CALENDER APTITUDE

Calendar aptitude problems short cuts

Calendar problems solving tricks - This is one of the reasoning question frequently asked in all the examination. Basically the calendar questions asked to find the leap year, how many days in the year, if first day of month is Tuesday what is the next last day. Very simple calendar problems short cuts is enough to solve this calendar aptitude question with in a 3 minutes. The people who felt that this one is difficult to solve, please see the calendar aptitude tricks formula below and practice it to do well. Firstly we given the calendar problems short cut methods formula, then some solved example are given. Not only that, at end of this post you will find useful calendar problems solving tricks video. But be true, practising this calendar aptitude concepts frequently to become familiar with it, without practising you can't do it faster in the exams.

Important Calendar Aptitude formula :

Odd days - Some question date and month will be given. You have to find the day regarding the given date. For these type of questions you can use odd day concept. Odd day concept is nothing but in the given period of time, number days more than the complete week.

Ordinary year - Only 365 days are present in the Ordinary year.

Leap year - A year which is divisible by 4, but it won't be century. But you have another option that every 4th century should be a leap year. It has 366 days.

Example for Leap year - The years 2004, 1948 are leap years. 2001, 2007, 2005 are not a leap year.

Counting odd days -

Rule 1 - Ordinary year = 365 days. It has 52 weeks plus 1 day. So ordinary years has 1 odd day.

Rule 2 - Leap year = 366 days. Totally 52 weeks plus 2 days. Leap year has two odd days.

Rule 3 - 50 years = 38 ordinary years + 12 leap years

Totally = (38x1 odd day) + 12x2 odd days)

Therefore 62 odd days, then separate it in to weeks.

So 8 weeks+ 6 odd days.

For 50 years there will be a 6 odd days.

Calendar aptitude solve examples :

It was Monday on February 1 2006, find what day will be on 2010 February 1?

Solutions -

On January 31st 2006 is a Tuesday.

No of odd day from year 2006 to 2009 = 5.

So count 5 days after Tuesday, there will be Wednesday, Thursday, Friday, Saturday, Sunday. The day which will come after Sunday, will be Monday this is the day of February 1 2010.

FOR  MORE SOLVED QUESTIONS GIVE A COMMENT IN COMMENT BOX.

BODMAS

For  Reasoning Section with Mathematical Operations, all calculations will be performed using BODMAS Rule. We can divide these problems in three types.Those are,

1. Symbol Substitution
2. Mathematical Logic
3. Interchange of Signs & Numbers

Now lets discuss these types with detailed examples.

1st Type : Symbol Substitution 

In this type, to find the value of the given expression you should replace the symbols by mathematical operations. Then apply the BODMAS Rule (i.e., Brackets, of, Division, Multiplication, Addition, Subtraction).

For the detailed understanding, lets have a look at an example

Example : 42% (9-2) + 6 x 3 - 4

Solution :
42 % (9-2) + 6 x 3 - 4
= 42 % 7 + 6 x 3 - 4
= 6 + 6 x 3 - 4
= 6 + 18 - 4
= 24 - 4 = 20


2nd Type : Mathematical Logic

In this type, they just give you some logical statements followed by some mathematical operations. You should solve the given mathematical operations with the help of given logical statements.

Example : If % stands for greater than, x stands for addition, + stands for division, - stands for equal to, > stands for multiplication, = stands for less than, < stands for minus, then which of the following alternatives is correct ?

1. 3 + 2 < 4 % 6 > 3 x 2 
2. 3 x 2 < 4 % 6 + 3 < 2
3. 3 > 2 < 4 - 6 x 3 x 2 

3 x 2 x 4 = 6 + 3 < 2 

Usually for these type of questions people tend to make direct decisions with the help of given statements. But confusion arises when you tried to  coordinate given mathematical symbols with the logical statements mentioned above. It is because lack of coordination between eyes and brain. So, to avoid confusion you just should make a note of given logical statements in simple mathematical equations format. So that it will be easier for your eye to pick the correct operation.

Solution : Using the symbols correctly as given in the above operations, for the option (2) you will get, 3+2-4>6%3-2. Using BODMAS rule in this inequality, we will get 5-4>2-2 or, 1>0, which is correct.

3rd Type : Interchange of Signs & Numbers

In this type they just give you some random set of numbers and mathematical symbols. And then ask you to choose the correct of symbols to fit the given equation.

Example : Select the correct set of symbols which will fit in the given equation 5  0  3  5  = 20

1. x, x, x
2. =, +, x
3. x, +, x
4. +, -, x

Solution : Using the set of symbols given in options, you will get the Option 2 as the correct answer.

