SSC CGL level Solution Set 66 on Time-Work Work-Wages Pipes-Cisterns 5

66th SSC CGL level Solution Set, 5th on topics Time-Work, Work-Wages, and Pipes-Cisterns

ssc cgl level solution set 66 time-work work-wages pipes-cisterns 5

This is the 66th solution set of 10 practice problem exercise for SSC CGL exam and 5th 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 level Question Set 66 on Time-Work Work-Wages Pipes-Cisterns 5.


66th solution set - 10 problems for SSC CGL exam: 5th on topics Time-Work Work-Wages Pipes-Cisterns - time 12 mins

Problem 1.

A pipe takes 36 hours more than its normal time of filling a cistern due to a leak at the bottom of the cistern. If the pipe inflow rate is double the leak outflow rate, how many hours would the pipe alone take to fill the cistern completely without the leak?

  1. 30
  2. 18
  3. 24
  4. 36

Solution 1: Based on mathematical reasoning

Effective rate of filling the cistern with both the pipe and and leak active is half the normal rate of filling the cistern by the pipe alone (as rate of filling is directly proportional to rate of effective inflow, and subtractive leak outflow is half the pipe inflow).

Given: at half the normal inflow rate, the cistern is filled up in normal filling time plus 36 hours.

As rate of inflow is inversely proportional to time to fill (the more the inflow, the less the time to fill, and less the inflow more the time to fill), at half the normal rate of inflow the time taken to fill must be double the normal time taken, which is, normal time plus 36 hours.

Thus normal time to fill = 36 hours.

Solution 1: Efficient conceptual deductive solution

If $P$ and $L$ be the portion of tank filled and emptied by the pipe and the leak per hour respectively,

$PH=T$,

where $H$ is the normal filling hours with the pipe working alone and $T$ is the capacity of the cistern.

With pipe and leak working together,

$(H+36)(P-L)=T$,

Or, $P(H+36)=2T$, as $L$ is half of $P$,

Or, $PH + 36P=2T$,

Or, $T+36P=2T$,

Or, $36P=T$.

Note: This also is an efficient solution based on efficient concept based simple deductions, but solution through mathematical reasoning in most cases is faster than any other path to the solution. Usually solution through mathematical reasoning needs more control on conceptual reasoning. If you have the confidence in conceptual reasoning, always take that path; it will be the fastest.

Answer: d: 36.

Key concepts used: Basic pipes and cisterns concepts -- Inflow to fill time inverse proportionality -- Mathematical reasoning -- Filling or emptying rate technique that are equivalent to Work rate technique -- Filling and emptying pipes working together concept that are equivalent to Working together per unit time concept.

Problem 2.

A and B working alone 7 hours a day can complete a job in 8 days and 6 days respectively. In how many days would they finish the job if they work together 8 hours per day?

  1. 2.5
  2. 3
  3. 3.6
  4. 4

Solution 2: Problem analysis and execution

Taking $A$ and $B$ as portion of the job completed by A and B respectively in 1 hour, the smallest time unit of work, we have,

$56A=42B=W$, where $W$ is the quantum of work in work units.

If working together 8 hours a day they finish the job in $d$ days,

$8d(A+B)=W$.

From the given conditions,

$A=\displaystyle\frac{W}{56}$, and

$B=\displaystyle\frac{W}{42}$.

So when working together,

$8d(A+B)=W$,

Or, $8d\left(\displaystyle\frac{W}{56}+\displaystyle\frac{W}{42}\right)=W$,

Or, $d\left(1+\displaystyle\frac{4}{3}\right)=7$, multiplying both sides by 7, taking 8 inside the brackets and cancelling $W$,

Or, $\displaystyle\frac{7d}{3}=7$,

Or, $d=3$.

Answer: b: 3.

Key concepts used: Basic time and work concepts -- Work rate technique -- Smallest work time unit concept -- Work time unit selection -- Work agents working together per unit time concept -- Basic algebra concepts.

Problem 3.

