Concept 3 - Time, Speed and Distance

The basic concepts of TSD are used in solving questions based on motion in a straight line, relative motion, circular motion, problems based on trains, problem based on boats, clocks, races, etc.

THEORY OF TSD

Concept of Motion and Mathematical Representation of Motion

Motion / movement occurs when a body of any shape or size changes its position with respect to any external stationary point. Thus, when a person travels from city A to city B, we say that he has moved from city A to city B. In general, whenever a body moves with respect to a stationary point, we say that the body has undergone a displacement / motion with respect to the starting point. Thus, for motion to have occurred, there must have been some displacement with respect to a stationary point on the ground.

The mathematical model that describes motion has three variables, namely: Speed, Time and Distance. The interrelationship between these three is also the most important formula namely: 

Speed × Time = Distance (Equation for the description of one motion of one body)

The above equation is the mathematical description of the movement of a body. In complex problems, we tend to get confused regarding the usage of this equation and often end up mixing up the speed, time and distance of different motions of different bodies.

It must be mentioned here that this formula is the cornerstone of the chapter Time, Speed and Distance. Besides, this formula is also the source of the various formulae applied to the problems on the applications of time, speed and distance — to trains, boats and streams, clocks and races, circular motion and straight line motion.

In the equation above, speed can be defined as the rate at which distance is covered during the motion. It is measured in terms of distance per unit time and may have any combination of units of distance and time in the numerator and the denominator respectively. (m/s, km/hour, m/min, km/min, km/day, etc.) When we say that the speed of a body is S kmph, we mean to say that the body moves with S kmph towards or away from a stationary point (as the case may be). 

Time (t) is the time duration over which the movement/motion occurs/has occurred. The unit used for measuring time is synchronous with the denominator of the unit used for measuring speed. Thus, if the speed is measured in terms of km/h then time is measured in hours. 

Distance (d) is the displacement of the body during the motion.

The above equation, as is self-evident, is such that the interrelationship between the three parameters defines the value of the third parameter if two of the three are known. Hence we can safely say that if we know two of the three variables describing the motion, then the motion is fully described and every aspect of it is known.

The Proportionalities Implicit in the Equation S × T = D

The above equation has three implicit proportionality dimensions each of which has its own critical bearing on the solving of time, speed and distance problems.

1. Direct proportionality between time and distance (when the speed is constant) time μ distance

Illustration

A car moves for 2 hours at a speed of 25 kmph and another car moves for 3 hours at the same speed. Find the ratio of distances covered by the two cars.

Solution: Since, the speed is constant, we can directly conclude that time μ distance.

Hence

Since, the times of travel are 2 and 3 hours respectively, the ratio of distances covered is also 2/3.

Note: This can be verified by looking at the actual distances travelled—being 50 km and 75 km in this case.

2. Direct Proportionality between speed and distance (when the time is constant) speed μ distance

(a) A body travels at S1 kmph for the first 2 hours and then travels at S2 kmph for the next two hours. Here two motions of one body are being described and between these two motions the time is constant hence speed will be proportional to the distance travelled.

(b) Two cars start simultaneously from A and B respectively towards each other with speeds of S1 kmph and S2 kmph. They meet at a point C…. Here again, the speed is directly proportional to the distance since two motions are described where the time of both the motions is the same, that is, it is evident here that the first and the second car travel for the same time.

In such a case the following ratios will be valid: S1/S2 = d1/d2.

Illustrations

(i) A car travels at 30 km/h for the first 2 hours of a journey and then travels at 40 km/h for the next 2 hours of the journey. Find the ratio of the distances travelled at the two speeds.

Solution: Since time is constant between the two motions described, we can use the proportionality between speed and distance.

Hence, d1/d2 = s1/s2 = 3/4

Alternatively, you can also think in terms of percentage as d2 will be 33.33% higher than d1 since S2 is 33.33% higher than S1 and time is constant.

(ii) Two cars leave simultaneously from points A and B on a straight line towards each other. The distance between A and B is 100 km. They meet at a point 40 km from A. Find the ratio of their speeds.

Solution: Since time is the same for both the motions described, we have ratio of speed = ratio of distance.

SA/SB = 40/60 = 2/3

(iii) Two cars move simultaneously from points A and B towards each other. The speeds of the two cars are 20 m/s and 25 m/s respectively. Find the meeting point if d(AB) = 900 km.

Solution: For the bodies to meet, the time of travel is constant (since the two cars have moved simultaneously).

Hence, speed ratio = distance ratio Æ 4/5 = distance ratio

Hence, the meeting point will be 400 km from A and 500 km from B.

3. Inverse proportionality between speed and time (when the distance is constant) Speed μ 1/time

(a) A body travels at S1 kmph for the first half of he journey and then travels at S2 kmph for the second half of the journey. Here two motions of one body are being described and between these two motions the distance travelled is constant. Hence he speed will be inversely proportional to the time travelled for.

(b) Two cars start simultaneously from A and B respectively towards each other. They meet at a point C and reach their respective destinations B and A in t1 and t2 hours respectively. Here again, the speed is inversely proportional to the time since two motions are described where the distance of both the motions is the same, that is, it is evident here that the first and the second car travel for the distance, viz., AB. In such a case, the following ratio will be valid:

S1/S2 = t2/t1 i.e. S1t1 = S2t2 = S3t3 = K

Illustrations

(i) A train meets with an accident and moves at 3/4 its original speed. Due to this, it is 20 minutes late. Find the original time for the journey beyond the point of accident.

Solution: Speed becomes 3/4 (Time becomes 4/3)

Extra time = 1/3 of normal time = 20 minutes

Normal time = 60 minutes

Alternatively, from the table on product constancy in the chapter of percentages, we get that a 25% reduction in speed leads to a 33.33% increase in time.

But, 33.33% increase in time is equal to 20 minutes increase in time.

Hence, total time (original) = 60 minutes.

(ii) A body travels half the journey at 20 kmph and the other half at 30 kmph. Find the average speed.

Solution: The short-cut process is elucidated in the chapter on ‘averages’. Answer = 24 kmph.

(iii) A man travels from his house to his office at 5 km/h and reaches his office 20 minutes late. If his speed had been 7.5 km/h, he would have reached his office 12 minutes early. Find the distance from his house to his office.

Solution: Notice that here the distance is constant. Hence, speed is inversely proportional to time.

Solving mathematically

S1/S2 = t2/(t2 + 32)

5/7.5 = t2/(t2 + 32)

5t2 + 160 = 7.5 t2

t2 = 160/2.5 = 64 minutes

Hence, the distance is given by 7.5 × 64/60 = 8 km.

