Time and Distance Problems with Solution: SSC CGL Tier 2 Set 13

Submitted by Atanu Chaudhuri on Thu, 05/07/2018 - 15:01

SSC CGL Tier 2 Set 13: Time and distance questions, train questions with solution

Learn from 10 selected Time and Distance Problems with Solution in SSC CGL Tier 2 Set 13 how to solve the problems easy and quick by basic concepts.

Contents are,

10 carefully selected relatively hard time and distance questions, train questions for SSC CGL Tier 2.
Answers to the time and distance problems, train problems.
Detailed and quick solution to time and distance problems and train problems.

Take the test first with timer on.

Time and distance problems, Train problems for SSC CGL Tier 2 set 13 - time to solve 15 mins

Problem 1.

A man walked $\displaystyle\frac{1}{3}$rd of a total distance at 5 km/hr, next $\displaystyle\frac{1}{3}$rd at a speed of 10 km/hr and the rest at 15 km/hr. His average speed over the distance is,

$8\displaystyle\frac{2}{11}$ km/hr
$7\displaystyle\frac{2}{11}$ km/hr
$8\displaystyle\frac{1}{11}$ km/hr
$7\displaystyle\frac{1}{11}$ km/hr

Problem 2.

On increasing the speed of a train by 10 km/hr, 30 mins is saved in a journey of 100 km. Initial speed of the train is,

42 km/hr
40 km/hr
44 km/hr
45 km/hr

Problem 3.

Driving to his office in his car from his home, covering the distance at a speed of 50 km/hr, a man is late by 20 mins. But when he increased the speed to 60 km/hr he reached office 10 minutes early. The distance of his office from his home is,

140 km
160 km
120 km
150 km

Problem 4.

The distance between place A and place B is 999 km. An express train leaves place A at 6 am and runs at a speed of 55.5 km/hr. The train stops on the way for 1 hour 20 minutes. It reaches place B at,

1.20 am
11 pm
12 pm
6 pm

Problem 5.

A train passes two bridges of lengths 800 m and 400 m in 100 secs and 60 secs respectively. The length of the train is,

80 m
90 m
150 m
200 m

Problem 6.

Two places P and Q are 162 km apart. A cyclist leaves P for Q and simultaneously a second cyclist leaves Q for P. They meet at the end of 6 hrs. If the first cyclist travels at a speed of 8 km/hr faster than the second cyclist, then the speed of the second cyclist is,

$9\displaystyle\frac{1}{2}$ km/hr
$10\displaystyle\frac{5}{6}$ km/hr
$12\displaystyle\frac{5}{6}$ km/hr
$8\displaystyle\frac{1}{2}$ km/hr

Problem 7.

Two trains A and B, start from stations X and Y towards Y and X respectively. After passing each other, they take 4 hrs 48 mins and 3 hrs 20 mins to reach Y and X respectively. If train A is moving at 45 km/hr, the speed of the train B is,

64.8 km/hr
60 km/hr
54 km/hr
37.5 km/hr

Problem 8.

A canon was fired twice at an interval of 12 mins from a fort. A passenger sitting in a train moving towards the fort heard the shots at an interval of 11 mins 40 secs. Assuming the speed of sound as 330 m/sec, what was the approximate speed of the train?

36 km/hr
38 km/hr
32 km/hr
34 km/hr

Problem 9.

If a man walks at the rate of 5 km/hr, he misses a train by 7 minutes. However, if he walks at the rate of 6 km/hr, he reaches the station 5 minutes before arrival of the train. The distance covered by him to reach the station is,

5 km
6 km
4 km
6.25 km

Problem 10.

Two trains start from stations A and B and travel towards each other at speeds of 50 km/hr and 60 km/hr respectively. At the time of their meeting, the second train has traveled 120 km more than the first. The distance between A and B is,

1320 km
1200 km
990 km
1440 km

Answers to the Time and distance problems, train problems SSC CGL Tier 2 set 13

Problem 1. Answer: Option a: $8\displaystyle\frac{2}{11}$ km/hr.

Problem 2. Answer: Option b: 40 km/hr.

Problem 3. Answer: Option d: 150 km.

Problem 4. Answer: Option a: 1.20 am.

Problem 5. Answer: Option d: 200 m.

Problem 6. Answer: Option a: $9\displaystyle\frac{1}{2}$ km/hr.

Problem 7. Answer: Option c: 54 km/hr.

Problem 8. Answer: Option d: 34 km/hr.

Problem 9. Answer: Option b: 6 km.

Problem 10. Answer: Option a: 1320 km.

Solution to Time and distance questions, train questions SSC CGL Tier 2 Set 13 - time to solve was 15 mins

Problem 1.

