## 44th SSC CGL level Solution Set, on topics Work-time, Pipes-cisterns and Speed-time-distance

This is the 44th solution set of 10 practice problem exercise for SSC CGL exam on topics Work-time, Pipes-cisterns and Speed-time-distance. 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.

To watch **video solutions** following is the video.

### 44th solution set- 10 problems for SSC CGL exam: topics Work-time, Pipes-cisterns and Speed-time-distance - time 12 mins

**Problem 1.**

A man and a boy can complete a work together in 24 days. If the man alone does the work for the last six days then it is completed in 26 days. How long will the boy take to complete the work alone?

- 20 days
- 72 days
- 36 days
- 24 days

** Solution 1: Problem analysis and basic and rich time and work concepts**

As a technique for solving work-time problems **we use two strategies,**

- Rate of work as a variable strategy, and
- Work amount expressed in terms of mandays or boydays or womandays strategy

Both of these are **Rich Time and Work concepts simplifying most problems on this topic significantly.**

The * third strategy is the classical one* where the

**work rate is expressed as an inverse of number of days an agent takes to complete the work.**For example if it is given that a man and a boy take 6 days and 12 days to complete a work respectively while working alone, we express the portion of work that a man and a boy can do in a day as $\frac{1}{6}$th and $\frac{1}{12}$th of total work $W$.

In this form it is always possible to * add the work* of the man and that of the boy

**for a day***when they work together*as,Â

$\frac{1}{6}+\frac{1}{12}=\frac{1}{4}$th of $W$.

This means, they complete one-fourth of the work in a day working together.

So we conclude, working together they will take 4 days to complete the work.

This is the * classical and conventional approach* and forms the

**basic time and work concept.**#### Rate of work as a variable strategy - rich time and work concept

The work portion that a work agent, say, a boy or a woman or a man, does in a day is assumed to be the variables, say, B, W or M. These are basically the respective **rates of work expressed in terms of portions of work done in a day.**

By this strategy, if it is given that a man can complete a work alone in 12 days, we express it as,

$12M=W$, where $W$ is the total amount of work.

The advantage of this strategy is, we don't have to deal with many fractions. The fractions are imposed primarily on the work amount $W$.

**Example problem 1:** A man can do a work in 6 days and a boy in 12 days each working alone. How many days will they take to complete work if they work together?

**Solution - example problem 1:** Let the rate of work of the boy and the man per day working alone be $B$ and $M$. So by the given statements,

$6M=W$ and,

$12B=W$.

Multiplying the first equation by 2 and adding them,

$12(M+B)=3W$,

Or, $4(M+B)=W$, where $W$ is the work amount.

It means that the man and the boy working together will complete the work in 4 days.

Here we can add $M$ with $B$ as these are rates of work per day which is the basic condition for addition when the work agents work together. As can be seen, we didn't have to deal with any fractions, just simple linear equation handling.

#### Work amount expressed in terms of mandays - rich time and work concept

The idea is, if it is given that a man can do a job alone in 6 days, we express the work amount $W$ as equivalent to 6 mandays. This not only gives a measure of the work in terms of rate of work of a particular work agent, it also introduces the flexibility of changing the number of days or number of work agents and get easily the number of men required or number of days required. For example if the work amount is 6 man days, we can very easily say, 2 men can complete the work in 3 days, as well as 6 men in 1 day.

Let us **return to our problem** and apply requisite strategies.

By the first statement,

$24(M+B)=W$, where $M$ is the portion of work the man does alone in a day and $B$ is the portion the boy does alone in a day,

Or, $24M +24B=W$

**Solution 1: Final Problem execution**

By the second statement we can conclude that the man and the boy worked together for the first 20 days and then the boy left and the man took another six more days to complete the work in a total of 26 days. This means,

$20(M+B)+6M=W$,

Or, $26M+20B=W$

Thus we get the second linear equation in $M$ and $B$.

One easy way to solve this problem from this state is to subtract one equation from the other getting,

$2M=4B$.

Substituting $2M$s with $4B$s in the first equation we then get,

$72B=W$, that is, the boy will take 72 days to do the work alone. Simple, isn't it?

