## 49th SSC CGL level Solution Set, 4th on Time and Work problems in simpler steps

This is the 49th solution set of 10 practice problem exercise for SSC CGL exam and 4th on Time and Work in simpler steps. Students must complete the corresponding question set in prescribed time first and then only refer to the solution set for extracting 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.

Before going through the solutions you should go through our sessions on * How to solve Time and work problems in simpler steps type-1* and

*for detailed explanation of the efficient time and work problem solving techniques that we will use for this present set of 10 problems.*

**How to solve more time and work problems in simpler steps type-2**### 49th solution set- 10 problems for SSC CGL exam: 4th on Time and Work in simpler steps - time 15 mins

**Problem 1.**

A alone can complete a job in 20 days while A and B together can complete it in 12 days. If B does the work only half a day daily then how many days would they take to complete the job working together?

- 10 days
- 15 days
- 11 days
- 20 days

** Solution 1 : Problem analysis and execution**

Assuming $A_w$ and $B_w$ to be the portion of the job $W$ done by A and B each working alone respectively in 1 day, from the first statement we have,

$A_w=\frac{1}{20}W$.

Similarly from the second statement we have,

$A_w+B_w=\frac{1}{12}W$.

Eliminating $A_w$ between the two equations,

$B_w=\left(\frac{1}{12}-\frac{1}{20}\right)W = \frac{1}{30}W$.

Half day's work of B is then,

$\frac{1}{2}B_w=\frac{1}{60}W$.

A and B together with B working half a day daily will then complete the portion of work in a day,

$A_w+\frac{1}{2}B_w=\left(\frac{1}{20}+\frac{1}{60}\right)W=\frac{1}{15}W$.

In other words they will take 15 days to complete the work.

**Answer:** b: 15 days.

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

*.*

**Working together per unit time concept****Problem 2.**

A and B together can do a piece of work in 12 days which B and C can do together in 16 days. After A has been working at it for 5 days and B for 7 days, C finishes it in 13 days. In how many days B would finish the job working alone?

- 24 days
- 48 days
- 12 days
- 16 days

**Solution 2 : Problem analysis**

From the first two statements in the first sentence we get two linear equations in $A_w$, $B_w$, $C_w$, the individual work rates of A, B, and C respectively in portion of work $W$ done in 1 day.

From the second sentence we will get a third linear equation in terms of the four quantities. From these three equations, it is easy to find $B_w$ and hence number of days to be taken by B to complete the work alone.

Though the problem is a straightforward one, there may be slight hesitation in forming the third equation. The doubt that may arise is, whether they worked sequentially one after the other, or started to work together at the same time. From the construction of the sentence, "After A has been working at it for 5 days and B for 7 days, C finishes it in 13 days." in the absence of any hint regarding whether they started to work together, we would assume, they worked one after the other - the simplest assumption.

**Solution 2 : Problem solving execution**

The three equations are,

$A_w+B_w=\frac{1}{12}W$,

$B_w+C_w=\frac{1}{16}W$, and

$5A_w+7B_w+13C_w=W$.

We will first convert $A_w$ in first equation to $5A_w$,

$5A_w+5B_w=\frac{5}{12}W$.

Then we will convert $C_w$ in second equation to $13C_w$,

$13B_w+13C_w=\frac{13}{16}W$.

Next we sum up the two,

$5A_w+18B_w+13C_w=\frac{59}{48}W$.

Lastly we will subtract the third equation from the last result thus eliminating $A_w$ and $C_w$,

$11B_w=\frac{11}{48}W$,

Or, $48B_w=W$.

B completes the work alone in 48 days.

**Answer:** Option b : 48 days

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

*--*

**Working together per unit time concept**

**Solving linear equations -- Efficient simplification.****Problem 3.**

If 90 men can do a certain job in 16 days working 12 hours a day, then the portion of the job that can be completed by 70 men working 8 hours day for 24 days is,

- $\displaystyle\frac{5}{8}$
- $\displaystyle\frac{1}{3}$
- $\displaystyle\frac{2}{3}$
- $\displaystyle\frac{7}{9}$

**Solution 3 : Problem analysis**

As an hour is the smallest unit of time and teams of men work for days and hours, we will convert all times into hours and will use the * manhours* concept instead of

*concept for designating work amount.*

**mandays****Solution 3 : Problem solving execution**

90 men can do a certain job in 16 days working 12 hours a day. So total work amount is,

$W=90\times{16}\times{12}$ manhours.

By the second statement, 70 men work 8 hours day for 24 days, thus completing an work amount,

$W_1=70\times{8}\times{24}$ manhours.

