Word Problems

(Speed, Distance & Time)

Complex Scenarios:

It’s time to start absorbing the complex concepts in the same topic that you’ve learned in yesterday lecture.

Train Scenarios:

One Train Scenario:

Let’s take train scenario, where the length of the train also come into calculation.

For instance a train of 50mile length is moving at 120miles/hr, has to cross signal, which is 40miles ahead of train. How much time will it take the train to cross this signal?

Solution:
First let’s imagine the figure below (You will only create a line to suppose train and a dot to suppose signal, during actual exam).

From the figure above, you should understand that the total distance that the train has to travel is not just 40 miles, rather length of the train would also be included into the distance. So in other words, we have to calculate the time, when the backside of the train cross the signal. So the distance between backside of the train and the signal is 50 miles + 40 miles = 90 miles.

Now, as the speed and distance is available, and we need to find time, so by applying equation of distance, speed & time, we’ll get

Time = Distance / Speed = ⁹⁰⁄₁₂₀= ⁹⁄₁₂ = ³⁄₄ hours

⇒ T = ³⁄₄ × 60 = 45 min Answer

Train crossing bridge scenario:

Suppose a train of 200 meter length has to cross a bridge of length 700 meter. If the train will take 10 minutes to cross the bridge, what is the speed of the train?

Solution:
Imagine a figure below, (in exam as I said earlier, just sketch lines to show train and bridge.)

Here,
Focus on the wording of question that has asked. This requires a common sense. It asked time required by the train to cross the bridge. This means that currently the train is in a position where its front part is just touching the start of the bridge. So what speed is required by the train to cross the bridge in 10 minutes? That again require common sense. It has said cross the bridge, which means the back side of train must pass the end side of bridge on finished time.

So all in all, we have to see the backside of the train that has to cross the bridge in order to fulfill the condition that the train must cross the bridge. And currently the backside of the train is 200 meter far from the starting point of the bridge, because question says train has to cross the bridge, which suggests train has not enter on the bridge yet. It’s just ready to enter at current moment. Therefore, the total distance that the backside of the train has to cover is 200 meter + 700 meter = 900 meter.

Now we have distance = 900 meter and time = 10 minutes. So by applying equation that signify relation of speed, distance and time:

⇒ Speed = ⁹⁰⁰⁄₁₀ = 90 meters/min
⇒ Speed = ⁹⁰⁄₆₀ meters/s = 1.5 meters/s Answer

Two trains scenario:

Moving towards eachother:

Suppose two trains of lengths 200 meter and 400 meter are moving towards each others with a speed of 10 miles/hr and 15 miles/hr. Approximately how much time will it take for the trains to cross each other? (1 mile ≈ 1600 meters)

Solution:

Here, you must assume that one of the train is not moving (suppose train on right side). And the other train must have net speed (more specifically relative speed) by adding the speed of both trains. That’s it. Remaining things you’ll have to do same as discussed in previous lecture. So remember when the two trains moving towards each other, you should add up their speed and assume one train to be fixed (i.e not in motion). While other train will move with speed of the sum of the two speed as follows:

Relative speed of the left-side train = 10 + 15 = 25 miles/hr.
Distance to be traveled = 400 + 200 = 600 meter.
Time is required

First you need to convert miles to meter to make sure same unites while calculating. It will be given in question regarding the conversion rate of complex or tough units. But conversion rate of basic units wouldn’t be given in question (i.e conversion of km to m, m to cm and vice versa etc.)

Relative speed of the train = 25 × 1600 = 40000 meters/hr
Total distance to be traveled = 600 meter

So,

Time = ⁶⁰⁰⁄₄₀₀₀₀ = ⁶⁄₄₀₀ = ³⁄₂₀₀ hours
⇒ Time = (³⁄₂₀₀) × 60 = ⁹⁄₁₀ = 0.9 minute

⇒ Time = 0.9 × 60 = 54 seconds Answer

Moving along side:

Similarly, when the two trains are moving in same direction in such a way that one train is ready to cross the other train while moving both at 15 miles/hr and 10 miles/hr respectively as shown below:

Again here you have to assume one of the train (which is ahead at start) fixed, while the following train must move with speed that is the positive difference between the two speeds.

Remember here that the net speed (more specifically relative speed) would be the difference between the speeds of the two trains, because trains are moving alongside of each other. So here relative speed would be 15 – 10 = 5 miles/hr and Total distance would the same as in previous scenario when trains were moving towards each other. Therefore, Total distance would be 400 + 200 = 600 meters. Similarly you need to convert miles into meters as follows:

Relative speed = 5 miles/hr = 5 × 1600 = 8000 meters/hr
total Distance = 600 meters
Time is required to find.

