How to solve a difficult SSC CGL level problem on percentage in seconds by conceptual reasoning

Solve a difficult percentage problem in a few seconds by number domain concepts

In this session of efficient math problem solving we will show how a difficult SSC CGL level percentage problem can be solved few seconds by using very basic number domain concepts along with basic percentage concepts.

Before taking up the problem solving exercise we will first try to clearly state the Basic percentage concepts. This part is taken from our detailed session on Concepts of Ratio and Proportion which also you may refer to for the few solved example problems on percentage which we are not including here.

Percentage concepts and its use

Percentage represented by the symbol ${\%}$ is a special kind of ratio in which the denominator is converted to 100.

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.

Use of percentage in daily life

In academic life we find frequent use of percentage concept 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 percentage score.

Here two important aspects of percentage use comes out,

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

The equivalent decimal value of $85\text{%}$ is,

$\displaystyle\frac{85}{100}=0.85$.

It is so because by definition of percentage, when the actual score is $85\text{%}$ of the total marks, $100\text{%}$ is the total marks. In other words, the actual marks is $0.85$ portion of the total marks. That is why to get the actual value expressed as percentage of a total value, the percentage is converted to its decimal equivalent form by dividing it by 100 first.

On many occasions in real life, the percentage is used to express a value without mentioning the reference anchor 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.

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 comparison tool. 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.

This brings us to the second way to use percentage when in multiplication. 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.

If the last year's profit were Rs.20000, this year's profit will be,

$20000\times{\displaystyle\frac{120}{100}}=20000\times{\displaystyle\frac{6}{5}}=\text{Rs.}24000$. 120 in the numerator indicates increase of 20% from 100%.

We can call these concepts as rich percentage concepts.

Though the percentage concept is simple it is one of the most universally used concepts specially in business environment.

With this grounding we will go straight into the process of solving our chosen percentage problem lightning fast by use of concepts only, bypassing any linear equation formation and solving.

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Chosen Problem 1.

A number is divided into two parts in such a way that 80% of the second part is 6 more than 90% of the first part and 80% of the first part is 3 more than 60% of the second part. The number is,

1. 130
2. 135
3. 125
4. 145

Problem solving in a few seconds by mathematical reasoning using basic number domain concepts for key pattern discovery

Deductive reasoning and choosing the right strategy

The conventional approach of forming two equations with each part as one of the two variables and solving for the variables from the two linear equations should take more time than we liked.

So we looked for a pattern by reading the problem carefully. In any case you have to read the problem you are going to solve and simultaneously analyze it. Primarily we searched for a pattern to use as a key to the solution.

It didn't surprise us much as we found the pattern quickly. It is there if you look for it. It should be there as these problems are meant to be solved quickly using such pattern identification and special strategies. Basically this approach is the problem solver's approach which is efficient and quite different from the conventional approach.

Key pattern discovery

There are two statements. We looked closely into the first statement first. We need to examine the potential of each parts of a problem. Two statements form two parts of the problem. This is problem breakdown for key pattern discovery.

First statement: 80% of the second part is 6 more than 90% of the first part.

The second portion of the statement, "90% of the first part" must be a multiple of 3 we decided.

It is $0.9$ multiplied by the anchor value as we call the original value on which percentage is expressed. This is basic percent concept and use. As one of the factors, $0.9$ is a multiple of 3, the result "90% of the first part" must also be a multiple of 3.

Automatically our eyes fell on the difference between the 90% of the first part and the 80% of the second part. The difference being 6, 80% of the second part also must be a multiple of 3. Primarily you are adding 6, a multiple of 3, to "90% of the first part", which is also a multiple of 3. So the result of addition, 80% of the second part, must also be a multiple of 3.

This reasoning is simple. If you add a multiple of 3 to another multiple of 3, the factor of 3 remains in the result which turns into a multiple of 3 with certainty. This is application of basic number domain concepts of factors and multiples.

This is first level reasoning but we have got the crucial key pattern we knew.

Second level reasoning: If 80% of the second part is a multiple of 3, the second part itself must be a multiple of 3, as obviously the percentage multiplier $0.8$ has no 3 as factor in it. Again we apply the basic factors and multiples concepts.

