Basic and Rich Percentage Concepts for solving difficult SSC CGL problems

Percentage is one of the most universally used but underrated mathematical concepts

percentage concepts

In business world, academics, Science and technology, or daily life, use of the concept of Percentage is ubiquitous and all-pervading. Because of its apparent simplicity though, this concept is taken for granted with not much emphasis on its elaboration and explanation.

In this session we will cover the basic and rich percentage concepts supported by selected examples from diverse problem topics.

At the end, a 10 question exercise problem set will be provided for trying out the concepts.

Basic Percentage concepts

Definition

Percentage represented by a value followed by the symbol ${\%}$ is a special kind of ratio in which the denominator is converted to 100. It is defined as a ratio between two quantities with the quantity in the numerator as portions of each 100 portions of the quantity in the denominator.

The statement, "Quantity A is $x$% of Quantity B", is mathematically equivalent to the statement, "Quantity A is $\displaystyle\frac{x}{100}$ times of B". 

For example, if the ratio of son's age to father's age is 2 : 5, we have,

\[ \frac{\text{Son's age}}{\text{Father's age}}=\frac{2}{5}=\frac{2\times{20}}{5\times{20}}=\frac{40}{100}=40\text{%} \]

We can then say, son's age is 40% of father's age. This is again equivalent to saying, son's age is 0.4 times father's age.

Thus percent symbol can be replaced by the number divided by 100.

If we reverse the ratio, we get,

\[ \frac{\text{Father's age}}{\text{Son's age}}=\frac{5}{2}=\frac{5\times{50}}{2\times{50}}=\frac{250}{100}=250\text{%} \]

Now we would have to say, father's age is 250% of son's age, or 2.5 times son's age.

The percent value more than 100 means the antecedent value is larger than the consequent.

Percentage reference concept and Decimal equivalence of percentage

In academic life, we find frequent use of percentage when we say, "My niece got 85% in Maths in the finals." In this statement nowhere are the information about the total or actual marks available. If the total marks were 200 then the student's score would have been, $200\times{85\text{%}}=170$.

How did we arrive at the actual marks? To get the actual marks the Total marks is just multiplied with the decimal equivalent of percentage score,

$\text{Actual marks}=\text{Total marks}\times{\text{Percent score}}$

$=200\times{\text{85%}}$

$=200\times{\displaystyle\frac{85}{100}}$

$=200\times{0.85}$

$=170$.

Two important basic concepts of percentage used are then the percentage reference concept and decimal equivalence of percentage.

When a percentage, say, $x$% is mentioned, it indicates a quantity as a portion of a second reference quantity. Without clarity on this reference quantity, the percentage statement would not be meaningful.

In the example, student score of 85% is a percentage of the total score which is the reference value.

Note: In actual use though often we find the actual reference value not mentioned. We will deal with such instances later on.

Secondly, the percentage of 85% is mathematically equivalent to, $\displaystyle\frac{85}{100}$ which again is simply the decimal, $0.85$.

As the formal score statement is, "85% of total score", when total score is given as 200, the actual score statement is transformed to, "85% of 200", or, "0.85 of 200".

In this last form only we are able to deduce the actual score mathematically. This equivalence of percentage to decimal form is essential for mathematical evaluation. We call this concept as, decimal equivalence of percentage.

A percentage value is to be always expressed with respect to a reference value. In this case the reference value is the Total score. When a percentage value is multiplied with the reference value, it is first converted to its equivalent decimal value and then multiplied.

Percentage to fraction conversion

Just as a percentage is equivalent to a decimal portion of reference quantity, it can also be expressed as a fraction of the reference quantity. This is the concept of Percentage to fraction conversion.

For example, if value of A is given as 25% of B, we can very well express value of A as $\displaystyle\frac{1}{4}$the of B. For further evaluations, whether we will use the decimal equivalent or the percentage equivalent, it depends purely on convenience and nature of the problem. The choice of conversion is determined by what we call as Percentage conversion technique. Later we will highlight this aspect through a selected problem.

Frequently used percentage to decimal and fractions

Often we need to to quickly convert a given percentage value to it decimal or fraction equivalent value for further solution of the problem. For fast and efficient simplification memorization of the following list of equivalent values should help.