If 8 men working together 8 hours per day for 8 days produce 8 units of work, the units of work produced by 4 men working together 4 hours per day for 4 days is,

  1. $1$
  2. $4$
  3. $\displaystyle\frac{1}{2}$
  4. $\displaystyle\frac{1}{4}$

Solution 3: Problem analysis and execution

Following smallest work unit concept, if work units produced by 1 man in 1 hour is $M$, according to the given information,

$8\times{8}\times{8}M=8$ work unit,

Or, $M=\displaystyle\frac{1}{8^2}$ work unit.

So when 4 men work together 4 hours per day for 4 days producing $x$ units of work (let's assume), we have,

$4\times{4}\times{4}M=x$,

Or, $x=\displaystyle\frac{4^3}{8^2}$,

Or, $x=\displaystyle\frac{64}{64}=1$.

Answer: Option a: 1.

Key concepts used: Work rate technique -- Work amount as Mandays concept (or manhours concept) -- Work time unit selection by which we select the smallest time unit for measuring the amount of work units in terms of mandays (or manhours here).

Problem 4.

A can do a job in 20 days, while A and B together do it in 15 days. If they are paid Rs.400 for the job, what is the share of B?

  1. Rs.100
  2. Rs.250
  3. Rs.50
  4. Rs.200

Solution 4: Problem analysis and execution

If $A$ and $B$ be the portions of work completed per day by A and B respectively, according to the first given condition,

$20A=W$, where $W$ is the amount of work,

Or, $A=\displaystyle\frac{1}{20}W$.

Similarly by the second given condition, A and B working together,

$15(A+B)=W$,

Or, $\displaystyle\frac{15W}{20}+15B=W$,

Or, $15B=W\left(1-\displaystyle\frac{15}{20}\right)=\displaystyle\frac{1}{4}W$,

Or, $60B=W=20A$.

Or, $A=3B$.

So A works three times of B; when A completes 3 units, B completes 1 unit, together completing 4 units of work.

Concept: Share of earning is directly proportional to share of work done.

So out of Rs.400 worth of work done, A will get 3 portions equal to Rs.300 and B will get 1 portion equal to Rs.100.

Answer: a: Rs.100.

Key concepts used: Basic work and wages concepts -- Work rate technique -- Working together per unit time concept -- Earning to work done proportionality.

Problem 5.

A cistern is filled up by two inlet pipes, working separately, in 30 mins and 20 mins respectively, while it is filled up in 18 mins with both pipes working together but the second pipe closed off after some time. After how much time is the second pipe closed?

  1. 12 mins
  2. 8 mins
  3. 5 mins
  4. 10 mins

Solution 5: Problem analysis and exceution

Assuming $A$ and $B$ to be the portion of cistern filled up per minute by the first and the second pipe respectively, we have from the first two conditions,

$30A=T$, where $T$ is the cistern capacity,

Or, $A=\displaystyle\frac{1}{30}T$, and

$20B=T$,

Or, $B=\displaystyle\frac{1}{20}T$.

If the second pipe is closed after $x$ mins from the start, first pipe works for full 18 mins while the second pipe works for $x$ mins only, in filling up the cistern in a total time of 18 mins. Thus in the case of two pipes working together,

$18A+xB=T$,

Or, $\displaystyle\frac{18}{30}T+\displaystyle\frac{x}{20}T=T$,

Or, $\displaystyle\frac{x}{20}=1-\displaystyle\frac{3}{5}=\displaystyle\frac{2}{5}$,

Or, $x=8$ mins.

Answer: Option b: 8 mins.

Key concepts used: Inlet pipe filling rate technique (in terms of portion of total cistern capacity) equivalent to work rate technique -- Filling pipes working together concept equivalent to Working together per unit time concept -- Event sequencing.

Problem 6.

Working 9 hours a day for 20 days a team of 8 men finish a job completely. How many days would a team of 7 men take to complete the same job working for 10 hours a day?

  1. $20\displaystyle\frac{4}{7}$
  2. $20\displaystyle\frac{1}{2}$
  3. $20\displaystyle\frac{3}{5}$
  4. $20\displaystyle\frac{3}{7}$

Solution 6: Problem analysis and execution

An hour being the smallest time unit we will define work rate of 1 man in terms of work portion done by 1 man in 1 hour, as well as work amount in terms of manhours.

