Test: Distance And Speed- 3 - GMAT MCQ

Let b be the number of hours Bob spends biking. Then (t – b) is the number of hours he spends walking. Let d be the distance in miles from his home to school. Since he had the flat tire halfway to school, he biked d/2 miles and he walked d/2 miles. Now we can set up the equations using the formula rate x time = distance. Remember that we want to solve for d, the total distance from Bob's home to school.

1) xb = d/2
2) y(t – b) = d/2

Solving equation 1) for b gives us:

3) b = d/2x                 Substituting this value of b into equation 2 gives:

4) y(t – d/2x) = d/2    Multiply both sides by 2x:

5) 2xy(t – d/2x) = dx  Distribute the 2xy

6) 2xyt – dy = dx       Add dy to both sides to collect the d's on one side.

7) 2xyt = dx + dy       Factor out the d

8) 2xyt = d(x + y)      Divide both sides by (x + y) to solve for d

9) 2xyt / (x + y) = d

The correct answer is C.

We begin by figuring out Lexy’s average speed. On her way from A to B, she travels 5 miles in one hour, so her speed is 5 miles per hour. On her way back from B to A, she travels the same 5 miles at 15 miles per hour. Her average speed for the round trip is NOT simply the average of these two speeds. Rather, her average speed must be computed using the formula RT = D, where R is rate, T is time and D is distance. Her average speed for the whole trip is the total distance of her trip divided by the total time of her trip. 

We already know that she spends 1 hour going from A to B. When she returns from B to A, Lexy travels 5 miles at a rate of 15 miles per hour, so our formula tells us that 15T = 5, or T = 1/3. In other words, it only takes Lexy 1/3 of an hour, or 20 minutes, to return from B to A. Her total distance traveled for the round trip is 5+5=10 miles and her total time is 1+1/3=4/3 of an hour, or 80 minutes.

We have to give our final answer in minutes, so it makes sense to find Lexy's average rate in miles per minute, rather than miles per hour. 10 miles / 80 minutes = 1/8 miles per minute. This is Lexy's average rate.
 
We are told that Ben's rate is half of Lexy's, so he must be traveling at 1/16 miles per minute. He also travels a total of 10 miles, so (1/16)T = 10, or T = 160. Ben's round trip takes 160 minutes.

Alternatively, we could use a shortcut for the last part of this problem. We know that Ben's rate is half of Lexy's average rate. This means that, for the entire trip, Ben will take twice as long as Lexy to travel the same distance. Once we determine that Lexy will take 80 minutes to complete the round trip, we can double the figure to get Ben's time. 80 × 2 = 160.

The correct answer is D.

There is an important key to answering this question correctly: this is not a simple average problem but a weighted average problem.  A weighted average is one in which the different parts to be averaged are not equally balanced.  One is "worth more" than the other and skews the "simple" average in one direction.  In addition, we must note a unit change in this problem: we are given rates in miles per hour but asked to solve for rates in miles per minute.
Average rate uses the same D = RT formula we use for rate problems but we have to figure out the different lengths of time it takes Dan to run and swim along the total 4-mile route. Then we have to take the 4 miles and divide by that total time. First, Dan runs 2 miles at the rate of 10 miles per hour.  10 miles per hour is equivalent to 1 mile every 6 minutes, so Dan takes 12 minutes to run the 2 miles.  Next, Dan swims 2 miles at the rate of 6 miles per hour.  6 miles per hour is equivalent to 1 mile every 10 minutes, so Dan takes 20 minutes to swim the two miles.
Dan's total time is 12 + 20 = 32 minutes.  Dan's total distance is 4 miles.  Distance / time = 4 miles / 32 minutes = 1/8 miles per minute.
Note that if you do not weight the averages but merely take a simple average, you will get 2/15, which corresponds to incorrect answer choice B.  6 mph and 10 mph average to 8mph.  (8mph)(1h/60min) = 8/60 miles/minute or 2/15 miles per minute.
The correct answer is A.