  1. $4\text{%}=\displaystyle\frac{1}{25}=0.04$
  2. $5\text{%}=\displaystyle\frac{1}{20}=0.05$
  3. $8\text{%}=\displaystyle\frac{2}{25}=0.08$
  4. $10\text{%}=\displaystyle\frac{1}{10}=0.10$
  5. $12.5\text{%}=\displaystyle\frac{1}{8}=0.125$
  6. $20\text{%}=\displaystyle\frac{1}{5}=0.20$
  7. $25\text{%}=\displaystyle\frac{1}{4}=0.25$
  8. $30\text{%}=\displaystyle\frac{3}{10}=0.30$
  9. $40\text{%}=\displaystyle\frac{2}{5}=0.40$
  10. $50\text{%}=\displaystyle\frac{1}{2}=0.50$
  11. $60\text{%}=\displaystyle\frac{3}{5}=0.60$
  12. $75\text{%}=\displaystyle\frac{3}{4}=0.75$
  13. $80\text{%}=\displaystyle\frac{4}{5}=0.80$
  14. $90\text{%}=\displaystyle\frac{9}{10}=0.90$

Percentage use in real life

On many occasions in real life, the percentage is used to express a value without mentioning the reference value. This is both an opportunity to hide information, a disadvantage of the concept, as well as a powerful means to express a value in its simplest form when the reference value is well known or universally accepted.

Percentage as a performance measurement tool

A very common example of the second intent of expressing value as a percentage is, "Profit this year has gone up by 20%."

Here it is generally known that the this year's profit percentage is expressed with reference to the Profit last year, whatever be that profit. This is a very apt use of percentage as a tool for comparing rate of increase. The expression of 20% increase in profit simply says,

This year profit is more than last year's profit by one-fifth of last year's profit.

By this we can compare the performance in terms of profit increase of same organization through the past years, or compare the performances of various organizations in the same year.

A generally accepted use of measuring performance of national economies is GDP or Gross Domestic Product which is expressed as a percentage.

As GDP is defined as a ratio with respect to the size of economy, this use of percentage neutralizes the variation of economy growth due to economy size and concentrates on relative growth.

Percentage as a rate of growth measuring tool

One of the most frequent use of percentage used as a growth measurement tool is the Interest rate on investment in banks or financial institutions. In this case also the reference value is not mentioned as it is defined and known by all as the invested amount.

In case of Simple interest, the interest or growth of invested amount remains same throughout the investment period.

For example, if Rs.1000 is invested for 2 years at an interest rate of 10% on Simple interest basis, each year the interest accrued will be,

$1000\times{0.1}=\text{Rs.}100$.

After two years, the invested amount will then grow to an amount,

$\text{Rs.}1000+\text{Rs.}200=\text{Rs.}1200$.

On the other hand, in case of Compound interest, the interest will be calculated on the amount at the beginning of the year, and not on the beginning of the investment period.

For example, when Rs.1000 is invested on Compound interest basis for 2 years at an interest rate of 10%, at the end of two years the amount will grow to,

$\text{Rs.}1000(1+0.1)^2=\text{Rs.}1210$.

The Rs.10 excess of simple interest case happens because in the second year, 10% interest rate acted not only on the invested amount of Rs.1000 but also on the interest of Rs.100 accrued during the first year.

In these cases, percentage is used as an indicator of rate of growth. This is the concept of Percentage as rate of growth.

The concept of Compound growth at a given percentage is not only used on money investments, but also on any kind of growth, say for example bacterial growth.

In this case also the reference of percentage is not mentioned as it is known by the definition of system of growth itself.

These are only a few examples of use of the concept of percentage in real life. We will now shift our attention to use of percentage in competitive test maths.

This is the area of rich percentage concepts.


Percentage concepts are heavily used in Compound and simple interest problems, Liquid or other types of mixing problems, Ratio and proportion problems, Age based problems, Profit and loss problems and potentially all types of Arithmetic problems. It is one of the most basic mathematical concepts in practical use.

Percentage use in Profit and loss problems solving

In profit and loss problems, a few pre-defined percentages form the conceptual base.

Two of the most important definitions are,

$\text{Profit % }= \displaystyle\frac{\text{Sale Price} -\text{Cost Price}}{\text{Cost Price}}\times{100}$,

where,

$\text{Actual Profit}=\text{Sale Price} -\text{Cost Price}$,

$\text{Discount % }= \displaystyle\frac{\text{Discount value}}{\text{Marked Price}}\times{100}$.

For profit, the percentage reference is Cost Price while for discount, the percentage reference is Marked Price.

Also, Profit is added to Cost Price to get Sale Price, and Discount is subtracted from Marked Price to get the Sale Price.

Together these special type of percentage concepts we call as, Percentage application for profit and loss.

By solving the first selected problem on profit and loss we will show how to use these concepts.

Without clarity on these simple concepts, solving profit and loss problems may be difficult. 

