## 32nd SSC CGL level Solution Set, 1st exclusively on topics Work-time, Work-wages & Pipes-cisterns

This is the 32nd solution set of 10 practice problem exercise for SSC CGL exam and 1st exclusively on topics Work-time, Work-wages & Pipes-cisterns that are related. 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.

### 32nd solution set- 10 problems for SSC CGL exam: 1st on topics Work-time, Work-wages & Pipes-cisterns - time 12 mins

**Problem 1.**

A team of 16 persons who can finish a job in 10 days start on the job together but after doing 80% of the job, 50% of the persons left the team. The number of additional days that will be required to complete the job will then be,

- 4
- 12
- 2
- 3

** Solution 1 : Problem analysis:**

The very first step that we take is to assess the total amount of work in terms of mandays.

$\text{Total amount of work} =16\times{10}=160$ mandays.

80% of this total amount has been completed and so 20% is left which is,

$\text{Work left = 20% of 160} = 32$ mandays.

This is the job left for the rest of the team.

In the team we have now,

Number of team members left $=$ 50% of $16 = 8$.

**Solution 1 : Final Problem execution**

So 32 mandays work is to be done by 8 persons which would take 4 days.

As the work would have been normally finished in 10 days out of which 8 days already have passed, now a total of $8 + 4 = 12$ days, that is, an additional 2 days extra will be required.

**Answer:** c: 2.

**Key concepts used:** Percentage concept -- mandays concept to assess total work and also how many days will be required by certain number of men, dual purpose concept.

**Problem 2.**

If 6 weavers can weave 14 sarees in 14 days how many more weavers will be required for weaving 28 sarees in 7 days?

- 20
- 18
- 16
- 12

**Solution 2 : Problem analysis **

Using weaver-days concept to assess the amount of work involved, we have from the given statement,

To weave 14 sarees, $6\times{14} = 84$ weaver-days are required.

So for weaving 28 sarees, $2\times{84}=168$ weaver-days will be required. This is the assessment of labour required in terms of labourer-days.

**Solution 2 : Final Problem execution**

To do this amount of labour in 7 days then, the number of labourers or weavers required will be,

$\displaystyle\frac{168}{7}=24$.

Original strength of weaver team was 6. So an extra $24-6=18$ numbers of weavers will be required for weaving 28 sarees in 7 days.

**Answer:** Option b : 18.

**Key concepts used:** Mandays concept to assess total amount of work involved as well as number of weavers required -- unitary method to assess actual amount of work as number of sarees to be weaved doubled, labour required in this case is directly proportional to the work involved. The proportionality of the number of labourers is absorbed in the labourer-days concept which is a measure of work in terms of workers as well as time required.

**Problem 3.**

Runu can do a piece of work in 12 days and Biju in 16 days. They begin together and with help of another man finish the work in 6 days. If they get a total wage of Rs.3200 for the work, how much Runu should get as his share of wage?

- Rs.1600
- Rs.1200
- Rs.800
- Rs.900

**Solution 3 : Problem analysis**

In practice three persons worked where we don't know the work capacity of the third man. Assuming that the third man take $x$ number of days to finish the work alone, we can now form a linear equation in $x$ easily,

$\displaystyle\frac{1}{12} + \displaystyle\frac{1}{16} + \displaystyle\frac{1}{x} = \displaystyle\frac{1}{6}$

As the three of them finished the work in 6 days, in one day they worked together to finish one-sixth portion of the work.

**Solution 3 : Final Problem execution**

So we have,

$\displaystyle\frac{1}{12} + \displaystyle\frac{1}{16} + \displaystyle\frac{1}{x} = \displaystyle\frac{1}{6}$,

Or, $\displaystyle\frac{1}{x} = \displaystyle\frac{1}{6} - \displaystyle\frac{7}{48}=\displaystyle\frac{1}{48}$,

Or, $x = 48$ days, that is, the third man can finish the work in 48 days.

Finally then the wage they will get will be in ratio of inverse of the number of days they take to finish the work individually which is,

Runu : Biju : other man = 1/12 : 1/16 : 1/48 = 4 : 3 : 1, a total number of 4 + 3 + 1 = 8 shares.