Time = ⁶⁰⁰⁄₈₀₀₀ = ⁶⁄₈₀ = ³⁄₄₀ hours
⇒ Time = (³⁄₄₀) × 60 = 4.5 minutes = 4 minutes + 0.5 minutes

Converting minutes into seconds require multiplying minutes with 60. So 0.5 minutes = 0.5 × 60 = 30 seconds

⇒ Time = 4 minutes and 30 seconds Answer

Unsteady motion scenarios (Medium-Level):

Move & Stop scenario:

An old man walks at 15 miles/hr. After every 10 miles, he rests for 8 minutes. How much time does he take to walk 50 miles?

Solution:
These short statement questions are sometimes very hard. Here you may quickly add 8 minutes to the time required to travel 10 miles (which is suppose t minutes) and conclude that the old man would take (t + 8) minutes to cover 10 miles distance. It’s OK, you are doing well. But problem would arise when you reach the last 10 miles distance, i.e distance from 40 miles to 50 miles. This distance should never add 8 minutes rest time, because the old man rests after covering 10 miles distance. So only here is the tricky point where many students would probably have huge chance to do mistake. Even some of the over confident students would do other way around and do similar mistake. They most probably assume after concluding the time which old man require is (t + 8) minutes to cover 10 miles distance therefore his average speed would be ¹⁰⁄_{(t + 8)}. Here again you would be trapped or mislead with the wrong understanding and technique. As the last 10 miles would no more have average speed of ¹⁰⁄_{(t + 8)}. In fact it would become ¹⁰⁄_t to cover last ten miles. So be careful! 🙂

So, let’s solve this.

First we need to find the number of intervals when the old man rests. So by analyzing you would find that there are 4 intervals for rest having duration of 8 minutes each. All in all, the old man rested for 8 × 4 = 32 minutes. While the walking time without rest can be found easily by using basic equation as below:

Time = ^Distance⁄_Speed = ⁵⁰⁄₁₅ = ¹⁰⁄₃ hours = 3 hrs and 20 min

Now by adding total resting time to this, we’ll get total time required to complete the journey. i.e

⇒ Total time required to complete journey = Total walking time + Total resting time
⇒ Total time required to complete journey = (3 hrs 20 minutes) + 32 minutes = 3 hrs and (20 + 32) minutes

⇒ Total time required to complete journey = 3 hrs and 52 minutes Answer

Fast & Slow motion scenario:

Similarly, When a biker has to travel a journey from City A to City B. During the day timings, his non-fractional speed is 1 mile/hr slower than his non-fractional speed during night timings. If his average speed of the entire trip is 15.75 miles/hr, what part of the total distance he covered during night timings?

Solution:
This question clearly looks difficult as well as tricky. But it’s not beyond perception. You can think easily that the two speeds have difference of 1 mile/hr, and average of the two speeds is 15.75 miles/hr. It is given that the two speeds must be non-fractional i.e must be integers; so clearly, one speed must be 15 miles/hr and the other one 16 miles/hr, so that the average speed comes in between the two integers. Therefore:

As we know,

Distance covered in day + Distance covered in night = Total Distance

⇒ d_d + d_n = D ————————— (eq. A)

⇒ (Day speed) × (Day time) + (Night Speed) × (Night time) = D

⇒ 15 t_d + 16 t_n = D ————————— (eq. 1)

As,

t_d + t_n = Total Time = T ————————— (eq. 2)

As we know that,

⇒ 15.75 = ^D⁄_T

By putting values of (equation 1) and (equation 2),

⇒ 15.75 = ^{(15 t_d + 16 t_n)}⁄_{(td + tn)}

⇒ 15.75 t_d + 15.75 t_n = 15 t_d + 16 t_n

⇒ 0.75 t_d = 0.25 t_n

⇒ 75 t_d = 25 t_n

⇒ 3 t_d = t_n

Remember, I’m saying you again, don’t worry on timing at this stage. I know you may start thinking that the steps and process is much time taking. For your information, the last eight steps in above solution steps is actually a one step. Later on in Arithmetic you will learn how to make it in one step (short-cut) and how that step can be done quickly. So at this stage just you need what I’m providing you. OK? So don’t be fast right now. Just try to become accurate at this level.

