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Numerical series in psychotechnical tests, how to overcome them

Numerical series in psychotechnical tests, how to overcome them

With this entry dedicated to numerical series, we inaugurated a new section in which we will talk about psychotechnical tests, and how to overcome them successfully.

We will see different types of questions, and some techniques that will help us find the solution in each case.

The numerical series they are the most common type of question that we will find in the psychotechnical tests, and it consists of a sequence of numbers, in which each element can be deduced, by means of a logical process or mathematical calculation.

Content

  • 1 Fixed Factor Arithmetic Series
  • 2 arithmetic variable factor series
  • 3 Geometric series with fixed factor
  • 4 Geometric series of variable factor
  • 5 Series with powers
  • 6 Alternative Series
  • 7 Series with fractions
  • 8 Series with compound factor
  • 9 batch series
  • 10 Multiple interleaved series
  • 11 Calculation of core values
  • 12 The 4 golden rules to pass the psychotechnical tests

Fixed Factor Arithmetic Series

Let's start with a very easy example, which will help us see how this type of series behaves.

Could you say what is the number that continues this series?

1  · 2 · 3 · 4 · 5 · ?

Obviously, the next element of the series is number 6. It is a growing series, since the increase between each element is positive, specifically: (+1). We will call this value the series factor.

It is a simple case but it already shows us the basis of this type of series, and that is: each element of the series, is obtained by adding a fixed value, to the previous element.

If the fixed value or factor is positive, the series will be increasing, and if it is negative, it will be decreasing.

This same idea can be used to create more complicated series, but that follow the same principle. Look at this other example:

27 · 38 · 49 · 60 · ?

Guess what is the number that continues the series?

In this case, the next value would be 71.

It is a series, of the same type as we have seen before, only, in this case, the increase between every two elements is +11 units.

In a psychotechnical test, to see if we are faced with a series of fixed factor, it is useful to subtract each pair of values, to see if it always matches.

Let's see it more graphically with this other example. Guess, what is the next element of this series?

4 · 1 · -2 · -5 · ?

Although we see that the factor is repeated in the first elements, it is important to subtract for ALL the terms of the series, since it could be the case that it was a series that evolves differently and the only way we have to make sure, it is calculating the difference between all the elements.

Let's place the value of this subtraction between each pair of numbers:

4   ·   (-3)   ·   1   ·   (-3)   ·   -2   ·   (-3)   ·   -5   ·   ? 

We will call the original series: main series. We will call the series formed by the differential between each two elements (numbers in brackets): secondary series.

We see that the difference is the same in all pairs of elements, so we can deduce that the next term in the main series is obtained by subtracting 3 from the last value, -5, so that we will have -8.

In this case, it is a decreasing series, with a fixed factor (-3), and with the added difficulty, that we have positive and negative values ​​in the series, since we cross zero, but the mechanism used continues to be exactly the same, as that of the first series we saw.

Normally, the psychotechnical tests are structured with increasing difficulty, so that the problems are becoming more complicated and it will take us more time to solve them as we go along.

Knowing this, it is very likely that the first series we meet are of this type and can be solved easily and quickly with a bit of agility in mental calculation.

Arithmetic series of variable factor

Watch this series and try to solve it:

1 · 2 · 4 · 7 · 11 · 16 · ?

Do you know how it continues?

At first glance it may not be obvious, so let's apply the technique we have learned before.

Let's subtract between each pair of consecutive numbers to see if we find out something:

Main Series: 1 · 2 · 4 · 7 · 11 · 16 · ?

Secondary series: 1 · 2 · 3 · 4 · 5

Secondary series differential: 1 · 1 · 1 · 1

When doing the subtraction, we clearly see that an incremental secondary series appears to us, as we saw in the previous section, so that the jump between each two values ​​of the main series is not a fixed factor, but is defined for a series with fixed increase +1.

Thus, the next value of the secondary series will be 6, and we only have to add it, to the last value of the main series, to obtain the result: 16 + 6 = 22.

Here we have had to work a little more, but we have only done the same method twice. First, to obtain the variable factor series and then to obtain the increment of this new series.

Let's consider another series that follows this same logic. Try to solve it:

6 · 9 · 15 · 24 · 36 · ?

We will follow the subtraction method we know to solve it:

Main Series: 6 · 9 · 15 · 24 · 36 · ?

Secondary series: 3 · 6 · 9 · 12

And we will reapply the subtraction method with the secondary series:

Tertiary series: 3 · 3 · 3 (Differential of the secondary series)

In other words, our main series is increased according to a secondary series, which increases three by three.