Selected problem example 1.

A dealer purchased an item for Rs.510. After giving a discount of 15% on its marked price, he still made a profit of 20%. The marked price was,

  1. Rs.720
  2. Rs.708
  3. Rs.696
  4. Rs.732

Solution 1

As per definition, applying profit percentage on given value of Cost Price or CP, we will get the Sale Price or SP as,

$\text{SP}=510+0.2\times{510}=612$.

Here we have directly converted the 20% value to its decimal equivalent of 0.2, and not its fraction equivalent as, further calculation with decimal equivalent form is more convenient in this case.

The percentage reference of profit is Cost Price by definition.

Turning our attention to the Marked Price or MP, we recall that by reducing the MP by the discount percentage operated on MP itself, we will get the Sale Price or SP. And the good thing is we already have evaluated SP.

So we directly express this relation as,

$\text{SP}=612=0.85\text{MP}$,

Or, $\text{MP}=\displaystyle\frac{612}{0.85}=\frac{3600}{5}=720$.

We take the first opportunity to transform the Sale Price in terms of Marked Price by first converting the Discount percentage to its decimal equivalent and subtracting it from 1. Marked Price or MP being the target value which we already have, this step immediately leads us to the solution.

Directly transforming SP in terms of MP by using Discount percentage, or transforming SP in terms of CP by using Profit percentage is what we call as, Transformation to percentage reference technique. In each of these two cases, target variable is the percentage reference variable.

This is a rich percentage concept.

The transformation of prices and relationships between the percentages and prices are clearly shown in the following graphic.

basic percentage concepts 1

Notice how profit percentage is first applied on its reference value of CP to get actual profit of Rs.102 and then is added to CP to arrive at SP. Effectively then,

$SP=(1+0.2)CP=1.2CP$.

Similarly Discount percentage is first applied on reference value of MP to get actual discount value of Rs.108 and then this actual discount is subtracted from MP to arrive at SP again. Effectively in this case,

$SP=(1-0.15)MP=0.85MP$.

Examine the reverse relationships of, CP=(83.3% of SP), and MP=(117.6% of SP).

Answer: Option a: Rs.720.

Key concepts used: Basic profit and loss concepts -- Basic percentage concepts -- Rich percentage concepts -- Percentage application concept for profit and loss, you need to apply the percentage profit on cost price and percentage discount on marked price -- Percentage reference concept -- Transformation to percentage reference technique -- Percentage to decimal conversion -- Efficient simplification.

We will solve a second problem on profit and loss to consolidate the percentage reference concept.

Selected problem example 2.

If a man were to sell a chair for Rs.720, he would lose 25%. For gaining 25%, he has to sell it for,

  1. Rs.960
  2. Rs.1000
  3. Rs.1200
  4. Rs.900

Solution 2.

In the first case loss is 25% on CP when SP is Rs.720. So,

$0.75CP=720$, we have expressed SP directly in terms of CP using Transformation to percentage reference technique.

Or, $\displaystyle\frac{3}{4}CP=720$, this is the case for expressing percentage as a fraction using Percentage to fraction conversion according to Percentage conversion technique,

Or, $CP=\displaystyle\frac{4}{3}\times{720}$.

To make a profit of 25%, the New Sale Price or NSP will then be 125% of CP, or,

$\text{NSP}=\displaystyle\frac{5}{4}CP=\frac{5}{3}\times{720}=\text{Rs.}1200$.

We have applied Delayed evaluation technique by not calculating the CP at the first stage, and instead carrying forward the fraction expression to simplify and evaluate it at the end once only.

Answer: Option c: Rs.1200.

Percentage conversion technique in selecting the right conversion form is a rich percentage concept. This speeds up the solution process.

Key concepts used: Basic profit and loss concepts -- Basic percentage concepts -- Rich percentage concepts -- Percentage reference concept -- Transformation to percentage reference technique -- Percentage to fraction conversion -- Percentage conversion technique -- Delayed evaluation technique -- Efficient simplification.

Rich percentage concepts in ratio and proportion problems

A ratio $p:q$ of two quantities represents the proportion of actual value $P$ with respect to actual value $Q$ and after cancelling out the HCF between $P$ and $Q$ the ratio is expressed as a proper fraction with no common factors between the two terms.

As the ratio terms are essentially in division relationship, we can always multiply or divide each ratio term by the same number without changing the ratio. This is according to Basic ratio concepts.

Furthermore in a ratio, each of the terms in the ratio represents portions of the whole. We use this as Portions use technique often. Conceptually though the portions are equivalent to percentages of the whole. This useful concept of Portions as percentages sometimes helps to solve a problem in a few easy steps.

