SSC CGL level Solution Set 62 on Speed-time-distance Train-running Boats-rivers 2

62nd SSC CGL level Solution Set, on topics Speed-time-distance, Train running and Boats in rivers

ssc cgl solution set 62 speed time distance train running boats in rivers 2

This is the 62nd solution set of 10 practice problem exercise for SSC CGL exam and 2nd 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 62 on speed-time-distance train running boats in rivers 2, take the test and then go through these solutions here.


62nd solution set- 10 problems for SSC CGL exam: topic Speed-time-distance, Train running, Boats in rivers - time 12 mins

Problem 1.

A horse rider rides at a speed of 18 km/hr but stops at the end of every 7th km for 6 mins to change horses. The time that he will take to reach his destination 90 km away is,

  1. 6 hrs 24 mins
  2. 6 hrs
  3. 6 hrs 18 mins
  4. 6 hrs 12 mins

Solution 1: Problem analysis

As $7\times{13}=91 \gt 90$, after 12 numbers of 7 km stops, the rider will reach his destination without any more stoppage. Then we need to deduce time taken for each such 7 km ride including 6 mins stoppage, multiply this duration by 12, and then add to it the time taken to cover the last, $90-7\times{12}=6$ km to the destination.

Solution 1: Problem solving execution

At 18 km per hour, for every 18 km the rider takes 1 hour,

So, for every 1 km the rider takes, $\displaystyle\frac{1}{18}$ hour,

Or, for every 7 km ride the rider takes, $\displaystyle\frac{7}{18}\text{ hour}=\displaystyle\frac{70}{3}$ mins

With 6 mins stoppage, the time taken for every 7 km becomes, $\displaystyle\frac{88}{3}$ mins,

For 12 such 7 km sections of ride, the time taken is, $352\text{ mins }=5 \text{ hours } 52\text { mins}$.

Add to it the last stretch of 6 km coverage time of one-third of 1 hour, that is, 20 mins.

The total time of ride is then, $6\text{ hours } 12\text{ mins}$.

Answer: Option d: 6 hours 12 mins.

Key concepts used: Event sequencing -- Boundary identification -- unitary method -- basic speed time and distance concepts -- boundary treatment.

Event sequencing

Let us state what event sequencing is,

In any problem where some entity goes through stages of action and accordingly the problem state changes, if we place each such event sequentially one after the other in a time line, we get what we call Event sequence.

Once we place the events in a sequence, the problem becomes clear to understand and a problem structure can then be formed easily. This leads to the clear and accurate solution.

In our problem, we can visualize the rider riding for the first 7 km, then stopping for 6 mins, and then starting for the next ride. If we go on this way, after 12th 7 km stoppage it becomes clear that no more stoppage will be required. Here the rider is the entity who goes through stages of problem state by his riding 7 km and then stopping for 6 mins.

For most of the arithmetic application topics such as, Simple interest and compound interest, Profit and loss, Time and work, Pipes and cistern, Work and wages, Speed time distance, Train running, Boats in rivers, sequencing of events helps to clarify the problem and makes it easy to solve it.

Experienced problem solvers automatically do the event sequencing in their minds, but it is good to know explicitly that such a greatly useful process exists and how to apply it.

Problem 2.

Bishnu and Ritesh start walking from the same place at the same time but in opposite directions. If Ritesh walks at a speed of $2\frac{1}{2}$ km/hr and Bishnu at a speed of 2 km/hr, in how much time will they be 18 km apart?

  1. 5.0 hrs
  2. 4.0 hrs
  3. 4.5 hrs
  4. 4.8 hrs

Solution 2: Problem analysis

The relative speed at which Bishnu and Ritesh will move away from each other will be the sum of theirs speeds,

$2+2\displaystyle\frac{1}{2}=\displaystyle\frac{9}{2}$ km per hour,

as they are walking on opposite directions starting at the same place and at the same time.

Efffectively then we have to find out the time required to cover a distance of 18 kms at a speed of $\displaystyle\frac{9}{2}$ km /hour.

As time required to cover a distance is directly proportional to the speed of travel, unitary method can be applied to produce the desired time as,

$\text{Time required}=\displaystyle\frac{\text{Distance}}{\text{Speed}}=\frac{18}{\displaystyle\frac{9}{2}}=4$ hours.

