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Test: Distance/Rate Problems - GMAT MCQ


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10 Questions MCQ Test - Test: Distance/Rate Problems

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Test: Distance/Rate Problems - Question 1

Julia drives for 4 hours to meet her fiancé's parents, stopping two times along the way. She drives 10 miles longer than 2/7 of the entire distance before she stops for the first time. Then she drives 4 miles less than 3/5 of the rest of the distance before stopping again. If after the second stop, she travels 80 more miles until her destination, what was her average speed over the whole trip?

Detailed Solution for Test: Distance/Rate Problems - Question 1

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.

Test: Distance/Rate Problems - Question 2

A motorist covers a distance of 39 km in 45 min by moving at a speed of x km/h for the first 15 min, then moving at double the speed for the next 20 min and then again moving at his original speed for the rest of the journey. Then, x is equal to:

Detailed Solution for Test: Distance/Rate Problems - Question 2

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.

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Test: Distance/Rate Problems - Question 3

John leaves from his home at 10 AM and starts driving towards a city that is 300 kilometres away, at a constant speed of 50 miles per hour. His brother Martin leaves the home at 11:30 AM and starts driving on the same route at a constant speed of 60 miles per hour. If they stop driving once they reach the city, which of the following statements must be true? (1 mile = 1.6 kilometres)

I. John reaches the city before Martin
II. At 1:30 PM, John is 55 kilometres ahead of Martin
III. Martin overtake John at 4 PM

Detailed Solution for Test: Distance/Rate Problems - Question 3

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.

Test: Distance/Rate Problems - Question 4

An international flight takes off from New Delhi in India for Washington DC in the United States of America via Doha, where the flight stops for 2 hours and 40 minutes. During flight, the average speed of the plane is 660 kilometres per hour and the flight reaches Washington DC, at 3:50 PM, Washington time. If the distance between New Delhi and Washington DC via Doha is 12000 kilometres and New Delhi is 9 hour 30 minutes ahead of Washington DC, what is the approximate time in New Delhi when the flight takes off from there?

Detailed Solution for Test: Distance/Rate Problems - Question 4

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.

Test: Distance/Rate Problems - Question 5

Two trains left from two stations P and Q towards station Q and station P respectively. 3 hours after they met, they were 675 Km apart. First train arrived at its destination 16 hours after their meeting and the second train arrived at its destination 25 hours after their meeting. How long did it take for the first train to make the whole trip?

Detailed Solution for Test: Distance/Rate Problems - Question 5

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 × SA
B = 3 × SB
3SA + 3SB = 675
SA + SB = 225
Remaining distance to be covered by 1st train = 

Remaining Distance covered by second train = 

Test: Distance/Rate Problems - Question 6

Half an hour after Car A started traveling from Newtown to Oldtown, a distance of 62 miles, Car B started traveling along the same road from Oldtown to Newtown. The cars met each other on the road 15 minutes after Car B started it's trip. If Car A traveled at a constant rate that was 8 miles per hour greater than Car B's constant rate, how many miles had Car B driven when they met?

Detailed Solution for Test: Distance/Rate Problems - Question 6

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

Test: Distance/Rate Problems - Question 7

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. 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?

Detailed Solution for Test: Distance/Rate Problems - Question 7

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 

Test: Distance/Rate Problems - Question 8

If the driver wants to repeat his ABCDA trip, what must be his speed on leg DA so that the trip would last 3 hours? Use the graph above. Assume that the speed on other legs remains unchanged.

Detailed Solution for Test: Distance/Rate Problems - Question 8

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)

Test: Distance/Rate Problems - Question 9

Nicole cycles at a constant rate of 20 kilometers per hour, and is passed by Jessica, who cycles at a constant rate of 30 kilometers per hour. If Jessica cycles at her constant rate for x minutes after passing Nicole, then stops to wait for her, how many minutes will Jessica have to wait for Nicole to catch up to her?

Detailed Solution for Test: Distance/Rate Problems - Question 9

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.

Test: Distance/Rate Problems - Question 10

Sam covers a total distance of 80km while travelling from a certain point X to a certain point Y and come back again to the start point X. He covers the first 40 km distance from point X to point Y at a speed of 25 km/hr. While travelling back to the point X from point Y, Sam drives his vehicle at a speed of 15 km/hr. Calculate the average speed of Sam for the entire journey.

Detailed Solution for Test: Distance/Rate Problems - Question 10

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.

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