## 10th SSC CGL tier II level Solution Set, 1st on topics Time-Work, Work-Wages, and Pipes-Cisterns

This is the 10th solution set of 10 practice problem exercise for SSC CGL tier II exam and 1st on related topics Time-Work, Work-Wages and Pipes-Cisterns. 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.

Before going through these solutions you should take the test by referring to * SSC CGL Tier II level Question Set 10 on Time-Work Work-Wages Pipes-Cisterns 1*.

### 10th solution set - 10 problems for SSC CGL Tier II exam: 1st on topics Time-Work Work-Wages Pipes-Cisterns - time 12 mins

**Problem 1.**

One pipe in a tank can fill it in 45 mins and a second one can empty it in 1 hr. In what time would the empty tank be filled if the pipes are opened one at a time in alternate minutes?

- 2 hrs 55 mins
- 4 hrs 48 mins
- 3 hrs 40 mins
- 5 hrs 53 mins

** Solution 1: Problem analysis**

The filling pipe must be taken as the first pipe opened in the sequence of the two pipes opened in alternate minutes. This is because, even if the emptying pipe is opened in the first minute, it will have no effect on the outcome as it would try to empty an already empty tank.

This is a problem involving identification of boundary conditions, and accordingly it needs to be treated with a suitable approach.

**Solution 1: Problem solving first stage: Finding effective fill every 2 mins**

Assuming $I$ as the filling rate of the first inlet pipe in terms of portion of tank filled in a minute and $D$ as the drain rate of the emptying pipe in terms of portion of tank drained per minute (with sufficient water to drain in the tank), from the first condition we have,

$45I=T$, where $T$ is the tank capacity

Or, $I=\displaystyle\frac{1}{45}T$.

Similarly from the second emptying condition,

$60D=T$, 1 hour is converted to 60 mins as per work time unit selection of the smallest time unit of mins

Or, $D=\displaystyle\frac{1}{60}T$.

This amount will be subtractive.

Together in every two mins then effective portion of the tank filled up will be,

$Fill_{2mins}=T\left(\displaystyle\frac{1}{45}-\displaystyle\frac{1}{60}\right)=\displaystyle\frac{1}{180}T$.

In every 2 mins, 4 out of 180 portions of tank will be filled up by the filling pipe and 3 out of 180 portions will be drained by the second emptying pipe, with resulting effective filling of 1 out of 180 portions of tank.

#### Solution 1: Problem solving second stage: Finding boundary condition

We can't say that in $2\times{180}=360\text{ mins}=6\text{ hrs}$, the tank will be fully filled, because near the end of the filling and emptying in alternate minute sequence, the tank will get fully filled before the emptying pipe gets any further chance of emptying it. We need to find this boundary condition for solving this problem.

As in each 2 mins the two pipes fill up 1 portion out of full 180 portions of the tank and 4 portions out of 180 portion of the tank is filled up in every odd min, we subtract 4 from 180 to get 176 as the number portions that will be safely filled up by,

$2\times{176}=352\text{ mins}=5\text{ hrs }52\text{ mins}$.

In the next min, that is, in 5 hrs 53rd min, the tank will then be fully filled up.

We could use * Working backwards approach* here to find the boundary by just subtracting the first and odd min fill portion from the total portion, because the effective fill is 1 portion every 2 mins. If it were any other figure such as 2 or 3 portions out of 180 total portions, the approach would have been different (number of multiples of effective fill portion less than the whole portion).

**Answer:** d: 5 hrs 53 mins.

**Key concepts used:** * Boundary condition problem -- Basic time and work concepts* --

*--*

**Fill rate or Work rate technique***--*

**Effective fill concept***--*

**Working backwards approach**

**Boundary determination -- Portion use technique***.*

**Problem 2.**

In a project, a job was targetted to be completed by 30 workers in 38 days. But after 25 days it was decided to add 5 more workers to the job to get it finished 1 day earlier. Had the extra number of workers not been added, what would have been the number of days taken by the existing team to finish the job?