Answer: Option b: 4 hours.

Key concepts used: Problem modelling -- Relative speed in opposite direction movement -- distance time proportionality -- unitary method.

Problem 3.

Walking at the rate of 4 km/hr a man covers a certain distance in 3 hrs 45 mins. If he covers the same distance on cycle, cycling at a speed of 16.5 km/hr, the time take by him will be,

  1. 45.55 mins
  2. 55.45 mins
  3. 54.55 mins
  4. 55.44 mins

Solution 3: Problem analysis and solving execution

There are two scenarios here.

In the first case the man walking at a speed of 4 km/hour covers a certain distance in 3 hours 45 minutes. So the disctance covered is equal to the product of the two,

$\text{Distance covered}=4\times{3\displaystyle\frac{3}{4}}=15$ km.

The second time, cycling speed being 16.5 km/hour, the man will take,

$\text{Time}=\displaystyle\frac{\text{Distance}}{\text{Speed}}$

$=\displaystyle\frac{15}{16.5}$ hours

$=\displaystyle\frac{600}{11}$ mins

$=54.54\overline{54}$ mins.

$=54.55$ mins, rounding off.

Answer: Option c: 54.55 mins.

Key concepts used: Basic speed, time and distance concepts -- Event sequencing, there are two scenarios here, first gives the distance which is used in the second to find time.

Problem 4.

A car driver leaves Mumbai at 8.30 am and expects to reach a place 300 km from Mumbai at 12.30 pm. At 10.30 he finds that he has covered only 40% of the total distance. By how much he has to increase his speed to keep his schedule?

  1. 45 km/hr
  2. 30 km/hr
  3. 40 km/hr
  4. 35 km/hr

Solution 4: Problem analysis

In the first part of the journey, 40% of the total distance of 300 km, that is, 120 km is covered by the car in a time duration of,

$\text{Time}_1=10.30-8.30=2$ hours.

So the speed at which the car travelled in this first part of the journey was,

$\text{Speed}_1=\displaystyle\frac{120}{2}=60$ km/hour.

The distance left is,

$\text{Distance left}=300-120=180$ km.

The time left to reach as per scedule is, 2 hours.

So the required speed is,

$\text{Speed}_2=\displaystyle\frac{\text{Distance}}{\text{Time}}$

$=\displaystyle\frac{180}{2}$

$=90$ km/hr, an increase of $90-60=30$ km/hr.

Answer: Option b: 30 km/hr.

Key concepts used: Basic Speed time disctance concepts -- Event sequencing.

Problem 5.

Keshav started his journey on scooter at a speed of 30 km/hr. After 30 mins, Kedar started on his scooter from the same place but with a speed of 40 km/hr. How much time will Kedar take to overtake Keshav?

  1. $2$ hrs
  2. $1$ hr
  3. $\displaystyle\frac{3}{2}$ hr
  4. $\displaystyle\frac{3}{8}$ hr

Solution 5: Problem analysis

This is a case of a Race.

Race concept

In a race between two moving objects, the race starts with the slower moving object a certain distance ahead of the faster object. Often this distance separation is called handicap. The concept is, the faster object starts the race with a handicap. As he moves faster, with every passing second he is to close this gap of handicap between the two racers till the point when the faster object catches up and overtakes the slower object.

During the whole duration of the race, both the racers will move ahead, the faster one behind the slower one but moving at faster speed and narrowing the gap between them gradually.

Easiest way to deal with races

As one can understand, because of the dynamically changing situation, it is not easy to form a conceptual and mathematical model for a race. The easiest and most elegant way to deal with such a situation is to eliminate dynamism of one of the two moving objects.

Conceptually we can always say,

The slower moving object stands still while the faster moving object covers the gap of handicap at their relative speed, that is, difference of their speeds.

This is a virtual situation but mathematically and conceptually valid for calculating the time that will be taken by the faster object to catch up with the slower object.

Once we get the time to catch up, we can always get the distance travelled by each during this time duration.

Actually both moves but the gap narrows down to zero.

Solution 5: Problem solving execution

Let us first calculate how far Keshav, the slower man got ahead of Kedar, the faster man when the race started with Kedar starting his scooter.