A man walked $\displaystyle\frac{1}{3}$rd of a total distance at 5 km/hr, next $\displaystyle\frac{1}{3}$rd at a speed of 10 km/hr and the rest at 15 km/hr. His average speed over the distance is,

$8\displaystyle\frac{2}{11}$ km/hr
$7\displaystyle\frac{2}{11}$ km/hr
$8\displaystyle\frac{1}{11}$ km/hr
$7\displaystyle\frac{1}{11}$ km/hr

Solution 1: Problem analysis

Each section of the journey is of same length, that is, $\displaystyle\frac{1}{3}$rd of the total distance. Let us assume it is $d$. The time to cover the total distance of $3d$ will then be,

$T=d\left(\displaystyle\frac{1}{5}+\displaystyle\frac{1}{10}+\displaystyle\frac{1}{15}\right)$

$=\displaystyle\frac{11d}{30}$.

So the average speed is,

$Speed_{av}=\displaystyle\frac{3d}{T}$

$=\displaystyle\frac{90}{11}$

$=8\displaystyle\frac{2}{11}$ km/hr.

Answer: Option a: $8\displaystyle\frac{2}{11}$ km/hr.

Key concepts used: Problem breakdown -- Speed time distance concept -- Average concept, all basic concepts.

Problem 2.

On increasing the speed of a train by 10 km/hr, 30 mins is saved in a journey of 100 km. Initial speed of the train is,

42 km/hr
40 km/hr
44 km/hr
45 km/hr

Solution 2: Problem analysis and intuitive solution

Knowing that if we follow the conventional method we would end up solving a quadratic equation, we decided to examine the choice values.

At a cursory glance at the choice values, the value of 40 could give us 2.5 hours travel time cleanly and the increased value of 50 resulted again in a round figure of 2 hours travel time which is half an hour less as required. Problem solved immediately.

Solution 2: Mathematical reasoning and choice values test

Examining the other values, we find that all of the other (original value, increased value) pairs—(42, 52), (44, 54) or (45, 55) would involve in non-terminating decimals because of presence of 7, 9, 11 or 13 in the denominator when dividing 100 to determine time to cover the distance, even after accounting for multiplication by 60 minutes for 1 hour.

Only (40, 50) pair of speeds results in round figure in minutes at the same time satisfies the problem statement.

This method is systematic and mathematical logic based.

Solution 2: Conventional deductive method

If $S$ is the original speed in km/hr, by problem statement,

$\displaystyle\frac{100}{S}-\displaystyle\frac{100}{S+10}=\displaystyle\frac{1}{2}$,

Or, $S^2+10S-2000=0$,

Or, $(S+50)(S-40)=0$

So $S=40$ km/hr, as it cannot be negative.

Answer: Option b: 40 km/hr.

Key concepts used: Intuitive problem solving -- Choice value test -- Mathematical reasoning -- Number system concepts -- Solving quadratic equation -- Train problems concepts -- Many ways problem solving.

Problem 3.

140 km
160 km
120 km
150 km

Solution 3: Problem analysis and solving execution

We know,

$ST=D$, where $S$ is speed, $T$ is time and $D$ is distance.

With distance $D$ remaining same, then,

$S \propto \displaystyle\frac{1}{T}$,

Or in other words, over a fixed distance, speed and time are inversely proportional.

This falls under basic speed time distance concept.

In this problem distance is fixed. So,

$\displaystyle\frac{5}{6}=\frac{T-10}{T+20}$, being a ratio we could convert scheduled duration $T$ in minutes,

Subtracting the equation from 1,

$\displaystyle\frac{1}{6}=\frac{30}{T+20}$, this is adapted componendo dividendo for efficient simplification,

Or, $180=T+20$,

It means, 3 hours is 20 minutes late from scheduled time of 160 minutes and the distance was covered at a speed of 50 km/hr to arrive late by 20 mins taking 3 hrs travel time.

So distance is,

$3\times{50}=150$ km.

On verifying, to cover the distance of 150 km at a speed of 60 km/hr, time taken would be 150 minutes, 10 minutes early from scheduled time of 160 minutes.

Answer: Option d: 150 km.

Key concepts used: Basic speed, time and distance concepts -- Speed time inverse proportionality over a fixed distance -- basic ratio concept -- efficient simplification by adapted componendo dividendo.

Problem 4.

The distance between place A and place B is 999 km. An express train leaves place A at 6 am and runs at a speed of 55.5 km/hr. The train stops on the way for 1 hour 20 minutes. It reaches place B at,

1.20 am
11 pm
12 pm
6 pm

Solution 4: Problem analysis and solving execution

The train at speed 55.5 km/hr takes,

$\displaystyle\frac{999}{55.5}=\frac{90}{5}=18$ hrs to cover the distance of 999 kms (numerator and denominator multiplied by 10 to eliminate decimal, and 111 cancelled out).