**Answer:** b: 72 days.

**Key concepts used:** * Rate of work as a variable* --

*-- basic time and work concept --*

**Mandays concept***-- linear algebraic equations -- efficient simplification.*

**rich time and work concept****Note:** We always want to use simple approaches that lead us to the solution quickly by the use of basic or rich concepts as the case may be. Both these concepts of rate of work as a variable and mandays concepts are **rich time and work concepts.**

**Problem 2.**

Two men $M_1$ and $M_2$ can complete a piece of work in 12 days and 18 days respectively. $M_1$ begins to do the work and they work alternately one at a time for one day each. The whole work will be completed in,

- $14\displaystyle\frac{1}{3}$ days
- $15\displaystyle\frac{2}{3}$ days
- $18\displaystyle\frac{2}{3}$ days
- $16\displaystyle\frac{1}{3}$ days

**Solution 2: Problem analysis - boundary condition problem**

As the two men work alternately, for each 2 days certain portion of the work will be completed till we reach the point where work of another 2 days will overdo the work, that is, work done will become more than work amount $W$. That is the point where we have to stop adding each 2 days work and analyze the remaining work and remaining days which will come as a fraction. This type of problem involves careful consideration of the * boundary condition*, that is, the point up to which we can safely add up the results of joint work. Rest of the work will have to be done individually.

**Solution 2: Problem execution**

In such problems we need to take the classical approach.

By this approach, the first man does $\frac{1}{12}$th and the second man $\frac{1}{18}$th of the work in a day while working alone individually.

So as per the statement in this problem, in 2 days they will complete,

$\frac{1}{12}+\frac{1}{18}=\frac{5}{36}$th part of the work.

So in 14 days they will complete, $\frac{35}{36}$th portion of the work. We have just stopped short of the whole work with only $\frac{1}{36}$th portion of work left.

On the 15th day it is the turn of the first man to work whose rate of work in a day is, $\frac{1}{12}$th of the work. By unitary method this man will then take another $\frac{1}{3}$rd of a day to complete the leftout $\frac{1}{36}$th portion of work.

Total days will then be, $14\frac{1}{3}$ days.

**Answer:** Option a : $14\frac{1}{3}$ days.

**Key concepts used:** * Boundary condition problem* --

*--*

**boundary determination***-- basic time and work concepts --*

**leftover analysis***.*

**mathematical reasoning****Note:** One needs to be careful with this type of problems, though with concept clarity these problems can be solved quickly.

**Problem 3.**

There are two pumps to fill a tank with water. First pump can fill the empty tank in 8 hours, while the second in 10 hours. If both the pumps are opened at the same time and kept open for 4 hours, the part of the tank that will be filled up is,

- $\displaystyle\frac{1}{10}$
- $\displaystyle\frac{1}{5}$
- $\displaystyle\frac{2}{3}$
- $\displaystyle\frac{9}{10}$

**Solution 3: Problem analysis and solving**

The first pump fills up the tank in 8 hours. So in 1 hour it fills up $\frac{1}{8}$th portion of the tank.

The second pump fills up the tank in 10 hours. So in 1 hour it fills up $\frac{1}{10}$th portion of the tank.

Adding the effect of filling up of the tank by the two pumps when both are working for 1 hour, the portion of the tank filled up in 1 hour is,

$\displaystyle\frac{1}{8}+\displaystyle\frac{1}{10}=\displaystyle\frac{9}{40}$.

In 4 hours then the two pumps working together will fill up the portion of the tank,

$4\times{\displaystyle\frac{9}{40}}=\displaystyle\frac{9}{10}$.

**Answer:** Option d: $\displaystyle\frac{9}{10}$.

**Key concepts used:** Pipes-cisterns basic concepts -- per hour work rate addition -- fraction arithmetic -- filling up as portion of the whole tank.

**Problem 4.**

A boy and a girl together fill a cistern with water. The boy pours 4 litres of water every 3 minutes while the girl pours 3 litres every 4 minutes. How much time will it take to fill 100 litres of water in the cistern?

- 44 minutes
- 42 minutes
- 36 minutes
- 48 minutes

**Solution 4: Problem analysis and conventional solution using per minute rate of pouring of water of the boy and the girl**

The rate of filling up of the boy and the girl are different. To add the rate, we need to add their per minute rate of filling up.

The boy fills up 4 litres every 3 minutes. So he fills up $\frac{4}{3}$ litres every minute.