In 30 mins travelling at a speed of 30 km/hr, Keshav will cover 15 kms. This is then the handicap distance that Kedar will have to cover at their relative speed of, $40-30=10$ km/hr.

The time to catch up will then be,

$\text{Time to catch up}=\displaystyle\frac{\text{handicap}}{\text{relative speed}}$

$=\displaystyle\frac{15}{10}$

$=1.5$ hr.

$=\displaystyle\frac{3}{2}$ hr.

Answer: Option c: $\displaystyle\frac{3}{2}$ hr.

Key concepts used: Basic speed time distance concepts -- Race concept -- Handicap -- Relative speed -- Reducing dynamism.

Problem 6.

Two cities P and Q are 500 km apart. A train starts at 8 am from P towards Q at a speed of 70 km/hr. At 10 am a second train starts from Q towards P at a speed of 110 km/hr. When will the two trains meet?

  1. 12.30 pm
  2. 1 pm
  3. 1.30 pm
  4. 12 noon

Solution 6: Problem analysis and solving execution

In this problem model, two moving objects start from two ends of a distance and moves towards each other at their relative speed which is sum of their speeds. The time taken to cover the distance will be the distance divided by the relative speed.

A variation is introduced in the problem with the first train starting from P first towards the other end Q and in $10-8=2$ hours at a speed of 70 km/hr covering a distance of 140 km, when the second train starts towards the first train from Q with a speed of 110 km/hr.

So in this case, the two trains will cover at their relative speed, $70+110=180$ km/hr a distance of not 500 km, but, $500-140=360$ km.

In $\displaystyle\frac{360}{180}=2$ hours from 10 am, that is, at 12 noon then the two trains will meet.

Answer: Option d: 12 noon.

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 -- Event sequencing.

Problem 7.

A boat goes 6 km/hr in still water but takes thrice as much time in going the same distance against the current. The speed of the current is,

  1. 4 km/hr
  2. 3 km/hr
  3. 2 km/hr
  4. 5 km/hr

Solution 7: Problem analysis

Basic relation between speed, time, and distance is,

$\text{Time}=\displaystyle\frac{\text{Distance}}{\text{Speed}}$.

In other words, if distance remains same, less speed means more time and vice versa.

Against the current the boat moves at a speed lesser than its still water speed by the current speed. Naturally then with this slower speed the boat will take more time to cover the same distance against the current compared to the time it would have taken in still water.

More formally, distance remaining same, we say, time taken to cover the distance is inversely proportional to the speed,

$T \propto \displaystyle\frac{1}{S}$, where $T$ is time in hours and $S$ speed in km/hr.

As the time against current is three times the time in still water, the speed against the current will be one-third of its speed in still water, that is,

$\displaystyle\frac{6}{3}=2$ km/hr.

Thus we have the speed of the current as,

$\text{Speed of the current}=\text{Still water speed}-\text{Speed against current}$

$=6-2$

$=4$ km/hr.

Answer: Option a: 4 km/hr.

Key concepts used: Speed time inverse proportionality -- against current speed and still water speed relation -- basic boats in rivers concepts.

Problem 8.

A man with $\displaystyle\frac{3}{5}$th of his usual speed reaches his destination $2\displaystyle\frac{1}{2}$ hrs late. His usual time to reach his destination is,

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

Solution 8: Problem analysis

If $T$ be the usual time to reach at his usual speed,

$T+\displaystyle\frac{5}{2}=\displaystyle\frac{\text{Distance}}{\displaystyle\frac{3}{5}\text{th of Usual speed}}$

$=\displaystyle\frac{5}{3}\times{\displaystyle\frac{\text{Distance}}{\text{Usual speed}}}$

$=\displaystyle\frac{5T}{3}$,

Or, $\displaystyle\frac{2T}{3}=\frac{5}{2}$,

Or, $T=3\displaystyle\frac{3}{4}$ hrs.

Answer: Option b: $3\displaystyle\frac{3}{4}$ hrs.

Key concepts used: Inverse proportionality of time and speed -- basic speed time distance concepts.

Problem 9.

A car travels from city A to city B at a constant speed. If its speed were increased by 10 km/hr it would have taken one hour less to reach its destination. Again if its speed were increased further by 10 km/hr it would have taken a further 45 mins lesser. The distance between the two cities is,

  1. 540 km
  2. 620 km
  3. 420 km
  4. 600 km

Solution 9: Problem analysis

Let the distance, and speed be $D$ and $S$, so that, usual time to reach,

$T=\displaystyle\frac{D}{S}$.