For example, in a ratio of two quantities, $2:3$, the first quantity is 2 portions out of total five portions, which is equivalent to 40% of the whole. The second quantity in this case is 60% of the whole 100%.

Selected problem example 3.

Number of students in three classes A, B and C in a school were in a ratio 2:3:4 in last year. This year the numbers increased by certain percentages to change the ratio to 11:18:26. If the percentage increase in class C were 30% what is the percentage increase in class B?

  1. 25%
  2. 20%
  3. 10%
  4. 15%

Solution 3.

Trying out multiples of 4 adding 30% of which to itself will result in the corresponding ratio term value of 26, we get immediately the factor 5 that results in,

$1.3\times{4\times{5}}=26$.

Here we have used the Transformation to Percentage reference technique and formed the problem relation,

$1.3\times{\text{Reference value}}=26$.

By using this rich percentage concept coupled with basic ratio concepts, we divide 26 by 1.3 to get 20 in a few seconds. From 20 we get the multiplicative factor 5 by which the ratio term 4 were to be multiplied. So the target term 3 for Class B also is to be multiplied by 5 to get 15.

Thus 15 is the percentage reference and we need to get 18 by applying a percentage on it.

Lastly getting the multiplying factor 1.2 of 15 to get 18 is easy, and this means an increase of 20%. Again we have used Transformation to percentage reference concept.

All deductions done by Mathematical reasoning in mind using rich concepts, and in quick time.

Alternate solution 3.

More deductive procedural method is to introduce cancelled out HCFs as $x$ and $y$ respectively in the two ratios according to HCF reintroduction technique to transform the ratios to ratios of actual values, $2x:3x:4x$ and $11y:18y:26y$ respectively.

From given percentage increase condition, applying Transformation to percentage reference concept again we get the relation,

$26y=1.3\times{4x}$,

Or, $x=5y$.

With this $x$ value we first get actual value for Class B as,

$3x=15y$.

This value is changed to $18y$ by an increment of $3y$ which is one-fifth or $20\text{%}$ of $15y$.

Answer: b: 20%.

Key concepts used: Basic ratio concepts -- HCF reintroduction technique -- Rich percentage concepts -- Percentage reference concept -- Transformation to percentage reference technique -- Mathematical reasoning.

Rich percentage concepts in proportional relations

Two types of proportional relations we encounter,

  • Direct proportionality
  • Inverse proportionality.

Direct proportionality

Two quantities $A$ and $B$ are directly proportional when,

$A=cB$, where $c$ is a constant.

Alternatively,

$\displaystyle\frac{A}{B}=c$, a constant.

This leads to Percentage in direct proportionality concept,

If a percentage is applied on one of two directly proportional quantities, the same percentage must be applied to the other quantity also.

This leads to the associated concept,

If one of two directly proportional quanties is changed by a certain percentage, the other quantity must also be changed by the same percentage.

For example, in proportional relation above, if $A$ is reduced by 20%, it becomes $80\text{% of } A$ or $0.8A$.

As, $\displaystyle\frac{A}{B}=c$, a constant, when $A$ is changed to $0.8A$, $B$ must also change to $0.8B$ whih means a 20% reduction.

This is a simple but powerful concept.

Inverse proportionality

Two quantities $A$ and $B$ are inversely proportional when,

$A=\displaystyle\frac{c}{B}$, where $c$ is a constant.

Alternatively,

$AB=c$, a constant.

This leads to Percentage in inverse proportionality concept,

If a percentage is applied on one of two inversely proportional quantities, inverse of the same percentage must be applied to the other quantity also.

In this case percentage change in one won't lead to same percentage change in the other.

We will showcase these rich concepts in solving two example problems.

Selected problem example 4.

Full tank petrol for Vijay's mobike lasts for 20 days. How many days a full tank will last for him if he increases his daily consumption of petrol by 25%?

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

Solution 4.

Petrol consumption per day $C$ and number of days $D$ a full tank of petrol lasts are inversely proportional as,

$CD=t$, where $t$ is the full tank capacity.

Daily consumption increasing by 25%, it is changed to $1.25C=\displaystyle\frac{5}{4}C$.

So original value 20 of $D$ will reduce to,

$\displaystyle\frac{4}{5}\times{20}=16$.

Here we have first used Transformation to percentage reference concept, and then Percentage to fraction conversion concept. By Percentage conversion technique we chose fraction equivalent of percentage because it would need to be inverted at the next stage.

