Important Points and Shortcuts for Clock problems

• Clock formula: 11/12. This is a very important point for solving aptitude questions on clocks.

• As minute hand covers one full circle i.e. 360 degrees in one hour, that means it travels 360/60 = 6 degrees/min.

• An hour hand covers one part of the 12 major parts of the circle which means it covers 360/12 = 30 degrees in one hour i.e. it travels 30/60 = 1/2 degree per min.

• Now, the relative speed of the minute hand is 6 – 1/2 = 11/2 or 5.5 degrees. This 11/2 degrees will be useful in finding the angle between the two hands of the clock.

• The complete circle has a total of 360 degrees and in terms of minute spaces, it has been divided into 60 minutes spaces, which means each minute space represents 360/60 = 6 degrees.

• The complete circle has been divided into 12 equal bigger units also, which we call as hours. This further implies that every hour space covers a total of 360/12 = 30 degrees.

• The hour hand and minute hand meet once every hour. But in a 12 hour period, they meet 11 times.

• There is one angle of 180 degrees in every hour i.e. both the hands are in the straight line in the opposite direction, but in a twelve-hour period, it happens 11 times.

• There are 2 right angles every hour, but in a 12 hour period, there are 22 such angles.

• If the two hands are moving at the normal speeds, they should meet after every 65 5/11 min.

• Clock Aptitude Tricks -Short Cut - Between x and (x + 1) O’clock, the 2 hands will be ‘t’ min apart at (5x t) 12/11 past x.

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Aptitude Trick#7(Multiple any number with 99)

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Aptitude Trick#6(Multiplication of number greater than 100)

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Aptitude Trick#5 (Multiplication of number less than 100)

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Aptitude Tricks#8(Multiply Quickly Two Number near A Power of 10)

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Aptitude Trick#4(Square of two digit number)

Square of two digit number using Formule 1

Consider An Example

Example: Square of 26

• Step 1:

           we separate the number into
            two parts(a & b)

• Step 2: Now using formulae

                 a２| 2ab | b２

                we get 2*2 | 2*2*6 | 6*6

Answer = 676

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Progression (AP & GP & HP)

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Area And Perimeter

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Logarithms

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Permutations and Combinations

Permutations and Combinations :

Permutations

1.nPr = n!/(n-r)!

2.nPn = n!

3.nP1 = n

Combinations

1.nCr = n!/(r! (n-r)!)

2.nC1 = n

3.nC0 = 1 = nCn

4.nCr = nCn-r

5.nCr = nPr/r!

Number of diagonals in a geometric figure of n sides = nC2-n

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VOLUME & SURFACE AREA

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INEQUALITIES

INEQUALITIES :

1. > First element is Greater than the Second Element.

2. < First element is Smaller than the Second Element.

3. = First element is Equals to Second element.

4. ≥ The first element is Greater than or Equals to the Second Element.

5. ≤ The first element is Smaller than or Equals to the Second Element.

6. ≠ The first element is either greater than or smaller than the Second Element.

Two signs opposite to each other will make the conclusion wrong. But again if the signs are in the same manner that will not make it wrong.

Example :

If A > B < C > D then A < C is False , C > A is False. But If E > F > G > H then E > G is True, F > H is True, E > H = True.

If A ≥ B ≤ C then A ≤ C = False, 

C ≥ A = False. But

If A ≥ B ≥ C then A ≥ C = True , 

C ≤ A = True

Points to Remember :

1.A > B > C

2.A > B ≥ C

3.A ≥ B > C

4.A = B > C

5.A > B = C

From the All above statement we conclude that conclusion is A>C

1. A < B < C

2.A < B ≤ C

3.A ≤ B< C

4.A = B < C

5.A < B = C

From the All above statement we conclude that conclusion is A< C

1. A ≥ B ≥ C

2. A = B ≥ C

3. A ≥ B = C

From the All above statement we conclude that conclusion is A≥C(Either A>C or A=C)

1. A ≤ B ≤ C

2. A = B ≤ C

3. A ≤ B = C

From the All above statement we conclude that conclusion is A ≤ C (Either A < C or A = C)

1. A < B > C

2.A ≤ B> C

3.A < B≥ C

4.A > B < C

5.A > B ≤ C

6.A ≥ B < C

From the All above statement we conclude that conclusion is that,

Either 1 or 2 follows if any of the following cases (a, b, c and d) are given as they form a complementary pair.

a) 1. A > C     2. A ≤ C

b) 1. A ≥ C     2. A < C

c) 1. A < C     2. A ≥ C

d) 1. A ≤ C      2. A > C  

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CALENDER

CALENDER :

ODD DAYS :

The number of days more than the complete weeks are called odd days in a given period.

LEAP YEAR :

A leap year has 366 days. In a leap year, the month of February has 29 days. Every year divisible by 4 is a leap year if it is not a century. Every 4th century is a leap year and no other century is a leap year.

Examples:

Each of the years 1948, 2004, 1676 etc. isa leap year.

Each of the years 400, 800, 1200, 1600, 2000 etc. is a leap year.

None of the years 2001, 2002, 2003, 2005,1800, 2100 is a leap year.

ORDINARY YEAR :

The year which is not a leap year is called an ordinary year. An ordinary year has 365 days.