Alternatively, using the Product Constancy Table from the chapter of percentages. If speed increases by 50%, then time will decrease by 33.33%.

But the decrease is equal to 32 minutes.

Hence, original time = 96 minutes and new time is 64 minutes.

Hence, the required distance = 5 × 96/60 km = 8 km.

or distance = 7.5 × 64/60 km = 8 km

[Note: The entire process can be worked out mentally while reading the problem.]

CONVERSION BETWEEN kmph to m/s
1 km/h = 1000 m/h = 1000/3600 m/s = 5/18 m/s.
Hence, to convert y km/h into m/s multiply by 5/18.
Thus, y km/h = m/s.
And vice versa : y m/s = 18 y/5 km/h. To convert from m/s to kmph, multiply by 18/5.

Relative Speed : Same Direction and Opposite Direction
Normally, when we talk about the movement of a body, we mean the movement of the body with respect to a stationary point. However, there are times when we need to determine the movement and its relationships with respect to a moving point/body. In such instances, we have to take into account the movement of the body/point with respect to which we are trying to determine relative motion.
Relative movement, therefore, can be viewed as the movement of one body relative to another
moving body. The following formulae apply for the relative speed of two independent bodies with respect to each other:
Case I: Two bodies are moving in opposite directions at speeds S1 and S2 respectively.
The relative speed is defined as S1 + S2
Case II: Two bodies are moving in the same direction.
The relative speed is defined as
(a) S1 – S2 when S1 is greater than S2.
(b) S2 – S1 when S1 is lesser than S2.
In other words, the relative speed can also be defined as the positive value of the difference between the two speeds, that is, | S1 – S2 |.

Motion in a Straight Line
Motion in a straight line is governed by the rules of relative speed enumerated above.
A. Two or more bodies starting from the same point and moving in the same direction: Their relative speed is S1 – S2.
(a) In the case of the bodies moving to and fro between two points A and B: The faster body will reach the end first and will meet the second body on its way back. The relative speed S1 – S2 will apply till the point of reversal of the faster body and after that the two bodies will start to move in the opposite directions at a relative speed of S1 + S2. The relative speed governing the movement of the two bodies will alternate between S1 – S2 and S1 + S2 every time any one of the bodies reverses directions. However, if both the bodies reverse their direction at the same instant, there will be no change in the relative speed equation.
In this case, the description of the motion of the two bodies between two consecutive meetings will also be governed by the proportionality between speed and distance (since the time of movement between any two meetings will be constant).
Distances covered in this case: For every meeting, he total distance covered by the two bodies will be 2D (where D is the distance between the extreme points). However, notice that the value of 2D would be applicable only if both the bodies reverse the direction between two meetings. In case only one body has reversed direction, the total distance would need to be calculated on a case-by-case basis. The respective coverage of the distance is in the ratio of the individual speeds.
Thus, for the 9th meeting (if both bodies have reversed direction between every 2 meetings) the total
distance covered will be 9 × 2D = 18D.
This will be useful for solving problems that require the calculation of a meeting point.
(b) In the case of the bodies continuing to move in the same direction without coming to an end point
and reversing directions: The faster body will take a lead and will keep increasing the lead and the movement of the two bodies will be governed by the relative speed equation: S1 – S2. Here again, if the two bodies start simultaneously, their movement will be governed by the direct proportionality between speed and distance.

B. Moving in the opposite direction: Their relative speed will be initially given by S1 + S2.
(a) In the case of the bodies moving to and fro between two points A and B starting from opposite ends of the path: The two bodies will move towards each other, meet at a point in between A and B, then move apart away from each other. The faster body will reach its extreme point first followed by the slower body reaching its extreme point next. Relative speed will change every time; one of the bodies reverses direction.
The position of the meeting point will be determined by the ratio of the speeds of the bodies (since the 2 movements can be described as having the time constant between them).
Distances covered in the above case: For the first meeting, the total distance covered by the two bodies will be D (the distance between the extreme points). The coverage of the distance is in the ratio of the individual speeds. Thereafter, as the bodies separate and start coming together, the combined distance to be covered is 2D. Note that if only one body is reversing direction between two meetings, this would not be the case and you will have to work it out.
Thus, for the 10th meeting (if both bodies have reverse direction between every 2 meetings) the total
distance covered will be D + 9 × 2D = 19D. This will be useful for solving of problems that require the calculation of a meeting point.

Illustrations
(i) Two bodies A and B start from opposite ends P and Q of a straight road. They meet at a point 0.6D from P. Find the point of their fourth meeting.
Solution: Since time is constant, we have ratio of speeds as 3 : 2.
Also, total distance to be covered by the two together for the fourth meeting is 7D. This distance is divided in a ratio of 3 : 2 and thus we have that A will cover 4.2D and B will cover 2.8D. The fourth meeting point can then be found out by tracking either A or B’s movement. A, having moved a distance of 4.2D, will be at a point 0.2D from P. This is the required answer.

(ii) A starts walking from a place at a uniform speed of 2 km/h in a particular direction. After half an hour, B starts from the same place and walks in the same direction as A at a uniform speed and overtakes A after 1 hour 48 minutes. Calculate the speed of B.
Solution: Start solving as you read the question. From the first two sentences you see that A is 1 km ahead of B when B starts moving.
This distance of 1 km is covered by B in 9/5 hours [1 hour 48 minutes = 1(4/5) = 9/5 hours].
The equation operational here (SB – SA) × T = initial distance
(SB – 2) × 9/5 = 1
Solving, we get SB = 23/9 km/h.

(b) In the case of the bodies continuing to move in the same direction without coming to an end point and reversing directions: The bodies will meet and following their meeting they will start separating and going away from each other. The relative speed will be given by S1 + S2 initially while approaching each other and, thereafter, it will be S1 + S2 while moving away from each other.
Try to visualise how two bodies separate and then come together. Also, clearly understand the three proportionalities in the equation s × t = d, since these are very important tools for problem solving.

Concept of Acceleration
Acceleration is defined as the rate of change of speed.
Acceleration can be positive (speed increases) or negative (speed decreases Æ also known as deceleration)
The unit of acceleration is speed per unit time (e.g. m/s2)
For instance, if a body has an initial speed of 5 m/s and a deceleration of 0.1 m/s2 it will take 50 seconds to come to rest.
Final speed = Initial speed + Acceleration × Time

Some more examples:
(i) Water flows into a cylindrical beaker at a constant rate. The base area of the beaker is 24 cm2. The water level rises by 10 cm every second. How quickly will the water level rise in a beaker with a base area of 30 cm2.
Solution: The flow of water in the beaker is 24 cm2 × 10 cm/s = 240 cm3/s.
If the base area is 30 cm2 then the rate of water level rise will be 240/30 = 8 cm/s.
Note: In case of confusion in such questions we are advised to use dimensional analysis to understand what to multiply and what to divide.