The formula to calculate distance is Distance = (Rate)(Time). So at any given moment Tom's distance (let's call it DT) can be expressed as DT  = 6T. So, at any given moment, Linda's distance (let's call it DL) can be expressed as DL = 2(T + 1) (remember, Linda's time is one hour more than Tom's). The question asks us to find the positive difference between the amount of time it takes Tom to cover half of Linda's distance and the time it takes him to cover twice her distance.  Let's find each time separately first.

When Tom has covered half of Linda's distance, the following equation will hold: 6T = (2(T + 1))/2. We can solve for T:
6T = (2(T + 1))/2
6T = (2T + 2)/2
6T = T +1
5T = 1
T = 1/5

So it will take Tom 1/5 hours, or 12 minutes, to cover half of Linda's distance. When Tom has covered twice Linda's distance, the following equation will hold: 6T = 2(2(T + 1). We can solve for T:

6T = 2(2(T + 1)
6T = 2(2T + 2)
6T = 4T + 4
2T = 4
T = 2

So it will take Tom 2 hours, or 120 minutes, to cover twice Linda's distance.
We need to find the positive difference between these times: 120 – 12 = 108.

The correct answer is E. 

and as high speed train takes x hours which will definitely be lesser than the time taken by a normal train which is y hours, So y>x

To determine Bill’s average rate of movement, first recall that Rate × Time = Distance. We are given that the moving walkway is 300 feet long, so we need only determine the time elapsed during Bill’s journey to determine his average rate.

There are two ways to find the time of Bill’s journey. First, we can break down Bill’s journey into two legs: walking and standing. While walking, Bill moves at 6 feet per second. Because the walkway moves at 3 feet per second, Bill’s foot speed along the walkway is 6 – 3 = 3 feet per second. Therefore, he covers the 120 feet between himself and the bottleneck in (120 feet)/(3 feet per second) = 40 seconds.

Now, how far along is Bill when he stops walking? While that 40 seconds elapsed, the crowd would have moved (40 seconds)(3 feet per second) = 120 feet. Because the crowd already had a 120 foot head start, Bill catches up to them at 120 + 120 = 240 feet. The final 60 feet are covered at the rate of the moving walkway, 3 feet per second, and therefore require (60 feet)/(3 feet per second) = 20 seconds. The total journey requires 40 + 20 = 60 seconds, and Bill’s rate of movement is (300 feet)/(60 seconds) = 5 feet per second. 

This problem may also be solved with a shortcut. Consider that Bill’s journey will end when the crowd reaches the end of the walkway (as long as he catches up with the crowd before the walkway ends). When he steps on the walkway, the crowd is 180 feet from the end. The walkway travels this distance in (180 feet)/(3 feet per second) = 60 seconds, and Bill’s average rate of movement is (300 feet)/(60 seconds) = 5 feet per second.

The correct answer is E. 
 

It is easier to break this motion up into different segments. Let's first consider the 40 minutes up until John stops to fix his flat. 

40 minutes is 2/3 of an hour. 
In 2/3 of an hour, John traveled 15 × 2/3 = 10 miles (rt = d)
In that same 2/3 of an hour, Jacob traveled 12 × 2/3 = 8 miles
John therefore had a two-mile lead when he stopped to fix his tire. 

It took John 1 hour to fix his tire, during which time Jacob traveled 12 miles. Since John began this 1-hour period 2 miles ahead, at the end of the period he is 12 – 2 = 10 miles behind Jacob. 
The question now becomes "how long does it take John to bridge the 10-mile gap between him and Jacob, plus whatever additional distance Jacob has covered, while traveling at 15 miles per hour while Jacob is traveling at 12 miles per hour?" We can set up an rt = d chart to solve this. 


