Numerical series in psychotechnical tests, how to overcome them
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- Glen Vandervort Sr.
With this entry dedicated to numerical series, We inaugurate a new section in which we will talk about psychotechnical test, 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 consists, in a sequence of numbers, in which each element can be deduced, through a Logical or mathematical calculation process.
Content
Toggle- Arithmetic fixed factor series
- Arithmetic series of variable factor
- Geometric series with fixed factor
- Geometric series of variable factor
- Series with powers
- Alternative series
- Fibonacci series
- Series with prime numbers
- Changes in the position and alteration of individual digits
- Increase or decrease in the number of figures
- Other cases
- Series with fractions
- Composite factor series
- Discontinuous series
- Multiple interspersed series
- Central values calculation
- The 4 gold rules to overcome psychotechnical tests
Arithmetic fixed factor series
Let's start with a very easy example, which will help us see how this type of series behaves.
Would you know how to say what is the number that this series continues?
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 it is that: Each element of the series is obtained by adding a fixed value, to the previous element.
If the fixed or factor value 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 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 following value would be 71.
This is a series, of the same type we have seen before, only that, in this case, the increase between every two elements is +11 units.
In a psychotechnical test, to see if we are facing a fixed factor series, it is useful to subtract each couple of values, to see if it always coincides.
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 make sure, it is calculating the difference between all the elements.
We will place the value of this subtraction between each couple of numbers:
4 · (-3) · 1 · (-3) · -2 · (-3) · -5 · ?
We will call the original series: main series. To the series formed by the differential between every two elements (numbers in parentheses) we will call it: Secondary series.
We see that the difference is the same in all couples of elements, so we can deduce that The following term of the main series is obtained by subtracting 3 at the last value, -5, with what will remain -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 the zero, but the mechanism used, continues to be exactly the same, that the first series we saw.
Normally, psychotechnical tests are structured with increasing difficulty, so that problems are increasingly complicated and will take more time to solve them as we are moving forward.
Knowing this, it is very likely, that the first series that we find are of this type and can be easily and quickly resolved with a little agility in mental calculation.
Arithmetic series of variable factor
Look at 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 evident, so we will apply the technique we have learned before.
We are going to do the subtraction between each couple 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 remains, we see clearly, that an incremental secondary series appears, such as those we saw in the previous section, so that the jump between every two values of the main series is not a fixed factor, but is defined For a series with fixed increase +1.
Therefore, The following secondary series value will be 6, and we have nothing more 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 followed the same method twice. First, to obtain the series of the variable factor and then to obtain the increase in this new series.
We are going to consider another series that follows this same logic. Try to solve it:
6 · 9 · 15 · 24 · 36 · ?
We will follow the method of the subtractions we know to solve it:
Main series: 6 · 9 · 15 · 24 · 36 · ?
Secondary series: 3 · 6 · 9 · 12
And we will apply the subtraction method again with the secondary series:
Tertiary series: 3 · 3 · 3 (Secondary series differential)
That is, our main series, increases according to a secondary series, which increases from 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 meet series, which need more than two levels of depth to find the solution, but the method we will use to solve them is the same.
Charles Spearman and Spearman's correlation coefficient Geometric series with fixed factor
Until now, in the series that we have seen, each new value, was calculated by sums or subtractions on 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 since their elements grow or decrease very quickly, according to 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 we look, that the series grows very quickly, we can assume that the increase is calculated with a multiplication operation, so what we will do is try Find a link, between each element, and the following, using the product.
Why do we have to multiply 1 to get 2? Well, obviously 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 following value of the series will be 16 x 2 = 32.
For this type of series, we do not have a method as mechanical as we used in the arithmetic series. Here we will have to try to multiply, each element, with different numbers, until the appropriate value.
Let's try this other example. Find the following element of this series:
2 · -6 · 18 · -54 · ?
In this example, the sign of each element is alternating 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 of the series, we get it by multiplying -54 × -3 = 162.