(ii) A 2 kilowatt heater can boil a given amount of water in 10 minutes. How long will it take for
(a) a less powerful heater of 1.2 kilowatts to boil the same amount of water?
(b) a less powerful heater of 1.2 kilowatts to boil double the amount of water?Solution:
(a) The heating required to boil the amount of water is 2 × 10 = 20 kilowatt minutes. At the rate of 1.2 kilowatt, this heat will be generated in 20/1.2 minutes = 16.66 minutes.
(b) When the water is doubled, the heating required is also doubled. Hence, heating required = 40
kilowatt minutes. At the rate of 1.2 kilowatt, this heat will be generated in 40/1.2 = 33.33 minutes.

AN APPLICATION OF ALLIGATION IN TIME, SPEED AND DISTANCE
Consider the following situation:
Suppose a car goes from A to B at an average speed of S1 and then comes back from B to A at an average speed of S2. If you had to find out the average speed of the whole journey, what would you do?
The normal short cut given for this situation gives the average speed as: (2S1S2) / (S1+S2)
However, this situation can be solved very conveniently using the process of alligation as explained
below:
Since, the two speeds are known to us, we will also know their ratio. The ratio of times for the two parts of the journey will then be the inverse ratio of the ratio of speeds. (Since the distance for the two journeys are equal). The answer will be the weighted average of the two speeds (weighted on the basis of the time travelled at each speed)
The process will become clear through an example:
A car travels at 60 km/h from Mumbai to Poona and at 120 km/h from Poona to Mumbai. What is the average speed of the car for the entire journey. 
Solution

The process of alligation, will be used here to give the answer as 80.
Note here, that since the speed ratio is 1:2, the value of the time ratio used for calculating the weighted average will be 2:1.
What will happen in case the distances are not constant?
For instance, if the car goes 100 km at a speed of 66kmph and 200 km at a speed of 110 kmph, what will be the average speed?
In this case the speed ratio being 6:10 i.e. 3:5 the inverse of the speed ratio will be 5:3. This would have been the ratio to be used for the time ratio in case the distances were the same (for both the speeds). But since the distances are different, we cannot use this ratio in this form. The problem is overcome by multiplying this ratio (5:3) by the distance ratio (in this case it is 1:2) to get a value of 5:6. This is the ratio which has to be applied for the respective weights. Hence, the alligation will look like:
Solution

Thus the required answer is 90 kmph.

APPLICATIONS OF TIME, SPEED AND DISTANCE

Trains
Trains are a special case in questions related to time, speed and distance because they have their own theory and distinct situations.
The basic relation for trains problems is the same: Speed × Time = Distance.
The following things need to be kept in mind before solving questions on trains:
(a) When the train is crossing a moving object, the speed has to be taken as the relative speed of the train with respect to the object. All the rules for relative speed will apply for calculating the relative speed.
(b) The distance to be covered when crossing an object whenever a train crosses an object will be equal to: Length of train + Length of object
Thus, the following cases will yield separate equations, which will govern the crossing of the object by the train:
For each of the following situations the following notations have been used:
ST = Speed of train SO = Speed of object t = time
LT = Length of train LO = Length of object
Case I: Train crossing a stationary object without length:
ST × t = LT
Case II: Train crossing a stationary object with length:
ST × t = (LT + LO)
Case III: Train crossing a moving object without length:
• In opposite direction: (ST + SO) × t = LT
• In same direction: (ST – SO) × t = LT
Case IV: Train crossing a moving object with length:
• In opposite direction: (ST + SO) × t = (LT + LO)
• In same direction: (ST – SO) × t = (LT + LO)
[Note: In order for a train to completely cross a stationary point on the ground, the train has to traverse a distance that is equal to its entire length.
This can be visualised by remembering yourself stationary on a railway platform and being crossed by a train. You would say that the train starts crossing you when the engine of the train comes in line with you. Also, you would say that you have been crossed by the train when the end of the guard’s compartment comes in line with you. Thus, the train would have travelled its own length in crossing you].

Illustrations
(i) A train crosses a pole in 8 seconds. If the length of the train is 200 metres, find the speed of the train.
Solution: In this case, it is evident that the situation is one of the train crossing a stationary object without length. Hence, Case I is applicable here.
Thus, ST = 200/8 = 25 m/s Æ 25 × 18/5 = 90 kmph.

(ii) A train crosses a man travelling in another train in he opposite direction in 8 seconds. However, the train requires 25 seconds to cross the same man if the trains are travelling in the same direction. If the length of the first train is 200 metres and that of the train in which the man is sitting is 160 metres, find the speed of the first train.
Solution: Here, the student should understand that the situation is one of the train crossing a moving object without length. Thus the length of the man’s train is useless or redundant data. Then applying the relevant formulae after considering the directions of the movements we get the equations:
ST + SM = 25
ST – SM = 8
ST = 33/2 = 33/2 * 18/5 = 59.4 kmph


Boats and Streams
The problems of boats and streams are also dependent on the basic equation of time, speed and distance : Speed × Time = Distance.
However, as in the case of trains the adjustments to be made for solving questions on boats and streams are:
The boat has a speed of its own, which is also called the speed of the boat in still water (SB).
Another variable that is used in boats and streams problems is the speed of the stream (SS).
The speed of the movement of the boat is dependent on whether the boat is moving:
(a) In still water the speed of movement is given by Æ SB.
(b) While moving upstream (or against the flow of the water), the speed of movement is given by Æ
SU = SB – SS
(c) While moving downstream (or with the flow of the water), the speed of movement is given by Æ
SD = SB + SS
The time of movement and the distance to be covered are to be judged by the content of the problem.

Circular motion: A special case of movement is when two or more bodies are moving around a circular track.
The relative speed of two bodies moving around a circle in the same direction is taken as S1 – S2.
Also, when two bodies are moving around a circle in the opposite direction, the speed of the two
bodies is taken to be S1 + S2.
The peculiarity inherent in moving around a circle in the same direction is that when the faster body overtakes the slower body it goes ahead of it. And for every unit time that elapses, the faster body keeps increasing the distance by which the slower body is behind the faster body. However, when the distanceby which the faster body is in front of the slower body becomes equal to the circumference of the circle
around which the two bodies are moving, the faster body again comes in line with the slower body. This event is called as overlapping or lapping of the slower body by the faster body. We say that the slower body has been lapped or overlapped by the faster body.