A percentage increase of 25% of C resulted in application of a percentage of 125% on C. Inverse of this percentage must then be applied on number of days $D$.

Answer: a: 16.

Key concepts used: Basic proportionality concepts -- Inverse proportionality concept -- Percentage in inverse proportionality -- Percentage reference concept -- Percentage to fraction conversion concept --Transformation to percentage reference technique -- Percentage conversion technique.

Selected problem example 5.

Due to price rise of rice by 25% per kg Pitambar could only purchase 20 kg less than what he had purchased at the expense of Rs.400 before the price rise. The increase in price of rice per kg (in Rs.) is,

  1. 2
  2. 1.5
  3. 1
  4. 3

Solution 5.

Price per kg $P$ and number of kgs purchased $K$ in Rs.400 are inversely proportional as,

$PK=400$.

By 25% increase, when $P$ increases to $1.25P$ or $\frac{5}{4}P$, $K$ must be reduced to $\frac{4}{5}K$, or decrease by 20%.

Here 20% is 20kgs. So 100% is 100kg, and so original price is,

$\displaystyle\frac{400}{100}=\text{Rs.}4$ per kg.

Increase in price is then 25% of Rs.4 or Rs.1 to a value of Rs.5 per kg.

Simple isn't it?

Answer: Option d: 1.

Key concepts used: Basic proportionality concepts -- Inverse proportionality concept -- Percentage in inverse proportionality -- Percentage change equivalence, 20% equivalent to 20kgs -- Unitary method -- Percentage change concept -- Percentage to fraction conversion concept -- Fraction to percentage conversion concept -- Transformation to percentage reference technique -- Percentage conversion technique.

Percent as actual value

Occasionally, percentage values are to be used in place of actual values for easy solution. We will explain this rich concept by solving a chosen problem example.

Selected problem example 6.

In a village each of the 60% of the families has a cow, each of the 30% of the families has a buffalo and each of the 15% of the families has both a cow and a buffalo. If there are 96 families in the village, how many families have neither a cow nor a buffalo?

  1. 24
  2. 28
  3. 20
  4. 26

Solution 6.

As 60% of the families each has a cow and 30% of the families each has a buffalo, when we add these two percentages, we have 90% of the families in which percentage of families each having both a cow and buffalo, that is, 15% is included twice.

Subtracting this 15% from 90% then we get 75% of the families each having either a cow or a buffalo or both.

The rest, that is, $100\text{%}-75\text{%}=25\text{%}$ of the families then each would have neither a cow nor a buffalo.

Till this point of evaluation, we have dealt with only percentages, and not actual values. The reason we could do that is, the percentage reference of total number of families is common for all these percentages.

Applying the 25% on total number of families at the end we get the desired number of families having neither a cow nor a buffalo as 24.

Instead, we could have converted 60%, 30% and 15% to actual numbers and then used set exclusion operation to get the answer. Operating only on percentages as actual values, we have saved two percentage conversions.

Answer. Option a: 24.

We state the rich concept of Percentage as actual values,

If all the variables in a problem are in the form of percentages of a common percentage reference, value of which is given, we can avoid calculating the actual values and use only the percentages to save extra steps.

While calculating actual value as a given percentage of a reference value, instead of converting the percentage to its equivalent decimal by dividing it by 100, often we convert it to a fraction by dividing it by 100 alright, but transforming the result to a fraction. We do this because usually fraction arithmetic is easier than decimal arithmetic.

Key concepts used: Basic percentage concepts -- Basic set theoretic concepts -- Percentage as actual value concept -- Efficient simplification.

An offshoot of this is the concept of Percentage of a percentage.

Following is a problem example highlighting this use of percentage.

Selected problem example 7.

Salary of an employee is first raised by 10% and then decreased by 10%. What is the effective increase or decrease percentage of salary?

  1. 10% increase
  2. 11% decrease
  3. 9% decrease
  4. 1% decrease

Solution 7.

It is a decrease because the first increase of 10% was on 100%, while the second action of reduction of same 10% amount was on the amount 110%. So, decrease is more than the increase. All percentage are with respect to the starting value, which is equivalent to 100%.

After the first increase, the whole is 110% of the initial starting value.

When this value is decreased by 10%, the percentage reference is this value of 110% of the starting value, and effectively this reduction is,

$10\text{% of }110\text{%}=11\text{%}$.

All these percentages have reference as the starting value.

So final value after the initial increase and subsequent decrease is,

$110\text{%}-11\text{%}=99\text{%}$.

This is equivalent to a decrease of 1% from the starting value.

Answer. Option d: 1%.

