Time and work problems with solutions: SSC CGL Set 67

Submitted by Atanu Chaudhuri on Sun, 21/05/2017 - 14:30

Solutions Time and Work Problems for SSC CGL Set 67

Solutions to SSC CGL Time and Work Problems Set 67

Learn to solve 10 Time and Work Problems in SSC CGL Solutions Set 67 in 15 mins. Time and work and Time work wages problems are selected to test you well.

Contents are,

Solutions to 10 selected Time and work problems for SSC CGL. These can easily be used for bank POs and other competitive exams.
The problems cover Time and work, Time work wages and Pipes and cisterns that all are essentially time and work problems.
All solutions target solving quickly in mind.

For best results take the test first at,

SSC CGL level Question Set 67 on Time-Work Work-Wages Pipes-Cisterns 6.

Solutions to SSC CGL Time and Work Problems Set 67 - time to solve was 15 mins

Problem 1.

A property developer decided to build a small property in 40 days. He employed 100 men in the beginning and 100 more after 35 days and completed the construction in stipulated time. If he had not employed the additional men, how many days behind schedule he would have finished the construction?

Solution 1:Problem analysis and execution

If $M$ be the work rate as a portion of total work done by 1 man in 1 day, completion of the construction in stipulated time can be represented by,

$35\times{100}M+5\times{200}M=W$, where $W$ is the quantum of total work.

Rearranging to make $100M$ a factor in LHS (the other factor will then be the number of days required by 100 men to complete the work),

$35\times{100}M+5\times{200}M=W$,

Or, $45\times{100}M=W$.

It means, with 100 men throughout the period of construction, without the addition of 100 more men after 35 days, the construction would have been finished in 45 days, that is, 5 days later than the stipulated period of 40 days.

We have used the mandays rearrangement technique in rearranging the mandays in the LHS suitably.

Answer: b: 5.

Key concepts used: Basic time and work concepts -- Work rate technique -- Time work scenario analysis -- Work amount as mandays concept -- Mandays rearrangement technique -- Mathematical reasoning -- Many ways technique.

Solution 1: Based on mathematical reasoning

For 35 days 100 men worked, but for the last 5 days 200 men worked, thus completing the work. The situation is then equivalent to existing 100 men working for 40 days, and another team of 100 men working for 5 days.

Naturally then, if the extra 100 men were not provided for the last 5 days, the existing 100 men would have had to work for another 5 days more than 40 days to complete the work.

Problem 2.

8 men and 12 women complete a certain piece of special job in 9 days. If each of the men takes twice the time taken by a woman to finish the special job, in how many days will 12 women finish the same job?

Solution 2: Problem analysis and execution

If $M$ and $F$ are the portion of the job done by a man and a woman respectively in 1 day, by the first condition,

$9(8M+12F)=W$, where $W$ is the quantum of the job in work units.

But as a man takes twice the time a woman takes to complete the job, effectively work rate of a woman is double that of a man,

$F=2M$,

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

Substituting,

$9(8M+12F)=W$,

Or, $9(4F+12F)=W$,

Or, $144F=W$,

Rearranging this total work amount expressed in terms of womandays,

$144F=W$,

Or, $12\times{12}F=W$.

So in 12 days 12 women will complete the job.

Answer: d: 12.

Key concepts used: Basic time and work concepts -- Work rate technique -- Work agents working together per unit time concept -- Work agent equivalence -- Work amount as mandays -- Mandays rearrangement technique -- Mathematical reasoning -- Many ways technique.

Solution 2: By mathematical reasoning

As work rate of a man is half the work rate of a woman, the team of 8 men is equivalent in this case to a team of 4 women. So we can say, effectively the work amount is $4+12=16$ women working for 9 days, that is, 144 womandays.

By manday rearrangement technique, rearranging the factors in the total womandays of 144 we can express, $144=12\times{12}$, that is, 12 women working for 12 days.

Problem 3.

A cistern is normally filled in 8 hrs. After a leak at the bottom has developed, it takes now 2 hrs longer to fill it. How many hours would it take for the cistern to become empty from a fully filled condition with all inlet pipes closed?

Solution 3: Problem analysis and execution

If $P$ and $L$ be the fill rate by inlet pipes and drain rate by the leak as portions of the cistern per hour respectively, when there was no leak,

$8P=T$, where $T$ is the cistern capacity.