Psychotechnical tests are normally. This can help us check if we have been wrong in our calculations, but you can also play against us, when we quickly answer the questions. Imagine that the answers available for the previous series are as follows:
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 confusion, the other possible answer, also has a negative sign, which can make us believe that we have been 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 it will probably try Create confusion with the available answers.
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 , since, we will not always have paper and pencil to make the calculations.
Geometric series of variable factor
We are going to 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 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 a multiplication.
So, we are going to look for the factor, for which we have to multiply each element to obtain the next one, to see if it gives us any clue:
Secondary series: × 1 · × 2 · × 3 · × 4 · ?
We see that, to achieve each element of the series, we must multiply by a factor, which is increasing, according to a growing arithmetic series.
If we calculate the following value of this secondary series, the 5, we have the factor, for 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 ourselves, with geometric or others, 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 multiplenders we find this:
× 2 · × 4 · × 8 · ?
That, clearly, it 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 have to multiply the last value of the main series, to obtain The result: 64 x 16 = 1024.
Series with powers
Until now, all the series that we have seen evolved according to sum, subtraction, multiplication or division operations 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 calculation skills, if not the ability to deduction, the discovery of patterns and logical rules.
That is why it is very useful, 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 that we have used so far, we will not reach any conclusion. But if we know the powers of two, (or squares), of the first natural numbers, we will see right away, that this series is the succession of the squares from zero to 4: 0² · 1² · 2² · 3² · 4²
Hence The next element will be 5² = 25.
Let's see a last example, let's see how these types of problems are given. Try to solve this series:
-1 · 0 · 1 · 8 · 27 · ?
This case perhaps is not so obvious, but it will help you know the powers of 3 (or cubes) since we will immediately recognize the values and we will see that the series is obtained when calculating the cubes from -1 to 3: -1³ · 0³ · 1³ · 2³ · 3³
Now we see clearly that The next element will be 4³ = 64.
What is the PFEiffer geriatric assessment scale (SPMSQ) Alternative series
In all series that we have seen so far, the way to get the next 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 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 announced this type of series, and although the original succession 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, product or any other type of mathematical operation.
Let's see an example. Look at 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 quickly, 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 couple of elements this secondary series appears: 1 2 3 5 8
We see that it is not a series with a fixed increase, so we will see if it is a series with a variable increase:
If we calculate the difference between every two elements of this new the series we get the following: 1 1 2 3
Nor is it an arithmetic series of variable increase! We have applied the methods we know and we have not reached any conclusion, so we will make use of our observation capacity.
If we look at The secondary series values, 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 the next element will be calculated by adding to the last number the one that precedes it in the series: 21 + 13 = 34. Get!
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 values increase very quickly, which gives us a track, that it is surely a geometric series in which we will have to use multiplication, but, clearly it is not a series with an 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 for a variable value we see the following: × 2 · × 1 · × 2 · × 2 · × 4
If we look, we see that again the main series values are repeated in the secondary series, so we can conclude that the following value of the secondary series will be the value that follows to 4 in the main series, that is, 8 and therefore to multiply 32 x 8 = 256 We will obtain the following series value.
We are going to do a last exercise on this type of series. Try to solve it:
-4 · 1 · -3 · -2 · -5 · -7 · ?
Knowing the type of series we are treating, we are very facilitated by things, since we can see right away, that, each value is obtained as the sum of the previous two by what The answer is -5 + (-7) = -12.
In the examples we have seen in this section, all calculations were based on using the previous two 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 get 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 last 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 tracks, you will have tried to add alternate numbers, and you may 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 the three positions beyond, that is to say:
1 + 1 = 2
1 + 2 = 3
1 + 3 = 4
2 + 4 = 6
Hence The next element will be 3 + 6 = 9.
Series with prime numbers
Look at 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 the 1 is not considered a prime number.
The elements of this series are the first prime numbers, so finding the following value does not depend on the fact that we perform any mathematical operation but that we have realized this.
In this case, The next element of the series will be 23 which is the following prime number.
