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# The cent per cent solution

**Introduction:**

Few mathematical concepts are as popular as per cent. Be it your marks in the examination, the discount at the local market or the growth of the Indian economy, it is a fair chance that the figures mentioned will be followed by the term 'per cent'. In fact, so ingrained is the concept of percentage in human society that even corrupt middlemen demand payments [or cuts] in terms of per cent!

Now, if per cent is so popular, you can scarcely expect it to steer clear of competitive examinations, can you? Sure enough, questions pertaining to percentage are an integral part of competitive examinations all over the world. If there is a question paper on quantitative aptitude or mathematics, you can be sure that an exercise on percentage will be lurking somewhere within it. What's more, per cent is at the core of many other concepts such as profit and loss, rate of growth, simple and compound interest - all of which play their role in competitive exams.

Fortunately, it does not take much to master percentage. If you know your fractions, elementary algebra, and basic mathematical operations [good old add, subtract, divide and multiply], you will be able to handle percentage problems in no time at all.

**Nature of exercises****What you need to solve them****Some strategies****Points to keep in mind****Sample questions with solutions****Send us a question****Feedback**

## Nature of exercises:

As I mentioned in the introduction, the concept of percentage is used in a variety of questions. However, for the sake of simplicity, we will here deal with questions that pertain directly to percentage. Fear not, we will be tackling profit and loss, interest and other related concepts too in the days that come. But now is the time for basics.

Questions on percentage could take the form of a simple equation or might be elaborately described.

Generally, questions on percentage are found in the following flavours:

- Calculating the percentage [yes, sometimes it is actually as simple as that]
- Measuring the change in percentage
- Determining the change in quantities involved
- Finding the original value of the quantities

## What you need to solve them:

The charm of percentage related questions is that most of them boil down to simply using the basic percentage formula.

But before we get cracking, let me just refresh your minds about per cent. The term per cent is derived from the Latin term 'per centum', which literally means 'per hundred' or 'out of hundred'. So any fraction with a denominator of a hundred is a percentage. Any fraction can be expressed in terms of percentage by multiplying it with 100. Thus ¼ can be expressed in per cent as ¼ x100 = 25%.

Similarly, you can convert a percentage figure into a fraction by simply putting 100 in the denominator and the per cent figure in the numerator and reducing it to its lowest terms. Thus 25% is 25/100 or ¼.

When you are dealing with an increase or decrease, the formula to use is: Per cent increase or decrease = [amount of increase or decrease/ earlier amount] x 100

A tricky use of percentage is when it is used to compare two quantities. For instance, if we say that Bob is earning 20% more than David, what we mean is that Bob's income exceeds David's by 20/100 x Bob's income. You can understand the concept better by looking at sample question 3.

Simple enough, isn't it? Believe it or not, that is all you need to solve questions on per cent.

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## Some strategies:

There are no secret methods to unravel the solutions of percentage exercises. One has to know one's mathematics and the formulae to use. It is as simple as that. Of course, there are ways in the process can be made a bit easier. In the case of descriptive questions, I used to make it a point to collect all the numerical information at one place before getting down to the operations. Collecting the information also help me make sure that I was not making any mistake with the figures and helped me understand the question more clearly.

Holmes. So I used to imagine that I was Sherlock Holmes, the information in the question were the clues and the answer was the solution to some terrific mystery. Call me childish To be honest, I used to be [and still am], a huge fan of the detective Sherlock but it seemed to work. It made me focus on the problem and if you can do that, you are on your way.

## Points to keep in mind:

When dealing with purely numerical questions involving different mathematical operations, remember to follow the BODMAS rule for the sequence of different mathematical operations. Quite simply, this is:

BO: Bracket Operations; do the mathematical operations within the brackets first of all.

D: Division; after the bracket operations, do the division.

M: Multiplication; once the division is done, you can proceed to multiplication

A: Addition; after multiplication, you can move on to addition.

S: Subtraction; when all else has been done, you can subtract.

If that sounds a bit complex, do try out this sample question given below:

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[Sample question for BODMAS rule:

What is 25x[10+2]/6+17-4?

Solution:

As per the BODMAS rule,

1. BO: Bracket Operations, so first we tackle the terms within the brackets which is [10+2]=12.

2. D: Division, we move on then to division which is 12/6=2

3. M: Multiplication, multiply 2 by 25=50

4. A: Addition, add 17, 50+17=67

5. S: top it all off by subtraction, 67-4=63>

All this may seem rather elementary but believe me, it is important. For instance, if you had carried out your addition before multiplication, you would have added 2 to 17 and got 19. Multiplying this with 25 would give you 475 and finally subtracting 4 would have made for a final answer of 471 - a solution that is nowhere near the correct answer of 63.