Therefore, the next element of the secondary series will be 12 + 3 = 15 and this will be the value that must be added to the last element of the main series to obtain the following element: 36 + 15 = 51.

We can find series that need more than two levels of depth to find the solution, but the method we will use to solve them is the same.

Geometric series with fixed factor

Until now, in the series that we have seen, each new value was calculated by adding or subtracting from the previous element of the series, but it is also possible that the increase in values ​​occurs, multiplying or dividing its elements by a fixed value.

The series of this type, they can be easily detected as their elements grow or decrease very quickly, depending on whether the operation applied is, a multiplication, or a division respectively.

Let's see an example:

1 · 2 · 4 · 8 · 16 · ?

If we apply to this series, the method we have seen before, we see that we do not reach any clear conclusion.

Secondary series: 1 · 2 · 4 · 8

Tertiary series: 1 · 2 · 4

But if you look, the series grows very quickly, we can assume that the increase is calculated with a multiplication operation, so what we will do is try look for a link, between each item, and the next one, using the product.

By what number do we have to multiply 1 to get 2? Well evidently by 2: 1 x 2 = 2.

And we see that, if we do it with all the elements of the series, each is the result of multiplying the previous value by 2, so the next value in the series will be 16 x 2 = 32.

For this type of series, we do not have such a mechanical method as we used in arithmetic series. Here we will have to try to multiply, each element, with different numbers, until we find the appropriate value.

Let's try this other example. Find the following item in this series:

2 · -6 · 18 · -54 · ?

In this example, the sign of each element alternates between positive and negative, which indicates that our multiplication factor will be a negative number. We have to:
2 × -3 = -6
-6 × -3 = 18
18 × -3 = -54

so, the next value in the series, we get it by multiplying -54 × -3 = 162.

Psychotechnical tests are usually of the test type, where we have to mark the correct answer, among which we have available. This can help us to verify if we have made a mistake in our calculations, but it can also play against us, when we quickly answer the questions. Imagine that the answers available for the previous series are the following:
a) -152
b) -162
c) None of the above

If we do not look, we can erroneously mark option b) in which the value is correct but the sign is wrong.

To increase the confusion, the other possible response also has a negative sign, which can lead us to believe that we were wrong with the sign. The correct answer would be option "c".

The examiner is aware that, having several results to choose from, simplifies the task of solving the problem, so he will probably try create confusion with available responses.

The difficulty associated with this type of series is that, if we have large numbers, we will have to make complicated calculations, so it is very important to master the multiplication tables and be able to perform mentally, operations with 2 or 3 digit numbers , since, we will not always have paper and pencil to do the calculations.

Geometric series of variable factor

Let's complicate a little more, the geometric series we had seen, making the multiplication factor a variable value. That is, the factor by which we will multiply each element will increase as if it were a numerical series.

Let's start with an example. Take some time to try to solve this series:

2 · 2 · 4 · 12 · 48 · ?

You've got it? This series cannot be solved with the methods we have seen so far, since we cannot find a fixed value, which allows us to obtain each element from the previous one through multiplication.

So, let's look for the factor, by which we have to multiply each element to get the next one, to see if it gives us any clues:

Secondary series: × 1 · × 2 · × 3 · × 4 ·?

We see that, to get each element of the series, we must multiply by a factor, which is increasing, according to a growing arithmetic series.

If we calculate the next value of this secondary series, 5, we have the factor, by which we must multiply, the last value of the main series, to obtain the result: 48 x 5 = 240.

In this case, the secondary series was an arithmetic series, but we can also find, with geometric or other series, which we will see later.

Try now, solve this series:

1 · 2 · 8 · 64 · ?

You got it? In this case, if we obtain the secondary series with the multiplicands we find this:

×2 · ×4 · ×8 · ?

That, clearly, is a geometric series, in which each element is calculated by multiplying the previous one by 2, so the next factor will be 16, and this is the number by which we must multiply the last value of the main series, to get the result: 64 x 16 = 1024.

Series with powers

Until now, all the series that we have seen evolved according to operations of addition, subtraction, multiplication or division but it is also possible that they use the powers or the roots.

Normally we will find powers of 2 or 3, if not, the numbers obtained are very large, and it is difficult to solve the problem with complex calculations, when What is sought with these types of problems is not so much the calculation skills, but the ability for deduction, the discovery of patterns and logical rules.

That is why it is very useful to memorize the powers of 2 and 3 of the first natural numbers, to easily detect this type of series.

Let's start with an example:

0 · 1 · 4 · 9 · 16 · ?

