Time and work problems with solutions: SSC CGL Set 48

Submitted by Atanu Chaudhuri on Tue, 14/02/2017 - 20:44

Solutions Time and Work Problems for SSC CGL Set 48

Solution to Time and work problems Set 48

Learn to solve 10 Time and Work Problems in SSC CGL Solutions Set 48 in 12 mins. For best results, take the test first and then go through the solutions.

As all the problems are solved in quick time by problem solving techniques, to appreciate the solutions, you should take the test first at,

SSC CGL Question set 48 on Time and work.

Solutions to time and work problem set 48 for SSC CGL - time to solve was 12 minutes

Problem 1.

A alone can do a job in 6 days and B alone in 8 days, each working 7 hours a day. If they work together for 8 hours a day, how many days would they take to complete the job?

4 days
3 days
3.6 days
3.5 days

Solution 1 : Problem analysis:

The work agents do the work per day and then days are further stated in terms of varying hours per day. So the final time unit must be taken as an hour and the rate of work capacity for A and B will be assumed as $A_w$ and $B_w$ portions of work $W$ per hour. For two given statements we will get two equations in terms of $A_w$, $B_w$ and $W$. Combining the two we will have to form an equation in terms of $A_w+B_w$.

Solution 1 : Problem solving execution

A alone can do the job in 6 days 7 hour per day, that is, in $6\times{7}=42$ hours. So,

$42A_w=W$,

Or, $A_w=\displaystyle\frac{1}{42}W$.

B alone does the same job in 8 days 7 hours per day, that is, in 56 hours. So,

$56B_w=W$,

Or, $B_w=\displaystyle\frac{1}{56}W$.

Then,

$A_w+B_w=\displaystyle\frac{1}{14}\left(\displaystyle\frac{1}{3}+\displaystyle\frac{1}{4}\right)W$

Or, $14(A_w+B_w)=\displaystyle\frac{7}{12}W$,

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

This means A and B together will take 24 hours, or 3 days with 8 hours per day to complete the job.

Answer: b: 3 days.

Key concepts used: Work rate technique -- Working together per unit time concept -- Time unit selection.

The smallest time unit of hours is selected and all times were converted to hours. After solving for the number of hours required for doing the job together the hours required is converted back to days easily on dividing the hours by hours per day.

Problem 2.

Ravi and Bittu are working on an assignment. Ravi takes 6 hours to type 32 pages while Bittu takes 5 hours to type 40 pages. How long would they take to complete the typing assignment totaling 110 pages if they work on two separate typing machines?

7 hours 30 minutes
8 hours 25 minutes
8 hours
8 hours 15 minutes

Solution 2 : Problem analysis

Here also we will work with per hour typing rate in terms of pages typed by Ravi as $R_w$ and Bittu as $B_w$ and then sum up the two to get the per hour page typing rate of Ravi and Bittu working together. Lastly applying Unitary method we will get the required solution.

Solution 2 : Problem solving execution

Ravi types 32 pages in 6 hours. So,

$R_w=\displaystyle\frac{32}{6}=\frac{16}{3}$.

Bittu types 40 pages in 5 hours. So,

$B_w=8$.

$R_w+B_w=\displaystyle\frac{16}{3}+8=\displaystyle\frac{40}{3}$.

Ravi and Bittu then types $\displaystyle\frac{40}{3}$ pages in 1 hour,

Or, Ravi and Bittu types 1 page in $\displaystyle\frac{3}{40}$ hours,

Or, Ravi and Bittu types 110 pages in $\displaystyle\frac{330}{40}$ hours = 8 hours 15 minutes.

This is application of Unitary method in three statement steps with to-be-found-out quantity on the right of each statement.

Answer: Option d : 8 hours and 15 minutes .

Key concepts used: Work rate technique -- Working together per unit time concept -- Unitary method.

Problem 3.

A can do as much work as B and C together can do. A and B together can do a job in 9 hours 36 minutes and C can do it in 48 hours. In how much time (in hours) would B do the job working alone?

Solution 3 : Problem analysis

We will designate $A_w$, $B_w$ and $C_w$ as the work portion done by A, B and C respectively in unit time, that is, 1 hour and $W$ as the whole work amount. Eliminating $A_w$ and $C_w$ from the resulting equations we will derive $B_w$ in terms of $W$ which will give us number of hours B will take to complete the job.

This is use of work rate for each agent technique.

Solution 3 : Problem solving execution

A can do as much work as B and C together can do (in same time period, that may be 1 hour also). So,

$A_w=B_w+C_w$.

A and B together can do a job in 9 hours 36 minutes, or $\displaystyle\frac{48}{5}$ hours. So,

$\displaystyle\frac{48}{5}\left(A_w+B_w\right)=W$,

Or, $48A_w + 48B_w=5W$.