Thus $W_1$ as a portion of $W$ is,

$\displaystyle\frac{W_1}{W}=\frac{70\times{8}\times{24}}{90\times{16}\times{12}}$

$=\displaystyle\frac{7}{9}$.

**Answer:** Option d: $\displaystyle\frac{7}{9}$.

**Key concepts used: *** Mandays technique *--

*-- Unitary method -- Delayed evaluation -- Efficient simplification.*

**Manhours technique****Note:** We have not carried out the two multiplications and waited for the common factors to cancel out in the last division operation. This is delayed evaluation technique in action which is very useful in saving precious seconds in calculation.

**Problem 4.**

A, B and C can do a job in 30, 20 and 10 days respectively. A is assisted by B on one day and by C on the next day alternately. How long would the work take to finish?

- $8\displaystyle\frac{4}{13}$ days
- $3\displaystyle\frac{9}{13}$ days
- $9\displaystyle\frac{3}{8}$ days
- $4\displaystyle\frac{3}{8}$ days

**Solution 4 : Problem analysis**

Assuming $A_w$, $B_w$, and $C_w$ to be the work rates or portion of work $W$ done in a day by A, B and C respectively, from the given values we have,

$A_w=\frac{1}{30}W$,

$B_w=\frac{1}{20}W$, and

$C_w=\frac{1}{10}W$.

A works here continuously throughout the period up to the completion whereas B and C work on alternate days. In the absence of any other specific information, as B's work is memtioned first, we will assume the work started with A and B working together on the first day and A and C worked on the second day. This pattern of working is repeated for every two days.

In first two days then the total work done is,

$2A_w+B_w+C_w=W\left(\frac{1}{15}+\frac{1}{20}+\frac{1}{10}\right)=\frac{13}{60}W$.

If 60 were a multiple of 13 the work would have been finished in an even number of days. As it is not the case here, we have to use the **Boundary condition evaluation technique.**

**Solution 4 : Problem solving execution**

As $52=4\times{13}$ is the largest multiple of 13 (portions out of 60 in each 2 days), we can safely go up to the work done in $4\times{2}=8$ days as,

$\displaystyle\frac{52}{60}W$.

After 8 days then leftover work will be,

$\displaystyle\frac{8}{60}W$.

On the 9th day A and B will complete,

$A_w+B_w=W\left(\frac{1}{30}+\frac{1}{20}\right)=\frac{5}{60}W$.

On the start of the 10th day then work left will be,

$\displaystyle\frac{3}{60}W$.

Now we will apply the unitary method on the leftover work and rate of work of A and C working together, that is,

$A_w+C_w=W\left(\frac{1}{30}+\frac{1}{10}\right)=\frac{8}{60}W$.

At this rate to finish the leftover job of $\frac{3}{60}W$, A and C will take,

$\displaystyle\frac{3}{8}$ days.

So total duration for the job to be completed is, $9\displaystyle\frac{3}{8}$ days.

**Answer:** c: $9\displaystyle\frac{3}{8}$ days.

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

**Leftover work analysis -- Boundary determination technique -- Unitary method.****Note: **On 9th day itself the cumulative work done could have exceeded the whole work $W$. In that case, whole number of days would have been 8 and unitary method would have been applied on the 9th day with A and B working together.

**Problem 5.**

A, B, and C can do a work separately in 16, 32 and 48 days respectively. They started the work together but B left 8 days and C six days before completion of the work. What was the duration of completion of the work?

- 14 days
- 9 days
- 10 days
- 12 days

**Solution 5 : Problem analysis**

As this is a problem of three work agents working together for different durations we will as usual use the work rate technique assuming $A_w$, $B_w$, and $C_w$ as the portion of work $W$ done by A, B, and C respectively in 1 day.

By given information these work rates are,

$A_w=\frac{1}{16}W$,

$B_w=\frac{1}{32}W$, and

$C_w=\frac{1}{48}W$.

As B stopped working 8 days and C stopped 6 days before completion of the work we need to assume the duration of completion as an unknown number of $n$ days and work back for number of days B and C worked.

**Solution 5 : Problem solving execution**

A worked for the whole duration, B worked for $(n-8)$ days and C for $(n-6)$ days. So,

$W=nA_w+(n-8)B_w+(n-6)C_w$,

Or, $W=W\left(\frac{1}{16}n +(n-8)\frac{1}{32}+(n-6)\frac{1}{48}\right)$,

Or, $\frac{1}{16}n +(n-8)\frac{1}{32}+(n-6)\frac{1}{48}=1$.

