SSC CGL level Solution Set 63 on Speed-time-distance Train-running Boats-rivers 3

63rd SSC CGL level Solution Set, 3rd on topics Speed-time-distance, Train running and Boats in rivers

ssc cgl solution set speed time distance train running boats in rivers 3

This is the 63rd solution set of 10 practice problem exercise for SSC CGL exam and 3rd on topic Speed-time-distance, Train running and Boats in rivers. Students must complete the corresponding question set in prescribed time first and then only refer to this solution set for gaining maximum benefits from this resource.

In MCQ test, you need to deduce the answer in shortest possible time and select the right choice.

Based on our analysis and experience we have seen that, for accurate and quick answering, the student

  • must have complete understanding of the basic concepts in the topic area
  • is adequately fast in mental math calculation
  • should try to solve each problem using the basic and rich concepts in the specific topic area and
  • does most of the deductive reasoning and calculation in his or her head rather than on paper.

Actual problem solving is done in the fourth layer. You need to use your problem solving abilities to gain an edge in competition.

If you have not taken the test yet first refer to SSC CGL level Question set 63 on speed-time-distance train running boats in rivers 3, take the test and then go through these solutions here.


63rd solution set - 10 problems for SSC CGL exam: 3rd on topics Speed-time-distance, Train running, Boats in rivers - time 12 mins

Problem 1.

The ratio of lengths of two trains is 5: 3 and the ratio of their speeds is 6 : 5. The ratio of time taken by them to cross a pole is,

  1. 11 : 8
  2. 27 : 16
  3. 5 : 6
  4. 25 : 18

Solution 1: Problem analysis and solving execution

To cross a pole a train needs to cover its own length.

The relation between Speed, time and distance is,

$T=\displaystyle\frac{D}{S}$, where $T$ is the time taken to cover the distance $D$ at speed $S$

So ratio of time taken to cover their lengths is,

$\displaystyle\frac{T_1}{T_2}=\displaystyle\frac{L_1\times{S_2}}{L_2\times{S_1}}$

$=\displaystyle\frac{5\times{5}}{3\times{6}}$

$=\displaystyle\frac{25}{18}$.

Answer. Option d: 25 : 18.

Key concepts used: Basic speed time distance relation -- basic proportionality concepts.

Problem 2.

A man can swim 3 km/hr in still water. If the velocity of the stream is 2 km/hr, the time taken by him to swim to a place 10 km upstream and back is,

  1. $12$ hrs
  2. $8\displaystyle\frac{1}{3}$ hrs
  3. $10$ hrs
  4. $9\displaystyle\frac{1}{3}$ hrs

Solution 2: Problem analysis

Effective upstream speed of the swimmer is,

$V_{up}=V-S=1$ km/hr, where $V$ is still water speed, $S$ is the stream speed and $V_{up}$ is the uspstream speed.

To swim 10 kms upstream at 1 km/hr he will then take 10 hrs.

Again, the downstream speed is,

$V_{down}=V+S=5$ km/hr., where $V_{down}$ is the downstream speed.

To swim 10 km downstream at 5 km/hr he will then take an additional 2 hrs, that is, a total of 12 hrs.

Answer: Option a: 12 hrs.

Key concepts used: Boats rivers concepts -- still water speed -- upstream speed -- downstream speed.

Problem 3.

A man standing on a platform finds that a train takes 3 seconds to pass him and another train of same length moving in the opposite direction takes 4 secs. The time taken by the trains to pass each other will be,

  1. $2\displaystyle\frac{3}{7}$ secs
  2. $5\displaystyle\frac{3}{7}$ secs
  3. $3\displaystyle\frac{3}{7}$ secs
  4. $4\displaystyle\frac{3}{7}$ secs

Solution 3: Problem analysis and solving execution

If $V_1$ km/hr is the velocity and $L$ metres is the length of the first train, in passing the man, the train will cover its length $L$ in 3 secs. So its velocity will be,

$V_1=\displaystyle\frac{L}{3}$ metre/sec

Similarly the speed of the second train will be,

$V_2=\displaystyle\frac{L}{4}$ metre/sec

To pass each other the trains have to cover $2L$ distance at relative speed,

$V_1+V_2=L\left(\displaystyle\frac{1}{3}+\displaystyle\frac{1}{4}\right)$ metre/sec

$=\displaystyle\frac{7L}{12}$ metre /sec

So the time taken to pass each other will be,

$T=\displaystyle\frac{2L}{\displaystyle\frac{7L}{12}}$ secs

$=\displaystyle\frac{24}{7}$ secs

$=3\displaystyle\frac{3}{7}$ secs.

The equal length $L$ is cancelled out.

Answer: Option c: $3\displaystyle\frac{3}{7}$ secs.

Key concepts used: Basic speed, time and distance concepts -- Train passing concept -- relative speeds.

Problem 4.