The girl fills up 3 litres every 4 minutes. So she fills up every minute, $\frac{3}{4}$ litres of water.

Adding the work of 1 minute for both we get litres filled up in 1 minute as,

$\displaystyle\frac{4}{3} + \displaystyle\frac{3}{4}=\displaystyle\frac{25}{12}$.

Applying unitary method, to pour 100 litres then the time taken will be,

$T=100\times{\displaystyle\frac{12}{25}}=48$ minutes.

**Answer:** d: 48 minutes.

**Key concepts used:** Individual work rate per minute added up to get combined work rate per minute -- basic time and work concept -- unitary method.

#### Second innovative solution to Problem 4: Using concept of domain mapping to two pouring frequencies synchronized at LCM

**Ignoring filling amounts,** first *concentrate on the two duration 3 minutes and 4 minutes in which the boy and the girl take their respective filling up actions.*

**Question is:** *what is the least time duration in which their two actions will coincide?*

It will simply be the LCM of the two, that is, 12 minutes. It is like two bells chiming every 3 minutes and every 4 minutes chiming together at every 12 minutes.

In every 12 minutes the boy will fill up $4\times{4}=16$ litres, and the girl will fill up $3\times{3}=9$ litres of waterâ€”together a total of 25 litres of water.

So to pour $4\times{25}=100$ litres of water it would take $4\times{12}=48$ minutes.

**Answer:** Option d: 48 minutes.

**Key concepts used:** **Domain mapping to synchronization of filling frequencies to the LCM -- Innovative math -- Solving in mind -- Conceptual solution.**

**This conceptual solution can easily be carried out mentally in least time.**

**Problem 5.**

You arrive at your school 5 minutes late if you walk at a speed of 4 km/hour. But if you walk at a speed of 5 km/hour you arrive 10 minutes before scheduled time. The distance of your school from your house (in km) is,

- 2
- 4
- 5
- 10

**Solution 5: Problem analysis and first conventional solution**

Let us assume the distance from school to house as $d$ km, and scheduled time duration to reach school $T$ hours.

In the first case the boy walking at speed 4 km/hr is 5 minutes late, that is, he takes $T\text{ hours } +5\text{ minutes}$ to cover distance $d$ at speed 4 km/hr. Algebraically we express this condition as,

$4\left(T+\displaystyle\frac{1}{12}\right)=d$,

Or, $4T +\displaystyle\frac{1}{3}=d$.

Again when the boy walks at speed 5 km/hr he reaches 10 minutes early, that is,

$5\left(T-\displaystyle\frac{1}{6}\right)=d$,

Or, $5T-\displaystyle\frac{5}{6}=d$.

We had to convert both given times in minutes to hours.

To eliminate $T$ and get the value of $d$, we multiply the second equation by the fraction, $\frac{4}{5}$ to get,

$4T-\displaystyle\frac{2}{3}=\displaystyle\frac{4}{5}d$.

Subtracting this equation from the first,

$1=\displaystyle\frac{1}{5}d$,

Or $d=5$ km.

**Answer:** Option c: 5 km.

**Key concepts used:** Basic time and distance concepts, as $\text{distance}=\text{speed}\times{\text{time}}$ -- solving two linear algebraic equations in two variables -- time unit conversion.

#### Innovative conceptual solution to Problem 5 by domain mapping to a race between a faster and a slower boy

Solving conceptually, we'll transform the problem to a race in scheduled time between two boys walking at 4 km/hr and 5 km/hr starting from house at same time.

In scheduled time the first boy is **behind the school by 5 minutes** or $\frac{1}{12}$th of an hour.

In this time, at 4 km/hr speed he would have covered,

$\displaystyle\frac{4}{12}=\frac{1}{3}$rd of a km.

Similarly, in scheduled time the second boy **crosses the school and walks for 10 more minutes** or $\frac{1}{6}$th of an hour. In this time, walking at speed 5 km/hr he would have covered,

$\displaystyle\frac{5}{6}$th of a km.

If the two boys started at the same time from the house, separation between them in scheduled time would have been then,

$\displaystyle\frac{1}{3}+\displaystyle\frac{5}{6}=\displaystyle\frac{7}{6}$th of a km.