Key concepts used: Basic percentage concepts -- Percentage as actual value concept -- Percentage reference concept -- Percentage of percentage concept.

We will end this session with use of percentage concepts in problems involving mixing of liquids and other materials that can be homogeneously mixed together. We call this set of concepts as Percentage in mixing concepts.

Proportional relations are multiplicative but in mixing, the variable relations are additive, and so are to be treated differently.

Percentage use in problems involving mixing of liquids or other materials

We will use two selected problems to highlight how percentage concepts are to be adapted to this new situation.

Selected problem example 8.

In 70 litres of milk diluted with water, milk is 90%. How much water is to be added to the diluted milk to make percentage of water as 12.5%.

  1. 1 litre
  2. 2.5 litres
  3. 1.5 litre
  4. 2 litres

Solution 8.

If milk is 90%, water is then 10%, that is 7 litres. Just by trial we can deduce that adding 2 to 7 will make it 9, as well as will increase the volume of the new mixture to 72, making the water portion as one-eighth or 12.5% of the total volume.

This is a problem of mixing where one component remains fixed in amount.

Formally we can form the relation after adding $x$ litres as,

$\displaystyle\frac{7+x}{70+x}=\frac{1}{8}$, for convenience we have converted 12.5% to its equivalent fraction.

Inverting and subtracting 1 from both sides,

$\displaystyle\frac{63}{7+x}=7$,

Or, $7+x=9$,

Or, $x=2$.

Answer. Option d: 2 litres.

Key concepts used: Basic mixing concepts -- Basic percentage concepts -- Percentage reference concept -- Percentage to fraction conversion.

Selected problem example 9.

In what ratio must a milk diluted with water containing 30% milk be mixed with a diluted milk containing 50% milk and rest water to get a mixture with milk content of 45%?

  1. $3 : 1$
  2. $1:3$
  3. $1:2$
  4. $2:1$

Solution 9.

This is a standard problem of mixing mixtures of two components with two different component ratios, to get a mixture of a specified component ratio.

Mark that we have used the phrase, "component ratios" instead of "component percentage ratios". It is so because in a ratio, the numeric ratio terms are replaceable by corresponding percentages.

The equivalent numerical term values in ratios of milk to water for the two mixtures in this case are respectively,

$30\text{%}:70\text{%}=3:7$, and

$50\text{%}:50\text{%}=1:1$.

Percentage ratios and numerical term value ratios are equivalent and interchangeable.

Let us start solving the problem.

Using concentration in mixture technique, we will find how much milk 1 litre of each mixture contains.

For the first mixture each 1 litre contains 0.3 litre of milk.

Similarly for the second mixture each 1 litre milk contains 0.5 litre of milk.

Assuming total volume of new mixture to be 100 litres, if $x$ litres of first mixture were taken for mixing, $100-x$ litres of second mixture would have been taken for the mixing.

Milk content in the new mixture would have been then,

$0.3x+0.5(100-x)=50-0.2x=45$.

So,

$0.2x=5$,

Or, $x=25$ litres is the volume of first mixture mixed with rest 75 litres of second mixture to get the new mixture of 100 litres.

The desired ratio would have been then, 1 : 3.

As in a homogeneous mixture any volume of the mixture would contain the milk and water in same ratio, we could have assumed first 1 litre of each of the mixtures and later 100 litres of new mixture.

Why 100 litres? Because it conveniently is equivalent to 100% of the total volume of the new mixture in which milk would have been 45% or 45 litres. This is Percentage as specific volume concept and quite effective in solving especially mixture problems elegantly.

Taking advantage of the same percentage of milk in any volume, 100 litre volume of new mixture is used thus eliminating any uncertain values and introducing convenient specific values.

A second advantage of assuming a specific new mixture volume has been the ability to define portion of one mixture mixed in terms of the portion of the second mixture. By this we restricted the number of variables to just one. In a linear equation on a single unknown variable, solution is quick and certain. This is what we call Reduction in number of variables technique. We take the first opportunity to reduce number of variables while simplifying an expression.

Answer. Option b: 1 : 3.

Key concepts used: Concentration of mixture technique -- Percentage to decimal conversion -- Percentage as specific volume concept -- Reduction in number of variables technique.

The powerful concept of Percentage change in product relation

The percentage change in product relation states,

In a product of factors, if the factors change by certain percentages, the decimal change coefficient of the product after change would always be equal to the product of the decimal change coefficients of the factors that changed.

Associated concept is, decimal change coefficient of a quantity $P$ is 1.2 if the quantity is increased by 20% and the quantity becomes $1.2P$ after change. Here Transformation to percentage reference concept is used along with Percentage to decimal conversion concept.