If we try to find a relationship that allows us to find each element with the methods we have used so far, we will not reach any conclusions. But if we know the powers of two, (or squares), of the first natural numbers, we will see immediately, that this series is the succession of the squares from zero to 4: 0² · 1² · 2² · 3² · 4²

So that the next element will be 5² = 25.

Let's see one last example, let's see how you get this kind of problem. Try to solve this series:

-1 · 0 · 1 · 8 · 27 · ?

This case may not be so obvious, but it will help you to know the powers of 3 (or cubes) since we will immediately recognize the values ​​and see that the series is obtained by calculating the cubes from -1 to 3: -1³ · 0³ · 1³ · 2³ · 3³

Now we clearly see that the next element will be 4³ = 64.

Alternative series

In all the series that we have seen so far, the way to achieve the following element has been applying mathematical calculations, but there are many cases in which it is not necessary to perform any mathematical operation to find the result.

Here, the limit is in the examiner's imagination, but we are going to give you enough guidelines so that you can solve most of the series of this type that you can find.

Fibonacci Series

They receive this name thanks to Fibonacci, who is the mathematician who disclosed this type of series, and although in the original sequence the sum is used to calculate the elements of the series, here we will group all the series whose elements are obtained only from of its own members, regardless of whether we need to use the sum, the product or any other type of mathematical operation.

Let's see an example. Watch this series:

2 · 3 · 5 · 8 · 13 · 21 · ?

Are you able to find the following term? We will try to solve it with the methods we know.

As the numbers do not grow very fast, we will assume that it is an arithmetic series and we will apply the method we know to try to reach some conclusion.

When calculating the subtraction between each pair of elements this secondary series appears: 1 2 3 5 8

We see that it is not a series with fixed increase, so let's see if it is a series with variable increase:

If we calculate again the difference between each two elements of this new series we get the following: 1 1 2 3

Nor is it an arithmetic series of variable increment! We have applied the methods we know and we have not reached any conclusions, so we are going to make use of our ability to observe.

If you look at the values ​​of the secondary series, we see that they are the same as those of the main series but displaced a position.

This means that the difference between an element of the series and the following is exactly the value of the element that precedes it or what is the same, each new value is calculated as the sum of the two previous elements. So we will calculate the following element by adding to the last number the one that precedes it in the series: 21 + 13 = 34. Got it!

Keep in mind that in this case, the first two terms of the series do not follow any defined pattern, they are simply necessary to calculate the following elements.

This is a simple case, but it is also possible to find series that use operations other than the sum. Let's complicate it a little more. Try to discover the value that follows in this series:

1 · 2 · 2 · 4 · 8 · 32 · ?

In this case, we do see that the values ​​increase very quickly, which gives us a clue, that surely it is a geometric series in which we will have to use multiplication, but, clearly it is not a series with increase by multiplication of a fixed value. If we try to obtain the multiplication factors, to see if the increase is calculated with a multiplication by a variable value we see the following: × 2 · × 1 · × 2 · × 2 · × 4

If we look, we see that again the values ​​of the main series are repeated in the secondary series, so we can conclude that the next value of the secondary series will be the value that follows the 4 in the main series, that is, the 8 and therefore when multiplying 32 x 8 = 256 we will get the next value of the series.

We will do one last exercise on this type of series. Try to solve it:

-4 · 1 · -3 · -2 · -5 · -7 · ?

Knowing the type of series that we are trying, makes things much easier for us, since we can immediately see, that, each value, is obtained as the sum of the previous two, so the answer is -5 + (-7) = -12.

In the examples we have seen in this section, all the calculations were based on using the two previous values ​​of the series, but, you can find cases in which more than 2 elements or even alternate elements are used. Let's see a couple of examples of this type. Try to solve them with the indications we have given you:

3 · 3 · 4 · 10 · 17 · 31 · ?

In this case, it is clear that it is not enough to add two terms to obtain the following, but, if we try to add three, we see that we achieve the expected result:

3 + 3 + 4 = 10
3 + 4 + 10 = 17
4 + 10 + 17 = 31

So, the following term will be equal to the sum of the last three elements: 10 + 17 + 31 = 58.

And now a final example of this type of series:

1 · 1 · 1 · 2 · 3 · 4 · 6 · ?

This series is not trivial, but if you have been attentive to the clues, you will have tried to add alternate numbers, and perhaps you have found the solution. The first three elements are needed to obtain the first calculated value, which is obtained as the sum of the previous element plus that found three positions beyond, that is to say:

1 + 1 = 2
1 + 2 = 3
1 + 3 = 4
2 + 4 = 6

So that the next element will be 3 + 6 = 9.

Series with prime numbers

Watch this series:

2 · 3 · 5 · 7 · 11 · 13 · 17 · 19 · ?