This gap would have been created because of faster boy travelling at relative speed of (5 km/hr - 4 km/ hr) = 1 km/hr.

At this speed of 1 km/hr to cover $\displaystyle\frac{7}{6}$th of a km, $\displaystyle\frac{7}{6}$th of an hr would elapse.

**This is the scheduled time** and is **10 minutes in excess of 1 hour.** As the faster boy walking at 5 km/hr reaches the school 10 minutes before scheduled time, **he walks for 1 hour and covers 5 km**, *the distance from his house to the school.*

**Answer:** Option c: 5 km.

**Key concepts used:** **Innovative math -- Domain mapping to a race between a faster and a slower boy -- Solving in mind -- Conceptual solution.**

#### Problem 6.

A train covers a distance of 10 km in 12 minutes. If its speed is decreased by 5 km/hr, the time taken by it to cover the same distance is equal to,

- $\displaystyle\frac{40}{3}$ minutes
- $40$ minutes
- $15$ minutes
- $20$ minutes

**Solution 6: Problem analysis**

We know,

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

So the original speed of the train was,

$S_1=\displaystyle\frac{10}{\displaystyle\frac{1}{5}}=50$ km/hr, we converted the minutes to hours,

After reduction by 5 km/hr the speed becomes,

$S_2=50-5=45$ km/hr, and the time taken will be,

$T_2=\displaystyle\frac{10}{45}$ hrs

$=\displaystyle\frac{10\times{60}}{45}$ minutes

$=\displaystyle\frac{40}{3}$ minutes

**Answer:** Option a : $\displaystyle\frac{40}{3}$ minutes.

**Key concepts used:** Basic time and distance concepts -- fraction arithmetic -- unit conversion.

** Problem 7.**

Walking at three-fourth of his usual speed, a man covers a certain distance in 2 hours more than the time he takes to cover the same distance at his usual speed. The time taken by him to cover the distance at his usual speed is,

- 4.5 hours
- 5 hours
- 5.5 hours
- 6 hours

**Solution 7: Problem analysis and solving**

As the speed is inversely proportional to the time taken when distance remains same,

$\displaystyle\frac{\frac{3}{4}S}{S}=\frac{T}{T+2}$, where the man covers the distance in $T$ hours at his usual speed of $S$ km/hr

Or, $3(T+2)=4T$,

Or, $T=6$ hours.

**Answer:** Option d: 6 hours.

** Key concepts used:** Basic speed time distance concepts -- speed varies inversely with time if distance remains same -- linear algebraic equation.

** Problem 8.**

In a river, the speed of current is 5 km/hour. A motorboat goes 10 km upstream and back again to the starting point in 50 minutes. The speed (in km/hour) of the motorboat in still water is,

- 20
- 25
- 26
- 28

**Solution 8: Problem analysis and execution - conventional method**

As time taken to cover a distance is, distance divided by speed, the total time taken may be expressed as,

$\displaystyle\frac{5}{6}=10\left(\frac{1}{S-5}+\frac{1}{S+5}\right)$, 50 minutes has been converted to hours and still water speed of the motorboat assumed to be $S$ km/hr

Or, $S^2-25=24S$,

Or, $S^2-24S-25=0$,

Or, $(S-25)(S+1)=0$.

So $S=25$ or $S=-1$.

As speed can't be negative, $S=25$

**Answer:** Option b: 25.

**Key concepts used:** Basic speed time distance concept -- basic boat and stream concept, upstream speed is still water speed of motorboat minus stream speed and downstream speed still water speed plus stream speed -- factorization of quadratic equation.

**Problem 9.**

A person travels 600 km by train at 80km/hour, 800 km by ship at 40 km/hour, 500 km by aeroplane at 400 km/hour and 100 km by car at 50 km/hour. What is the average speed over the entire distance?

- $65\frac{5}{123}$ km/hr
- $60\frac{5}{123}$ km/hr
- $60$ km/hr
- $62$ km/hr

**Solution 9: Problem analysis**

To get the average speed, we need to evaluate time taken at each leg of the journey, add those up and divide the total distance covered by the total time taken.

**Solution 9: Problem execution**

Time taken by train, $\displaystyle\frac{600}{80}=\frac{15}{2}$ hours.

Time taken by ship, $\displaystyle\frac{800}{40}=20$ hours.