As total wage they get is Rs.3200, the value of 1 share is, $\frac{3200}{8}=Rs.400$.

Runu will then get 4 shares equivalent to $4\times{400}=Rs.1600$.

**Answer:** Option a: Rs.1600.

**Key concepts used:** Work - time concepts -- wage as work contribution share concept.

**Problem 4.**

Out of 120 poles to be erected 40 labourers could erect 40 poles in 4 hours. But after that 8 labourers fell ill and left the job. How many hours more will be required to finish erecting all the poles?

- 4
- 6
- 10
- 8

**Solution 4 : Problem analysis and execution**

40 labourers could erect 40 poles in 4 hours.

So to erect 40 poles, $40\times{4}=160$ labourer hours were required.

As total number of poles to be erected is 120 out of which $120-40=80$ are still left to be erected, a total amount of work left is as follows.

To erect 40 poles 160 labourer hours were required,

So to erect 80 poles, $160\times{2}=320$ labourer hours will be required.

From 40 labourers 8 labourers left leaving 32 still working. So this 32 labourers will erect the rest 80 poles in $\frac{320}{32} = 10$ more hours.

**Answer:** c: 10.

**Key concepts used:** Work amount to time proportionality -- work assessment in terms of labourer hours.

**Problem 5.**

7 men and 7 women can do a work in 2 days. How much time (in days) will 1 man and 1 woman take to do the same work?

- 7
- 1
- 14
- 49

**Solution 5 : Problem analysis and exceution**

The amount of work in terms of mandays and womandays is,

$Work = 7\times{2}$ mandays $+$ $7\times{2}$ womandays = 14 mandays + 14 womandays.

So if this work is to be done by 1 man and 1 woman, they will take together 14 days to complete it, the total work amount completed would then be = 14 mandays + 14 womandays.

**Answer:** Option c: 14.

**Key concepts used:** Careful assessment of total work in terms of number of labourer days and inverse proportionality of days to labourers concept**.**

#### Problem 6.

A piece of work can be done by $B$ in 12 days and by $A$ in 20 days. $A$ first worked on the job for a few days after which $B$ started the work replacing him. If the work was finally completed in total number of 14 days, after how many days from start did $A$ leave?

- 6
- 4
- 8
- 5

**Solution 6 : Problem analysis**

Assumption of number of days that $A$ worked as $x$ will quickly give us the results here.

We will express then the number of days $A$ worked as $x$ and days for which $B$ worked as $14-x$.

As we know their rate of work, we will be able to calculate each person's share of work done. Summing up the two portions of work done should be the whole work, that is, 1.

#### Solution 6 : Problem execution

Work portion done by $A$ working forĀ $x$ number of days is,

$\displaystyle\frac{x}{20}$.

Similarly work portion done by $B$ in $14-x$ number of days is,

$\displaystyle\frac{14-x}{12}$.

Sum of these two portions should be 1 as they finally finished the whole work. So,

$\displaystyle\frac{x}{20} + \displaystyle\frac{14-x}{12} = 1$,

Or, $\displaystyle\frac{3x + 70 - 5x}{60} = 1$,

Or, $70 - 2x = 60$,

Or, $2x = 10$,

Or, $x=5$.

**Answer:** Option d : 5.

**Key concepts used:** Work rate and work portion.

** Problem 7.**

$A$, $B$ and $C$ can do a piece of work in 20, 30 and 12 days respectively. $A$ and $B$ begin to do the work together and taking help of $C$ every third day complete the job. If they receive Rs.6000 as wages for the work, how much did $C$ receive (in Rs.)?

- 1800
- 1200
- 1500
- 1600

**Solution 7 : Problem analysis**

Everyday $A$ and $B$ work together to complete,

$\frac{1}{20} + \frac{1}{30} = \frac{1}{12}$ portion of the work.

Every third day $C$ contributes by completing $\frac{1}{12}$ portion of work.

**Solution 7 : Problem execution**

So in two days total work done is,

$\frac{1}{12} + \frac{1}{12}=\frac{1}{6}$ portion of work.