Now we need to find distance covered in night time as a fraction of total distance i.e ^d_n⁄_D,

By putting this value of the above t_n in (eq. 1)

⇒ D = 15 t_d + 16 (3 t_d)

⇒ D = 15 t_d + 48 t_d

⇒ D = 63 t_d ————————— (eq. 3)

From (eq. A) and (eq. 1),

^{d_n = 16 t_n = 16 (3 t_d) = 48 t_d ————————— (eq. 4)}

Now,

By putting values of (3) and (4) in required fraction,

⇒ ^d_n⁄_D = ^{48 t_d}⁄_{63 td} = ⁴⁸⁄₆₃

⇒ ^d_n⁄_D = ¹⁶⁄₂₁ Answer

Move & Drag scenario (Hard-Level) :

An insect has to climb a wall of 0.9 meter tall. The insect moves constantly for 3 minutes and then rest for one minute. During rest, it drags down by one quarter the distance it cover in 3 minutes. If insect was moving at 0.2 cm per second, how much time will it take the insect to climb the wall?

Solution:
This is the hardest type of question scenario in word problem topic of ‘distance, speed and time’.

The solution of this problem is similar to the ‘Move & Stop scenario ‘ that we have discuss earlier. But here, we’ll do some little different thing, because resting and dragging makes different remaining distance. Let’s solve it by skipping some steps.

As it is given that, the insect drag down ¹⁄₄, of the distance it move upward in 3 minutes, at 0.2 cm per second that is 12 cm per minute.

So 12 cm × 3 = 36 cm

Therefore, the insect moves 36 cm in 3 minutes.

So he must drag by (¹⁄₄) × 36 cm = 9 cm in a minute.

At this stage, many of the students were doing fine perhaps. But here’s the tough part come.

Now, we need to find the number of times (i.e intervals) when the insect drag downward. How we can find this? Any idea? Think again and again to help advance your level of scope of your mental process. And BE CAREFUL here!

If you have tried and few of you perhaps reached the answer, let’s solve to see whether you have done it perfect.

We were in need to find number of times the insect drag till it reach at the top. For that purpose, similar to previous case of “Move & Stop scenario”, the insect will not drag at its last move upward to reach at top, because it will reach at the top finally. So what we need to do is to think smartly, and do the follow step:

The figure above shows how the split of time is done. Let’s understand it know. First, the questions said that insect moved upward for 3 minutes, and covered 36 cm distance. And after every 3 minutes, the insect drag downward for 1 minute, thereby dragging 9 cm downward. So, all in all, the insect moved 27 cm every 4 minutes. Now this trend will be same till the insect will cover the greatest distance in multiple of 27 (but less than 90). That can be found as below:

⁹⁰⁄₂₇ ≅ 3.33

So the closest multiple of 27 would be 27 × 3 = 81

Now the pattern will become:

                27 cm                             27 cm                               27 cm
|——————————|——————————|——————————|———–|
              4 minutes                       4 minutes                         4 minutes

As it is given that, wall is 0.9 meter tall, i.e 0.9 x 100 = 90 cm tall, so the remaining distance after making last times stop (i.e drag back) will become 90 — 81 = 9 cm.
thus,

                27 cm                             27 cm                               27 cm                     9 cm
|——————————|——————————|——————————|———–|
              4 minutes                       4 minutes                         4 minutes                 x minutes

Now in these intervals till the insect cover distance of 81 (i.e 27 × 3), the rate will be applied as 9 cm per minute {i.e (36 — 9) every 3 minutes} while taking dragging distance into count. But after the distance of 81, the insect will never drag any further, rather it will climb the wall. So at this last part of distance 9 cm, the insect will move with rate of 36 cm every 3 minutes, that is 12 cm every minute. So by applying unity method:

In last movement after stopping,

Insect covers 12 cm ————– every 1 minute

By dividing 12 on both sides.

Insect covers 1 cm ————– every ¹⁄₁₂ minute

Multiplying by 9 on both sides,

Insect covers 9 cm ————– every ⁹⁄₁₂ minute = ³⁄₄ minutes

So, x minutes = ³⁄₄ minutes

So, total time it take for insect to climb the wall will be:

= 4 minutes + 4 minutes + 4 minutes + ³⁄₄ minutes

= 12 minutes and 45 sec Answer

Other scenarios (Hard-Level) :

In a 100 mile race, Mr A can beat Mr B by 4 miles, while Mr B can beat Mr C by 25 miles. By how many miles can Mr A beat Mr C?

