Test: Distance/Rate Problems - GMAT MCQ

Assuming the total distance Julia traveled is 700 miles, we can easily calculate the distances covered before each stop.

Before the first stop, Julia covered 2/7 of the total distance plus 10 miles: 700 x 2/7 + 10 = 210 miles.

For the second stop, Julia covered 3/5 of the remaining distance minus 4 miles: (700 - 210) x 3/5 - 4 = 98 x 3 - 4 = 290 miles.

The distance left at this point is 200 miles, but the problem states that Julia traveled an additional 80 miles until her destination, which corresponds to 2/5 of the assumed distance.

To find the actual distance Julia traveled, we calculate 700 x 2/5 = 280 miles.

Therefore, Julia traveled a total distance of 280 miles over 4 hours, resulting in an average speed of 70 miles per hour.

In the first phase, the motorist travels for 15 minutes at a speed of x km/h. Since distance = speed × time, the distance covered in this phase is given by:
Distance = speed × time = x km/h × (15/60) h = (x/4) km

In the second phase, the motorist travels for 20 minutes at double the speed of the first phase. So the speed in this phase is 2x km/h. The distance covered in this phase is given by:
Distance = speed × time = 2x km/h × (20/60) h = (2x/3) km

In the third phase, the motorist travels for the remaining time, which is 45 minutes - 15 minutes - 20 minutes = 10 minutes. The speed in this phase is x km/h, and the distance covered in this phase is:
Distance = speed × time = x km/h × (10/60) h = (x/6) km

The total distance covered is given as 39 km. So we can add up the distances covered in each phase and set it equal to 39 km:
(x/4) + (2x/3) + (x/6) = 39

To simplify the equation, we can find a common denominator of 12:
3x + 8x + 2x = 39 × 12
13x = 468
x = 468/13

Calculating this value, we find:
x ≈ 36

Therefore, the value of x is approximately 36 km/h, which corresponds to option B.

John's travel time:
Distance = Speed × Time
300 km = (50 miles/hour) × (1.6 km/mile) × Time
Time = 300 km / (50 × 1.6) hour
Time = 3 hours

Martin's travel time:
Distance = Speed × Time
300 km = (60 miles/hour) × (1.6 km/mile) × Time
Time = 300 km / (60 × 1.6) hour
Time = 2.5 hours

Now, let's evaluate each statement:

I. John reaches the city before Martin.
John's travel time is 3 hours, while Martin's travel time is 2.5 hours. Since John started earlier, he will reach the city before Martin. So, statement I is true.

II. At 1:30 PM, John is 55 kilometers ahead of Martin.
John started at 10 AM and travels for 3 hours. At 1:30 PM, he has been driving for 3.5 hours. The distance he has covered is:
Distance = Speed × Time
Distance = (50 miles/hour) × (1.6 km/mile) × (3.5 hours)
Distance = 280 km

Martin started at 11:30 AM and travels for 2.5 hours. At 1:30 PM, he has been driving for 2 hours. The distance he has covered is:
Distance = Speed × Time
Distance = (60 miles/hour) × (1.6 km/mile) × (2 hours)
Distance = 192 km

The difference in distance between John and Martin at 1:30 PM is:
280 km - 192 km = 88 km

So, statement II is false.

III. Martin overtakes John at 4 PM.
John reaches the city in 3 hours, so he arrives at 1 PM. Martin's travel time is 2.5 hours, so he arrives at 2 PM. Martin cannot overtake John at 4 PM because John has already reached the city by then. So, statement III is false.

Based on our analysis, statement I is the only true statement. Therefore, the correct answer is option A: I only.

Given:

• Distance between New Delhi and Washington DC via Doha: 12,000 kilometers
• Average speed of the plane: 660 kilometers per hour
• Flight reaches Washington DC at 3:50 PM, Washington time
• New Delhi is 9 hours and 30 minutes ahead of Washington DC

First, let's calculate the total travel time of the flight:
Total travel time = Distance / Speed
Total travel time = 12,000 km / 660 km/h
Total travel time ≈ 18.18 hours

Next, we need to find the time in New Delhi when the flight takes off. Since New Delhi is 9 hours and 30 minutes ahead of Washington DC, we subtract this time difference from the scheduled arrival time in Washington DC.

Scheduled arrival time in New Delhi = 3:50 PM - 9 hours 30 minutes
Scheduled arrival time in New Delhi = 6:20 AM

Since the flight stops in Doha for 2 hours and 40 minutes, we subtract this time from the scheduled departure time in New Delhi to find the approximate time when the flight takes off.

Approximate departure time in New Delhi = 6:20 AM - 2 hours 40 minutes
Approximate departure time in New Delhi ≈ 3:40 AM

Therefore, the approximate time in New Delhi when the flight takes off is 3:40 AM.

The correct answer is (B) 4:30 AM, since New Delhi is 1 hour ahead of Washington DC due to Daylight Saving Time.