For example, if two quantities $P$ and $Q$ are increased by 10% and 20% respectively, their product after these changes in the factors will change from $PQ$ to,

$1.1P\times{1.2Q}=1.1\times{1.2}PQ=1.32PQ$.

The decimal change coefficient after change 1.32, equals the product of the decimal change coefficients of the factors, 1.1 and 1.2.

This change in the product is equivalent to an increase of 32%.

Let us clarify this concept by solving a selected problem.

Problem 10.

An airline having daily flights between Kolkata and Delhi, increased its daily average fare by 30%. This caused a fall in daily average number of passengers by 20%. The overall effect of fare increase on the daily average fare revenue is then,

  1. 10% increase
  2. 12% decrease
  3. 12% increase
  4. 10% decrease

Solution 10 : Problem analysis and execution: Conventional solution

If $N_1$ and $F_1$ are the daily average number of passengers and daily average fare before fare increase, the daily average fare revenue before fare increase is,

$R_1=N_1\times{F_1}$.

Similarly, if $N_2$ and $F_2$ are the daily average number of passengers and daily average fare after fare increase, the daily average fare revenue after fare increase is,

$R_2=N_2\times{F_2}$.

By given conditions then,

$1.3F_1=F_2$, and

$0.8N_1=N_2$.

Both percentages are on initial values.

Using these relations in the two expressions,

$R_1=N_1\times{F_1}$, and

$R_2=N_2\times{F_2}=0.8N_1\times{1.3F_1}$=0.8\times{1.3}N_1\times{F_1}=1.04R_1$.

Taking the ratio,

$R_2 : R_1=1.04 : 1$.

This means, daily average fare revenue after fare increase has increased by 4% from the original corresponding amount. The 20% drop in daily average number of passengers is over-compensated by the 30% fare increase by 4%.

Answer: Option a: 4% increase.

Alternate solution 10: Elegant solution applying the concept of Percentage change in product relation

Daily average fare revenue is equal to the product of daily average number of passengers and daily average fare.

As fare is increased by 30% to 1.3 times of initial fare, and number of passengers reduced by 20% to 0.8 times of intial number of passengers, the Daily average fare revenue after fare rise effectively becomes $1.3\times{0.8}=1.04$ times intial Daily average fare revenue.

This is equivalent to a 4% increase in Daily average fare revenue.

In this conceptual solution we have used first the concept of Transformation to percentage reference concept to express the changed values in terms of initial values and then formed the product of the decimal change coefficients of the two changed values to get the decimal change coefficient of changed product value in terms of initial product value.

This is use of the concept of Percentage change in product relation.

The same concept will be applicable for product or division of any number of factors.

For example, if a quantity is a product of two factors A and B with a third factor C dividing, when A is increased by 20%, B is decreased by 25% and C is increased by 50%, the change coefficients of the three will be, 1.2, 0.75 and 1.5 respectively. The change coefficient after the change would then be, 

$\displaystyle\frac{1.2\times{0.75}}{1.5}=0.6$.

This is equivalent to a 40% reduction in the quantity after change. This is a good example of how use of a rich concept can help to solve a complex problem very easily.

Transforming the decimal change coefficient after change to percentage and getting the change percentage from 100% gives us the solution.

Key concepts used: Problem modelling to form the two Daily average fare revenue expressions before and after change of fare -- Percentage to decimal conversion -- Transformation to percentage reference concept --  Decimal change coefficient -- Percentage change in product relation -- Decimal to percentage conversion -- Basic percentage concepts.

Mark that all these rich percentage concepts and techniques are firmly based on the basic concepts of relevant topics such as profit and loss, ratio and proportion or mixing liquids. We use these rich concepts for faster and more elegant solution of problems which otherwise also could have been solved using conventional procedural methods.

Lastly it is true that we have formed and used these rich concepts while solving new type of problems. This set of concepts naturally is then open-ended and extendable.


Problems as exercise

Problem 1.

Instead of giving a discount, for every 19 kites sold, a kite-seller gives 1 kite extra free. In order to give 5% more discount, the number of extra kites the kite-seller needs to give free in a sale of 27 kites is,

  1. 3
  2. 6
  3. 8
  4. 7

Problem 2.

Two fixed points A and B are 15 cm apart. A third point C is located between the two points on the straight line joining them so that length of AC is 9 cm. If the point C is moved towards point B so that the length of AC is increased by 6%, the length of BC must have been decreased by,

  1. 6%
  2. 8%
  3. 7%
  4. 9%

Problem 3.