A boat covers 12 km upstream and 18 km downstream in 3 hrs, while it covers 36 km upstream and 24 km downstream in $6\displaystyle\frac{1}{2}$ hrs. What is the speed of the current?

  1. 2.5 km/hr
  2. 2 km/hr
  3. 1.5 km/hr
  4. 1 km/hr

Solution 4: Problem analysis

In the first journey,

$\displaystyle\frac{12}{V-S}+\displaystyle\frac{18}{V+S}=3$, where $V$ is the still water speed and $S$ is the stream speed

Similarly in the second journey,

$\displaystyle\frac{36}{V-S}+\displaystyle\frac{24}{V+S}=\displaystyle\frac{13}{2}$

Multiplying the first equation by 3 and subtracting the second equation from the result we get,

$\displaystyle\frac{30}{V+S}=\displaystyle\frac{5}{2}$, eliminating $\displaystyle\frac{1}{V-S}$

Or, $V+S=6\times{2}=12$.

Substituting value of $(V+S)$ in the first equation,

$\displaystyle\frac{12}{V-S}+\displaystyle\frac{18}{12}=3$,

Or, $\displaystyle\frac{12}{V-S}=3-\displaystyle\frac{3}{2}=\displaystyle\frac{3}{2}$,

Or, $V-S=4\times{2}=8$.

So,

$V=\displaystyle\frac{12+8}{2}=10$ km/hr, and

$S=10-8=2$ km/hr.

Answer: Option b: 2 km/hr.

Key concepts used: Basic spped time distance concepts -- boats in streams concepts -- upstream speed concept -- downstream speed concept -- key pattern identification of inverse of $(V-S)$ and inverse of $(V+S)$ appearing unchanged in two linear equations so that we solved not for $V$ and $S$ but first solved for inverses of $(V-S)$ and $(V+S)$ and then for $V$ and $S$ both from two linear equations thus avoiding solving quadratic equations. This is equivalent to Component expression substitution and substitution back or reverse substitution according to principle of representative.

Problem 5.

In covering a certain distance, the speed of A and B are in the ratio 3 : 4. A takes 30 mins more than B to reach the destination. The time taken by A to reach the destination is,

  1. $1$ hrs
  2. $2\displaystyle\frac{1}{2}$ hr
  3. $1\displaystyle\frac{1}{2}$ hr
  4. $2$ hrs

Solution 5: Problem analysis and solving execution

Distance being same, speed is inversely proportional to time taken. So if time taken by B is $T$ mins,

$\displaystyle\frac{S_A}{S_B}=\frac{T}{T+30}$, where $S_A$ and $S_B$ are the speeds of A and B respectively

Or, $\displaystyle\frac{T+30}{T}=\frac{4}{3}$

Subtracting 1,

$\displaystyle\frac{30}{T}=\frac{1}{3}$.

So $T=90$ mins which is the time taken by B to cover the distance.

As A takes 30 mins more than B, the time taken by A is 120 mins or 2 hrs.

Answer: Option d: 2 hrs.

Key concepts used: Basic speed time distance concepts -- Speed time inverse proportionality with distance constant -- Efficient simplification by subtracting 1.

Problem 6.

A man completes 30 km of a journey at a speed of 6 km/hr and the remaining 40 km journey in 5 hrs. His average speed for the whole journey is,

  1. $7$ km/hr
  2. $7.5$ km/hr
  3. $6\displaystyle\frac{4}{11}$ km/hr
  4. $8$ km/hr

Solution 6: Problem analysis and solving execution

For the first part of 30 kms the man takes,

$\displaystyle\frac{30}{6}=5$ hrs.

For the second part of 40 kms as the man takes same 5 hrs he moves faster during this second part of journey.

Overall he takes $5+5=10$ hrs for covering a total distance of $30+40=70$ kms.

His average speed is then,

$S_{avg}=\displaystyle\frac{70}{10}=7$ km/hr

Answer: Option a: 7 km/hr.

Key concepts used: Basic speed time distance concepts -- Average speed concept.

Problem 7.

A cyclist after cycling a distance of 70 kms on the second day finds that the ratio of distance covered by him on the first two days is 4 : 5. If he travels a distance of 42 kms on the third day, the ratio of distances travelled on the third day and the first day is,

  1. 4 : 3
  2. 3 : 4
  3. 3 : 2
  4. 2 : 3

Solution 7: Problem analysis

This is purely a ratio problem.

From the given information we have the first ratio as,

$\displaystyle\frac{d_{first}}{d_{second}}=\frac{4}{5}$, where distance covered on the second day is, $d_{second}=70$ kms.

so the distance covered on the first day is,

$d_{first}=70\times{\displaystyle\frac{4}{5}}=56$ kms.

Distance covered on the third day,

$d_{third}=42$ kms

So the ratio of distance covered on the third day to the distance covered on the first day is,

$\displaystyle\frac{d_{third}}{d_{first}}=\frac{42}{56}=\frac{3}{4}$.

Answer: Option b: 3 : 4.