COUNTING OF ODD DAYS :

• 1 ordinary year = 365 days = (52 weeks + 1 day)
  
• 1 ordinary year has 1 odd day
• 1 leap year = 366 days = (52 weeks + 2 days)
  1 leap year has 2 odd days
  
• 1 leap year has 2 odd days
• 100 years = 76 ordinary years + 24 leap years.
• = (76 *1 + 24 * 2) odd days = 124 odd days
• = (17 weeks + 5 days) 5 odd days

Day Codes :

• 0 = Sunday

• 1 = Monday

• 2 = Tuesday

• 3 = Wednesday

• 4 = Thursday

• 5 = Friday

• 6 = Saturday

Month Codes

• January = 0

• February =3

• March = 3

• April = 6

• May = 1

• June = 4

• July = 6

• August = 2

• September = 5

• October = 0

• November = 3

• December = 5

• 1 year = 12 months
• 12 months = 52 weeks
• 1 week = 7 days
• Ordinary Year = 52 weeks + 1 odd day
• Leap Year = 52 weeks + 2 odd days

Odd days are the number of days more than the complete week or extra days in a year.

• Number of odd days in 100 years = 5
• Number of odd days in 200 years = (5 * 2) = 3 odd days
• Number of odd days in 300 years = (5 * 3) = 1 odd day
• Number of odd days in 400 years = (5 * 4 +1) = 0 odd day

Similarly, each one of 800 years, 1200 years, 1600 years, 2000 years etc  has 0 odd days

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SYLLOGISM

SYLLOGISM :

The syllogism is known as Statements and Logical Conclusions with Logical Arguments.
In all the statement of syllogism questions,first-term is called subject and second is called predicate.

Example :

1. All pigs are ducks – here pigs is subject and ducks is the predicate

2. All readers are commentators – here readers are subject and commentators are the predicate

When a statement starting with All, every, any, none, not a single, only is a universal statement.

Some, many, a few, quite a few, not many very little, most of, almost, generally, often, frequently, which are all indicates Particular one.

The easy way to identify the conclusion is that,

• All + All =All

• All + No = No

• All + Some = No Conclusion

• Some + All = Some

• Some + Some = No Conclusion

• Some + No = Some Not

• No + No = No Conclusion

• No + All = Some not reversed

• No + Some = Some not reversed.

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NEW WORD FORMATION

NEW WORD FORMATION :

Write down the given word on a rough sheet paper and number the individual letters according to alphabetical series

Then round off the given words as given in the question for easy identification without getting confused

Then take the round of words and form sequence one after another letter and count the number of meaningful words formed.

In some questions, it is asked to find if more than one meaningful word can be formed from the given keyword. In such type of questions, if you could make 2 words with the given keyword, and mark the option which has more than one word as the answer.

Example :

How many meaningful English words can be formed with letters NPA using each letter only once.


a) Three  b) Two  c) One  d) Five


Answer: b) Two
Explanation :

There are two words can be formed with letters NPA.
i.e., NAP and PAN

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RANKING

 
RANKING :

First of all, we do,
Read the statement line by line carefully.

Position can be either side of row and rank is top to bottom of a row.

Rank of a person from lower – total persons in the row - rank of that person from upper+1

Rank of a person from right = total persons in the row- rank of that person from left + 1

Rank of a person from upper = total persons in the row - rank of that person from lower + 1

Rank of a person from left = total persons in the row- rank of that person from right+ 1

The total person in row = rank of a person from the upper/ left + rank of a person from lower/right – 1

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ALPHANUMERIC SEQUENCE PUZZLE

ALPHANUMERIC

SEQUENCE PUZZLE :

1. Alphanumeric is a combination of alphabetic and numeric characters. In some cases, it may include upper and lower case letters, punctuation marks, and symbols (such as @,&, and *,).


2. Few keywords that are generally used in such questions like "following, followed by, preceding, and preceded by".


3. Let us take two successive alphabets Y and Z. Here, Y is preceding Z and Z is preceded by Y. Also, Z is following Y and Y is 
followed by z.

4. Now, take three successive alphabets X, Y, and Z. Here X and Y are preceding Z, but Y is immediately preceding Z whereas, X is not immediately preceding Z. Similarly, Y and Z are following X, but Y is immediately following X whereas, Z is not immediately
following X.

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DIRECTION SENSE TEST

DIRECTION SENSE TEST :

1. The shortcut for solving direction sense problems is to draw a schematic or block diagram of the given data and thus we can solve all the questions through it.

2. North, South, East, and West are the four directions.

3. North-East (N-E or NE), South-West (SW), North-West(NW) and South-East(SE) are the four main cardinal directions.

4. If a person A is standing at the time sunrise, his shadow point towards west.

5. At the time of sunset, the shadow of an object is always to the east.

6. If a man stands facing the North, at the time of sunrise his shadow will be towards his left and at the time of sunset, it will be towards his right.

7. At 12:00 noon, the rays of the sun are vertically downward hence there will be no shadow.


8. The angle between any two main (or cardinal) direction is 900 but the angle between one main and one cardinal direction is 450.

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INPUT OUTPUT IN REASONING

INPUT OUTPUT IN REASONING:

1. Combination of Numbers and words series in Input changing their patterns in a different manner and finally produces an output.

2. Arranging the Numbers and words either in Ascending or Descending order as well as Dictionary order if English words are given.

3. All words at one End and all numbers at one end or vice versa.

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SEQUENTIAL OUTPUT TRACING

SEQUENTIAL   OUTPUT TRACING :

Tips About Number Of Steps :

1. If there are "n" words/digits in the input then at most "n-1" steps are required to rearrange it completely.

2. The number of words/digits arranged until the present step is greater than or equal to the present step number.

3. If the input is not given we cannot determine the previous step from the given step number or we cannot determine input from given step number.

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