In a three hour test, half way through, out of a total number of 200 questions Abhi answered 40% of the questions. What should be the percentage increase in average answer rate during the rest of the time so that Abhi can answer all the questions?

  1. $60$%
  2. $50$%
  3. $40$%
  4. $33\frac{1}{3}$%

Problem 4.

In a final exam, Vikas scores 30% but fails to pass by 5 marks, while Chandra who got 10 more marks than the pass marks scores 40%. The pass marks is then,

  1. 100
  2. 70
  3. 50
  4. 150

Problem 5.

The price of an item was increased by 10%. This caused a reduction in monthly total item sales by 20%. The overall effect of price increase on the monthly total sale value is then,

  1. 10% increase
  2. 12% decrease
  3. 12% increase
  4. 10% decrease

Problem 6.

A number is increased by 20% and then again by 20%. By what percent should the inreased number be reduced to get back to the original number?

  1. $30\displaystyle\frac{5}{9}$%
  2. $44$%
  3. $19\displaystyle\frac{11}{31}$%
  4. $40$%

Problem 7.

In a basic science library 25% of the books are on Bioscience, 200 are on Physics and the rest are on Chemistry. If the ratio of number of books on Chemistry, Physics and Bioscience is, 5 : 4 : 3, how many new books only on Bioscience are to be added to the library collection to make the percentage of Bioscience books as 40%?

  1. 75
  2. 150
  3. 100
  4. 125

Problem 8.

The sum of two numbers is 520. If the larger number is decreased by 4% and the smaller number is increased by 12%, the numbers thus obtained become equal. The smaller of the two original numbers is then,

  1. 300
  2. 240
  3. 280
  4. 210

Problem 9.

In an examination 65% of the students passed in Mathematics, 48% passed in Physics and 30% passed in both. How much percentage of students then failed in both subjects?

  1. 13%
  2. 17%
  3. 43%
  4. 47%

Problem 10.

A's salary is 40% of B's salary and B's salary is 25% of C's salary. What percent of C's salary is A's salary?

  1. 15%
  2. 5%
  3. 20%
  4. 10%

Answers to these question are given below and refer to SSC CGL level Solution Set 69, Percentage 3 for detailed conceptual solutions.


Answers to problems

Problem 1. Answer: Option a: 3.

Problem 2. Answer: Option d : 9%.

Problem 3. Answer: Option b: $50$%.

Problem 4. Answer: Option c: 50.

Problem 5. Answer: Option b: 12% decrease.

Problem 6. Answer: Option a : $30\displaystyle\frac{5}{9}$%.

Problem 7. Answer: Option b: 150.

Problem 8. Answer: Option b: 240.

Problem 9. Answer: Option b: 17%.

Problem 10. Answer: Option d: 10%.


Resources that should be useful for you

7 steps for sure success in SSC CGL tier 1 and tier 2 competitive tests or section on SSC CGL to access all the valuable student resources that we have created specifically for SSC CGL, but generally for any hard MCQ test.

Concept Tutorials on related topics

Basic concepts on fractions and decimals part 1

Basic concepts on Ratio and Proportion

How to solve problems on mixing liquids and based on ages

Basic and Rich Percentage Concepts for solving difficult SSC CGL problems

Efficient solution techniques on related topics

How to solve SSC CGL level arithmetic mixture problems in a few simple steps 1

How to solve SSC CGL level arithmetic mixture problems in a few simple steps 2

How to solve SSC CGL level number and age ratio problems lightning fast

Other SSC CGL question and solution sets on Ratio & Proportion and Percentage

SSC CGL level Solution Set 69, Percentage 3

SSC CGL level Question Set 69, Percentage 3

SSC CGL level Solution Set 68, Ratio proportion 6

SSC CGL level Question Set 68, Ratio Proportion 6

SSC CGL level Solution Set 31, Ratio and Proportion 5

SSC CGL level Question Set 31, Ratio and Proportion 5

SSC CGL Level Solution Set 25, Percentage, Ratio and Proportion 4

SSC CGL level Question Set 25, Percentage, Ratio and Proportion 4

SSC CGL level Solution Set 24, Arithmetic Ratio & Proportion 3

SSC CGL level Question Set 24, Arithmetic Ratio & Proportion 3

SSC CGL level Solution Set 5, Arithmetic Ratio & Proportion 2

SSC CGL level Question Set 5, Arithmetic Ratio & Proportion 2

SSC CGL level Solution Set 4, Arithmetic Ratio & Proportion 1

SSC CGL level Question Set 4, Arithmetic Ratio & Proportion 1

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