Adding stoppage time of 1 hour 20 minutes, total travel time is, 19 hours 20 minutes.

Adding it to starting time of 6 am result is 25 hours 20 minutes, which is equaivalent to 1.20 am.

Answer: Option a: 1.20 am.

Key concepts used: Basic Speed time distance concepts -- clock time concept -- Train problems concepts -- Efficient simplification by decimal elimination.

Problem 5.

A train passes two bridges of lengths 800 m and 400 m in 100 secs and 60 secs respectively. The length of the train is,

80 m
90 m
150 m
200 m

Solution 5: Problem analysis and execution

Taking train length as $d$ m, to pass the bridges, the train has to traverse, $d+800$ m and $d+400$ m in two cases taking 100 secs and 60 secs respectively.

Assuming constant speed of the train as $S$ then,

$S=\displaystyle\frac{d+800}{100}=\frac{d+400}{60}$,

Or, $\displaystyle\frac{d+400}{d+800}=\frac{3}{5}$, minimizing the fraction

Subtracting the equation from 1 to simplify the numerator,

$\displaystyle\frac{400}{d+800}=\frac{2}{5}$, this is efficient simplification by adapted componendo dividendo.

Simplifying we get, $d=200$ m.

Answer: Option d: 200 m.

Key concepts used: Basic speed time distance concepts -- train traversing platform or bridge concept -- Ratio concept -- Efficient simplification by adapted componendo dividendo -- Train problems concepts.

Problem 6.

$9\displaystyle\frac{1}{2}$ km/hr
$10\displaystyle\frac{5}{6}$ km/hr
$12\displaystyle\frac{5}{6}$ km/hr
$8\displaystyle\frac{1}{2}$ km/hr

Solution 6: Problem analysis and solving execution

Assuming the speed of the second cyclist as $S$ km/hr, the two cyclists will cover the total distance of 162 km in 6 hrs at a relative speed of $S +(S+8)=2S+8$ km/hr.

So,

$\displaystyle\frac{162}{2S+8}=6$,

Or, $27 = 2S +8$,

Or, $S=\displaystyle\frac{19}{2}=9\displaystyle\frac{1}{2}$ km/hr

Answer: Option a: $9\displaystyle\frac{1}{2}$ km/hr.

Key concepts used: Basic speed time distance concepts -- relative speed -- two objects meet when they cover distance between them while moving towards each other -- two moving objects meeting concept.

Problem 7.

64.8 km/hr
60 km/hr
54 km/hr
37.5 km/hr

Solution 7: Problem analysis

After starting simultaneously from station X and station Y, the two trains A and B meet at a point P and then move on to reach their destinations Y and X respectively.

To travel from meeting point P to destination Y train A at speed 45 km/hr takes 4 hr 48 mins, that is, $4\displaystyle\frac{4}{5}=\frac{24}{5}$ hrs and train B takes 3 hrs 20 mins, that is, $3\displaystyle\frac{1}{3}=\frac{10}{3}$ hrs to reach its destination X.

As a general principle we evaluate what we can, immediately, and so get the distance from meeting point P to destination Y as,

$d_2=45\times{\displaystyle\frac{24}{5}}=216$ km

The distance $d_1$ from P to X that B covered is unknown as well as the target speed of train B, say $S$ km/hr.

The following figure shows the situation.

trains-reaching-destination-after-crossing

Sensing complications we decide to proceed using very basic concepts of trains meeting while covering a fixed distance $d$.

Solution 7: Problem solving execution

When A and B meet at point P, both have taken the same time, say $T$ after starting.

So for train A and B,

$T=\displaystyle\frac{d_1}{45}=\frac{216}{S}$, the distance $d_2=216$ km was covered by B at a speed S to reach point P.

With two unknowns we can't solve this equation. |Something more we should know. Then we remember, yes we know that distance $d_1$ was covered by train B at speed $S$ in $\displaystyle\frac{10}{3}$ hrs. So we get a relation between $d_1$ and $S$ as,

$\displaystyle\frac{d_1}{S}=\frac{10}{3}$.

Can we use it in the first equation? Yes, we can just substitute the value of $d_1$ in terms of $S$,

$\displaystyle\frac{d_1}{45}=\frac{216}{S}$,

Or, $\displaystyle\frac{10S}{3\times{45}}=\frac{216}{S}$,

Or, $S^2=\displaystyle\frac{3\times{45}\times{216}}{10}=3\times{9}\times{108}=(54)^2$.

Answer: Option c: 54 km/hr.