Key concepts used: Basic ratio concepts -- Event sequencing.

Problem 8.

A and B run a 5 km race on a round course of 400 m. If their speeds are in the ratio of 5 : 4, the number of times the winner passes the other is,

  1. 5
  2. 1
  3. 2
  4. 3

Solution 8: Analysis of Race on a round course problem

The first basic concept of two runners racing on a round course and the faster passing the slower is,

To pass once, the faster runner must cover the intial gap of the length of the course running at their relative speed.

Based on race concept, the slower runner is held still at the starting point and the faster runner passing him after covering the full length of the course and coming round at their relative speed.

To know how many times the faster runner passes the slower one we have to know how much distance the two actually cover running at their actual speeds when the faster runners covers the gap of full length of the course.

This is the second basic concept of actual distance covered by the faster runner in passing the slower and is given by the product of actual speed of the faster runner and the time taken to pass.

Solution 8: Problem solving execution

Applying basic ratio concepts of reintroducing the cancelled out HCF in a ratio, the actual speeds of the faster and the slower runner are assumed to be $5x$ and $4x$ where $x$ is the common HCF cancelled out while forming the ratio.

Relative speed is then $x$ and time taken to cover 400 metres is,

$T_{400}=\displaystyle\frac{0.4}{x}$ hr.

In this time the faster runner covers,

$d_{pass}=5\times{0.4}=2$ kms

In other words, to pass once the faster runner has to cover 2 kms.

As the course length is 5 kms, in completing the race the faster runner will pass the slower runner 2 times in 4 kms. In the last 1 km he won't be able to pass the slower runner any further.

Counting one more passing at the start itself, the winner will pass the slower runner a total of 3 times in completing the race.

Answer: Option d: 3.

Key concepts used: Race concepts -- Race on a round course concepts -- Relative speed concept -- Handicap on a round course concept.

Problem 9.

If a boy walks from his house to his school at the rate of 4 km/hr, he reaches the school 10 mins earlier than the scheduled time. However if he walks at the rate of 3 km/hr he reaches 10 mins late. Find the distance of the school from his house.

  1. 4 km
  2. 5 km
  3. 4.5 km
  4. 6 km

Solution 9: Problem analysis

Distance being same, time taken at two different speeds is inversely proportional to the speeds,

So,

$\displaystyle\frac{T+10}{T-10}=\frac{4}{3}$, where $T$ mins is the scheduled time to reach the school.

Applying componendo dividendo technique on the relation we have,

$\displaystyle\frac{T}{10}=\frac{4+3}{4-3}=7$.

So,

$T=70$ mins.

Taking the first case then he covers the distance to school at 4 km/hr in $70-10=60$ mins or in 1 hr. So the distance to school is, 4 kms.

Answer: Option a: 4 kms.

Key concepts used: Basic speed, time and distance concepts -- inverse proportionality of time and speed with distance constant -- basic algebra concepts -- Componendo dividendo technique -- efficient simplification.

Problem 10.

In a 100 m race, Kanu defeats Ritwik by 5 secs. If the speed of Kanu is 18 km/hr, the speed of Ritwik is,

  1. 14.4 km/hr
  2. 14 km/hr
  3. 14.5 km/hr
  4. 15.4 km/hr

Solution 10: Problem analysis

Kanu covers 100 m at a speed of 18 km/hr. So the time taken by Kanu is,

$T_{K}=\displaystyle\frac{D}{S_K}=\displaystyle\frac{100\times{3600}}{18000}=20$ secs, where $T_K$ and $S_K$ are the time and speed of Kanu.

Ritwik takes 5 secs more, that is, $T_R=25$ secs to cover the same distance of 100 metres.

So the speed of Ritwik is,

$S_R=\displaystyle\frac{D}{T_R}=\displaystyle\frac{100\times{3600}}{25\times{1000}}=14.4$ km/hr

Answer: Option a: 14.4 km/hr.

Key concepts used: Basic speed time distance concepts -- Speed time inverse proportionality with distance constant.

Note: Though the problem refers to a race, for solving the problem use of race concept and techniques are not involved.


Related important reading list:

Guidelines

7 steps for sure success in SSC CGL Tier 1 and Tier 2 competitive tests

Tutorial

Basic concepts on Arithmetic problems on Speed-time-distance Train-running Boat-rivers

Question and answer sets

SSC CGL Tier II

SSC CGL Tier II Solved questions Set 13 on Time and distance problems Train problems 1

SSC CGL

SSC CGL level Question Set 63 on Speed-time-distance Train Running Boats in Rivers 3

SSC CGL level Solution Set 62 on Speed-time-distance Train Running Boats in Rivers 2

SSC CGL level Question Set 62 on Speed-time-distance Train running Boats in Rivers 2

SSC CGL level Solution Set 44 on Work-time Pipes-cisterns Speed-time-distance

SSC CGL level Question Set 44 on Work-time Pipes-cisterns Speed-time-distance