First meeting: Three or more bodies start moving simultaneously from the same point on the circumference of the circle, in the same direction around the circle. They will first meet again in the LCM of the times that the fastest runner takes in totally overlapping each of the slower runners.
For instance, if A, B, C and D start clockwise from a point X on the circle such that A is the fastest runner then we can define TAB as the time in which A completely overlaps B, TAC as the time in which A completely overlaps C and TAD as the time in which A completely overlaps D. Then the LCM of TAB, TAC and TAD will be the time in which A, B, C and D will be together again for the first time.

First meeting at starting point: Three or more bodies start moving simultaneously from the same point
on the circumference of a circle, in the same direction around the circle. Their first meeting at the starting point will occur after a time that is got by the LCM of the times that each of the bodies takes to complete one full round.
For instance, if A, B and C start from a point X on the circle such that TA, TB and TC are the times in which A, B and C respectively cover one complete round around the circle, then they will all meet together at the starting point in the LCM of TA, TB and TC.


Clocks
Problems on clocks are based on the movement of the minute hand and that of the hour hand as well as on the relative movement between the two. In my opinion, it is best to solve problems on clocks by considering a clock to be a circular track having a circumference of 60 km and each kilometre being represented by one minute on the dial of the clock. Then, we can look at the minute hand as a runner running at the speed of 60 kmph while we can also look at the hour hand as a runner running at an average speed of 5 kmph.
Since, the minute hand and the hour hand are both moving in the same direction, the relative speed of the minute hand with respect to the hour hand is 55 kmph, that is, for every hour elapsed, the minute hand goes 55 km (minute) more than the hour hand.
(Beyond this slight adjustment, the problems of clocks require a good understanding of unitary method. This will be well illustrated through the solved example below.)

Important Information
Number of right angles formed by a clock: A clock makes 2 right angles between any 2 hours. Thus, for instance, there are 2 right angles formed between 12 to 1 or between 1 and 2 or between 2 and 3 or between 3 and 4 and so on.
However, contrary to expectations, the clock does not make 48 right angles in a day. This happens because whenever the clock passes between the time period 2–4 or between the time period 8–10 there are not 4 but only 3 right angles.
This happens because the second right angle between 2–3 (or 8–9) and the first right angle between 3–4 (or 9–10) are one and the same, occurring at 3 or 9.
Right angles are formed when the distance between the minute hand and the hour hand is equal to 15 minutes.
Exactly the same situation holds true for the formation of straight lines. There are 2 straight lines in every hour. However, the second straight line between 5–6 (or 11–12) and the first straight line between 6–7 (or 12–1) coincide with each other and are represented by the straight line formed at 6 (or 12).
Straight lines are formed when the distance between he minute hand and the hour hand is equal to either 0 minutes or 30 minutes.

Illustration
At what time between 2–3 p.m. is the first right angle in that time formed by the hands of the clock?
Solution: At 2 p.m. the minute hand can be visualised as being 10 kilometres behind the hour hand. (considering the clock dial to be a race track of circumference 60 km such that each minute represents a kilometre).
Also, the first right angle between 2–3 is formed when the minute hand is 15 kilometres ahead of the hour hand.
Thus, the minute hand has to cover 25 kilometres over the hour hand.
This can be written using the unitary method:
Distance covered by the minute hand over the hour hand.
55 kilometres in 1 hour
25 kilometres in what time?
Æ 5/11 of an hour.
Thus, the first right angle between 2–3 is formed at 5/11 hours past 2 o’clock.
This can be converted into minutes and seconds using unitary method again as:
1 hour 60 minutes
5/11 hours ? minutes
Æ 300/11 minutes = 27 (3/11) minutes
1 minute 60 seconds
3/11 minutes ? seconds Æ 180/11 seconds = 16.3636 seconds.
Hence, the required answer is: 2 : 27 : 16.36 seconds.

Concept 2 - Percentages

BASIC DEFINITION AND UTILITY OF PERCENTAGE

Percent literally, means ‘for every 100’ and is derived from the French word ‘cent’, which is French for 100. The basic utility of Percentage arises from the fact that it is one of the most powerful tools for comparison of numerical data and information. It is also one of the simplest tools for comparison of data. In the context of business and economic performance, it is specifically useful for comparing data such as profits, growth rates, performance, magnitudes and so on.

Mathematical definition of percentage 

The concept of percentage mainly applies to ratios, and the percentage value of a ratio is arrived at by multiplying by 100 the decimal value of the ratio.

For example, a student scores 20 marks out of a maximum possible 30 marks. His marks can then be denoted as 20 out of 30 = (20/30) or (20/30) × 100% = 66.66%.

The process for getting this is perfectly illustrated through the unitary method:

Then the value of x × 30 = 20 × 100

x = (20/30) × 100 Æ the percentage equivalent of a ratio.

Now, let us consider a classic example of the application of percentage:

Example: Student A scores 20 marks in an examination out of 30 while another student B scores 40 marks out of 70. Who has performed better?

Solution: Just by considering the marks as 20 and 40, we do not a get clear picture of the actual performance of the two students. In order to get a clearer picture, we consider the percentage of marks.

Thus, A gets (20/30) × 100 = 66.66%

While B gets (40/70) × 100 = 57.14%

Now, it is clear that the performance of A is better.

Consider another example:

Example: Company A increases its sales by 1 crore rupees while company B increases its sales by 10 crore rupees. Which company has grown more?

Solution: Apparently, the answer to the question seems to be company B. The question cannot be answered since we don’t know the previous year’s sales figure (although on the face of it Company B

seems to have grown more). If we had further information saying that company A had a sales turnover of ` 1 crore in the previous year and company B had a sales turnover of ` 100 crore in the previous year, we can compare growth rates and say that it is company A that has grown by 100%. Hence, company A has a higher growth rate, even though in terms of absolute value increase of sales, company B has grown much more.

IMPORTANCE OF BASE / DENOMINATOR FOR PERCENTAGE CALCULATIONS

Mathematically, the percentage value can only be calculated for ratios that, by definition, must have a

denominator. Hence, one of the most critical aspects of the percentage is the denominator, which in other words is also called the base value of the percentage. No percentage calculation is possible without knowing the base to which the percentage is to be calculated.

Hence, whenever faced with the question ‘What is the percentage …?’ always try first to find out the

answer to the question ‘Percentage to what base?’

CONCEPT OF PERCENTAGE CHANGE

Whenever the value of a measured quantity changes, the change can be captured through

(a)Absolute value change or

(b)Percentage change.

Both these measurements have their own advantages and disadvantages.

Absolute value change: It is the actual change in the measured quantity. For instance, if sales in year 1 is ` 2500 crore and the sales in year 2 is ` 2600 crore, then the absolute value of the change is ` 100 crore.