After the leak has developed, it takes 10 hrs to fill the cistern. So,

$10(P-L)=T$.

From the first equation,

$P=\displaystyle\frac{1}{8}T$.

Substituting this in the second equation,

$\displaystyle\frac{5}{4}T-10L=T$,

Or, $10L=\displaystyle\frac{1}{4}T$,

Or, $40L=T$.

It means, the leak working alone will empty the full cistern in 40 hours.

Answer: Option a: 40.

Key concepts used: Fill rate equivalent to Work rate technique -- Cistern capacity as pipe-hours equivalent to Work amount as Mandays concept -- Mathematical reasoning -- Many ways technique.

Solution 3: By Mathematical reasoning

Normally the inlet pipes take 8 hrs to fill the cistern without leak. In 10 hrs then the inlet pipes would have filled up $\displaystyle\frac{10}{8}=\frac{5}{4}$th of the cistern without the leak. The extra filling would then be one-fourth of the cistern. This amount must have been drained by the leak in 10 hrs, so that at the end of 10 hrs, we get exactly filled cistern.

So to empty the full tank, the leak will take 4 times 10 hrs, that is, 40 hrs.

Note: This is what we call concept based mathematical reasoning. This path creates more pressure on the mind as you are not writing down any intermediate steps, and also because in general carrying out logical reasoning in mind creates an extra pressure on mind.

The first efficient solution is method based, partly conceptual partly mechanical, while the second one is fully conceptual and fastest if you have control.

With concept and method, large and complex problems can be solved with ease and efficiency, though intuitional or conceptual strength may enable one to skip many steps to the solution.

Problem 4.

If two pipes are opened together, a tank is filled in 12 hrs. If one pipe fills the tank 10 hrs faster than the other, how many hrs would it take for the faster pipe to fill the tank?

Solution 4 : Problem analysis and execution

If $A$ and $B$ be the portions of tank filled per hr by the faster and the slower pipe respectively, according to the first given condition,

$12(A+B)=T$, where $T$ is the tank capacity (equivalent to amount of work),

Or, $A+B=\displaystyle\frac{1}{12}T$.

By the second condition,

$xA=T$, where the faster pipe is assumed to fill the tank in $x$ hrs

Or, $A=\displaystyle\frac{1}{x}T$.

Similarly,

$(x+10)B=T$,

Or, $B=\displaystyle\frac{1}{x+10}T$.

Substituting values $A$ and $B$ in first equation,

$A+B=\displaystyle\frac{1}{12}T$

Or, $\displaystyle\frac{1}{x}T+\displaystyle\frac{1}{x+10}T=\displaystyle\frac{1}{12}T$.

Cancelling out $T$, transposing and rearranging we have,

$x^2+10x=24x+120$,

Or, $x^2-14x-120=0$,

Or, $(x-20)(x+6)=0$.

As $x \neq -6$,

$x=20$.

The faster pipe fills the tank alone in 20 hrs.

Answer: c: 20.

Key concepts used: Basic pipes and cisterns concepts -- Fill rate equivalent to Work rate technique -- Working together per unit time concept -- Basic algebra concepts -- Factorization of quadratic equation -- Free resource use principle -- Choice value set test -- Many ways technique.

Solution 4: By use of choice values for trials, thus avoiding solving a quadratic equation

The two pipes fill the tank in 12 hrs and so, in 1 hr fill $\displaystyle\frac{1}{12}$th of the tank. This fraction must then be equal to sum of two fractions, $\displaystyle\frac{1}{x}$ and $\displaystyle\frac{1}{x+10}$, where $x$ is the hrs in which the faster pipe fills the tank.

We can form the required relation in mind,

$\displaystyle\frac{1}{x}+\displaystyle\frac{1}{x+10}=\displaystyle\frac{1}{12}$.

Using the free resource of the choice values, we find 35 and 45 to form a pair with no factor of 2, so sum of inverses of these two numbers cannot form $\displaystyle\frac{1}{12}$. The same reasoning will hold for 45 and 55.

For the third value of 20 as $x$ when we try 20 and 30, sum of inverses formed is $\displaystyle\frac{1}{12}$.

Problem 5.

