How to solve arithmetic boundary condition problems in a few simple steps

Boundary condition problems occur in Time and Work, Pipes and Cisterns, Average and Speed time Distance

efficient-solution-of-boundary-condition-problems-aritmetic

In a boundary condition problem,

A simple or compound event (consisting of more than one event) goes on happening systematically at regular intervals till a boundary point just before the end. With each regular occurrence of the event, problem parameters change. After crossing the boundary though, end is reached before the full impact of the event along with the change can happen. To solve such a problem, this boundary is to be determined. Occasionally, boundary coincides with the end.

Boundary condition problem example 1 from Time and Work

In 42 days 40 men complete a work. As it happened, instead of all of them working together to finish the job, they started working together, but at the end of every 10th day 5 men left. In how many days would then the work be completed?

  1. 61 days
  2. 63 days
  3. 60 days
  4. 65 days

Solution example 1

This is a Boundary condition problem in which we need to carefully detect the boundary, because no straightforward formula can produce the answer directly. By boundary we mean the last event point up to which a regular event goes on happening unchanged. In this case, the regular event is reducing workforce by 5 men every 10 days.

Crossing over the last point, the nature of event changes. We will be clear about this boundary concept as we proceed.

Conventional solution shows nature of the problem clearly

1st 10 days, 40 men worked, work done = 400 mandays,

2nd 10 days, 35 men worked, work done = 350 mandays,

3rd 10 days, 30 men worked, work done = 300 mandays,

4th 10 days, 25 men worked, work done = 250 mandays, till now in 40 days 1300 mandays work is done, while total work is 1680 mandays. We can proceed safely to the 5th segment of 10 days. This step takes care of unwittingly overshooting the target.

5th 10 days, 20 workers worked, work done = 200 mandays, till now total of 1500 mandays work done, work left is 180 mandays which will be worked upon by 15 men. We have to consider then the 6th segment of 10 days.

6th 10 days, 15 workers worked, work done = 150 mandays, a total of 1650 mandays work done till now, 30 mandays work left with number of workers 10.

This 6th and last full segment of 10 days is the boundary in this problem. Till this boundary, worker teams, even in regularly reduced numbers, worked for the full segment period of 10 days. Crossing this boundary, the full period is broken and job is finished in 3 more days.

Answer: b: In 63 days the work will then be done.

Work measure in mandays is useful here. But solution is procedural, step by step and is slow, but it shows the difficulty in arriving at the answer directly by a formula.

Faster solution by mathematical reasoning applying average of natural number concept

As 40 men working together all through take 42 days to complete the job, and in the second case, every 10 days worker force reduces by 5, we can safely test for 50 days work portion completion. This is called an Educated guess.

In MCQ based tests, such an estimate based on Educated guess, on which a trial will be made, is very useful.

In 50 days, as 5 persons are going off every 10 days, we have number of men working as 40, 35, 30, 25, and 20 over first five consecutive 10 day segments. Average number of men working over first 50 days will be then 30 and work portion completed $30\times{50}=1500 \text{ mandays}$.

We have used the concept of average of first 'n' natural numbers, where 'n' is odd. This is a basic number system concept. The middle number 30 is the average with which the number of days can safely be multiplied.

Total mandays work being $42\times{40}=1680 \text{ mandays}$, after 50 days, 180 mandays work will be left and 15 men will have to work for the full 6th segment of 10 days producing 150 mandays work. In 60 days then, $1500+150=1650 \text{ mandays}$ work will be finished.

Work left will be 30 mandays in the seventh 10 day segment when 10 men will be working.

In further 3 days, that is, in a total of 63 days the work will be finished.

Answer: b: 63 days.

No on paper calculation is necessary in this approach.

This speedy solution was possible because of the use of natural number concept.

More you use effective concepts from one topic area in solving a problem in another topic area, the faster you would be able to solve problems.

Boundary condition problem example 2 from Speed Time and Distance

A monkey is climbing a greased bamboo pole trying to reach the top. The monkey climbs the pole at a uniform speed of 5 ft every minute, but the next minute it slides down by 2 ft. If the bamboo pole is 30 ft high, how long would the monkey take to reach the top?

  1. 19 mins
  2. 20 mins
  3. 18 mins 36 secs
  4. 19 mins 30 secs

Solution example 2.

At the outset we decide that every two minutes the monkey climbs 3 ft effectively. So immediate answer would be, "To climb 30 ft then, the monkey will take $30\div{3}\times{2}=20\text{ minutes}$.