Take few minutes and try this by yourself. Lets see who can solve this. 🙂

It is highly advised to try yourself first, even it will take 15 minutes. Trust me, brain storming will help enhance ability to think and analyze with much broader angles.

Explanation:

If you still unable to solve it, you may proceed for explanation. This question seems quite easy, but practically it’s one of the trickiest question that you may encounter in your exam. Almost 99 percent people of you would have made quick answer in their mind before going for explanation. Unfortunately, simply 25 + 4 = 29 is not the right answer. In fact, in exam there was no as such choice available with option 29. At that point you may start panic, and loose control over your nerves. Because you’ll say yourself that its how easy question with just simple one statement. I have to answer it, but I cannot reach the answer that is available in answer choices. 🙁

Let’s learn how to deal with such situation.

First and for most point is that, how did you thought this question with very simple answer could be a part of exam like GMAT or GRE? So commonsense tell if such question come in exam like GMAT or GRE, that would be a tricky question for sure. And there you should be very very careful! 🙂

For those who are big fan of cricket or at least know this sports or game, have you every know spin bowl? This question is asked from females especially who have perhaps low interest in cricket. Spin ball is rather slower bowl thrown by the bowler to an overconfident batsman. Batsman can easily be trapped whether this slow ball would swing? It may swing either side or perhaps not swing. So batsman especially from England Cricket Team are quite comfortable to play fast pace ball rather than a slower ball.

Similarly, questions like this looks like slower ball. So don’t be fool with overconfidence. BE CAREFUL! I’m repeating this word again and again. You have to understand the importance of this word, while taking actual exam. After you complete a full practice test, you think perhaps there’s hardly any mistake. But upon looking the result, you see really different and shocking story at front of your screen. WHY? Because you just ignored the advise of BE CAREFUL! while making decision (i.e selecting answer choice). Therefore, if you see questions like this, never think only from that easy way that you think could be right answer. BE CAREFUL!

Let’s come to solution.

Theoretical Solution:

Firstly, let me make you clear what was the hidden trick inside this question and how you have deceived. First look at word beat. A can beat B by 4 miles means the moment when A is on finish point, B would have covered distance of 96 miles. That was the easy part, but on second sentence, it is given that B can beat C by 25 miles. Here’s the point where majority of you are doing mistake. ‘B can beat C by 25 miles’ means that at the moment when B will reach finish point, C will cover distance of 75 miles. But question asks the distance of C when B will cover 96 miles.

Secondly, notice that at start of the race, there will be no distance among all of the three persons. But as the race proceed, the distance among the three persons will begin to increase. So the moment when A will reach on finished point (i.e cover distance of 100 miles), B will cover distance of 96 miles. Now question is asking that at this moment (i.e when A covers 100 miles and B covers 96 miles), how much distance C would cover, rather than asking that at the moment when B covers 100 miles, how much C would cover. That’s the point where your mind couldn’t see the thing other way around and your mind was stuck only to how is this possible if answer is not 29. 😀

So what’s the answer then? Let’s solve this in two different methods. Second method is recommend, being the shortest.

Unity Method:

It’s given that if Mr A will covered 100 miles, and Mr B will covered 96 miles, we need to find distance covered by Mr C at this moment.

When B would be at 100 miles from start, C would have at 75 miles from start. But at start, both B and C were at same point. And we need the position of C when B is at 96 miles. Right!

If B will cover 100 miles,   ⇒   C would have covered 75 miles.

If B …………….. 1 mile,   ⇒   C ……………………. ⁷⁵⁄₁₀₀ = ³⁄₄ miles                 (Dividing by 100)

If B ……….. 1 × 96 mile,   ⇒   C ……………………. (³⁄₄) × 96 miles                 (Multiplying by 96)

So, Distance covered by C = 72 miles

⇒ A can beat C by = 100 – 72 = 28 miles Answer

It is recommended to avoid using unity method as much as you can; because in some scenarios, unity method is not effective and gives wrong result. And there’s other short way to solve on that situation. We’ll discuss where it gives wrong results and what is the latest short way to solve these in later lectures.

Ratio Method (Recommended):

The ratio method is discussed in next lecture of quantitative section. You may use that technique and solve this within 30 seconds. 🙂

Other Mixed topics scenarios (Hard Level):

Questions from Speed-Distance-Time also came by mixing with other topics such as in geometry with rotational speed or fractions with sequential impact and many other ways. Those scenarios you’ll learn in topic named as Miscellaneous later-on.