If $M$ be the work portion done by 1 man in 1 hour and $W$ be the total work amount in manhours, by the first condition,

$8\times{20}\times{9}M=W$,

Or, $M=\displaystyle\frac{W}{8\times{20}\times{9}}$

And for the target condition with $x$ as the desired number of days,

$7\times{x}\times{10}M=W$,

Or, $W\displaystyle\frac{7\times{x}\times{10}}{8\times{20}\times{9}}=W$, value of $M$ substituted

Or, $x=\displaystyle\frac{144}{7}=20\displaystyle\frac{4}{7}$ days.

Answer: Option a : $20\displaystyle\frac{4}{7}$.

Key concepts used: Work rate technique -- Work time unit selection; work rate for work agents as well as work amount is to be defined in terms of smallest time unit -- Work amount in terms of man hours; a variation of mandays concept.

Problem 7.

Three inlet pipes to a cistern with diameters $2$cm, $1$cm and $1\displaystyle\frac{1}{3}$cm fill the cistern with inflow rate of each pipe poroportional to the square of its diameter. If the pipe with the largest diameter can fill the cistern working alone in 61 mins, how much time will be taken by the three pipes to fill the cistern working together?

  1. 36 mins
  2. 32 mins
  3. 38 mins
  4. 34 mins

Solution 7: Problem analysis and execution

First let us get the proportion of rates of inflow of the three pipes as proportions of the square of their diameters as,

$2^2 : 1^2 : \displaystyle\frac{4^2}{3^2}$,

Or, $4 : 1 : \displaystyle\frac{16}{9}$,

Or, $36 : 9 : 16$.

Here, the first pipe of diameter 2cm is the largest pipe and it fills up the cistern in 61 mins.

If $A$, $B$ and $C$ be the filling rates of the three pipes respectively in terms of portion of cistern capacity filled up per minute,

$A: B : C=36 : 9 : 16=36p : 9p : 16p$,

where $p$ is the canceled out HCF reintrodued as per basic ratio concepts and $36p$, $9p$ and $16p$ are the actual values of $A$, $B$ and $C$.

The ratio of fill rates $A:B:C$ could be directly equated to the ratio of inflow rates $36:9:16$ because, inflow rate is directly proportional to the fill rate.

In the first instance the largest pipe with per minute filling rate $A$ fills up the cistern in 61 mins,

$61A=T$, where $T$ is the cistern capacity,

Or, $61\times{36p}=T$.

In the second case of all three pipes working together, assuming the three pipes fill up the whole cistern in $x$ mins,

$x(A+B+C)=T$,

Or, $x(36p+9p+16p)=61\times{36p}$,

Or, $x(61p)=61\times{36p}$,

Or, $x=36$ mins.

Answer: Option a: 36 mins.

Key concepts used: Rate of inflow proportional to square of diameter of pipe (circular area of cross-section being, $\text{Area} =\pi{r^2}=\pi{\frac{d^2}{4}}$, where $r$ is the radius and $d$ the diameter of the pipe) -- Inflow to fill rate proportionality -- The rates of inflow being in ratio, and each rate of inflow being proportional to the filling rate as portion of cistern filled up per min, the ratio of filling rates can be directly equated to the ratio of rates of inflow -- getting actual values of filling rates from the ratio by reintroduction of canceled out HCF according to Basic ratio concepts -- Pipes working together per unit time concept with sum of filling rates multiplied by the time of filling producing the total capacity of the cistern.

Note: Using a basic and rich concept based efficient methodology, complex problems can be solved quickly without any confusion. Here both the concepts and the methodology processes are important.

Problem 8.

In 18 days, working together, A, B and C earn a sum of Rs.2700. If on the same job, B and C together earn Rs.1520 in 20 days and, C and A together earn Rs.940 in 10 days, how much does C earn in a day?