If you are more cautious, you would instead take care to see what happens near the end. With this precautionary measure, you would assume a safe point up to which the regular event of effective climbing of 3 ft every 2 minutes can happen as 27 ft. To climb 27 ft, without any hesitation whatsoever you can decide that the monkey would take, $27\div{3}\times{2}=18\text{ minutes}$.

Now when you think of the 19th minute, you are not very surprised to find that the monkey easily climbs to the top in 19th minute.

But again, is it the right answer?

Being a careful thinker, you notice the phrase, "The monkey climbs the pole at a uniform speed of 5 ft every minute." If the pole were of height 32 ft, the monkey would certainly have taken full 19 minutes to reach the top. But pole height being 30 ft, it would actually take three-fifth of a minute, that is 36 seconds more than 18 minutes to reach the top.

In this case, boundary is at pole height 27ft or time 18mins.

Answer: c: 18 mins 36 secs.

Boundary condition problem example 3 from Average

Players in a team are weighed one by one. After each weighing the average weight increases by 1 kg. After weighing of which player the difference in weights in kg of the first and last player will be equal to or more than 23 kgs?

  1. 11th
  2. 14th
  3. 12th
  4. 13th

Solution example 3.

Let $x$ kg be the weight of the first member. So after weighing the first member, the total is $x$ and average is $x$.

In second weighing, average is $x+1$ and number of players 2, So total is $2x+2$ and weight of second player, $x+2$. Notice how we start with finding the average first (this increases at each step by 1kg as specified and thus is known) and then deduce all other values to get finally the weight of the player added at this step.

In third weighing, average is $x+2$, number of players 3, total weight $3x+6$ and weight of the player added in this third step is thus $x+4$.

The weights of the players then follow an AP series, x, x+2, x+4, x+6, x+8, x+10, x+12..... The difference in weight between the first player and the player weighed last will also follow an AP series, 0, 2, 4, 6, 8, 10, 12....

Without knowing the formula for $n$th term of an AP series, by observation of the series we find, if we leave the first player, the AP series will have 24 as its value on the weighing of the 12th player. This is when the answer gets clear and weighing stops. This is the boundary. We have detected a useful pattern in the numbers 0,2,4,6,8,10,12....; excluding the first number every term is double its position in the series. This is application of pattern identification technique which gives you results fastest.

Including the first player, answer is then 13.

Answer: Option d: 13th.

Boundary condition problem example 4 from Pipes and Cisterns

One pipe in a tank can fill it in 45 mins and a second one can empty it in 1 hr. In what time would the empty tank be filled if the pipes are opened one at a time in alternate minutes?

  1. 2 hrs 55 mins
  2. 4 hrs 48 mins
  3. 3 hrs 40 mins
  4. 5 hrs 53 mins

Solution example 4

First step: identifying which pipe works first in the alternate sequence

The filling pipe must be taken as the first pipe opened in the sequence of the two pipes opened in alternate minutes. This is because, even if the emptying pipe is opened in the first minute, it will have no effect on the outcome as it would try to empty an already empty tank.

Second step: Finding effective fill every 2 mins

In every 2 minutes, effective fill of the tank will be, 

$Fill_{2mins}=\text{fill portion in 1 min } - \text{ drain portion in 1 min}$, these portions are actually fill rate and drain rate in terms of portion of tank filled or drained in 1 minute,

$=T\left(\displaystyle\frac{1}{45}-\displaystyle\frac{1}{60}\right)=\displaystyle\frac{1}{180}T$, where $T$ is the tank capacity.

Effectively in every 2 mins, 1 out of 180 portions of tank will then be filled up.

Third step: Finding boundary condition

Now is the time to be careful and without jumping into the conclusion that in $2\times{180}=360\text{ mins}=6\text{ hrs}$, the tank will be fully filled, let's carefully examine what happens near the end.

It is generally true for boundary condition problems that,

The positive element, in this case the filling pipe, acts at the last segment of time to reach the goal before the negative element can get active.

Being aware of this property and knowing that the filling pipe will act last and fill up last 4 portions out of 180 in 1 minute, we work backwards and deduce—just before complete filling of the tank, 176 portions of the tank was filled in $2\times{176}=352\text{ minutes}$.

So in 353rd minute or 5 hours 53 mins the tank will be completely filled up.

Boundary in this case is when 176 out of 180 portions of the tank is filled up.

Answer: d: 5 hrs 53 mins.

Key concepts used: Boundary condition problem -- Basic time and work concepts -- Fill rate or Work rate technique -- Effective fill concept -- Working backwards approach -- Boundary determination -- Portion use technique.