- 42
- 39
- 40
- 41

** Solution 2: Problem analysis and execution: By Mathematical reasoning**

After 25 days, 5 workers were added to the team of 30 workers. With this increased strength of 35 workers after 25 days, the job could be finished in 37 days, 1 day less than targetted 38 days.

So for 37 days 30 workers worked and for 12 days an additional 5 workers worked to finish the job in 37 days. These 5 workers working for 12 days did an additional work amounting to, $5\times{12}=60$ mandays.

Thus if the additional 5 workers were not added, the existing team of 30 workers would have had to work for an additional, $\displaystyle\frac{60}{30}=2$ days to do this additional work of 60 mandays.

The period of completion of the job would then be, $37+2=39$ days.

**Answer:** b: 39.

**Key concepts used:** * Basic time and work concepts* --

**Work amount as Mandays concept***.*

**-- Mandays rearrangement technique -- Mathematical reasoning****Problem 3.**

Three pipes fill a tank individually in 12 hrs, 15 hrs and 20 hrs. With the first pipe open throughout and the second and third pipe open in alternate hrs starting with the second, how long would it take to get the tank filled fully?

- $6$ hrs
- $7\frac{1}{2}$ hrs
- $6\frac{2}{3}$ hrs
- $7$ hrs

**Solution 3: Problem analysis and execution**

If $A$, $B$ and $C$ be fill rates in terms of portion of tank filled up in 1 hr for the three pipes mentioned in sequence respectively, from the first three conditions we have,

$A=\displaystyle\frac{1}{12}T$, where $T$ is the tank capacity,

$B=\displaystyle\frac{1}{15}T$, and

$C=\displaystyle\frac{1}{20}T$.

In every 2 hrs then the portion of tank filled up will be, twice of fill rate of A, and sum of fill rates of B and C as,

$Fill_{2hrs}=2A+B+C=T\left(\displaystyle\frac{2}{12}+\displaystyle\frac{1}{15}+\displaystyle\frac{1}{20}\right)=\displaystyle\frac{17}{60}T$.

In three times 2 hrs, that is, 6 hrs, three times 17, that is 51 portions out of total 60 portions of the tank would be filled.

.Here 51 is the largest multiple of 17 less than 60 and is the boundary

This * largest multiple boundary determination* method is to be applied when the

*in*

**effective portions completed***(which may be 2 or 3 or 4 or any number of time units over which the workers complete*

**effective cycle time period***) is more than 1.*

**one cycle of combined work**During 7th hr, first two pipes will be active and will fill the rest of the 9 tank portions out of 60,

$A+B=T\left(\displaystyle\frac{1}{12}+\displaystyle\frac{1}{15}\right)=\displaystyle\frac{9}{60}T$

To fill the tank then 7 hrs will be required.

**Answer:** Option d: 7 hrs.

**Key concepts used:** * Fill rate* equivalent to

*--*

**Work rate technique**

**Boundary condition problem -- Boundary determination --***.*

**Portions use technique -- Largest multiple cycle portion boundary -- Effective cycle portions completed -- Effective cycle time -- Cycle of combined work****Problem 4.**

Two men start doing a job for Rs.200. One of them could do it alone in 6 days and the other can do it alone in 8 days. Together with a third man, the first two men finish the job in 3 days. How much money should the third man get?

- Rs.25
- Rs.20
- Rs.45
- Rs.65

**Solution 4 : Problem analysis and execution**

The first man can finish the job alone in 6 days. So in 3 days he finishes half of the job.

The second man alone can finish the job in 8 days. So in 3 days, he does, $\displaystyle\frac{3}{8}$ portion of the job.

In 3 days, the first and the second man then finish together,

$\displaystyle\frac{1}{2}+\displaystyle\frac{3}{8}=\displaystyle\frac{7}{8}$ portion of the job.

The third man then must have done the rest of $\displaystyle\frac{1}{8}$th of the job.