You can try to solve it, using any of the methods we have seen so far and you will not get anything. In this case, the secret is in the prime numbers, which are those that are only divisible by themselves and by the unit, taking into account that 1 is not considered a prime number.

The elements of this series are the first prime numbers, so finding the next value does not depend on whether we perform any mathematical operation, but rather that we have realized this.

In this case, the next element in the series will be 23 That is the next prime number.

Just as it is useful for us to memorize the first powers of natural numbers to more easily solve some series, it is also important to know the prime numbers to detect this type of series more quickly.

Changes in the position and alteration of the individual digits

We know that the digits are the individual figures that make up each number. For example, the value 354 is made up of three digits: 3, 5 and 4.

In this type of series, the elements are obtained by modifying the digits individually. Let's see an example. Try to solve this series:

7489 · 4897 · 8974 · 9748 · ?

This series does not follow any clear mathematical pattern, but, if we look closely, we can see that the digits of each of the elements of the series are always the same but changed in order. Now we just need to see what is the pattern of movement that the figures follow.

There are no universal laws here, it's about trial and error. Normally, the digits rotate or are exchanged. It can also happen that the digits increase or decrease cyclically or oscillate between several values.

In this specific case, we can see that the numbers seem to shift to the left and the end number goes to the position of the units. Thus the next value in the series will be the initial number again: 7489.

Increase or decrease in the number of figures

It is usual to meet sometimes with series that have very large numbers. It is unlikely that the examiner intends that we perform operations with a number of 5 or more figures, so in these cases, alternative behaviors must be sought.

In this type of series, what changes is the number of digits of each element. Let's see an example. Try to find the following element of this series:

1 · 12 · 312 · 3124 · 53124 · ?

In many cases, the visual aspect of the numbers will help us find the solution. In this series we see that one more digit appears, with each new element and that the digits of the previous element also appear as part of the value.

The digit that appears in each new element follows an incremental series and appears alternately to the right and left. The series begins with 1, then 2 appears on the right, in the next term 3 appears on the left and so on, so to obtain the last term we will have to add the number 6 to the right of the last element of the series and it will be: 531246.

Other cases

The limit on the complexity of the series is limited only by the examiner's imagination. In the most complex questions of the test we can find anything that can occur to us. We are going to propose a somewhat peculiar exercise as an example. Try to find the term that follows in this series:

1 · 11 · 21 · 1211 · 111221 · ?

The truth is that this series, there is nowhere to catch it. We can assume that it is not a conventional series, since the growth of numbers is very strange. This can give us a clue that the solution will not be obtained by calculating but seeing how the numbers progress.

Let's see the solution. The first value is the seed of the series and it is normally imposed on us, so we will begin with the following term, 11. The secret of this series is that, each element is a numerical representation of the digits that appear in the previous term.

The first element is a one: 11
The second element consists of two ones: 21
The third element contains a two and a one: 1211
The room has a one, a two and two ones: 111221
Therefore, the following element will be: three ones, two doses and one one: 312211

We cannot prepare you for everything you can find but if we want to help you open your mind and your imagination so that you consider all kinds of possibilities.

Series with fractions

Fractions are expressions, which indicate a quantity of portions that are taken from a whole. They are expressed as two numbers separated by a bar that symbolizes division. In the upper part (to the left in our examples), called numerator, the number of portions is indicated and in the lower part (to the right in our examples), called denominator, the quantity that forms the whole is indicated. For example, fraction 1/4 represents a quarter of something (1 portion of a total of 4) and results in 0.25.

The series with fractions will be similar to those we have seen so far with the proviso that in many cases, the examiners, play with the position of the digits when obtaining the elements of the series.

Let's look at a simple example series:

1/3 · 1/4 · 1/5 · ?

You don't have to know much about fractions or be a lynx to discover that the next element in the series will be 1/6, right?

The difficulty of the series with fractions is that sometimes we can have a series for the numerator and a different one for the denominator or we can find a series that treats both parts of the fraction as a whole. Simplification of fractions also increases the difficulty since the same value can be expressed in several different ways, for example ½ = 2/4. Let's look at a case of each type:

1/2 · 1 · 3/2 · 2 · ?

If you are not used to working with fractions, you may have to do a bit of recycling to get fluent with basic operations: addition, subtraction, multiplication and division with fractions.

In this example, each term is the result of adding the fraction ½ to the previous value. If we add ½ to the first value we get 2/2 which is equal to 1 and so on until the end, so that the last element will be 2 + ½ = 5/2.

Well, we have seen a simple case that is nothing more than an arithmetic series with a fixed increase but using fractions. Let's complicate it a little more. Try to find the following term in this series:

1/3 · 4/6 · 7/9 · 10/12 · ?