If 28 women complete $\frac{7}{8}$ portion of a job in 7 days, then the number of women required to get the remaining work completed in another 7 days is,

Solution 5: Problem analysis and exceution

Assuming a woman completes $F$ portion of the job in 1 day, by the first condition,

$7\times{28F}=\displaystyle\frac{7}{8}W$, where $W$ is the quantum of total job.

Applying mandays rearrangement technique, rearranging the factors in the equation,

$7\times{28F}=\displaystyle\frac{7}{8}W$,

Or, $7\times{4F}=\displaystyle\frac{1}{8}W$, the rest of the work.

4 women then will complete the rest of $\displaystyle\frac{1}{8}$th of the work in another 7 days.

Answer: Option b: 4.

Key concepts used: Work rate technique -- Mandays rearrangement technique -- Mathematical reasoning -- Many ways technique -- Work to worker proportionality.

Solution 5: By mathematical reasoning

Rest of the job yet to be done is $\displaystyle\frac{1}{8}$th of the whole job and $\displaystyle\frac{1}{7}$th of the job done by 28 women in 7 days.

As amount of job done is directly proportional to number of workers if period of work is fixed, to complete this one-seventh of job done by 28 women in same period of 7 days, one-seventh of 28, that is, 4 women will be required.

Problem 6.

Two pipes A and B fill a tank in 24 hrs and 32 hrs respectively. If both pipes are opened together, when should the first pipe be turned off so that the tank is fully filled in exactly 16hrs?

14 hrs
10 hrs
8 hrs
12 hrs

Solution 6 : Problem analysis and execution

If $A$ and $B$ be the portion of tank filled up by pipe A and B respectively in 1 hr, by the first two conditions,

$24A=T$, where $T$ is the tank capacity,

Or, $A=\displaystyle\frac{1}{24}T$.

And,

$32B=T$,

Or, $B=\displaystyle\frac{1}{32}T$.

With $x$ as the desired number of hrs after which the first pipe is turned off, by the third condition,

$xA+16B=T$,

Or, $\displaystyle\frac{xT}{24}+\displaystyle\frac{16T}{32}=T$, value of $A$ and $B$ substituted

Or, $x=24\left(1-\displaystyle\frac{1}{2}\right)=12$ hrs.

The first pipe is to be closed after 12 hrs.

Answer: Option d: 12 hrs.

Key concepts used: Fill rate equivalent to Work rate technique -- Working together per unit time concept -- Mathematical reasoning -- Many ways technique -- Work time to work done proportionality.

Solution 6: By Mathematical reasoning

For 16 hrs the pipe B runs, and this period being half of 32 hrs, the total tank filling time, pipe B fills up half of the tank. To fill up the rest half of the tank, the first pipe must take then, half of 24 hrs, that is, 12 hrs.

We have used here the concept of Work time to work done direct proportionality concept, with number of workers fixed (here, fill time to fill done direct proportionality)

If a pipe takes 24 hrs to fill a tank, it will take 8 hrs to fill one-third of the same tank.

Problem 7.

A tank is normally filled in $3\displaystyle\frac{1}{2}$ hrs, whereas with a leak it takes half an hour longer to fill it up. How long would it take to get the full tank emptied by the leak with all inlet pipes closed?

24 hrs
32 hrs
28 hrs
36 hrs

Solution 7: By Mathematical reasoning

It takes normally $3\frac{1}{2}=\frac{7}{2}$ hrs to fill the tank without leak, and 8 hrs to fill the tank with leak, which is half an hour more than the normal fill time.

In this extra time of half an hour, which is one-seventh of $\frac{7}{2}$ hrs, one-seventh more of tank capacity will be filled up by the inlet pipes if the leak were not there.

As the inlet pipes together with the leak fill the tank exactly in 4 hrs, in this duration of 4 hrs, the leak must have emptied the extra fill of one-seventh tank capacity by the inlet pipes.

By the concept of work time to work done proportionality, (and by unitary method), the leak will then take seven times of duration of 4 hrs, that is, 28 hrs to empty a full tank with all inlet pipes closed.

Solution 7: By Work rate (fill rate) technique

Assuming fill rate $P$ as portion of tank filled by the inlet pipes in 1 hr and drain rate $L$ as portion of tank drained in 1 hr by the leak, in the case of no leak we have,

$\displaystyle\frac{7}{2}P=T$, where $T$ is the tank capacity,

Or, $P=\displaystyle\frac{2}{7}T$.