As * earning is proportional to the work done*, the third man will get,

$\displaystyle\frac{1}{8}\text{th of Rs.}200=\text{Rs.}25$

**Answer:** a: Rs.25.

**Key concepts used:** * Basic work and wages concepts* --

*--*

**Work rate technique***--*

**Working together per unit time concept**

**Earning share concept -- Earning to work done proportionality.****Problem 5.**

Two pipes together fill a tank in 6 hrs. If one pipe fills the tank 5 hrs faster than the other, how long does the faster pipe take to fill the tank alone?

- 12
- 9
- 8
- 10

**Solution 5: Problem analysis and exceution**

If the faster pipe fill the tank alone in $x$ hrs, the other pipe fills it in $(x+5)$ hrs and together in 6 hrs,

$6T\left(\displaystyle\frac{1}{x}+\displaystyle\frac{1}{x+5}\right)=T$, where $T$ is the tank capacity,

Or, $6\left(\displaystyle\frac{1}{x}+\displaystyle\frac{1}{x+5}\right)=1$.

Trying out the choice values, the value of $x=10$ easily satisfies the expression.

Otherwise also the expression can be simplified into a quadratic equation in $x$,

$6\left(\displaystyle\frac{1}{x}+\displaystyle\frac{1}{x+5}\right)=1$,

Or, $x^2+5x=12x+30$,

Or, $x^2-7x-30=0$,

Or, $(x-10)(x+3)=0$.

As $x$ can't have negative value, $x=10$.

**Answer:** Option d: 10.

**Key concepts used:** * Fill rate equivalent to Work rate technique* --

**Working together per unit time concept -- Principle of free resource use -- Choice value set test -- Mathematical reasoning -- Many ways technique -- Basic algebra concepts**

**.**#### Problem 6.

In the same time duration A works three times as B. Together if they finish a job in 15 days, in how many days would A finish the job working alone?

- 20
- 40
- 50
- 60

**Solution 6 : Problem analysis and execution**

If $A$ and $B$ be the portion of work done by A and B respectively in 1 day, by the first condition,

$A=3B$.

When they finish the job together in 15 days,

$15(A+B)=W$, where $W$ is the total quantum of the work,

Or, $15A+5A=20A=W$, substituting value of B in terms of A.

Thus A finishes the work in 20 days working alone and B does it in 60 days working alone.

**Answer:** Option a: 20 days.

**Key concepts used:** * Work rate technique* --

*.*

**Working together per unit time concept -- Work amount as mandays concept**** Problem 7.**

Two men A and B complete a job working alone in 12 hrs and 8 hrs respectively. If they work for 1 hr each alternately, with B starting the work at 10 am, when will the work be finished?

- 8 pm
- 7.30 pm
- 6.30 pm
- 6 pm

**Solution 7: By Mathematical reasoning**

Working alone, in 1 hr, A does $\displaystyle\frac{1}{12}$th of the work and B does $\displaystyle\frac{1}{8}$th of the work.

Together in each 2 hrs then they do,

$\displaystyle\frac{1}{12}+\displaystyle\frac{1}{8}=\displaystyle\frac{5}{24}$ portion of the work.

In * effective cycle time of 2 hrs* they together working alternately complete

**effective 5 portions of work.**Largest multiple of 5 portions less than total 24 portions is 20, which is 4 times 5. *This is the boundary.*

So in $4\times{2}=8$ hrs A and B together will finish 20 portions of total 24 portions.

As B started the work, in the 9th hour B works at the rate of 1 out of 8 portions or 3 out of 24 portions per hour. So in 9th hour B will finish 3 out of rest 4 portions left. Remaining work will be 1 out of 24 portions.

In 10th hour A will finish this 1 out of 24 portions in half an hr working at the rate of 2 out of 24 portions per hour.

Total time required to finish the work will then be 9.30 hrs. As start time is 10.30 am, end time will be 7.30 pm.

**Answer:** Option b: 7.30 pm.

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

*--*

**Work rate technique***--*

**Working together per unit time concept, unit time in this case is 2 hrs**

**Boundary determination --***--***Portions use technique -- Largest multiple portion boundary -- Effective portions completed in a cycle -- Effective cycle time -- Cycle of combined work***.*

**Sequencing of events, after finding the boundary we needed to find wok done in each subsequent hr step by step**** Problem 8.**

If 5 women and 7 men earn Rs.3825 in 6 days and 2 women and 3 men earn Rs.1050 in 4 days, in how many days would 7 women and 6 men earn Rs.22500?

- 15
- 25
- 30
- 20

**Solution 8: By Mathematical reasoning**

Earning is proportional to work done and so is proportional to daily wage.

If $W$ and $M$ be the daily wage of 1 woman and 1 man respectively, by the first condition we have,

$6(5W+7M)=30W+42M=3825$.

Similarly by the second condition we have,

$4(2W+3M)=8W+12M=1050$,

Or, $4W+6M=525$, dividing by 2

Or, $28W+42M=3675$, multiplying by 7 with the objective of eliminating $M$.

Subtracting,

$2W=150$,

Or, $W=75$,

From $4W+6M=525$,

$300+6M=525$,

Or, $6M=225$,

Or, $M=37.5$ which is half of $W$.

If $x$ be the number of days for 7 women and 6 men to earn Rs.22500,

$x(7W+6M)=22500$,

Or, $10Wx=22500$, as $W=2M$,

Or, $x=30$, as $W=75$.

In 30 days the team of 7 women and 6 men will earn Rs.22500.

**Answer:** Option c: 30.

**Key concepts used:** * Earning share concept* --

**Earning to work done proportionality -- Daily wage concept -- Earning together concept -- Basic algebra concept -- Efficient simplification**

**.**** Problem 9.**

If 45 men earn Rs.15525 in 48 days and daily wage of a woman is double the daily wage of a man, how many women must work for 16 days to receive a total wage of Rs.5750?

- 15
- 25
- 30
- 20

**Solution 9: Problem analysis and solving**

Earning is proportional to work done and so is proportional to daily wage.

If $W$ and $M$ be the daily wage of 1 woman and 1 man respectively, by the first condition we have,

$48\times{45M}=15525$.

By the second condition we have again,

$W=2M$,

Or, $M=\frac{1}{2}W$.

Substituting,

$48\times{45M}=15525$,

Or, $24\times{45W}=15525$.

Or, $W=\displaystyle\frac{15525}{24\times{45}}=\displaystyle\frac{345}{24}=\displaystyle\frac{115}{8}$.

If $x$ be the number women who work for 16 days to earn Rs.5750,

$16(xW)=5750$,

Or, $x=\displaystyle\frac{5750\times{8}}{16\times{115}}$

$=\displaystyle\frac{2875}{115}$,

$=\displaystyle\frac{575}{23}$,

$=25$.

**Answer:** Option b: 25.

**Key concepts used:** * Earning share concept* --

**Earning to work done proportionality -- Daily wage concept -- Basic algebra concept -- Efficient simplification**

**.**** Problem 10.**

A finishes 7 portions out of 10 portions of a job in 15 days and finishes the rest of the job with the help of B in 4 more days. In how any days then A and B working together will be able to finish the entire job?

- $10\frac{2}{3}$
- $12\frac{2}{5}$
- $13\frac{1}{3}$
- $8\frac{1}{4}$

**Solution 10: By Mathematical reasoning using work time to work amount direct proportionality concept**

A does,

$\displaystyle\frac{7}{10}$th of the job in 15 days.

The job left is,

$\displaystyle\frac{3}{10}$th of the total job.

This portion is done by A and B together in 4 days.

As * work time is directly proportional to work amount with number of workers fixed*, by

*then, the whole work will be done by A and B together in,*

**unitary method**$\displaystyle\frac{4\times{10}}{3}=13\frac{1}{3}$ days

This is the elegant solution in which we didn't need to evaluate individual work rates of A and B.

**Answer:** Option c: $13\frac{1}{3}$.

**Key concepts used:** **Work time to work amount proportionality -- Context awareness -- Work left concept -- Unitary method****.**

**Note:** You will observe that in many instances we have used proportionality concepts to quickly solve the problems. As equations and dedutions are based on proportionality concepts, direct use of more basic proportionality concepts usually will yield faster solution.

### 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**

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

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

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

#### SSC CGL level Work Time Question and solution sets

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