Percentage change: It is the percentage change got by the formula

As seen earlier, this often gives us a better picture of the effect of the change.

Note: The base used for the sake of percentage change calculations is always the original quantity unless otherwise stated.

Example: The population of a city grew from 20 lakh to 22 lakh. Find the

(a) percentage change

(b) percentage change based on the final value of population

Solution:

(a) percentage change = (2/20) × 100 = 10%

(b) percentage change on the final value = (2/22) × 100 = 9.09%

Difference between the Percentage Point Change and the Percentage Change

The difference between the percentage point change and the percentage change is best illustrated through an example. Consider this:

The savings rate as a percentage of the GDP was 25% in the first year and 30% in the second year.

Assume that there is no change in the GDP between the two years. Then:

Percentage point change in savings rate = 30% – 25% = 5 percentage points.

Percentage change in savings rate = [(30-25) / 25] x 100 = 25%

PERCENTAGE RULE FOR CALCULATING PERCENTAGE VALUES THROUGH ADDITIONS

Illustrated below is a powerful method of calculating percentages. In my opinion, the ability to calculate percentage through this method depends on your ability to handle 2 digit additions. Unless you develop the skill to add 2 digit additions in your mind, you are always likely to face problems in calculating percentage through the method illustrated below. In fact, trying this method without being strong at 2-digit additions/subtractions (including 2 digits after decimal point) would prove to be a disadvantage in your attempt at calculating percentages fast.

This process, essentially being a commonsense process, is best illustrated through a few examples:

Example: What is the percentage value of the ratio: 53/81?

Solution: The process involves removing all the 100%, 50%, 10%, 1%, 0.1% and so forth of the

denominator from the numerator.

Thus, 53/81 can be rewritten as: (40.5 + 12.5)/81 = 40.5/81 + 12.5/81 = 50% + 12.5/81

= 50% + (8.1 + 4.4)/81 = 50% + 10% + 4.4/81

= 60% + 4.4/81

At this stage you know that the answer to the question lies between 60 – 70% (Since 4.4 is less than 10% of 81) At this stage, you know that the answer to the calculation will be in the form: 6a.bcde ….

All you need to do is find out the value of the missing digits.

In order to do this, calculate the percentage value of 4.4/81 through the normal process of multiplying the numerator by 100.

Thus the % value of = 4.4/81 = (4.4 x 100)/81 = 440/81

[Note: Use the multiplication by 100, once you have the 10% range. This step reduces the decimal

calculations.]

Thus 440/81= 5% with a remainder of 35

Our answer is now refined to 65.bcde. (1% Range)

Next, in order to find the next digit (first one after the decimal add a zero to the remainder;

Hence, the value of ‘b’ will be the quotient of b Æ 350/81 = 4 Remainder 26

Answer: 65.4cde (0.1% Range)

c Æ 260/81 = 3 Remainder 17

Answer: 65.43 (0.01% Range)

and so forth.

The advantages of this process are two fold:

(1) You only calculate as long as you need to in order to eliminate the options. Thus, in case there was only a single option between 60 – 70% in the above question, you could have stopped your

calculations right there.

(2) This process allows you to go through with the calculations as long as you need to. However, remember what I had advised you right at the start: Strong Addition skills are a primary requirement for using this method properly.

To illustrate another example:

What is the percentage value of the ratio 223/72 ?

223/72 Æ 300 – 310% Remainder 7

700/72 Æ 9. Hence 309 – 310%, Remainder 52

520/72 Æ 7. Hence, 309.7, Remainder 16

160/72 Æ 2. Hence, 309.72 Remainder 16

Hence, 309.7222 (2 recurs since we enter an infinite loop of 160/72 calculations).

In my view, percentage rule (as I call it) is one of the best ways to calculate percentages since it gives you the flexibility to calculate the percentage value up to as many digits after decimals as you are required to and at the same time allows you to stop the moment you attain the required accuracy range.

Effect of a Percent Change in the Numerator on a Ratio’s Value

The numerator has a direct relationship with the ratio, that is, if the numerator increases the ratio increases. The percentage increase in the ratio is the same as the percentage increase in the numerator, if the denominator is constant.

Thus, 22/40 is exactly 10% more than 20/40. (in terms of percentage change)

Percentage Change Graphic and its Applications

In mathematics there are many situations where one is required to work with percentage changes. In such situations the following thought structure (Something I call Percentage Change Graphic) is a very useful tool:

What I call Percentage Change Graphic (PCG) is best illustrated through an example:

Suppose you have to increase the number 20 by 20%. Visualise this as follows:

The PCG has 6 major applications listed and explained below: PCG applied to:

1. Successive changes

2. Product change application

3. Product constancy application

4. A Æ BÆA application

5. Denominator change to Ratio Change application

6. Use of PCG to calculate Ratio Changes

Application 1: PCG Applied to Successive Changes

This is a very common situation in most questions.

Suppose you have to solve a question in which a number 30 has two successive percentage increases

(20% and 10% respectively). The situation is handled in the following way using PCG:

Illustration

A’s salary increases by 20% and then decreases by 20%. What is the net percentage change in A’s salary?

Solution:

Hence, A’s salary has gone down by 4%

Illustration

A trader gives successive discounts of 10%, 20% and 10% respectively. The percentage of the original cost price he will recover is:

Solution:

Hence the overall discount is 35.2% and the answer is 64.8%.

Illustration

A trader marks up the price of his goods by 20%, but to a particularly haggling customer he ends up giving a discount of 10% on the marked price. What is the percentage profit he makes?

Solution:

Hence, the percentage profit is 8%.

Application 2: PCG applied to Product Change

Suppose you have a product of two variables say 10 × 10.

If the first variable changes to 11 and the second variable changes to 12, what will be the percentage

change in the product? [Note there is a 10% increase in one part of the product and a 20% increase in the other part.]

The formula given for this situation goes as: (a + b + ab/100)

Hence, Required % change = 10 + 20 + (10 x 20) / 100

(Where 10 and 20 are the respective percentage changes in the two parts of the product)

However, a much easier solution for this case can be visualized as:

Hence, the final product shows a 32% increase.

Similarly suppose 10 × 10 × 10 becomes 11 × 12 × 13

In such a case the following PCG will be used:

Hence, the final product sees a 71.6 percent increase

(Since, the product changes from 100 to 171.6)

Note: You will get the same result irrespective of the order in which you use the respective percentage changes. Also note that this process is very similar to the one used for calculating successive percentage change.

Application for DI:

Suppose you have two pie charts as follows:

If you are asked to calculate the percentage change in the sales revenue of scooters for the company from year one to year two, what would you do?