In mathematical operations, order is sacrosanct. ]

I have always found that breaking down a descriptive question into the form of a mathematical equation is the toughest part of a per cent exercise. The sooner you can do this, the quicker you will on your way to a solution. For instance, in the first sample question, we broke down the question to a simple equation of 33m /100 = 231. Solving the equations is a matter of basic mathematics but if you get the equation wrong, you can bid your marks farewell.

Always try to reduce fractions to their lowest level before doing any operations. Let us face it, it is easier to deal with 1/3 than with 25/75.

When percentages are given in decimal points, remember that each decimal point adds a further zero to the denominator. So, while 6% is 6/100, 6.5% is NOT 65/100 but 65/1000. Similarly, 6.52 will 652/10000.

Because in most questions pertaining to percentage can be broken down into an equation, some candidates try to fit in the alternatives given as answers into the equation and choose the correct fit as the answer. I would not recommend trying this as it takes up way too much time.

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## Sample questions with solutions

1. 33% marks are required to pass an examination. A candidate who gets 210 marks fails by 21 marks. The total marks in the examination are:

a. 500 b. 600 c.700 d.800

[RBI examination, 1985]

**Solution:**

1.The candidate got 210 marks and failed by 21 marks. This means that the marks needed to pass the exam are [210+21]=231.

2.As the 33% marks are needed to pass the exam, it follows that 33% of the total marks is 231.

3.Assuming the total marks to be m, we get

33/100 x m = 231

33m/100=231

m=231x 100/33

m=700

Therefore the total number of marks in the examination are 700 and the correct answer is c.

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## Send us a question

Is there a percentage-based question that you cannot solve? Or a question that you would like to share with other candidates? If so, then do mail them to us, along with details of the examination paper / sample paper in which you found them. We will try to solve them and send the solution to you. Any questions that you send will also be uploaded to the Sample Questions section of the web site and you will be given credit for contributing them.

## Feedback

Did this section help you in your preparation for competitive examinations? Do let us know if we are doing a decent job or an inept one. Mail us your opinions and suggestions at editor@enableall.org.

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Alternative approach:

We can also solve this question by resorting to algebra at an earlier stage. Here is how it is done:

1. Let the total number of marks be m.

2. Then in order to pass, a person has to obtain 33% or 33/100 x m marks, which amounts to 33m/100

3. As per the question, the candidate got 210 marks and was 21 marks short of passing. So it follows that

33m/100 - 21 = 210

33m/100 = 210+21

33m/100 = 231

m= 231x 100/33

m =700

As you can see, the solution is the same and even the equations are the same. It is only the start that is different.

Both approaches work. Pick the one you are most comfortable with.

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2. A's salary is 20% below B's salary. By how much per cent is B's salary above A's salary?

a. 16 2/3 b. 25 c. 20 d. 33 1/3

Solution:

1. First of all, resist the temptation to say that as A is earning 20% less than B, B must be earning 20% more than him. He does not.

2. Now, assume that B earns Rs. 100 [it is simpler than assuming than using algebra, trust me. Also notice that here we are concerned with the percentage and not the actual amount]

3. So, A's salary is 20% less than B, or in other words is 80% [100-20] of what B gets.

4. So if B's salary is 100, A's salary will be 80% of Rs.100, which will be Rs.80.

5. Then it follows that B's salary will [100-80]/80 more than A's salary.

6. [100-80] / 80 gives us 20/80 or ¼ or 25%.

7. The answer is 25% or b.

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3. A reduction of 20% in the price of oranges enables a man to buy 5 oranges more for Rs. 10. The price of an orange before reduction was

a. 20 paise b.40 paise c. 50 paise d. 60 paise

[Auditors', 1982]

Solution:

1. Do not head for the algebra yet. Just call up your reserves of common sense. Consider the facts:

There has been a 20% cut in the price of oranges.

The man spends Rs 10 on oranges.

2. Therefore after the price cut, the man needs to spend 20% less to get the same number of oranges that he used to get before the price cut. 20% of Rs. 10 is Rs. 2. The man saves Rs 2 due to the price cut,

3. These Rs.2 enable him to buy 5 additional oranges.

4. So the price of each orange right now will be 2/5 = Rs 0.40, or 40 paise.

5. Right, now let us go for the algebra. Let the price before the reduction be p.

6. Then p - 20%= 40 paise

7. p - 20/100 = 40 paise

p - 1/5=40 paise

4p/5 = 40 paise

p = [40x5/4 ] paise

p = 50 paise.

The price of oranges before the reduction in price was 50 paise per orange. The correct answer is c.