Using the first equation we substitute $A_w$ to eliminate it,

$48B_w+48C_w+48B_w=5W$,

Or, $96B_w+48C_w=5W$.

From third statement "C can do the job in 48 hours" we get now,

$48C_w=W$.

Using this in last equation,

$96B_w=4W$,

Or, $24B_w=W$.

So B does the job alone in 24 hours.

Answer: Option a: 24.

Key concepts used:Work rate technique -- Working together per unit time concept -- Linear equation solving.

Note: With a little care you should be able to solve the problem using the same concepts wholly in mind in a few tens of seconds.

Problem 4.

A can do a job in 12 days and B in 15 days. They work together for 5 days and then B leaves. The days taken by A to finish the remaining part of the job is,

Solution 4 : Problem analysis and execution

Assuming $A_w$ and $B_w$ to be the work portions done by A and B in 1 day and $W$ to be the whole amount of work,

$12A_w=W$,

Or, $A_w=\frac{1}{12}W$.

$15B_w=W$,

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

In 5 days work done by A and B working together then is,

$5(A_w+B_w)=5W\left(\frac{1}{12}+\frac{1}{15}\right)=\frac{3}{4}W$.

So leftover work is one-fourth the whole work $W$. A should do this in one-fourth of 12, that is, 3 days.

Answer: b: 3.

Key concepts used: Work rate technique -- Working together per unit time concept -- Leftover work analysis.

Problem 5.

A can do a piece of work in 8 days which B can destroy in 3 days. A has worked for 6 days during the last two days of which B has been destroying the work. How many days A must work alone to complete the job?

$8$
$7\frac{1}{3}$
$7$
$7\frac{2}{3}$

Solution 5 : Problem analysis

An interesting aspect in this problem is "B destroys the piece of work in 3 days". This is unusual. How to deal with this novel situation?

This reminds us of Pipes and Cisterns problems that are very similar to the Time and Work problems, the main difference being absence of draining pipe equivalent in Time and Work problems. In Pipes and Cisterns problems, a tank filling pipe fills up a portion of the tank in one time unit (which is the equivalent to portion of work done by a work agent in one unit of time). Usually in such problems, instead of tank filling rate, the number of time units a filling pipe requires to fill up the whole tank is given.

Additionally there are draining pipes that empty the tank at certan rates. It means that the positive work done by a filling pipe is destroyed or nullified by a draining pipe. When a filling pipe and a draining pipe work together, we would then need to subtract the per unit time drain from the per unit time fill to get the effective filling per unit time.

In the same way, in Time and work problems, when a work agent does positive work and another agent destroys the work, to get the effect of these two agents working together, we need to subtract the per unit time destruction of work portion from per unit time work portion done to arrive at effective per unit time work done. The concept of negative work is involved here.

Solution 5 : Problem solving execution

A can complete the piece of work in 8 days. So per day work portion done by A is,

$A_w=\frac{1}{8}W$, where $W$ is the whole work amount.

B destroys the same work in 3 days. So per day work portion destroyed by B, or negative work done by B is,

$B_w=\frac{1}{3}W$.

In 6 days A does work portion of,

$6A_w=\frac{3}{4}W$.

In the last 2 days negative work done by B is,

$2B_w=\frac{2}{3}W$.

Effective positive work done in 6 days is then,

$6A_w-2B_w=W\left(\frac{3}{4}-\frac{2}{3}\right)=\frac{1}{12}W$.

Leftover work is then,

$\displaystyle\frac{11}{12}W$.

Now we will apply unitary method in part.

Whole work $W$ is done by A in 8 days,

So, $\frac{11}{12}W$ work is done by A in $\frac{22}{3}$ or, $7\frac{1}{3}$ days.

Answer: Option b: $7\frac{1}{3}$ days.

Key concepts used: Work rate technique -- Working together per unit time concept -- Leftover work analysis -- Negative work concept.

Problem 6.

8 men can do a work in 12 days. After 6 days of work, 4 more men were engaged to finish the job. In how many days would the remaining work be completed?

Solution 6 : Problem analysis

Where a team of men do a job in days and number of men changes, we will use the mandays technique which is a rich time and work technique. This problem is eminently suitable for applying the simple but powerful mandays technique.

Solution 6 : Problem solving execution

8 men can do a work in 12 days. So the work amount done in 12 days is,

$W=8\times{12}=96$ mandays.

In 6 days 8 men do half the work (as, in 12 days the team does the whole work) and one half work, that is, 48 mandays work is left.

After 6 days total number men rises to 12. So to finish 48 mandays work left, 12 men will require an additional number of,

$\displaystyle\frac{48}{12}=4$ days.

Answer: Option c : 4.

Key concepts used: Mandays technique -- Work amount as Mandays concept -- Rich Time and Work concept -- Leftover work analysis.