Using the set of symbols given in option 2, you will get the equation,

5-0+3x5 = 20

Using BODMAS rule you will get,

5-0+15=20 or 20-0 = 20 which is correct

Now lets have a look at some examples,

Q1. '<' means subtraction, '>' means Addition, '+' means Multiplication and '$' means Division. Then what will be the value of 27 > 81 $ 9 < 6 

27+81%9-6 = 27+9-6 = 36-6 = 30

Q2. If % means +, - means x, x means -, + means %, then what will be the value of 15 - 2 % 900 + 90 x 100 ?

Solution :

15x2+900%90-100 = 30 + 10 - 100 = -60

Q3. Q means Addition, J means Multiplication, T means Subtraction, K means Division then what is the value of 30K2Q3J6T5 ?

Solution :

3%2+3x6-5  =  15 + 18 - 5 = 28

Q4. P means x, R means +, T means %, S means -, then what will be the value of 18T3P9S8R6 ?

Solution :

By substituting the given symbols, you will get the answer 52

Q5. P means Multiplication, T means Subtraction, M means Addition, B means Division then what will be the value of 28B7P8T6M4 ?

Solution :

28 % 7 x8 - 6 + 4 

= 4 x 8 - 6 + 4 

= 32 - 6 + 4 

= 36 - 6 = 30

Alphabetical Series

Almost every test on reasoning contains questions on alphabetical series. In such a question, if it consists of a single series of alphabets/combination, the alphabets/combinations are arranged in a particular manner and each alphabet/combination is related to the  earlier and the following alphabets in a particular way. The examinees is supposed to decode the logic involved in the sequence and then fill in the space containing the question mark with a suitable choice out of those given. But before we proceed to discuss the various types of questions related to alphabetical series, we will talk of some basic facts which are essential to an understanding of these types of questions.

A B C D E F G H I J K L M N O P Q R S T U V W X Y Z

1. THE ALPHABET: The normal English alphabet contains 26 letters in all, as shown above

(Usually, questions on alphabet are accompanied by this normal alphabet). From A to M, the alphabet completes its first half, while the other half starts from N and ends at Z.

A-M - 1-13 (First Alphabetical Half)

N-Z –14-26 (Second Alphabetical Half)

2. EJOTY: For purpose of convenience, it is helpful to remember this simple formula called EJOTY, with the help of which you can easily find the position of any letter without much effort. But for practical purposes, you need to learn by heart the positions of different letters in the alphabet.

E   J    O    T    Y

5     10   15   20  25

Now, for instance, we wish to find the position of, say, the 17^th letter from the left side. You already know that the 15^th letter from the left side is O, therefore, the only thing you have to do is to find a letter which is two positions ahead of O, which is Q (The Answer). Using this simple formula, you can quickly find the position of any letter from the left side without much brain-rattling. Remembering the positions of different alphabets is basic to solving any question on alphabetical series. One of the best ways to achieve it is to practice EJOTY. 

For example, let’s say the name of the person imagined is ZUBINA. Now from EJOTY, we know that Z stands for 26, U stands for 21, B stands for 2, I stands for 9, N stands for 14 and A stands for 1. Now add up all these positions (26+21+2+9+14+1). What you get on addition does not have any significance, but it can be a very good way to try to make out and remember the individual positions of letters in the alphabet.

3. FINDING POSITIONS: Much more commonly, you get questions in the tests which provide you alphabetical positions from the right side. Since we are used to counting from the left side i.e. A, B, C… and not Z, Y, X…, the formula we discussed earlier will be applicable with a bit of modification.  But before we proceed to discuss it, it is essential to remember one simple mathematical fact.

*Let’s say there is a row of 7 boys in which a boy is standing 3^rd from left. We want to know his position from the right side.

 I       I       I        I        I       I       I  

                        1st     2^nd    3^rd    4^th    5^th    6^th    7^th   

You can see for yourself that the boy who was 3^rd from the left is placed 5^thfrom the right side.

The sum of both the positions is 8 (3+5), while the total number of boys is 7. This happens because we are counting a single boy twice in the calculation process. If we had subtracted 3 from 7 (as some of us might do), we would have got 4, which is obviously not the correct position from the right side. An important conclusion emerges from this discussion. If we are dealing with an alphabet and we have been given the position of any letter from either side, we will add 1 to the total no. of letters and then subtract its position from one side to get its position from the other side. For example, let’s find the position from the right of a letter which is the 9^th from the left side.

