 50 mph she would reach half an hr early.
NOTE: BECAUSE SHE REACHES AIRPORT ½ HR EARLY WE SUBTRACT ½ (t–1/2)
(t – 1/2) = d/50 = d/(5*10)
From (1) and (2) d/5 = (t+1)*7 = (t-1/2)*10
7t + 7 = 10t – 5
12 = 3t
t = 4 hrs.
Therefore, d = (t + 1)*35 = 5*35 = 175 miles.
Total miles = 175 + 35 = 210 miles | C

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Hope it helps.

I used the below approach:

35+35(t+1) = 35+50(t - 0.5) {Since distance is the same in both the cases, I equated them}
Solving this for time t gives the below:
35t+70=50t+10
5t=60, t=4
So the distance is 35+35(4+1) = 210

35(t+2)=35+50(t-0,5)
solving, we get t=4 and after substituting t with 4 in any of the above equations, we get D=210 km, Answer C.

Method 1:

Let T = the travel time it would take going at 35 mph in which she is +1 hour late

Let D = Distance from instant she decides to switch speeds to airport (after the 1st 35 miles traveled)

T = D miles / 35 mph —(equation 1)


If she increases speed to 50 mph, she cuts down +1 hour late and shows up 1/2 hour early. This is -(3/2) hours less than T

T - (3/2) = D miles / 50 mph

—Substituting equation 1 in for T—

D / 35 - (3/2) = D / 50

—remove DEN’s by multiplying by LCD = 350—

10D - 525 = 7D

3D = 525

D = 175 miles

175 miles + 35 miles already driven = 210 miles

-C-

Method 2:

Let the Original Time to travel from the 35th mile marker to the airport = T hours

When she increases her speed to 50 mph, she cuts down 3/2 hours and the time traveled over this same distance = (T - 3/2) hours

The Distance in the 2 Scenarios remains Constant. Speed traveled at over the constant distance will be INVERSELY Proportional to Time traveled

If Speed INCREASES by (n/d)

Then Time traveled DECREASES by (n) / (d + n)

From Original Scenario 1 at 35 mph to Scenario 2 at 50 mph, her Speed Increases by:

(50 - 35) / 35 = 15/35 = 3/7

Thus, the Time traveled will DECREASE by (3/10) from the Original Scenario to New Scenario.


T - (3/2) hours = T - (3/10) * T

3/2 = (3/10) * T

T = 5 hours = Time it took her to get to the airport over the last stretch when she was +1 hour late


If she continues at 35 mph, she would have driven for 5 more hours to arrive +1 hour late.

Total Distance driven = 35 m + (35 mph * 5 h) = 35 + 175 = 210

-Answer-

210 miles total



Posted from my mobile device

Solution

Given:

• Sarah is driving to the airport
• Sarah drives for one hour at 35 mph

o She realizes that at the same rate, she will be an hour late for her flight.
• She then travels at 50 mph.

o And, arrives 30 minutes before her flight departs

To find:
• Total miles Sarah drove to reach the airport

Approach and Working:

• Let’s assume that Sarah was supposed to reach the Airport at time ‘t’.

Case- 1) Time Sarah would take reach the Airport at 35 MPH after she realizes

• \(\frac{Distance\ of\ Airport\ from\ the\ point }{ 35 }\)= (t + 1) h ---------- eq(1)

Case- 2) Time Sarah took to reach the Airport at 50 MPH after she realizes

• \(\frac{Distance\ of\ Airport\ from\ the\ point }{ 50}\) = (t -\(\frac{1}{2}\)) h-------------eq(2)

Subtracting eq(2) from eq(1)

• \(\frac{Distance\ of\ Airport\ from\ the\ point }{ 35} \)-\( \frac{Distance\ of\ Airport\ from\ the\ point }{ 50}\) = (t + 1) – (t -\(\frac{1}{2}\))

Let Distance of airport from the point of realization = D

• \(\frac{D}{ 35}\) - \(\frac{D}{ 50}\) = t +1 – t +1/2 = 3/2
• \(\frac{3D}{350}\) = \(\frac{3}{2}\)
• \(\frac{D}{350}\) = \(\frac{1}{2}\)
• D = 175

Total Distance = 175 + Distance covered in 1 hour by travelling at 35 MPH

• 175 + 35 = 210 miles

Hence, the correct answer is option C.

Answer: C

Alternate approach:

After driving at 35 miles per hour for one hour, she realizes that if she continues at that same average rate she will be an hour late for her flight. She then travels 50 miles per hour for the rest of the trip, and arrives 30 minutes before her flight departs.
Implication:
When Sarah's speed increases to 50 mph from 35 mph, the time decreases by 3/2 hours.

Let Sarah's normal speed = 35 mph.

Since 50/35 = 10/7, the higher speed (50 mph) is 10/7 of the normal speed (35).
Rate and time have a RECIPROCAL RELATIONSHIP.
Traveling at 10/7 of the normal speed, Sarah will take 7/10 of her normal time, shaving 3/10 of an hour from her normal time.
Implication:
The time decrease of 3/2 hours must be equal to 3/10 of Sarah's normal time to complete the remainder of the trip:
\(\frac{3}{2} = \frac{3}{10}t\)
\(30 = 6t\)
\(t = 5\)

Since Sarah's normal 35 mph speed requires a total of 6 hours -- the initial hour of travel plus the 5 hours yielded above for the remainder of the trip -- we get:
d = rt = 35*6 = 210

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