Multiplying both sides by 96, the LCM of the denominators of the three fractions, 16, 32 and 48,

$6n + 3n - 24+2n-12 = 96$,

Or, $11n=132$,

Or, $n=12$.

In 12 days the work will be completed.

**Answer:** Option d: 12 days.

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

*.*

**Working backwards approach**#### Problem 6.

A can do one-fourth of a work in 10 days and B can do one-third of the work in 20 days. In how many days would A and B be able to complete the work while working together?

- 32 days
- 30 days
- 25 days
- 24 days

**Solution 6 : Problem analysis and execution**

If $A_w$ and $B_w$ be the per day portion of work $W$ done by A and B, we have,

$10A_w=\frac{1}{4}W$,

Or, $A_w=\frac{1}{40}W$

Similarly,

$20B_w=\frac{1}{3}W$,

Or, $B_w=\frac{1}{60}W$.

So working together portion of work done by A and B in 1 day will be,

$A_w+B_w=\left(\frac{1}{40}+\frac{1}{60}\right)W=\frac{1}{24}W$,

Or, $24(A_w+B_w)=W$.

So A and B working together will be able to complete the work in 24 days while working together.

**Answer:** Option d : 24 days.

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

**Working together per unit time concept**

**.**** Problem 7.**

A and B working separately can do a job in 9 and 15 days respectively. If they work for a day alternately, with A beginning, then the work will be completed in,

- 11 days
- 9 days
- 12 days
- 10 days

**Solution 7 : Problem analysis and execution**

By the given information portion of work $W$ done by A in 1 day is,

$A_w=\frac{1}{9}W$.

Similarly portion of work done by B in 1 day is,

$B_w=\frac{1}{15}W$.

As A and B work together with A beginning the work, every 2 days work done by the two will be,

$A_w+B_w=\left(\frac{1}{9}+\frac{1}{15}\right)W=\frac{8}{45}W$.

So after 10 days work done will be, $\frac{40}{45}W$ with $\frac{1}{9}W$ work left which will be completed by A on the next day.

So total duration for completion of the work is $10+1=11$ days.

**Answer:** Option a: 11 days.

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

**Working together per unit time concept -- Boundary determination technique.**** Problem 8.**

A takes 3 times as long as B and C together take to complete a job. B takes 4 times as long as A and C together take to do the same job. If all three working together complete the job in 24 days, A alone can do it in,

- 100 days
- 90 days
- 95 days
- 96 days

**Solution 8 : Problem analysis and execution**

With $A_w$, $B_w$ and $C_w$ as the work rates per day of A, B and C respectively we have,

$3A_w=B_w+C_w$,

$4B_w=A_w+C_w$, and

$24(A_w+B_w+C_w)=W$.

Substituting $3A_W$ for $B_w+C_w$ in equation 3,

$24(A_w+3A_w)=96A_w=W$.

So A alone will take 96 days to complete the work.

Useful pattern recognition makes second equation superfluous.

**Answer:** Option d: 96 days.

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

*--*

**useful pattern identification**

**efficient simplification.**

**Problem 9.**

15 men take 20 days to complete a job working 8 hours a day. The number of hours a day should 20 men take to complete the job in 12 days will be,

- 5 hours
- 10 hours
- 18 hours
- 15 hours

**Solution 9 : Problem analysis and execution:**

Total manhours required for the job is,

$W=15\times{20}\times{8}=2400$ manhours.

This will be equal to,

$W=20\times{12}=240$ mandays.

So the team of 20 men will have to work 10 hours a day for 12 days to complete the work.

**Answer:** Option b: 10 hours.

**Key concepts used:** **Mandays technique.**

** Problem 10.**

A and B can do a piece of work in 15 days. B and C together can do the same job in 12 days and C and A in 10 days. How many days will A take to complete the job?

- 8
- 13
- 24
- 40

**Solution 10 : Problem analysis and execution:**

Assuming $A_w$, $B_w$ and $C_w$ to be the work rates of A, B and C respectively in terms of portion of work $W$ done in a day,

$A_w+B_w=\frac{1}{15}W$,

$B_w+C_w=\frac{1}{12}W$, and

$C_w+A_w=\frac{1}{10}W$.

Subtracting first equation from second,

$C_w-A_w=\frac{1}{60}W$.

Now subtracting this result from the third equation,

$2A_w=W\left(\frac{1}{10}-\frac{1}{60}\right)=\frac{1}{12}W$,

Or, $24A_w=W$.

A completes the work alone in 24 days.

**Answer:** Option c: 24.

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

**Solving linear equations**

**.**### Useful resources to refer to

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

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