   

    A   B    C   D   E    F    G    H    I    J   K     L  M    N      O     P     Q   R    S    T     U  V    W   X    Y    Z

    1    2     3    4    5    6     7    8     9  10  11    12  13    14    15    16    17   18   19  20  21  22    23   24   25   26

26 25    24  23  22  21   20   19  18 17  16    15  14    13    12     11    10    9    8   7    6     5     4    3     2    1

As you can see for yourself, the 9^th letter from the left side, I, comes out to be the 18^thfrom the right side. Their sum (9+18=27) is again one more than the total number of alphabets i.e. 26. We can do this operation easily by adding one to the total number of letters (26+1=27) and then subtracting 9 from it. It gives you the letter position 18^thfrom the right, which you can verify yourself from the above alphabet. The same procedure will be applicable if we are given an initial right position and are supposed to find it from the left side. Take for example, a letter which is placed 11^th from the right side. If we want to locate its position from the left side, we will add 1 to total no. of letters and then subtract the right position from it to get its position from the left side. 27 – 11 gives you 16. Using EJOTY, you can easily conclude that the letter is P (16^thfrom left, 11^th from right).

The same logic is applicable if we are dealing with a situation in which the position of an item from the top is given to us and we want to find it from the bottom side or vice-versa.

IV. Still another type of question concerns finding the midpoint between two letters in the alphabet. For instance, let’s talk of a case which requires us to find the mid-point between the 11^th letter and the 17^th letter from the left side.

 A   B    C   D   E    F    G    H   I     J   K      L   M    N      O     P     Q    R    S    T    U  V    W   X    Y    Z

 1    2     3    4    5    6    7      8   9    10  11    12  13    14    15    16    17   18   19   20 21  22    23   24   25   26

26    25  24  23  22  21   20    19 18   17  16     15   14   13    12     11    10    9      8  7     6     5     4    3   2    1

   

You can see that there are five letters between these two positions i.e.  L, M, N, O and P. Obviously, the midpoint of 5 items is the third item from either side, whether counted from the left or the right. It comes out to be N, which is the correct answer. But frankly speaking, so much labour is not exactly required in solving such questions. Let’s let the cat out of the bag. In such questions, if the positions are given from the same side (i.e. either both are from left or both are from right), simply add up the two positions, get their average and you have the answer. In this case, the two positions are 11 and 17 from left. Adding them and averaging them gives you 14. Recollect the EJOTY formula and you immediately come up with the letter which is 14^th from the left side (preceding O). The same procedure will be applicable if you are given a case in which both the positions are counted from the right side. Remember that the answer you get will be from the same sides which you have been given. Let’s make this thing clearer by taking a practical example.

*Consider a case in which we have to find the mid-point between the 13^th and the 19^th letter from the right side. Adding the two positions gives us 32, the average of which is 16. So we get the mid-point, which is 16^th from the right side (the same as the sides given in the question). Now we have to convert this position into a position from the left.

Applying the logic discussed earlier, we subtract 16 from 27 and get 11^th from the left, which is obviously K. You can verify this answer by looking up the above alphabet. In fact, for such questions, one should have so much practice that one does not need to look up the alphabet, which proves to be time-consuming.

A   B    C   D   E    F    G    H     I      J    K     L    M    N     O      P     Q    R   S    T    U   V   W   X    Y    Z

1    2      3   4    5    6      7    8      9    10  11    12    13    14    15    16    17   18   19  20  21  22    23   24   25   26

26 25     24  23  22  21    20  19    18   17  16    15   14    13    12   11     10    9      8   7    6     5     4    3     2    1

Now let us consider the third case in which we have to find the mid-point between two alphabetical positions, one of which is given from the left and the other from the right.

*Take, for instance, a case in which we have to find the mid-point between the 6^thposition from the left side and the 11^th position from the right side. The first thing we have to do is to convert the right position into a left position to make the data comparable in nature. Doing so gives us 16. Now add up 16 and 6 (because now both are from the same side), average them, apply EJOTY and you get the correct answer. 

Coding Decoding

Solved Examples

Q1. If HARISH is coded as ITJSBI, then how would REEMA be coded?

ANS: In this case, the immediate next letter in the alphabet is taken and the code is written in the reverse order i.e. the code for the letter H is I and it is in the end, the code for A is B and it is second from the end, similarly the code for R is S and it is third from the end and so on. While coding the word REEMA, the code for R will be S and it would be written in the end, similarly the codes for E, E, M and A would be F, F, N and B respectively. And the final code would be BNFFS. This type is called as+1 and reverse order.