Similarly when the tank is filled in 4 hrs with leak active,

$4(P-L)=T$,

Or, $\displaystyle\frac{8}{7}T-4L=T$, substituting value of $P$,

Or, $4L=T\left(\displaystyle\frac{8}{7}-1\right)=\displaystyle\frac{T}{7}$,

Or, $28L=T$,

which means the leak will take 28 hrs to empty the full tank with all inlet pipes closed.

Answer: Option c: 28 hrs.

Problem 8.

Three persons agree to do a job in exchange of Rs.660. The first two complete $\displaystyle\frac{7}{11}$th portion of the job and the third person then finshes the rest of the job working alone. How much should the third person get?

Rs.200
Rs.240
Rs.160
Rs.220

Solution 8: By Mathematical reasoning

As share of earning is proportional to share of work done, the third person having done $\frac{4}{11}$th of the job should get $\frac{4}{11}$th of the total earning of Rs.660, that is, Rs.240.

The statement fragment regarding other two persons have little bearing in the solution except that they had finished some portion of the job.

Answer: Option b: Rs.240.

Key concepts used: Earning share concept -- Earning to work done proportionality.

Problem 9.

A person can do a work alone in 10 days, but can do it in 6 days with the help of a second person. In completing the job together, the two persons earned Rs.50. What is the share of the second person?

Rs.40
Rs.10
Rs.20
Rs.30

Solution 9: By Mathematical reasoning

The first person does the whole job alone in 10 days, and so in 6 days he does, $\frac{6}{10}=\frac{3}{5}$th of the work while the second person does the rest $\frac{2}{5}$th of the work.

We have used the work time to work done proportionality concept.

Earning share being proportional to work done, the second person then gets $\frac{2}{5}$th of the total earning of Rs.50, that is, Rs.20.

Solution 9: Concept based deductive

Let $A$ and $B$ be the work rates as portion of work done per day for the first and second person respectively. When the first person thus completes the job in 10 days,

$10A=W$, where $W$ is the work amount,

Or, $A=\displaystyle\frac{1}{10}W$.

When the two together complete the job in 6 days,

$6(A+B)=W$,

Or, $\displaystyle\frac{6}{10}W+6B=W$,

Or, $6B=\displaystyle\frac{2}{5}W$.

This is the work portion done by the second person in 6 days.

As earning share is proportional to the work done, the second person will get an amount two-fifth of total earning of Rs.50,

$\text{Amount B gets}=\displaystyle\frac{2}{5}\times{50}=\text{Rs.}20$.

Answer: Option c: Rs.20.

Key concepts used: Work time to work done proportionality -- Earning share concept -- Earning to work done proportionality -- Mathematical reasoning.

Problem 10.

A job is normally completed in 100 days by a team of workers. But as 10 workers were absent, the rest took 10 more days to complete the job. What was the number of workers in the original team?

Solution 10: By Mathematical reasoning using work time to worker inverse proportionality and basic ratio concepts

By work time to number of worker inverse proportionality concept, the ratio of,

$\displaystyle\frac{\text{Original number of workers}}{\text{Reduced number of workers}}=\frac{\text{Increased completion time}}{\text{Original completion time}}$

$=\displaystyle\frac{110}{100}$.

It means, if original number of workers were 55, 66, 110, 220, 330 or any multiple of 11, reduced number of workers must be 50, 60, 100, 200, 300 or the same multiple of 10, with the ratio maintained at 110 : 100 = 11 : 10.

With the additional information that diffference in number of workers of two teams is exactly 10, we can conclude the values of original nyumber of workers as 110 and reduced number of workers as 100.

This is use of basic number system concepts along with basic ratio concepts.

Solution 10: Concept based deductive

Assuming number of original workers as $x$ and work rate of 1 worker as portion of job done in 1 day as $M$, in the original case,

$100xM=W$, where $W$ is the work amount.

With original team strength reduced by 10 workers the work is completed in 110 days. So,

$110(x-10)M=W$.

Dividing and transposing we get,

$10x-1100=0$,

Or, $x=110$.

A safe and steady way to reach the solution without much waste of time.

Answer: Option b: 110.

Key concepts used: Work rate technique -- Work amount as mandays concept -- efficient simplification -- Mathematical reasoning -- Worker to work time inverse proportionality -- Basic number domain concepts -- Basic ratio concepts.

Note: To us the first method is more attractive, but you choose your path.