As we find useful, 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 individual digits
We know that digits are the individual figures that make up each number. For example, 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 look at 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 the movement pattern is followed by the figures.
There are no universal laws here, it is essay and error. Normally, digits are rotating or exchanging. It can also happen that digits increase or decrease cyclically or that range between several values.
In this specific case, we can see that the numbers seem to move to the left and the end number goes to the position of the units. Therefore The following value of the series will be the initial number again: 7489.
Increase or decrease in the number of figures
It is common to sometimes meet series that have very large numbers. It is unlikely that the examiner intends to carry out operations with number of 5 or more figures, so in these cases, we must look for alternative behaviors.
In this type of series, what changes is the amount 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 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 the 2nd right appears, in the next term appears on the 3rd 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 we will have: 531246.
Other cases
The limit in 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 take 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 get it by making calculations 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 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 one: 11
The second element consists of two about: 21
The third element contains a two and one: 1211
The room has a one, a two and two about: 111221
Therefore, the next element will be: three, two two and one: 312211
We cannot prepare for everything you can find but if we want to help you open your mind and imagination to consider all kinds of possibilities.
Series with fractions
The fractions are expressions, which indicate a number of portions that are taken from a whole. They express themselves as two numbers separated by a bar that symbolizes the division. In the upper part (to the left in our examples), called numerator, the number of portions and at the bottom (right in our examples), called denominator, indicates the amount that forms the whole. For example, fraction 1/4 represents a quarter of something (1 portion of a total of 4) and has as a result 0.25.
The series with fractions will be similar to those we have seen so far with the proviso that on many occasions, 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 · ?
It is not necessary to know much about fractions or be a lynx to discover that the next element of 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 deals both of the fraction as a whole. The 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 accustomed to work with fractions you may have to do some recycling to take ease with basic operations: sum, 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 2/2 to the first value that is equal to 1 and so on 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 fixed increase but using fractions. Let's complicate it a little more. Try to find the following term of 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 so much about a fraction and a unique numerical value if not as two independent values separated by a line. The next term will be 13/15.
When we have fractions series, much of the difficulty is to discern whether fractions are treated as unique values or as independent numerator and denominator values.
Returning to the last series we have seen, he thinks that also You can find the series of simplified fractions which greatly hinders its resolution. Look how the previous series would be 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 more difficult to solve.
Let's see another much more complicated case. I'll give you a clue. Fractions are treated as two independent values of numerator and denominator:
6/3 · 3/4 · 18/15 · 7/8 · ?
And these are the possible answers:
a) 14/11
b) 27/30
c) 10/9
Have you tried to solve it? Have you reached any conclusion? View like this, this series seems that it does not 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 said, 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 obtain the denominator and thus, making a species of Zigzag with the numbers until reaching the last term so The value we are looking for is 30/27. But if we look possible, we see that option b) invests the values of numerator and denominator so it is a different value but we try to simplify the fraction 30/27, we get 10/9 that is The answer c).
Apart from everything seen, we must bear in mind that as in the series with entire numbers, 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 advance by test 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 no longer serve to obtain the third element. We will assume then 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
It does not seem that there is any clear pattern, so we are going to 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 we seem to get anywhere, so we are going to treat our series as a geometric series. We will now calculate the value for 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 increases by one and the denominator in two, we see that the second term can be rewritten as 3/3 = 1, and following the same criteria we have that the first issue It should be 2/1 And so it is!
This would be the series without simplifying to see it clearer:
Secondary series: × 2/1 · × 3/3 · × 4/5 · × 5/7
Therefore, we have concluded that it is a geometric series, in which, the fraction used to obtain each element, increases in a 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.
All the concepts we have seen in this section, you can also apply them in the dominoes of dominoes, since they can be treated as fractions, with the only proviso that the numbers range from zero to six cyclically for what is considered that after six the zero goes and before zero goes the six.
Composite factor series
In all the series that we have seen so far, the factor we used to calculate the following 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 little more, those factors can also be composed of more than one operation. We are going to 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 whole values or powers that generate exactly 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 are exactly a unit of distance so we can deduce that The values of this series will be obtained by starting with zero and calculating the square of each whole number and adding 1.