If you look closely, you will see that in this case the fraction is being treated as two different series, one that advances in the numerator adding 3 to the previous one and another in the denominator that also adds 3 to the previous denominator. In this case we do not have to think of a fraction as a single numerical value, but rather as two independent values ​​separated by a line. The next term will be 13/15.

When we have series of fractions, much of the difficulty is in discerning whether the fractions are treated as single values ​​or as independent numerator and denominator values.

Going back to the last series we've seen, think that too you can find the series of simplified fractions which greatly hinders its resolution. Look how the previous series would look with the simplified terms:

1/3 · 2/3 · 7/9 · 5/6 · ?

The series is exactly the same and the solution too, but it is much harder to solve.

We are going to see another much more complicated case. I'll give you a clue. Fractions are treated as two independent numerator and denominator values:

6/3 · 3/4 · 18/15 · 7/8 · ?

And these are the possible answers:

a) 11/14
b) 27/30
c) 10/9

Have you tried to solve it? Have you reached any conclusion? Seen this way, this series seems not to follow a clear criterion. The terms increase and decrease almost randomly.

Now we are going to rewrite the series with the terms without simplifying:

6/3 · 9/12 · 18/15 · 21/24 · ?

What about now? You see some pattern. As we have commented, in this case, the numbers of the fractions are treated as independent values. If you look you will see that starting with the denominator of the first term, add 3 to get the numerator and add 3 again, to get the numerator of the second term, to which we add again 3 to get the denominator and thus, making a kind of zigzag with the numbers until you reach the last term so the value we are looking for is 30/27. But if we look at the possible solutions, we see that option b) inverts the numerator and denominator values ​​so it is a different value but if we try to simplify the fraction 30/27, we get 10/9 which is the answer c).

Apart from all that has been seen, it should be taken into account that, as in series with integers, it is possible that the increase is achieved by multiplying by a value or with a factor that increases or decreases in each term. Let's see a complex example to close this section:

1 · 2 · 2 · 8/5 · 40/35 · ?

In this case, we will proceed by trial and error: To get 2 from 1, we can add 1 or multiply by 2. If we try to obtain the rest of the values ​​with these fixed terms we see that they are no longer useful to obtain the third element. We will then assume that it is an arithmetic series so we will calculate the difference between every two terms to see if we reach any conclusion:

Secondary series: 1 · 0 · -2/5 · -16/35

There seems to be no clear pattern, so let's rewrite these fractions with a common denominator that will be 35. We would have this:

Secondary series: 35/35 · 0/35 · -14/35 · -16/35

Nor does it seem that we get anywhere, so let's treat our series as a geometric series. We will now calculate the value by which each term must be multiplied to obtain the following:

Secondary series: × 2 · × 1 · × 4/5 · × 5/7

These numbers already seem more affordable but do not give us a clear sequence. Maybe they are simplified. Following the progress of the last two elements of this secondary series where the numerator is increased by one and the denominator by two, we see that the second term can be rewritten as 3/3 = 1, and following the same criteria we have that the first number it should be 2/1 and so it is!

This would be the series without simplifying to see it more clearly:

Secondary series: × 2/1 · × 3/3 · × 4/5 · × 5/7

Therefore, we have come to the conclusion that it is a geometric series, in which, the fraction that is used to obtain each element, increases by one unit in the numerator, and in two units in the denominator, so the next term will be 6/9 and if we multiply it by the last term of the main series we have to 40/35 x 6/9 = 240/315 that simplified, we have 48/63 left.

All the concepts that we have seen in this section, you can also apply in the series of dominoes, since they can be treated as fractions, with the only caveat that the numbers go from zero to six in a cyclical way for what is considered that after six goes zero and before zero goes six.

Series with compound factor

In all the series we have seen so far, the factor we used to calculate the next term was a single value, or series of values, on which we performed a single operation to obtain each element. But to complicate things a bit more, those factors can also be composed of more than one operation. Let's solve this example to see it more clearly:

1 · 2 · 5 · 10 · 17 · ?

These are numbers that grow very quickly, so we can think of a geometric series or a power, but we do not find integer values ​​or powers that exactly generate the values ​​of the series. If we look a little, we see that the values ​​of the series are suspiciously close to the squares of the first natural numbers: 1, 4, 9, 16 in fact they are exactly one unit away so we can deduce that the values ​​of this series will be obtained starting with zero and calculating the square of each integer and adding 1.

This is a concrete case that uses addition and power but we could have any combination of addition / subtraction with product / division and power.

Batch series

So far, in all the series, in the