 

John's travel during this "catch-up period" can be represented as 15t = d + 10
Jacob's travel during this "catch-up period" can be represented as 12t = d

If we solve these two simultaneous equations, we get:
15t = 12t + 10
3t = 10
t = 3 1/3 hours
Another way to approach this question is to note that when John begins to ride again, Jacob is 10 miles ahead. So John must make up those first 10 miles plus whatever additional distance Jacob has covered while both are riding. Since Jacob's additional distance at any given moment is 12t (measuring from the moment when John begins riding again) we can represent the distance that John has to make up as 12t + 10. We can also represent John's distance at any given moment as 15t. Therefore, 15t = 12t + 10, when John catches up to Jacob. We can solve this question as outlined above.

The correct answer is B.

This standard rate problem will rely heavily on the formula RT=D, where R is the rate, T is the time and D is the distance traveled.
First, we should find the driving and biking distances:
If Deb drives for 45 minutes, or 0.75 hours, at a rate of 40mph, she drives a total distance of 
(0.75)(40) = 30 miles.
If the bike route is 20% shorter than the driving route, the bike route is 30 – 30(0.2) = 30 – 6 = 24 miles.
Next, we need to determine how long it will take Deb to travel the route by bike.  She wants to ensure that she'll get to work by a particular time, so we want to calculate the longest possible time it could take her; therefore, we have to assume she will bike at the slowest end of the range of the speeds given: 12mph.  If she travels 24 miles at 12mph, it will take her 24/12 = 2 hours or 120 minutes.
If Deb normally takes 45 minutes to drive to work but could take up to 120 minutes to bike to work, then she must leave 120 – 45 = 75 minutes earlier than she normally does to ensure that she will arrive at work at the same time.
The correct answer is D.
 

If we want Brenda's distance to be twice as great as Alex's distance, we can set up the following equation: 

2(4T) = R(T – 1), where 4T is Alex's distance (rate × time) and R(T – 1) is Brenda's distance (since Brenda has been traveling for one hour less). 

If we simplify this equation to isolate the T (which represents Alex's total time), we get: 
2(4T) = R(T – 1)
8T = RT – R
R = RT – 8T
R = T(R – 8)

This is choice C

The key to solving this question lies in understanding the mathematical relationship that exists between the speed (s), the circumference of the tires (c) and the number of revolutions the tires make per second (r).  It makes sense that if you keep the speed of the car the same but increase the size of the tires, the number of revolutions that the new tires make per second should go down.  What, however, is the exact relationship?
Sometimes the best way to come up with a formula expressing the relationship between different variables is to analyze the labels (or units) that are associated with those variables.  Let’s use the following units for the variables in this question (even though they are slightly different in the question):
      c: inches/revolution   
      s: inches/sec
      r: revolutions/sec
The labels suggest:  (rev/sec) x (inches/rev) = (inches/sec), which means that rc = s. 
When the speed is held constant, as it is in this question, the relationship rc = s becomes rc = k.  r and c are inversely proportional to one another.  When two variables are inversely proportional, it means that whatever factor you multiply one of the variables by, the other one changes by the inverse of that factor.  For example if you keep the speed constant and you double the circumference of the tires, the rev/sec will be halved.  In this way the product of c and r is kept constant.    
In this question the circumferences of the tires are given in inches, the speed in miles per hour, and the rotational speed in revolutions per second.  However, the discrepancies here don’t affect the fundamental mathematical relationship of inverse proportionality: if the speed is kept constant, the rev/sec of the tires will change in an inverse manner to the circumference of the tires.    
Let’s assign     c₁ = initial circumference; c₂ = new circumference
r₁ = initial rev/sec ; r₂ = new rev/sec
Since the speeds are held constant:
c₁r₁ = c₂r₂
r₂ = (c₁/c₂)r₁
r₂ = (28/32)r₁  
r₂ = (7/8)r₁

If the new rev/sec is 7/8 of the previous rev/sec, this represents a 1/8 or 12.5% decrease and the correct answer is (B).  Relationships of inverse proportionality are important in any word problem on the GMAT involving a formula in the form of xy = z and in which z is held constant.