The formula for percentage change would give us:

Obviously this calculation is easier said than done.

However, the Product change application of PCG allows us to execute this calculation with a lot of ease comparatively. Consider the following solution:

Product for year one is: 0.2347 × 17342.34

Product for year two is: 0.2655 × 19443.56

These can be approximated into:

234 × 173 and 265 × 194 respectively (Note that by moving into three digits we do not end up losing any accuracy. We have elaborated this point in the chapter on Ratio and Proportions.)

The overall percentage change depends on two individual percentage changes:

234 increases to 265: A % change of 31/234 = 13.2 % approx. This calculation has to be done using the percentage rule for calculating the percentage value of the ratio 173 increases to 194 – A percentage change of approximately 12%.

Thus PCG will give the answer as follows:

Hence, 26.76 % increase in the product’s value. (Note that the value on the calculator for the full calculation sans any approximations is 26.82 %, and given the fact that we have come extremely close to the answer—the method is good enough to solve the question with a reasonable degree of accuracy.)

Application 3 of PCG: Product Constancy Application (Inverse proportionality)

Suppose you have a situation wherein the price of a commodity has gone up by 25%. In case you are required to keep the total expenditure on the commodity constant, you would obviously need to cut down on the consumption. By what percentage? Well, PCG gives you the answer as follows:

Hence, the percentage drop in consumption to offset the price increase is 20%.

I leave it to the student to discover the percentage drop required in the second part of the product if one part increases by 50 percent.

Note: Product constancy is just another name for Inverse proportionality.

Application 4 of PCG: A Æ B Æ A.

Very often we are faced with a situation where we compare two numbers say A and B. In such cases, if we are given a relationship from A to B, then the reverse relationship can be determined by using PCG in much the same way as the product constancy use shown above.

B’s salary is 25% more than A’s salary. By what percent is A’s salary less than B’s salary?

A drop of 25 on 125 gives a 20% drop.

Hence A’s salary is 20% less than B’s.

Note: The values which applied for Product Constancy also apply here.

Application 5 of PCG Æ Effect of change in Denominator on the Value of the Ratio

The denominator has an inverse relationship with the value of a ratio.

Hence the process used for product constancy (and explained above) can be used for calculating

percentage change in the denominator.

For instance, suppose you have to evaluate the difference between two ratios:

Ratio 1 : 10/20

Ratio 2 : 10/25

As is evident the denominator is increasing from 20 to 25 by 25%.

If we calculate the value of the two ratios we will get:

Ratio 1 = 0.5, Ratio 2 = 0.4.

% change between the two ratios = 0.1 / 0.5 × 100 = 20% Drop

This value can be got through PCG as:

100 --> 125 --> 100

Hence, 20% drop.

Note: This is exactly the same as Product constancy and works here because the numerator is constant.

Hence, R1 = N/D1 and R2 = N/D2

i.e. R1 × D1 = N and R2 × D2 = N, which is the product constancy situation.

Direct process for calculation

To find out the percentage change in the ratio due to a change in the denominator follow the following process:

In order to find the percentage change from 10/20 to 10/25, calculate the percentage change in the denominator in the reverse fashion.

i.e. The required percentage change from R1 to R2 will be given by calculating the percentage change in the denominators from 25 to 20 (i.e. in a reverse fashion) & not from 20 to 25.

Product Constancy Table, Inverse Proportionality Table, A Æ B Æ A table, Ratio Change to Denominator table

Application 6: Use of PCG to Calculate Ratio Changes:

Under normal situations, you will be faced with ratios where both numerator and denominator change.

The process to handle and calculate such changes is also quite convenient if you go through PCG.

Calculate the percentage change between the Ratios.

Ratio 1 = 10/20 Ratio 2 = 15/25

The answer in this case is 0.5 Æ 0.6 (20% increase). However, in most cases calculating the values of

the ratio will not be easy. The following PCG process can be used to get the answer:

When 10/20 changes to 15/25, the change occurs primarily due to two reasons:

(A) Change in the numerator (Numerator effect)

(B) Change in the denominator (Denominator effect)

By segregating the two effects and calculating the effect due to each separately, we can get the answer easily as follows:

Numerator Effect 

The numerator effect on the value of the ratio is the same as the change in the numerator. Hence, to calculate the numerator effect, just calculate the percentage change in the numerator: 

In this case the numerator is clearly changing from 10 to 15 (i.e. a 50% increase.) This signifies that the numerator effect is also 50%.

Denominator Effect 

As we have just seen above, the effect of a percentage change in the denominator on the value of the ratio is seen by calculating the denominator’s percentage change in the reverse order.

In this case, the denominator is changing from 20 to 25. Hence the denominator effect will be seen by

going reverse from 25 to 20 i.e. 20% drop.

With these two values, the overall percentage change in the Ratio is seen by:

This means that the ratio has increased by 20%.

Implications for Data Interpretation

Percentage is perhaps one of the most critical links between QA and Data Interpretation. In the chapter theory mentioned above, the Percentage Rule for Percentage Calculations and the PCG applied to product change and ratio change are the most critical.

As already shown, the use of PCG to calculate the percentage change in a product (as exhibited through the pie chart example above) as well as the use of PCG to calculate ratio changes are two extremely useful applications of the concepts of percentages into DI.

Applying Percentages for the special case of comparing two ratios to find the larger one.

Suppose you have two ratios to compare. Say R1 = N1/D1 and R2 = N2/D2

The first step is to find the ten percent ranges for each of these ratios. In case, they belong to different ranges of 10% (say R1 lies between 50-60 while R2 lies between 70 to 80), it becomes pretty simple to say which one will be higher.

In case, both of these values for percentage of the ratios belong to the same ten percent range, then we can use the following process

Step 1: Calculate the percentage change in the numerator

Step 2: Calculate the percentage change in the denominator. There could be four cases in this situation, when we move from Ratio1 to Ratio2:

Case 1: Numerator is increasing while denominator is decreasing Æ obviously the net effect of the two changes will be an increase in the ratio. Hence, R2 will be greater.

Case 2: Numerator is decreasing while denominator is increasing Æ obviously the net effect of the two changes will be a decrease in the ratio. Hence, R1 will be greater. It is only in the following cases that we need to look at the respective changes in the numerator and denominator.

Case 3: Numerator and denominator are both increasing Calculate the percentage value of the respective increases. If the numerator is increasing more than the denominator the ratio will go up. On the other hand, if the denominator is increasing more than the numerator, Ratio2 will be smaller than Ratio1.  (Note: Compare in percentage values)

Case 4: Numerator and denominator are both decreasing Æ Calculate the percentage value of the respective decreases. If the numerator is decreasing more than the denominator the ratio will go down. On the other hand, if the denominator is decreasing more than the numerator, Ratio2 will be greater than Ratio1.