Total distance travelled by both the trains before meeting = D
This distance will be covered in proportion of their speeds.
3 hours after meeting distance travelled by
A = 3 × S_A
B = 3 × S_B
3S_A + 3S_B = 675
S_A + S_B = 225
Remaining distance to be covered by 1st train = 

Remaining Distance covered by second train = 

The speed of Car B is B mph; thus, the speed of Car A = (B + 8) mph

In 1/2 hour, Car A would travel 1/2*(B + 8) = (B + 8)/2 miles

Thus, both cars together have to cover a distance of [62 - (B + 8)/2] miles

Since Car A and Car B are traveling in opposite directions, their relative speed would be the addition of their speeds.

Relative speed = A + B = (B + 8) + B = (2B + 8) mph

Time to meet = Distance covered / Rel speed = [62 - (B + 8)/2] / (2B + 8) hour

⇒ [62 - (B + 8)/2] / (2B + 8) = 15 minutes = 15/60 hours

⇒ B = 56 mph

Thus, Car B would meet Car A after traveling 56 mph * 15 min = 56*15/60 = 14 miles

Let t be the actual time and r be the actual rate.

"If a motorist had driven 1 hour longer on a certain day and at an average rate of 5 miles per hour faster, he would have covered 70 more miles than he actually did":

(t+1)(r+5) − 70 = tr
tr + 5t + r + 5 − 70 = tr
5t + r = 65

"How many more miles would he have covered than he actually did if he had driven 2 hours longer and at an average rate of 10 miles per hour faster on that day?"

(t+2)(r+10) − x = tr
tr+10t + 2r + 20 − x = tr
2(5t + r) + 20 = x

Since from above 5t +  r = 65 , then 2(5t+r) + 20 = 2 ∗ 65 + 20 = 150 = x. Therefore, x = 150 

Use t = d / r. The original journey was:

• 30 km at 90 km/h = 30 / 90 = 1/3 hour
• 50 km at 50 km/h = 50 / 50 = 1 hour (Note: The distance of 50 is easy to find since it is the hypotenuse of a 3 : 4 : 5 right triangle.)
• 60 km at 60 km/h = 60 / 60 = 1 hour
• 40 km at 40 km/h = 40 / 40 = 1 hour

So the time for the original journey was 1/3 + 1 + 1 + 1 = 3 1/3 hours

To complete the trip in 3 hours, the driver needs to take 1/3 hour less. Specifically, he needs to take 1/3 hour less on the last leg, DA.
DA took 1 hour, so it needs to take 1 − 1/3 = 2/3 of an hour.

For the time to decrease, the speed needs to increase. Use r = d / t.

40 km in 2/3 hour = r km/h

r = 40 / (2 / 3) = 40 × (3 / 2) = 60

He will need to drive the last leg at 60 km/h. The correct answer choice is (E)

Since Nicole cycles at a constant rate of 20 kilometers per hour, in x minutes, she would cover a distance of (20/60) * x kilometers. Simplifying this, we have (1/3) * x kilometers.

Now, let's consider Jessica. When she passes Nicole, she is traveling at a rate of 30 kilometers per hour. In x minutes, she would cover a distance of (30/60) * x kilometers. Simplifying this, we have (1/2) * x kilometers.

To find the distance Nicole travels before Jessica catches up, we subtract the distance covered by Jessica from the distance covered by Nicole:

(1/3) * x - (1/2) * x = (2/6) * x - (3/6) * x = -(1/6) * x kilometers.

Since Jessica is ahead of Nicole, the distance between them is negative. However, since we are interested in the time it takes for Nicole to catch up, we can ignore the negative sign.

Now, we need to determine the time it takes for Nicole to cover the distance of (1/6) * x kilometers while cycling at a rate of 20 kilometers per hour.

Time = Distance / Rate = ((1/6) * x) / 20 = (1/6) * (1/20) * x = (1/120) * x hours.

To convert this to minutes, we multiply by 60:

(1/120) * x * 60 = (1/2) * x minutes.

Therefore, Jessica will have to wait for (1/2) * x minutes for Nicole to catch up to her.

Hence, the correct answer is option b) x/2 minutes.

Let's break down the journey into two parts: from X to Y and from Y back to X.

Part 1: X to Y
Distance: 40 km
Speed: 25 km/hr
Time taken: Distance / Speed = 40 km / 25 km/hr = 1.6 hours

Part 2: Y to X
Distance: 40 km
Speed: 15 km/hr
Time taken: Distance / Speed = 40 km / 15 km/hr = 2.67 hours

Now, let's calculate the total time taken for the entire journey:

Total time = Time taken for part 1 + Time taken for part 2
= 1.6 hours + 2.67 hours
= 4.27 hours

Next, we calculate the total distance covered for the entire journey:

Total distance = Distance covered in part 1 + Distance covered in part 2
= 40 km + 40 km
= 80 km

Finally, we calculate the average speed:

Average speed = Total distance / Total time
= 80 km / 4.27 hours
≈ 18.77 km/hr

Therefore, the average speed of Sam for the entire journey is approximately 18.77 km/hr.

The correct answer is option B) 18.77 km/hr.