This is a specific case that uses sum and power but we could have any sum/subtraction combination with product/division and power.
The differences between the human brain and artificial intelligence Discontinuous series
Until now, in all series, in which we made some calculation on natural numbers, to obtain the elements of the series, we have used consecutive numbers, but it is also possible that the way to build the series is applying a calculation on the numbers pairs (2, 4, 6, ...), for example or on odd numbers (1, 3, 5, ...) or about one in three numbers (1, 3, 5, 6, ...) or even that this separation increases in each element (1, 2, 4, 7, 11, ...).
Let's look at a case. Try to find the following element of this series:
2 · 10 · 26 · 50 · ?
Knowing the type of series that we are trying, it is clear that it is obtained from some type of calculation, on a subset of natural numbers.
Seeing that values grow rapidly, we can deduce that it will be a geometric progression, either by multiplication or power, and if we have in mind the square numbers we will see right away that it is about 2 + 1 powers.
But here, the calculation does not apply to all natural numbers, if not only to the odd. We can rewrite the series in this way, to see it more clearly:
1²+1 · 3²+1 · 5²+1 · 7²+1 · ?
Hence The next element will be 9²+1 = 82.
Multiple interspersed series
To complicate things a little more, some examiners intersperate two or more different series, to form a single. Try to solve this series:
1 · 2 · 3 · 4 · 5 · 8 · 7 · 16 · 9 · ?
We promised them happy, since the first numbers seem consecutive, but after 5, everything falls apart. We can try all the methods seen so far, but we will not succeed, since in this case what we have are two different series interspersed, one formed by the elements of the odd positions (1 · 3 · 5 · 7 · 9) and another formed by the elements of the even positions (2 · 4 · 8 · 16 · ?).
If we write them separately, we easily see that we have an arithmetic series with factor 2 that begins with value 1, interspersed with another geometric series with factor 2 and that begins with value 2.
Seen this way, it is easy to realize that the next value of the complete series will be the following value of the geometric series. As each element is obtained from multiplying by 2 the previous, The solution will be 16 × 2 = 32.
It is unusual that there are more than two interspersed series, but obviously, it is possible. A track that can help us detect multiple series, is that they are usually longer than conventional series, since we need more information to obtain the factors.
Let's see a last year in this section:
2 · 1 · 5 · 2 · 8 · 9 · 11 · 28 · 14 · ?
We have the first track that the series is very long, which is indicative that it is probably a multiple series so we will separate the terms to try to solve it: (2 · 5 · 8 · 11 · 14) This first Part is an arithmetic series with fixed factor +3, although it does not help us to calculate the result since the next term is of the other series: (1 · 2 · 9 · 28 · ?). This partial series grows very quickly so it will probably be a geometric series of some kind. If we have in mind the powers to the cube of the first whole numbers (0, 1, 8, 27) we see that there is only one unit of distance with the numbers of the series, so we deduce that The elements are calculated by raising the entire numbers to the cube and adding 1, so the following term of the series will be 4³ + 1 = 65.
Central values calculation
Normally, in psychotechnical tests, they ask us to find the last term of a series, but it can also happen that the element they ask us is one of the centrals or even the first.
The way of acting here is in essence, the same that until now, only that when an intermediate term is missing, when we look for the factors we will have two questions in the secondary series. Let's look at some cases to clarify this. Let's start with a simple case:
5 · 8 · ? · 14 · 17
The elements grow slowly, so we will assume that it is an arithmetic series, and we will look for the difference between each couple of terms:
Secondary series: 3 · ? · ? · 3
In this case, when we miss a central element in the main series we have two unknowns in the secondary series, so we will look at the elements that we have been able to obtain. Interestingly they are the same number, so we will try what happens if we replace the two unknowns of the secondary series by 3. We have that the term sought would be 8 + 3 = 11 and now we would only have to calculate the following term to confirm that our assumption was correct: 11 + 3 = 14. Perfect! It is an arithmetic series with fixed factor equal to 3.