By the first statement,

$T-1=\displaystyle\frac{D}{S+10}$,

So,

$D=(T-1)(S+10)$

Or, $D=TS-S+10T-10$

Or, $D=D-S+10T-10$, as $D=TS$

Or, $10T-10=S$.

Similarly, by the second statement,

$T-1-\displaystyle\frac{3}{4}=\displaystyle\frac{D}{S+20}$,

Or, $D=\left(T-1-\displaystyle\frac{3}{4}\right)\left(S+20\right)$

Or, $D=\left(T-\displaystyle\frac{7}{4}\right)\left(S+20\right)$

Or, $D=TS -\displaystyle\frac{7S}{4}+20T-35$

Or, $20T-35=\displaystyle\frac{7S}{4}$,

Or, $80T-140=7S=70T-70$, as $S=10T-10$

Or, $10T=70$,

Or, $T=7$,

Or, $S=10T-10=70-10=60$,

So,

$D=TS=7\times{60}=420$ kms.

Answer: Option c: 420 kms.

Key concepts used: Basic speed, time and distance concepts -- inverse proportionality of time and speed with distance constant -- basic algebra concepts -- efficient simplification, in each equation we have eliminated distance to get two linear equations in speed and time solving which easily gives us distance. In this way we have avoided dealing with quadratic equations -- Speed time change relationship.

Elegant solution by applying the rich concept of speed time change relationship

The rich concept of speed time change relationship says,

The product of increase in speed and decreased time will be equal to the product of original speed and decrease in time.

Mathematically,

$S_{increase}(T_{decreased})=S_{original}(T_{reduction})$.

Let us see the mechanism of the powerful rich concept.

Let distance be $D$, original speed be $S$, original time taken be $T$, speed increase be $S_{increase}$ and time reduction be $T_{reduction}$.

Originally,

$T=\displaystyle\frac{D}{S}$.

Afrer speed increase and time reduction the situation is,

$T-T_{reduction}=\displaystyle\frac{D}{S+S_{increase}}$,

Or, $(T-T_{reduction})(S+S_{increase})=D$,

Or, $S_{inccrease}(T-T_{reduction})+TS-S(T_{reduction})=D$,

Or, $S_{inccrease}(T-T_{reduction})=S(T_{reduction})$, as $D=TS$

Or, $S_{increase}(T_{decreased})=S_{original}(T_{reduction})$.

Applying this rich concept to our problem, we have in the first situation,

$10(T-1)=S$, as time reduction is 1 hour and speed increase is 10 km/hr.

In the second instance applying the same rich concept, we get the second relation in $T$ and $S$ as,

$10\left(T-\displaystyle\frac{7}{4}\right)=\displaystyle\frac{3}{4}(S+10)$, as original speed here is $(S+10)$, speed increase 10 km/hr and time reduction $\displaystyle\frac{3}{4}$ hr from $(T-1)$ hr.

Or, $10T-\displaystyle\frac{35}{2}=\displaystyle\frac{15}{2}+\displaystyle\frac{3}{4}S$,

Or, $40T=100+3S=100+30T-30$, as $10(T-1)=S$

Or, $10T=70$,

Or, $T=7$.

We get the same result as before.

Note: Even if you are not able to understand, remember and apply this rich concept, you can always carry out the simple deduction as is done in the first solution without going into cumbersome quadratic equation handling.

Problem 10.

Two men are standing at the opposite ends of a bridge 1200 metres long. If they walk towards each other at speeds of 5m/min and 10 m/min respectively, in how much time would they meet each other?

  1. 90 mins
  2. 60 mins
  3. 85 mins
  4. 80 mins

Solution 10: Problem analysis

This again is a simple case of two moving objects approaching each other from two ends of a distance 1200 metres at their relative speed,

$5+10=15$ metres/min.

So total time to meet will be,

$T=\displaystyle\frac{1200}{15}$

$=80$ mins.

Answer: Option d: 80 mins

Key concepts used: Basic speed time distance concepts -- two objects moving towards each other to meet at an intermediate point -- relative speed.


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

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