Second statement analysis with knowledge of key pattern

Applying similar reasoning to the second statement, "80% of the first part is 3 more than 60% of the second part", we found immediately that 3 more than 60% of the second part is a multiple of 3 and as it equals 80% of the first part, the 80% of the first part, and so the first part itself must also be a multiple of 3.

Final conclusion

Together then the whole number must be a multiple of 3. If two multiples of 3 are added, the result must also be a multiple of 3, that is the reason.

This again is application of basic factors and multiples concepts.

Use of free resources of choice values

Now armed with the nature of the solution, we looked at the choice values, a use of free resources. Only 135 met the criterion.

Here we have used the method of first adding the digits of a number and check whether the integer sum thus formed is a multiple of 3. This is the fastest method for checking multiple of 3 condition and all know it, we think. This is application of Divisibility concepts for 3 divisibility.

The four solution choice numbers, 130, 135, 125, 145 have the integer sums, 4, 9, 8, and 10 of which only 9 is divisible by 3, so the corresponding number 135 must also be divisible by 3, and must be the answer.

Answer: Option b: 135.

Note: We feel, the clues were left to be discovered. The reason for our belief comes from the experience of solving numerous problems very fast using conceptual strategies, deductive reasoning and key pattern discovery. The fact remains that if one goes in for solving such problems using formula or statement based formal deductions, time taken will invariably be much more than the conceptual problem solver's path and will generally exceed the apportioned time limit of 1 minute per problem.

Key Concepts used: Deductive reasoning -- Problem breakdown technique -- Pattern recognition after judging the longer path of conventional solution -- Use of number domain concepts -- Basic factors and multiples concepts -- Divisibility concepts -- Use of basic parcentage concepts -- Anchor value of percentage concept -- Mathematical reasoning -- Concepual reasoning -- Principle of free resource use.

Important: Even though the conceptual solution is elaborately explained, it actually takes less than 20 seconds for a conceptually aware problem solver. Inherently this solution path is easy and natural involving no complicated calculations as a bonus.

To contrast this solution with the deductive solution we will now expose the deductive solution.

Conventional deductive solution based on linear equation solving in two variables

First statement, "80% of the second part is 6 more than 90% of the first part", gives us the linear equation,

$0.8B=0.9A + 6$, where $A$ and $B$ are the first and second parts and we have duly converted the percentages to their equivalent decimals.

Or, $8B=9A+60$, we have removed two decimals with multiplication by 10. This is efficient simplification.

Caution: from the statement "80% of the second part is 6 more than 90% of the first part", when you form the equation, the apparently harmless operation signified by the word "more" is error prone (just the way some particularly unstable tricky portion of a road is marked as "accident prone"). Unless you are a bit cautious, the 6 might be added to "0.8B" rather than to "0.9A". We have seen this happen. The unfailing rule is,

Whichever quantity is more by an amount from a second quantity, in the equivalent equation, "more" amount will go the other side of the first quantity and will fit in cozily with the second quantity. Example, "if 7 is 3 more than 4", we write, $7=4+3$.

In contrast to Efficient simplification principles, this is a part of what we call, Safe simplification principles.

From the second statement, "80% of the first part is 3 more than 60% of the second part" similarly we get,

$0.8A=0.6B+3$,

Or, $8A=6B+30$.

Multiply the second equation by 4,

$32A=24B+120$,

and multiply the first by 3,

$24B=27A+180$,

Substitute the value of $24B$ to the earlier equation,

$32A=27A+180+120=27A+300$,

Or $5A=300$,

Or, $A=60$,

Or, $8A=480=6B+30$,

Or, $6B=450$,

Or, $B=75$.

So, the number is, $A+B=60+75=135$.

Mathematical peace. Took a bit more time and it was not so interesting, even then we are home.

Concepts used: Basic percentage concepts -- Efficient simplification -- Safe simplification principles -- Solving linear equations in two variables.

When we solve a problem in two ways we apply the much favored Many ways technique that improves problem solving skill.

Basic concepts application approach

In conceptual solution, we have used very basic percentage concepts and number domain concepts, nothing complicated or difficult at all. This approach belongs to the broader and more powerful Basic concepts application approach which says,

The more basic a concept you use for solving a problem satisfactorily, the more efficient your solution will be, particularly so if you combine basic concepts from multiple domains or topic areas.

The method uses mathematical reasoning and is not conventional procedure based.