Let's give a more complicated example, let's see if you can solve it:
5 · 9 · ? · 21 · 25 · 33 · 37
We can start looking for a difference between every two terms, since the series grows slowly and could be an arithmetic series, but we quickly see that this does not lead us to anything. Nor will we find anything looking for a factor that multiplying the elements since the difference between values is small. We could have two different series interspersed but after a few attempts we will not find anything. So ... how about we try the prime numbers? It is clear that the numbers we see are not cousins but perhaps they are multiplied by some factor, so we are going to write the first prime numbers and we will try to turn them into these: 2 · 3 · 5 · 7 · 11 · 13 · 17 · 19
To convert the 2 into 5, we can multiply by 3 and subtract 1 or multiply by two and add 1. Let's see if with any of these options we manage to obtain the second element of the series, but it is impossible to obtain 9 from 3 using the aforementioned operations.
What else can we try? What if the first element of the series corresponds to another prime number? Let's try with 3. To make it 5 you have to multiply by 2 and subtract 1. Okay, we are going to do this same operation with the following prime number: 5 * 2 - 1 = 9, coincides! If we calculate The term that we need using this factor we get the value 13, But we have to make sure, calculating the rest of the values, and we see that everyone can be obtained, with the factor we have calculated, from the list of prime numbers.
Calculate series in which they ask us for the initial value is easier since it is enough to turn all the numbers to have a series with the unknown in the end.
Eidetic memory or photographic memory The 4 gold rules to overcome psychotechnical tests
It is a set of unwritten norms that must always be taken into account when answering the questions of a psycho-technical test And that we collect in this section:
1.- The logical process, which allows us to deduce the following value of a series, must be repeated at least twice in the statement series.
Let's explain it a little better. Look at this series:
2 · 4 · ?
These are the possible answers:
a) 8
b) 6
c) 16
Which is the right answer?
We could assume that each term is calculated by multiplying by 2 the previous value, so the answer would be 8, or we could assume that it is the first natural numbers multiplied by 2 with what the result would be 6. With the first option, we only have a repetition of our logical process, since the first value would be imposed and we would multiply by two to obtain the second value. With the second option, both the first value of the series and the second are obtained using the same factor (natural numbers multiplied by two) so we have two repetitions of our logical process, one to calculate the first value and another to calculate the second, so this should be the valid answer.
2.- If there are several possible solutions, the correct answer is the simplest.
Imagine you have the following series:
1 · 2 · 3 · ?
After all the possibilities we have seen, we can continue the series in several different ways. The most obvious is with 4, but we could also answer that it is the Fibonacci series so the answer would be 5. In general, the correct answer will always be the one that follows the simplest logical process, in this case on 4.
In the case of fractions, if there are several possible answers that symbolize the same value, for example 2/3 and 8/12, in general, the correct answer will be the simplified fraction, in this case 2/3.
3.- If you get stuck with a question leave it for the end.
This is a universal norm of psychotechnical test. It is possible that some questions are resist, so we should leave them for later and continue with the following. Once we arrive at the last question, it is time to review what we have not answered, preferably, in order of appearance in the test, since the questions usually be ordered by difficulty.
4.- Practice is your best ally.
Practicing with real psychotechnical test is the best way to improve, and get the necessary cognitive processes to solve these types of problems, they are almost mechanical.
Only practice will help us discover, what type of series we are facing, in order to apply the corresponding resolution method.
Try to memorize powers of 2, the powers of 3, the prime numbers and practices the mental calculation, to achieve agility when solving the operations.
Here are some links in which you will find evidence of this type to practice:
https: // www.psychoactive.com/tests/test-numeric.PHP
https: // ci-training.com/test-series-numeric.PHP
All the techniques we have seen, will also be useful in many other types of questions, such as dominoes or letters, in which the series construction mechanism is, in essence, the same.
You also have this video material available:
Test for Practice for oppositions