Time taken by aeroplane, $\displaystyle\frac{500}{400}=\frac{5}{4}$ hours.

Time taken by car, $\displaystyle\frac{100}{50}=2$ hours.

Total distance, $D=600+800+500+100=2000$ kms

Total time taken, $T=\displaystyle\frac{15}{2}+20+\displaystyle\frac{5}{4}+2=\displaystyle\frac{123}{4}$ hours.

So average speed over the total journey is,

$S=\displaystyle\frac{D}{T}=\frac{2000}{\displaystyle\frac{123}{4}} = \frac{8000}{123}=65\frac{5}{123}$ km/hr.

**Answer:** Option a: $65\frac{5}{123}$ km/hr

**Key concepts used:** Basic speed time distance concepts -- average speed concept -- fraction arithmetic.

** Problem 10.**

A man goes from Delhi to Agra at a uniform speed of 40 km/hr and comes back to Delhi at a uniform speed of 60 km/hr. His average speed for the whole journey is,

- 55 km/hr
- 48 km/hr
- 50 km/hr
- 54 km/hr

**Solution 10: Problem analysis and execution**

If $d$ be the distance between Delhi and Agra, the total time taken to cover the total distance of $2d$ by the man is,

$d\left(\displaystyle\frac{1}{40}+\displaystyle\frac{1}{60}\right)=\displaystyle\frac{d}{24}$ hours

So his average speed is,

$S_{average}=\displaystyle\frac{2d}{\displaystyle\frac{d}{24}}=48$ km/hr.

**Answer:** Option b: 48 km/hr.

**Key concepts used:** Basic speed time and distance concept -- average speed concept -- fraction arithmetic.

### Useful resources to refer to

#### Guidelines, Tutorials and Quick methods to solve Work Time problems

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

**How to solve Arithmetic problems on Work-time, Work-wages and Pipes-cisterns**

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

**How to solve a hard CAT level Time and Work problem in a few confident steps 3**

**How to solve a hard CAT level Time and Work problem in a few confident steps 2**

**How to solve a hard CAT level Time and Work problem in few confident steps 1**

**How to solve Work-time problems in simpler steps type 1**

**How to solve Work-time problem in simpler steps type 2 **

**How to solve a GATE level long Work Time problem analytically in a few steps 1**

**How to solve difficult Work time problems in simpler steps, type 3**

#### SSC CGL Tier II level Work Time, Work wages and Pipes cisterns Question and solution sets

**SSC CGL Tier II level Solution set 26 on Time-work Work-wages 2**

**SSC CGL Tier II level Question set 26 on Time-work Work-wages 2**

**SSC CGL Tier II level Solution Set 10 on Time-work Work-wages Pipes-cisterns 1**

**SSC CGL Tier II level Question Set 10 on Time-work Work-wages Pipes-cisterns 1**

#### SSC CGL level Work time, Work wages and Pipes cisterns Question and solution sets

**SSC CGL level Solution Set 72 on Work time problems 7**

**SSC CGL level Question Set 72 on Work time problems 7**

**SSC CGL level Solution Set 67 on Time-work Work-wages Pipes-cisterns 6 **

**SSC CGL level Question Set 67 on Time-work Work-wages Pipes-cisterns 6**

**SSC CGL level Solution Set 66 on Time-Work Work-Wages Pipes-Cisterns 5**

**SSC CGL level Question Set 66 on Time-Work Work-Wages Pipes-Cisterns 5**

**SSC CGL level Solution Set 49 on Time and work in simpler steps 4**

**SSC CGL level Question Set 49 on Time and work in simpler steps 4**

**SSC CGL level Solution Set 48 on Time and work in simpler steps 3**

**SSC CGL level Question Set 48 on Time and work in simpler steps 3**

**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**

**SSC CGL level Solution Set 32 on work-time, work-wage, pipes-cisterns**

*SSC CGL level Question Set 32 on work-time, work-wages, pipes-cisterns*

#### SSC CHSL level Solved question sets on Work time

**SSC CHSL Solved question set 2 Work time 2**

**SSC CHSL Solved question set 1 Work time 1**

#### Bank clerk level Solved question sets on Work time

**Bank clerk level solved question set 2 work time 2**

**Bank clerk level solved question set 1 work time 1**