FRACTION TO PERCENTAGE CONVERSION TABLE

The following percentage values appear repeatedly over the entire area where questions can be framed on the topic of percentage. Further, it would be of great help to you if you are able to recognize these values separately from values that do not appear in the following table. 

Some Utilisations of the Table

• The values that appear in the table are all percentage values. These can be converted into decimals by just shifting the decimal point by two places to the left. Thus, 83.33% = 0.8333 in decimal value.

• A second learning from this table is in the process of division by any of the numbers such as 2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 15, 16, 24 and so on, students normally face problems in calculating the decimal values of these divisions. However, if one gets used to the decimal values that appear in the Table, calculation of decimals in divisions will become very simple. For instance, when an integer is divided by 7, the decimal values can only be .14, .28, .42, .57, .71, .85 or .00. (There are approximate values)

• This also means that the difference between two ratios like can be integral if and only if x is divisible by both 6 and 7. This principle is very useful as an advanced short cut for option based solution of some questions.

Percentage Conversion Table

Calculation of Multiplication by Numbers like 1.21, 0.83 and so on

In my opinion, the calculation of multiplication of any number by a number of the form 0.xy or of the form 1.ab should be viewed as a subtraction/addition situation and not as a multiplication situation. This can be explained as follows.

Example: Calculate 1.23 × 473.

Solution: If we try to calculate this by multiplying, we will end up going through a very time taking process, which will yield the final value at the end but nothing before that (i.e. you will have no clue

about the answer’s range till you reach the end of the calculation). Instead, one should view this multiplication as an addition of 23% to the original number. This means, the answer can be got by adding 23% of the number to itself.

Thus 473 × 1.23 = 473 + 23% of 473 = 473 + 94.6 + 3% of 473 = 567.6 + 14.19 = 581.79

(The percentage rule can be used to calculate the addition and get the answer.)

The similar process can be utilised for the calculation of multiplication by a number such as 0.87

(Answer can be got by subtracting 13% of the number from itself and this calculation can again be done by percentage rule.)

Concept 1 - Time and Work

If A does a work in a days, then in 1 day, A does 1/a of the work.

If B does a work in a days, then in 1 day, A does 1/b of the work.

Then, in 1 day, if A and B work together, then their combined work is 1/a + 1/b = (a+b) / ab.

In the above case, we take the total work to be done as “1 unit of work”. Hence, the work will be completed when 1 unit of work is completed.

For example, if A can do a work in 10 days and B can do the same work in 12 days, then the work will be completed in how many days.

One day’s work = 1/10 + 1/12 = (12 + 10)/120 = 22/120

Then the number of days required to complete the work is 120/22.

Note that this is a reciprocal of the fraction of work done in one day. This is a benefit associated with solving time and work through fractions. It can be stated as—the number of time periods required to complete the full work will be the reciprocal of the fraction of the work done in one time period.

Alternative Approach:

Instead of taking the value of the total work as 1 unit of work, we can also look at the total work as 100 per cent work. In such a case, the following rule applies:

If A does a work in a days, then in one day A does 100/a % of the work.

If B does a work in b days, then in one day B does 100/b % of the work.

Then, in one day, if A and B work together, then their combined work is 100/a + 100/b.

This is often a very useful approach to look at the concept of time and work because thinking in terms of percentages gives a direct and clear picture of the actual quantum of work done.

What I mean to say is that even though we can think in either a percentage or a fractional value to solve the problem, there will be a thought process difference between the two.

Thinking about work done as a percentage value gives us a linear picture of the quantum of the work that has been done and the quantum of the work that is to be done. On the other hand, if we think of the work done as a fractional value, the thought process will have to be slightly longer to get a full understanding of the work done.

For instance, we can think of work done as 7/9 or 77.77%. The percentage value makes it clear as to how much quantum is left. The percentage value can be visualised on the number line, while the fractional value requires a mental inversion to fully understand the quantum.

An additional advantage of the percentage method of solving time and work problems would be the elimination of the need to perform cumbersome fraction additions involving LCMs of denominators.

However, you should realise that this would work only if you are able to handle basic percentage calculations involving standard decimal values. If you have really internalised the techniques of percentage calculations given in the chapter of percentages, then you can reap the benefits for this chapter. The benefit of using this concept will become abundantly clear by solving through percentages the same example that was solved above using fractions.

Example:

If A can do a work in 10 days and B can do the same work in 12 days, then the work will be completed in how many days.

One day’s work = 10% + 8.33% = 18.33% (Note, no LCMs required here)

Hence, to do 100% work, it will require: 100/18.33.

This can be solved by adding 18.33 mentally to get between 5–6 days. Then on you can go through options and mark the closest answer.

The process of solving through percentages will yield rich dividends if and only if you have adequate practice on adding standard percentage values. Thus, 18.33 × 5 = 91.66 should not give you any headaches and should be done while reading for the first time. 

Thus a thought process chart for this question should look like this.

If A can do a work in 10 days (Æ means 10% work) and B can do the same work in 12 days (Æ 8.33% work Æ 18.33% work in a day in 5 days 91.66% work Æ leaves 8.33% work to be done Æ which can be done in 8.33/18.33 of a day = 5/11 of a day (since both the numerator and the denominator are divisible by 1.66), then the work will be completed in 5 days. 

The entire process can be done mentally.

The Concept of Negative Work

Suppose, that A and B are working to build a wall while C is working to break the wall. In such a case, the wall is being built by A and B while it is being broken by C. Here, if we consider the work as the building of the wall, we can say that C is doing negative work.

Example: A can build a wall in 10 days and B can build it in 5 days, while C can completely destroy the wall in 20 days. If they start working at the same time, in how many days will the work be completed.

Solution: The net combined work per day here is:

A’s work + B’s work – C’s work = 10% + 20% – 5% = 25% work in one day.

Hence, the work will get completed (100% work) in 4 days.

The concept of negative work commonly appears as a problem based on pipes and cisterns, where there are inlet pipes and outlet pipes/leaks which are working against each other.

If we consider the work to be filling a tank, the inlet pipe does positive work while the outlet pipe/leak does negative work.

Application of Product Constancy Table of Time and Work

The equation that applies to Time and Work problems is

Work Rate × Time = Work done

This equation means that if the work done is constant, then Work rate is inversely proportional to time. 

Hence, the Product Constancy Table will be directly applicable to time and work questions.

[Notice the parallelism between this formula and the formula of time speed and distance, where again

there is product constancy between speed and time if the distance is constant.]