But in every third day the work done is also,

$\frac{1}{12} + \frac{1}{12} = \frac{1}{6}$

Then in every three days the total work done is,

$\frac{1}{6} + \frac{1}{6} = \frac{1}{3}$.

In $3\times{3} = 9$ days then the work will be finished.

In 9 days the portion of work done by $A$ will then be,

$9\times{\frac{1}{20}} = \frac{27}{60}$,

the portion of work done by $B$ will be,

$9\times{\frac{1}{30}} = \frac{18}{60}$.

The rest, $\frac{15}{60}$ will be the contribution of $C$ in 3 out of 9 days.

Thus the ratio of contribution to the total work by $A$, $B$ and $C$ is,

$\frac{27}{60} : \frac{18}{60} : \frac{15}{60} = 27 : 18 : 15$, where total portions is 60 of value Rs.6000. So each portion value is Rs.100 and share of $C$ is 15 portions, that is, Rs.1500.

**Answer:** Option c: 1500.

** Key concepts used:** Share of wage is in proportion to share in work contribution -- work rate per day in terms of work portion completed -- enumerating the events accurately.

** Problem 8.**

A cistern can be filled up by two pipes $A$ and $B$ in 30 minutes and 20 minutes respectively. Both the pipes were opened together when the cistern was empty, but after a little while $B$ was stopped and the cistern got filled up in a total duration of 18 minutes. When was the pipe $B$ stopped (in minutes afer start)?

- 6
- 8
- 10
- 4

**Solution 8 : Problem analysis**

We need to assume the time pipe $B$ was active as $x$ minutes. This assumption will enable us to form the requisite linear equation in $x$ from the problem statement.

**Solution 8 : Problem execution**

The portion of cistern filled up till the time both $A$ and $B$ were open is,

$\displaystyle\frac{x}{20} + \displaystyle\frac{x}{30} = \displaystyle\frac{x}{12}$.

And the portion filled up by $A$ alone after $B$ was stopped is,

$\displaystyle\frac{18-x}{30}$.

Sum of these two portions should be 1,

$\displaystyle\frac{x}{12} + \displaystyle\frac{18-x}{30} = 1$,

Or, $5x + 36 - 2x = 60$,

Or, $3x=24$,

Or, $x=8$.

**Answer:** Option b: 8.

**Key concepts used:** Contribution of each pipe in filling up the cistern according to its capacity per unit time.

**Problem 9.**

If 30 men work they take 6 more days than 40 men would have taken to complete a work. How many days would 60 men take to complete the work?

- 8
- 12
- 10
- 6

**Solution 9 : Problem analysis**

Here also assuming that 40 men take $x$ number of days to finish the work will make solution come easy.

**Solution 9 : Problem exceution**

By the first statement,

$30(x + 6) = 40x$, is the amount of work.

Or, $10x = 180$,

Or, $x=18$,

Work amount $=40x=720$ mandays.

So 60 men will take,

$\frac{720}{60} = 12$ days to complete the work.

**Answer:** Option b: 12.

**Key concepts used:** Mandays as amount of work.

** Problem 10.**

Ravi is thrice as fast as Mintu and is able to finish a work in 60 days less than Mintu. How long would they take (in days) to complete the work together?

- $18$
- $22\frac{1}{2}$
- $24\frac{1}{6}$
- $20\frac{2}{3}$

**Solution 10 : Problem analysis and execution**

Let's assume Mintu can do the work in $x$ number of days.

So by problem statement,

$3(x-60)=x$,

Or, $2x=180$

Or, $x=90$.

Thus Mintu finishes the work in 90 days and Ravi in 30 days (three times faster means finishes same work in one-third time) and together they will finsh portion of work in a day as,

$\frac{1}{30} + \frac{1}{90} = \frac{4}{90}$.

And the number of days they will take to finish the work together is the inverse of the daily portion achieved, that is,

$\frac{90}{4} = 22\frac{1}{2}$ days.

**Answer:** Option b: $22\frac{1}{2}$.

**Key concepts used:** Work capacity to daily work contribution -- careful problem modeling.

### Useful resources to refer to

**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 **

**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**

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

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