Q2. If CALENDAR = QZCMDKZB, then NEELAM = ?

ANS: This type is just a bit different from the previous type. This is called – 1 and reverse order. Now C – 1 = B and it is the last letter in the code, code for A, L, E are Z, K and D respectively and these are written as the second, third and fourth letter respectively from the end of the word in the code. Similarly the code for the word NEELAM would be LZKDDM.

Q3. If the code for the number 84219643 is 24139864,then what is the code for 97658256 ?

ANS: In this case, the eight-lettered number has been broken into two parts i.e. 84219643, the underlined part is 1st part starting with 2 and ending with 6,the second part is 4384, starting with 4 and ending with 4. So while making code 1 number from each part is taken. 2 is taken from part 1, then 4 is taken from part 2, 1 is taken from part 1, 3 is taken from part 2 and so on.

Q4. If SHIVANI = 574, then GANESH =  ?

ANS: in this case, EJOTY of all the words has been added,multiplied by the number of letters in that word. SHIVANI => S = 19, H = 8, I = 9, V = 22, A = 1, N = 14, I = 9 => 19 + 8 + 9 + 22 + 1 + 14 + 9 = 82 × 7 ( .: there are 7 letters in the word SHIVANI), Similarly while making code for GANESH => 7 + 1+ 14 + 5 + 19 + 8 = 54 × 6 ( .: there are six letters) = 324 would be the code.

Q5. If ACTIVITY = 24315137, Then ELEPHANT = ?

ANS: It involves the position of alphabet in the alphabetic order + 1. If it becomes a single digit number,  write it and if it is a two digit number then add it to get a single digit number as in question no. 13. Similarly ELEPHANT => E = 5 + 1 = 6, L = 12 +1 = 13 => 1 + 3 = 4, E = 5 + 1 =6, P = 16 + 1 = 17 => 1 + 7 = 8 and so on. The code will become 64689263.

Q6. If EVITCDNIV = HZNZJLWSG, then ABDICTION = ?

ANS: In this question, firstly + 2, then + 3, +4, +5, +6,+7 and so on. Similarly when the code for the word ABDICTION is made firstly+2, then +3, +4, +5, + 6 and so on. The code will finally become DFIOJBNYY.

Q7. If RELATION = 95312965, then MANAGEMENT = ?

ANS: In this case, the code for every letter is its position in the alphabetic order, represented as a single digit. If its position is already a single number, then it is simply assigned as the code,and if there are two digits in its position, then those digits are added to get the code for that letter. In this case R is the 18^th letter so itscode would be 1 + 8 = 9, E is the 5^th letter so its code is 5, L isthe 12^th letter so its code is 1 + 2 = 3, A is 1 and so on.Similarly when the code for the word MANAGEMENT is made. M is 13, 1 + 3 = 4. A= 1, N is 14 1 + 4 = 5, A = 1, G = 7 and so on. The final code for the word MANAGEMENT would be 4151754552

Q8. If PERSON = THTWRP, then ENGINE = ?

ANS: Here P + 4 = T, E + 3 = H, R + 2 = T, Then S + 4 =W, O + 3 = R, N + 2 = R. The method for coding is +4, +3, +2. The code forENGINE will become E + 4 = I, N + 3 = Q, G + 2 = I, I + 4 = M, N + 3= Q and E +2 = G. Therefore the code is IQIMQG.

 Q9. If SILVER = HROEVI, then MEENAKSHI = ?

ANS : In this case, if the letter is at ‘nth’ position from the beginning then the letter at ‘nth’ position from the end is written. This can always be checked, whenever the sum of the number and its respective code is 27. Then the method applied for the coding would be this only. As in SILVER, S is 19 and its code H is 8 and the sum is 27. I is 9 and its code R is 18 and sum is 27. While coding MEENAKSHI, the same method coding will be applied. M is 13, so what should be added in 13 to make it 27 ( that is14), write the 14^th letter which is N as the code for M. Similarly E is 5, find 22^nd letter to make sum as 27 ( V is 22^ndletter) and that is the code and so on. The code for the word MEENAKSHI will be NVVMZPSR.

Q10. If A = E, B = F, C = G and H = L, how will the word COME BACK be coded in this code?

ANS: Here in this case, it can be checked that 4^thletter has been made the code for every letter, in the same way, you can the 4^thletter as the code for every letter i.e. COME BACK would become as C + 4 = G, O will be S and the code would become – GSQI FEGO.

* Sometimes the complete pattern of numbers/letters codes are given for coding and based upon that you have to code the words/numbers.