Time is usually in days or hours although any standard unit of time can be used. The unit of time that has to be used in a question is usually decided by the denominator of the unit of work rate.

Here, there are two ways of defining the Work rate.

(a) In the context of situations where individual working efficiencies or individual time requirements are given in the problem, the work rate is defined by the unit: Work done per unit time.

In this case, the total work to be done is normally considered to be 1 (if we solve through fractions) or 100% (if we solve through percentages).

Thus, in the solved problem above, when we calculated that A and B together do 18.33% work in a day, this was essentially a statement of the rate of work of A and B together.

Then the solution proceeded as:

18.33% work per day × No. of days required = 100% work

Giving us: the no. of days required = 100/18.33 = 5 (5/11).

(b) In certain types of problems (typically those involving projects that are to be completed), where a certain category of worker has the same rate of working, the Work rate will be defined as the number of workers of a particular category working on the project.

For instance, questions where all men work at a certain rate, the work rate when 2 men are working together will be double the work rate when 1 man is working alone. Similarly, the work rate when 10 men are working together will be 10 times the work rate when 1 man is working alone.

In such cases, the work to be done is taken as the number of man-days required to finish the work. Note, for future reference, that the work to be done can also be measured in terms of the volume of work defined in the context of day-to-day life. For example, the volume of a wall to be built, the number of people to be interviewed, the number of chapattis to be made and so on.

WORK EQUIVALENCE METHOD

(To Solve Time and Work Problems)

The work equivalence method is nothing but an application of the formula:

Work rate × Time = Work done (or work to be done)

Thus, if the work to be done is doubled, the product of work rate × time also has to be doubled.

Similarly, if the work to be done increases by 20%, the product of work rate × time also has to be

increased by 20% and so on.

This method is best explained by an example:

A contractor estimates that he will finish the road construction project in 100 days by employing 50 men. However, at the end of the 50th day, when as per his estimation half the work should have been completed, he finds that only 40% of his work is done.

(a) How many more days will be required to complete the work?

(b) How many more men should he employ in order to complete the work in time?

Solution:

(a) The contractor has completed 40% of the work in 50 days.

If the number of men working on the project remains constant, the rate of work also remains constant. Hence, to complete 100% work, he will have to complete the remaining 60% of the work. For this he would require 75 more days. (This calculation is done using the unitary method.)

(b) In order to complete the work on time, it is obvious that he will have to increase the number of men working on the project. This can be solved as:

50 men working for 50 days Æ 50 × 50 = 2500 man-days.

2500 man-days has resulted in 40% work completion. Hence, the total work to be done in terms of the number of man-days is got by using unitary method:

Work left = 60% = 2500 × 1.5 = 3750 man-days.

This has to be completed in 50 days. Hence, the number of men required per day is 3750/50 = 75 men. Since, 50 men are already working on the project, the contractor needs to hire 25 more men.

[Note, this can be done using the percentage change graphic for product change. Since, the number of days is constant at 50, the 50% increase in work from 40% to 60% is solely to be met by increasing the number of men. Hence, the number of men to be increased is 50% of the original number of men = 25 men.]

The Specific Case of Building a Wall (Work as volume of work)

As already mentioned, in certain cases, the unit of work can also be considered to be in terms of the

volume of work. For example, building of a wall of a certain length, breadth and height.

In such cases, the following formula applies:

where L, B and H are respectively the length, breadth and height of the wall to be built, while m, t and d are respectively the number of men, the amount of time per day and the number of days. Further, the suffix

1 is for the first work situation, while the suffix 2 is for the second work situation.

Consider the following problem:

Example: 20 men working 8 hours a day can completely build a wall of length 200 meters, breadth 10 metres and height 20 metres in 10 days. How many days will 25 men working 12 hours a day require to build a wall of length 400 meters, breadth 10 metres and height of 15 metres.

This question can be solved directly by using the formula above.

Here, 

L1 is 200 metres 

L2 is 400 metres

B1 is 10 metres 

B2 is 10 metres

H1 is 20 metres 

H2 is 15 metres

while 

m1 is 20 men 

m2 is 25 men

d1 is 10 days 

d2 is unknown

and 

t1 is 8 hours a day

t2 is 12 hours a day

Then we get (200 × 10 × 20)/(400 × 10 × 15) = (20 × 8 × 10)/(25 × 12 × d2)

d2 = 5.333/0.6666 = 8 days

Alternatively, we can also directly write the equation as follows:

d2 = 10 × (400/200) × (10/10) × (20/15) × (20/25) × (8/12)

This can be done by thinking of the problem as follows:

The number of days have to be found out in the second case. Hence, on the LHS of the equation write

down the unknown and on the RHS of the equation write down the corresponding knowns.

Then, the length of the wall has to be factored in. There are only two options for doing so. viz:

Multiplying by 200/400 (< 1, which will reduce the number of days) or multiplying by 400/200 (>1, which will increase the number of days).

The decision of which one of these is to be done is made on the basis of the fact that when the length of the wall is increasing, the number of days required will also increase.

Hence, we take the value of the fraction greater than 1 to get

d2 = 10 × (400/200)

We continue in the same way to get

No change in the breadth of the wall Æ hence, multiply by 10/10 (no change in d2)

Height of the wall is decreasing Æ hence, multiply by 15/20 (< 1 to reduce d2)

Number of men working is increasing Æ hence, multiply by 20/25 (< 1 to reduce d2)

Number of hours per day is increasing Æ hence, multiply by 8/12 (< 1 to reduce the number of days)

The Concept of Efficiency

The concept of efficiency is closely related to the concept of work rate.

When we make a statement saying A is twice as efficient as B, we mean to say that A does twice the work as B in the same time. In other words, we can also understand this as A will require half the time required by B to do the same work.

In the context of efficiency, another statement that you might come across is A is two times more efficient than B. This is the same as A is thrice as efficient as B or A does the same work as B in 1/3rd of the time.

Equating Men, Women and Children This is directly derived from the concept of efficiencies.

Example: 8 men can do a work in 12 days while 20 women can do it in 10 days. In how many days can 12 men and 15 women complete the same work.

Solution: Total work to be done = 8 × 12 = 96 man-days.

or total work to be done = 20 × 10 = 200 woman-days.

Since, the work is the same, we can equate 96 man-days = 200 woman-days.

Hence, 1 man-day = 2.08333 woman-days.

Now, if 12 men and 15 women are working on the work we get

12 men are equal to 12 × 2.08333 = 25 women

Hence, the work done per day is equivalent to 25 + 15 women working per day.

That is, 40 women working per day.

Hence, 40 × no. of days = 200 woman days

Number of days = 5 days.