Key concepts used: Speed time distance concepts -- Event sequencing -- Efficient simplification -- Trains moving towards each other take same time to meet concept -- Train problems concepts.

Problem 8.

36 km/hr
38 km/hr
32 km/hr
34 km/hr

Solution 8: Problem analysis

The distance traversed by the first shot has been,

$d=12\times{60}\times{330}\text{ m}=72\times{3.3}\text{ km}$

when the second shot was fired.

Keeping this separation, the sound of two shots moved towards the train.

After the passenger heard the first sound, the situation was exactly like,

The passenger in the train and the sound of second shot traveling towards each other at speeds $S$, say, and 330 m/sec respectively, covering the distance $d$ when the passenger met the second shot, or rather heard the second shot.

This is a situation of two objects traveling at different speeds moving towards each other and meeting at an intermediate point.

Solution 8: Problem solving execution

The meeting time was 11 mins 40 secs, 20 secs less than the time of 12 minutes the second shot would have taken to reach the passenger if the train stood still. In this 20 secs, sound would have traversed at speed 330 m/sec, a distance of 6600 m.

The following figure explains the situation.

canon-shot-sound-and-train-approaching-each-other

This 6600 metres the train must have traversed in 11 mins 40 secs, the meeting time.

So the speed of the train was,

$S=\displaystyle\frac{6.6\times{3600}}{700}$ km/hr, 6600 m coverted to 6.6 km and 11 min 40 secs to $\displaystyle\frac{700}{3600}$ hr

$=36\times{0.94}=36-2.16=34$ km/hr approximately.

Answer: Option d: 34 km/hr.

Key concepts used: Domain mapping, we have mapped sound of canon shots and passenger in the train moving towards each other to two objects moving towards each other by precise event visualization and sequencing -- basic speed time distance concepts -- Efficient simplification by not calculating and using the value of $d$, as well as quick approximation -- Train problems concepts.

Note: Though the train moves towards the sound of shots (not the canon balls), for the mathematics of the problem it is actually the passenger sitting on the train moves towards the two sounds at the speed of the train. A passenger in such train problems is considered as a no dimension point.

Problem 9.

5 km
6 km
4 km
6.25 km

Solution 9: Problem analysis and execution

If $T$ is the time span between the train arrival time and the time the man started from home, at walking speed of 5 km/hr he takes time,

$T+7=\displaystyle\frac{D}{5U}$, where $D$ is the distance to the station and $U$ is the unit conversion factor for taking care of minutes on the LHS.

Or, $D=(T+7)\times{5U}$.

When he walks at a speed of 6 km/hr, as he arrives 5 minutes earlier than scheduled time covering same distance, similarly,

$D=(T-5)\times{6U}$,

So,

$5(T+7)=6(T-5)$, $U$ canceled out and $T$ is in minutes

Or, $T=65$ minutes

So, the man reaching the station 5 minutes earlier than 65 minutes, that is arriving in 1 hour, at a speed of 6 km/hr must be covering the distance of 6 km.

Answer: Option b: 6 km.

Key concepts used: Basic speed, time and distance concepts -- inverse proportionality of time and speed with distance constant -- basic algebra concepts -- efficient simplification, we have equated the two distance expressions eliminating the need of unit conversion.

Problem 10.

1320 km
1200 km
990 km
1440 km

Solution 10: Problem analysis and execution

This is a case of two moving objects approaching each other from two ends of a fixed distance and meeting at an intermediate point. In such cases the fact that is always true is,

To reach the meeting point, each moving object has taken exactly the same time.

This is though obvious, is a very important outcome of two moving objects meeting at an intermediate point.

A corollary to this fact is,

As time of travel is constant, the distances traversed till the meeting point are directly poroportional to the speeds of the two moving objects,

$S_1T=d_1$ and $S_2T=d_2$, so, $S_1 : S_2 = d_1 : d_2$

Applying this result to our problem, if $d_1$ and $d_2$ are the distances traversed by A and B upto the meeting point at speeds 50 km/hr and 60 km/hr speeds respectively,

$\displaystyle\frac{6}{5}=\frac{d_2}{d_1}=\frac{d_1+120}{d_1}$.

Subtracting 1 from both sides,

$\displaystyle\frac{1}{5}=\frac{120}{d_1}$, efficient simplification by adapted componendo dividendo,

Or, $d_1=600$,

So, $d_2=600+120=720$, and,

$d=d_1+d_2=1320$ km.

Answer: Option a: 1320 km.

Key concepts used: Basic speed time distance concepts -- two objects moving towards each other to meet at an intermediate point taking same time with speeds and distances covered directly proportional -- Efficient simplification by adapted componendo dividendo.

Useful resources to refer to: