## 2nd SSC CHSL Solved Question Set, 2nd on Work time

This is the 2nd solved question set of 10 practice problem exercise for SSC CHSL exam and the 2nd on topic Work Time. It contains,

- 2nd
**question set on Work time**for SSC CHSL to be answered in 15 minutes (10 chosen questions) **Answers**to the questions, and- Detailed
**conceptual solutions**to the questions.

For maximum gains, the test should be taken first, and then the solutions are to be referred to. But more importantly, to absorb the concepts, techniques and reasoning explained in the solutions, one must solve many problems in a systematic manner using the conceptual analytical approach.

Learning by doing is the best learning. There is no other alternative towards achieving excellence.

### 2nd Question set - 10 problems for SSC CHSL exam: 2nd on topic Work time - answering time 15 mins

**Q1. **If the work done by $(x-1)$ men in $(x+1)$ days is to the work done by $(x+2)$ men in $(x-1)$ days are in the ratio $9:10$, then the value of $x$ is,

- 8
- 7
- 6
- 5

**Q2.** 3 men and 7 women can do a job in 5 days, while 4 men and 6 women do it in 4 days. The number of days required for a group of 10 women working together at the same rate as before to finish the job is,

- 30 days
- 40 days
- 36 days
- 20 days

**Q3. **If there is a reduction in the number of workers in a factory in the ratio of $15:11$ and an increment of wages in the ratio of $22:25$, the ratio by which the total wages of the workers would be decreased is,

- $3:7$
- $5:6$
- $6:5$
- $3:5$

**Q4. **Debu and Monu take a piece of work for Rs.28,800. One alone could do it in 36 days and the other in 48 days. With the assistance of an expert they finish it in 12 days. How much remuneration the expert should get?

- Rs.10,000
- Rs.12,000
- Rs.18,000
- Rs.16,000

**Q5. **It was planned to construct a road 5 km long in 100 days and 280 workers were employed for the job. But after 80 days it was found that only $3\displaystyle\frac{1}{2}$ Km of the road was completed. How many more men need to be employed now to complete the work in planned time?

- 200
- 80
- 100
- 480

**Q6.** P can complete $\displaystyle\frac{1}{4}$th of a work in 10 days. Q can complete 40% of the same work in 15 days. R completes $\displaystyle\frac{1}{3}$rd of the work in 13 days and S completes $\displaystyle\frac{1}{6}$th of the work in 7 days. Who will be able to complete the work earliest?

- S
- P
- Q
- R

**Q7.** A and B can do a piece of work in 72 days, B and C can do it in 120 days and A and C can do it in 90 days.. When A, B and C work together, how much work will be finished by them in 3 days?

- $\displaystyle\frac{1}{40}$
- $\displaystyle\frac{1}{10}$
- $\displaystyle\frac{1}{20}$
- $\displaystyle\frac{1}{30}$

**Q8.** 40 men can complete a work in 18 days. Eight days after they started working together, 10 more men joined them. How many days will they now take to complete the remaining work?

- 12 days
- 8 days
- 10 days
- 6 days

**Q9.** A can do as much work as B and C together can do. A and B can do a piece of work in 9 hours 36 mins and C can do it in 48 hours. The time that B will take to complete the work alone is,

- 18 hours
- 12 hours
- 30 hours
- 24 hours

**Q10.** A and B can do a work in 18 and 24 days respectively. They worked together for 8 days and then A left. The remaining work was finished by B in,

- $8$ days
- $5\displaystyle\frac{1}{3}$ days
- $10$ days
- $5$ days

### Answers to the questions

Q1. Answer: Option a: 8.

Q2. Answer: Option d : 20 days.

Q3. Answer: Option c: $6:5$.

Q4. Answer: Option b: Rs.12,000.

Q5. Answer: Option a: 200.

Q6. Answer: Option c : Q.

Q7. Answer: Option c: $\displaystyle\frac{1}{20}$.

Q8. Answer: Option b: 8 days.

Q9. Answer: Option d: 24 hours.

Q10. Answer: Option b: $5\displaystyle\frac{1}{3}$ days.

### 2nd solution set - 10 problems for SSC CHSL exam: 2nd on topic Work time - answering time 15 mins

**Q1. **If the work done by $(x-1)$ men in $(x+1)$ days is to the work done by $(x+2)$ men in $(x-1)$ days are in the ratio $9:10$, then the value of $x$ is,

- 8
- 7
- 6
- 5

** Solution 1: Problem analysis and solving execution by using mandays as work measure**

By the first condition, work done by $(x-1)$ men in $(x+1)$ days is,

$W_1=(x-1)(x+1)$ mandays.

We have expressed the work amount in terms of mandays.

Similarly in the second case, the work done,

$W_2=(x+2)(x-1)$ mandays.

By the third condition,

The two amounts of work done are in the ratio $9:10$. So,

$\displaystyle\frac{W_1}{W_2}=\frac{(x-1)(x+1)}{(x+2)(x-1)}=\frac{9}{10}$,

Or, $\displaystyle\frac{x+1}{x+2}=\frac{9}{10}$.

Cross-multiplying and simplifying,

$x=8$.

**Answer:** Option a: 8.

**Key concepts used:** * Work time concept* --

*--*

**Mandays concept***.*

**Ratio and proportion****Q2.** 3 men and 7 women can do a job in 5 days, while 4 men and 6 women do it in 4 days. The number of days required for a group of 10 women working together at the same rate as before to finish the job is,

- 30 days
- 40 days
- 36 days
- 20 days

**Solution 2: Problem solving execution using Work rate technique**

Here two cases of work completion with two different combinations of workers are given. We will assume **work portion done by a man and a woman in a day** as, $M$ and $W$ respectively. These are the corresponding work rates using Work rate technique.

In Work rate technique, we express the work variables themselves as work rates. In this technique, the fraction arithmetic is transferred to the RHS on total work amount which we assume here as $W_A$.

By the first two conditions then,

$5(3M+7W)=W_A$, assuming $W_A$ as the whole work amount

Or, $15M+35W=W_A$, and

$4(4M+6W)=W_A$,

Or, $16M+24W=W_A$.

We need to eliminate $M$ and get the value of work rate of a woman in terms of work portion done in a day.

Multiplying the first equation by 16, the second by 15 and subtracting the second result from the first we get,

$W_A=W(16\times{35}-15\times{24})$

$=20W(28-18)$

$=200W$.

As 10 women will be working in the third case,

$W_A=20(10W)$.

10 women working together will then take 20 days to complete the work.

**Answer:** Option d: 20 days.

**Key concepts used:** **Work rate technique -- Work time concept -- Linear algebraic equation -- Efficient simplification****.**

As a part of efficient simplification, we have not evaluated the two terms, $16\times{35}$ and $15\times{24}$. Only when the second is subtracted from the first we have taken out the HCF of the two and then carried out the simple multiplication and subtraction.

After writing down the first two equations, rest of the calculations could be carried out in mind quickly.

**Q3. **If there is a reduction in the number of workers in a factory in the ratio of $15:11$ and an increment of wages in the ratio of $22:25$, the ratio by which the total wages of the workers would be decreased is,

- $3:7$
- $5:6$
- $6:5$
- $3:5$

**Solution 3: Problem analysis and execution by Ration and proportion and Product of ratios concepts**

$\text{Total wage }=\text{Number of workers}\times{\text{Wage per worker}}$.

So multiplying the two given ratios we will get the ratio of total wages before and after,

$\text{Ratio of total wage of workers}=\displaystyle\frac{15}{11}\times{\displaystyle\frac{22}{25}}$

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

**Answer:** Option c: $6:5$.

**Key concepts used: Work wage concept -- Ratio proportion concept -- Product of ratios -- HCF reintroduction technique**

**.**

#### Mathematical explanation of ratio multiplication

Let total wages in the before and after situations be $TW_1$ and $TW_2$.

In the first case, the before and after ratio of number of workers is,

$\displaystyle\frac{15}{11}=\frac{15x}{11x}$, where $15x$ and $11x$ are the actual number of workers and $x$ is the canceled out HCF reintroduced.

Similarly in the second case, the before and after ratio of wage per worker is,

$\displaystyle\frac{22}{25}=\frac{22y}{25y}$, where $22y$ and $25y$ are the actual wage per worker and $y$ is the canceled out HCF reintroduced.

As total wage is the product of number of workers and wage per worker,

$TW_1=xy(15\times{22})$, and

$TW_2=xy(11\times{25})$.

Taking ratio of the two total wages we get,

$\displaystyle\frac{TW_1}{TW_2}=\frac{15}{11}\times{\frac{22}{25}}$

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

You don't have to deduce the answer this way. This explanation shows how a product of two ratios can result in a third ratio. In fact, with clear concepts, the problem can easily be solved in mind.

**Q4. **Debu and Monu take a piece of work for Rs.28,800. One alone could do it in 36 days and the other in 48 days. With the assistance of an expert they finish it in 12 days. How much remuneration the expert should get?

- Rs.10,000
- Rs.12,000
- Rs.18,000
- Rs.16,000

**Solution 4: Problem solving using the concept of earning and work portion completed direct proportionality**

By earning to work portion done proportionality concept,

Total earning of a worker is directly proportional to the portion of work done by the worker.

The portion of work completed by Debu and Monu in 12 days of working together is,

$\displaystyle\frac{12}{36}+\displaystyle\frac{12}{48}=\displaystyle\frac{1}{3}+\displaystyle\frac{1}{4}=\displaystyle\frac{7}{12}$ portion of the work.

As total earning is directly proportional to the portion of total work done which is worth a total earning of Rs.28,800, in 12 days of working together Debu and Monu will earn,

$\displaystyle\frac{7}{12}$th of the total earning, and the expert will earn the rest as,

$\displaystyle\frac{5}{12}$th of total earning

$=\displaystyle\frac{5}{12}\times{\text{Rs.28,800}}$

$=\text{Rs.12,000}$.

**Answer:** Option b: Rs.12,000.

**Key concepts used: Work wage concept -- Work rate concept -- Earning proportional to work portion done concept -- Working together concept****.**

**Q5. **It was planned to construct a road 5 km long in 100 days and 280 workers were employed for the job. But after 80 days it was found that only $3\displaystyle\frac{1}{2}$ km of the road was completed. How many more men need to be employed now to complete the work in planned time?

- 200
- 80
- 100
- 480

**Solution 5: Problem analysis and execution using mandays concept**

We will use mandays as work measure.

280 workers working for 80 days results in,

$280\times{80}\text{ mandays}=3\displaystyle\frac{1}{2}=\displaystyle\frac{7}{2}$ km of road work

Simplifying we get the work equivalence of,

$80\times{80}\text{ mandays}=1$ km of road work.

Leftover work is, $5-3\displaystyle\frac{1}{2}=\displaystyle\frac{3}{2}$ km of road work

If $x$ be the number of workers completing this leftover work in 20 days we have the mandays equivalence as,

$20x\text{ mandays}=\displaystyle\frac{3}{2}$km of road work,

$=\displaystyle\frac{3\times{80}\times{80}}{2}$ mandays of work, substituting mandays equivalent to 1 km of road work,

Or, $x=480$ men

Extra number of worker required would then be,

$480-280=200$.

**Answer:** Option a: 200.

*Key concepts used:* **Mandays as work measure concept****.**

**Q6.** P can complete $\displaystyle\frac{1}{4}$th of a work in 10 days. Q can complete 40% of the same work in 15 days. R completes $\displaystyle\frac{1}{3}$rd of the work in 13 days and S completes $\displaystyle\frac{1}{6}$th of the work in 7 days. Who will be able to complete the work earliest?

- S
- P
- Q
- R

**Solution 6 - Problem analysis and execution by comparing number of days required to complete the work**

P can complete $\displaystyle\frac{1}{4}$th of a work in 10 days,

So, P completes the whole work in 40 days.

Q can complete 40% or $\displaystyle\frac{2}{5}$th of the same work in 15 days,

So, Q completes the whole work in $\displaystyle\frac{75}{2}=37.5$ days.

R completes $\displaystyle\frac{1}{3}$rd of the work in 13 days,

So, R completes the whole work in 39 days.

S completes $\displaystyle\frac{1}{6}$th of the work in 7 days,

So, S completes the whole work in 42 days.

Out of the four then Q takes minimum time of 37.5 days to complete work.

**Answer:** Option c : Q.

**Key concepts used:** ** Working time concept -- Work completion time as product of inverse of work portion and time to do the portion -- unitary method**.

**Q7.** A and B can do a piece of work in 72 days, B and C can do it in 120 days and A and C can do it in 90 days. When A, B and C work together, how much work will be finished by them in 3 days?

- $\displaystyle\frac{1}{40}$
- $\displaystyle\frac{1}{10}$
- $\displaystyle\frac{1}{20}$
- $\displaystyle\frac{1}{30}$

**Solution 7 - Problem analysis and excution by Work rate technique and working together concepts**

Assuming $A_w$, $B_w$ and $C_w$ as the work rate or work portion done in a day by A, B and C respectively, by the three given conditions,

$A_w+B_w=\displaystyle\frac{1}{72}W$, assuming total work as $W$,

$B_w+C_w=\displaystyle\frac{1}{120}W$, and

$C_w+A_w=\displaystyle\frac{1}{90}W$.

Adding the three we get work portion done in a day when A, B and C work together as,

$A_w+B_w+C_w=\displaystyle\frac{1}{2}\left(\displaystyle\frac{1}{72}+ \displaystyle\frac{1}{120}+\displaystyle\frac{1}{90}\right)W$

$=\displaystyle\frac{1}{60}W$.

So A, B, and C working together, will complete $\displaystyle\frac{1}{60}$th of the work in 1 day and $\displaystyle\frac{1}{20}$ th of the work in 3 days.

**Answer:** Option c: $\displaystyle\frac{1}{20}$.

** Key concepts used:** ** Work rate of each worker in terms of work portion done in a day **--

*--*

**Work rate technique**

**Working together concept, work portion done by workers working together is sum of their work rates -- Summing up three equations gives us the work portion done by all three together in a day.****Q8.** 40 men can complete a work in 18 days. Eight days after they started working together, 10 more men joined them. How many days will they now take to complete the remaining work?

- 12 days
- 8 days
- 10 days
- 6 days

** Solution 8 - Problem analysis and execution by mandays concept**

40 men complete a work in 18 days, and so the work amount is,

$40\times{18}=720$ mandays.

In first 8 days 40 men complete, 320 mandays of work leaving 400 mandays work yet to be done.

On 9th day 10 more men joining, the team strength becomes 50 men.

To complete the leftout work of 400 mandays, this team of 50 men will require then,

$\displaystyle\frac{400}{50}=8$ days.

**Answer:** Option b: 8 days.

**Key concepts used:** **Mandays as work measure concept -- leftout work concept.**

**Q9.** A can do as much work as B and C together can do. A and B can do a piece of work in 9 hours 36 mins and C can do it in 48 hours. The time that B will take to complete the work alone is,

- 18 hours
- 12 hours
- 30 hours
- 24 hours

**Solution 9 - Problem analysis and execution by Work rate as work portion done in a day and Working together concept by adding work rates**

A and B working together can complete the work in 9 hours 36 minutes or $9\displaystyle\frac{3}{5}=\frac{48}{5}$ hours.

Adopting **Work rate technique**, we assume work rates $a$, $b$ and $c$ as portion of work done in a day by A, B and C respectively, and $W$ as the whole work amount.

So by the second condition of A and B together doing the whole work in $\displaystyle\frac{48}{5}$ hours, we get **one day work equivalence** as,

$a+b=W\div{\displaystyle\frac{48}{5}}=\displaystyle\frac{5W}{48}$, Assuming $W$ as the whole work amount.

C does the work in 48 hours. So,

$c=\displaystyle\frac{W}{48}$

Also given, $a=b+c$. Substituting in the first equation,

$a+b=2b+c=\displaystyle\frac{5W}{48}$,

Or, $2b+\displaystyle\frac{W}{48}=\displaystyle\frac{5W}{48}$,

Or, $2b=\displaystyle\frac{W}{12}$,

Or, $b=\displaystyle\frac{W}{24}$.

It means B will complete the work in 24 hours.

**Answer:** Option d: 24 hours.

**Key concepts used:** * Work rate technique -- Working together concept* --

**Efficient simplification.****Q10.** A and B can do a work in 18 and 24 days respectively. They worked together for 8 days and then A left. The remaining work was finished by B in,

- $8$ days
- $5\displaystyle\frac{1}{3}$ days
- $10$ days
- $5$ days

**Solution 10 - Problem analysis and execution by work rate and working together concepts**

In 8 days, A and B working together complete portion of total work,

$8\left(\displaystyle\frac{1}{18}+\displaystyle\frac{1}{24}\right)=\displaystyle\frac{7}{9}$

So the leftout work is $\displaystyle\frac{2}{9}$th of the whole wotk.

B does the whole work in 24 days, so he will do $\displaystyle\frac{2}{9}$th of the whole work in,

$24\times{\displaystyle\frac{2}{9}}=\displaystyle\frac{16}{3}=5\displaystyle\frac{1}{3}$ days.

**Answer: **Option b: $5\displaystyle\frac{1}{3}$ days.

**Key concepts used:** **Work rate concept -- Working together concept.**

### 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*

#### 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**

### SSC CHSL level Question and Solution sets

#### Work and Time, Pipes and Cisterns

**SSC CHSL level Solved Question set 1 on Work time 1**

**SSC CHSL level Solved Question set 2 on Work time 2**

#### Number System, HCF and LCM

**SSC CHSL level Solved Question set 3 on Number system 1**

**SSC CHSL level Solved Question set 4 on Number system 2**

**SSC CHSL level Solved Question set 5 on HCF and LCM 1**

**SSC CHSL level Solved Question set 6 on HCF and LCM 2**

#### Surds and Indices

**SSC CHSL level Solved Question set 7 on Surds and Indices 1**

**SSC CHSL level Solved Question set 8 on Surds and Indices 2**

#### Mixture or Alligation

**SSC CHSL level Solved Question set 9 on Mixture or Alligation 1**

**SSC CHSL level Solved Question set 10 on Mixture or Alligation 2**

#### Algebra

**SSC CHSL level Solved Question set 11 on Algebra 1**

**SSC CHSL level Solved Question set 12 on Algebra 2**

**SSC CHSL level Solved Question set 13 on Algebra 3**