  1. Rs.40
  2. Rs.15
  3. Rs.10
  4. Rs.20

Solution 8: Problem analysis and execution based on concept of daily wage

Concept of Daily wage

Earning per day by a worker is the daily wage and it is directly proportional to the worker's work capacity, or the work done as a portion of total work per day. Just as Work capacity or work rate, Daily wage of a worker is fixed.

Concept: Earning together: When a few workers work together to get a job done and earn a total amount, each worker earns an amount proportional to work done by him (or her). Earning of a worker is a product of daily wage and number of days of work (daily wage for the worker is fixed just as the work rate and it is proportional to work rate).

Concept: Earning share is directly proportional to share of work done and so is proportional to the work capacity of the worker and also to the daily wage.

When A, B an C earns Rs.2700 working together for 18 days, assuming $A$, $B$ and $C$ to be the earning per day (or the daily wage) of A, B and C respectively,

$18(A+B+C)=2700$,

Or, $(A+B+C)=150$.

Similarly,

$20(B+C)=1520$, and

$10(C+A)=940$,

Or, $20(C+A)=1880$.

Adding the last two,

$20(A+B+2C)=3400$,

Or, $(A+B+2C)=170$.

Subtracting $(A+B+C)=150$, from this last result we get,

$C=20$.

Answer: Option d: 20.

Key concepts used: Daily wage concept -- Earning share concept -- Earning together concept -- Efficient simplification.

Problem 9.

Working 6 hours a day, 12 pumps can empty a completely filled reservoir in 15 days. How many such pumps working 9 hours a day can empty the same completely filled reservoir in 12 days?

  1. 10
  2. 9
  3. 15
  4. 12

Solution 9: Problem analysis and execution

Smallest work time unit is 1 hour and worker unit is 1 pump.

So assuming $P$ to be the portion of reservoir emptied by 1 pump in 1 hour or in 1 pump-hour, by the first given condition,

$15\times{6}\times{12}P=R$, where $R$ is the reservoir capacity,

Or, $P=\displaystyle\frac{R}{15\times{6}\times{12}}$.

If $x$ be the number of pumps required, by the second condition we have,

$12\times{9}\times{x}P=R$,

Or, $Rx\displaystyle\frac{12\times{9}}{15\times{6}\times{12}}=R$,

Or, $x=\displaystyle\frac{15\times{6}\times{12}}{12\times{9}}=10$.

So 10 pumps will be required to meet the given conditions.

Answer: Option a: 10.

Key concepts used: Smallest Work time unit selection -- pump-hours concept equivalent to work amount as mandays concept.

Problem 10.

A cistern is filled by three inlet pipes A, B, and C working together for 6 hours. Once, after the three work together for 2 hours, the pipe C is closed and then on, the pipes A and B fill the remaining part in 7 hours. How much is then the time taken by C alone to fill the cistern?

  1. 16 hours
  2. 10 hours
  3. 12 hours
  4. 14 hours

Solution 10: Problem analysis and execution

Let's assume $A$, $B$ and $C$ to be the portion of cistern filled in 1 hour by A, B and C respectively.

By the first statement then,

$6(A+B+C)=T$, where $T$ is the capacity of the cistern,

Or, $3A+3B+3C=\displaystyle\frac{1}{2}T$

By the second statement,

$2C+9(A+B)=T$, A and B worked for $2+7=9$ hours

Or, $3A +3B+\displaystyle\frac{2}{3}C=\displaystyle\frac{1}{3}T$.

Subtracting,

$C\left(3-\displaystyle\frac{2}{3}\right)=T\left(\displaystyle\frac{1}{2}-\displaystyle\frac{1}{3}\right)$,

Or, $\displaystyle\frac{7}{3}C=\displaystyle\frac{1}{6}T$,

Or, $14C=T$.

So C alone fills the cistern in 14 hours.

Answer: Option d: 14 hours.

Key concepts used: Filling rate or work rate in terms of portion of cistern filled up in 1 hour equivalent to Work rate technique -- Filling or work agents working together per unit time -- efficient simplification.


Useful resources to refer to

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