Problem 7.

A, B and C can complete a work in 10, 12 and 15 days respectively. A left the work 5 days before the work was completed and B left 2 days after A had left. Number of days required to complete the whole work is,

$8\frac{2}{3}$
$6$
$7$
$6\frac{2}{3}$

Solution 7 : Problem analysis

In this problem as individual work agents complete the job, we will use the work rate technique and specify the work rates of A, B and C as, $A_w$, $B_w$ and $C_w$.

Solution 7 : Problem definition

In usual Time and work problems we do not need to think twice regarding what is actually wanted in the problem. The requirements normally are straightforward so that the steps to efficient solutions can be mapped out and executed without any delay.

In this problem though we feel the need to straighten out the goal requirement that will aid the process of quick and simple solution.

We find that 5 days before the completion of the work A leaves and 3 days before completion of the work B leaves. Unspoken is the fact that A, B, and C started working together and only C works up to the end.

So, if days to complete the work is $n$, A worked for $(n-5)$ days, B worked for $(n-3)$ days and C worked for $n$ days with their respective work rates as,

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

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

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

Solution 7 : Problem solving execution

In $n$ days the work $W$ is completed with A, B and C working for the above-mentioned number of days at their respective work rates. So,

$W\left(\displaystyle\frac{n-5}{10}+\displaystyle\frac{n-3}{12}+\displaystyle\frac{n}{15}\right)=W$,

Or, $n\left(\frac{1}{10}+\frac{1}{12}+\frac{1}{15}\right)=1+\frac{1}{2}+\frac{1}{4}=\frac{7}{4}$,

Or, $n\times{\frac{15}{60}}=\frac{7}{4}$,

Or, $n=7$.

Answer: Option c: $7$.

Key concepts used: Problem definition -- Working backwards -- Work rate technique -- Working together per unit time concept -- Efficient simplification.

Problem 8.

A can do a piece of work in 25 days and B can do the same work in 30 days. They work together for 5 days after which how much work would be left?

$\displaystyle\frac{11}{30}$
$\displaystyle\frac{12}{30}$
$\displaystyle\frac{19}{30}$
$\displaystyle\frac{1}{2}$

Solution 8 : Problem analysis and execution

In 5 days A will complete one-fifth, $\left(5\times{\displaystyle\frac{1}{25}}=\displaystyle\frac{1}{5}\right)$ and B one-sixth, $\left(5\times{\displaystyle\frac{1}{30}}=\displaystyle\frac{1}{6}\right)$ of the work; and together,

$\displaystyle\frac{1}{5}+\displaystyle\frac{1}{6}=\displaystyle\frac{11}{30}$th of the work.

Leftover will then be, $\displaystyle\frac{19}{30}$th of the work.

Answer: Option c: $\displaystyle\frac{19}{30}$.

Key concepts used: Work rate technique -- Working together per unit time concept -- useful pattern identification -- efficient simplification -- deductive reasoning.

Problem 9.

A's 2 days' work is equal to B's 3 days' work. If A can complete the work in 8 days, then to complete the work B will take,

16 days
14 days
15 days
12 days

Solution 9 : Problem analysis and execution:

A completes the work in 8 days. So A's 2 days' work is one-fourth the whole work. This work is done by B in 3 days. Now we will invoke unitary method based on Time and Work proportionality concept.

B does one-fourth of the work in 3 days.

So B does the whole work in $4\times{3}=12$ days.

Note: With work agents (and corresponding work rates) remaining constant, time on work is directly proportional to the amount of work completed and vice-versa.

Answer: Option d: 12 days.

Key concepts used: Work rate technique -- Time and Work proportionality -- Unitary method.

Problem 10.

A and B can do a piece of work in 20 days and 12 days respectively. A started the work alone and then after 4 days B joined to work together till the completion of the work. How long did the work last?

10 days
20 days
6 days
15 days

Solution 10 : Problem analysis and execution:

Till B joined, in 4 days A finished one-fifth of the work (time-work proportionality) so that four-fifths of work was left.

As A and B completes the work alone in 20 days and 12 days respectively, together in 1 day they will be able to complete,

$\frac{1}{20}+\frac{1}{12}=\frac{8}{60}=\frac{2}{15}$ portion of the work.

So to complete the remaining four-fifths of the work they will require an additional,

$\frac{15}{2}\times{\frac{4}{5}}=6$ days.

Total duration of the work is then, $4+6=10$ days

Answer: Option a: 10 days.

Key concepts used: Work rate technique -- Working together per unit time concept -- Time and Work proportionality -- Unitary method.

Note: For efficient solution to time and work problems clear understanding of four concepts, namely, Work rate concept, Working together concept, Time and work proportionality concept and Mandays concept and their uses should prove to be invaluable.