Number Series Method Shortcut Tricks

Important info

Number Series shortcut tricks are very important thing to know for your exams. Time takes a huge part in competitive exams. If you manage your time then you can do well in those exams. Most of us miss that part. Few examples on number series shortcuts is given in this page below. These shortcut tricks cover all sorts of tricks on Number Series. We request all visitors to read all examples carefully. These examples will help you to understand shortcut tricks on Number Series.

First of all do a practice set on math of any exam. Choose any twenty math problems and write it down on a page. Solve first ten math problems according to basic math formula. You also need to keep track of timing. After solving all ten math questions write down total time taken by you to solve those questions. Now practice our shortcut tricks on number series and read examples carefully. After finishing this do remaining questions using Number Series shortcut tricks. Again keep track of the time. The timing will be surely improved this time. But this is not enough. If you need to improve your timing more then you need to practice more.

Math section in a competitive exam is the most important part of the exam. It doesn’t mean that other topics are not so important. But if you need a good score in exam then you have to score good in maths. You can get good score only by practicing more and more. All you need to do is to do math problems correctly within time, and only shortcut tricks can give you that success. But it doesn’t mean that you can’t do math problems without using any shortcut tricks. You may have that potential to do maths within time without using any shortcut tricks. But other peoples may not do the same. For those we prepared this number series shortcut tricks. Here in this page we try to put all types of shortcut tricks on Number Series. But it possible we miss any. We appreciate if you share that with us. Your little help will help so many need .

What is Number Series ?

Number series is a form of numbers in a certain sequence, where some numbers are mistakenly put into the series of numbers and some number is missing in that series, we need to observe first and then find the accurate number to that series of numbers.

Anything we learn in our school days was basics and that is well enough for passing our school exams. Now the time has come to learn for our competitive exams. For this we need our basics but also we have to learn something new. That’s where shortcut tricks are comes into action.

In competitive exams number series are given and where you need to find missing numbers and mistakenly put into the series numbers. The number series are come in different types. At first you have to decided what type of series are given in papers then according with this you have to use shortcut tricks as fast as you can.

Perfect Square Series

Perfect square series is a arrangement of numbers in a certain order, where some numbers Series are based on square of a number which is in same order you need to place one square number that is missing in that given series, we need to observe and find the accurate number to the series of numbers.

This type of problem are given in Quantitative Aptitude which is a very essential in banking exam. It is simple to work on perfect square root numbers you can easily obtain the result of perfect square number. How you get easily by Perfect square numbers missing term by memorize square and square root numbers shortcut tricks .

The square of same number and the square result of a number which is equal to the square of another same element. In mathematical world, a square number or perfect square is number of an integer positive integer that is the square of an same integer number always and the numbers are non-negative.

In other words, we say it is the result of product of multiplication of some positive integer numbers with itself always. For example, we consider 4 is a result of square numbers, since it as 2 × 2 in normal way.

The normal representation of square numbers is n²and that is similar with products of n × n, but it is similar with exponentiation of n² ,

In Square numbers are positive number. So we can explain it that a positive number is a square number, where its square roots are always integers positive numbers. so For example, √4 = ±2, so 4 is a square number.

Perfect Square Series
Here we see the some examples that how the perfect square are arranged how the missing square series are arranged.

Example 1: 441, 484, 529, 576, ?,
Answer: 441 = 21², 484 = 22², 529 = 23², 576 = 24²,625 = 25².

Example 2: 121, 144, 169, ?, 225
Answer: 121 = 11², 144 = 12², 169 = 13², 196 = 14², 225 = 15².

Example 3: ?, 2116, 2209, 2304, 2401, 2500
Answer: 2025 = 45², 2116 = 46², 2304 = 48², 2401 = 49², 2500 = 50²

Example 4: 961, 1024, ?, 1156, 1225
Answer: 961 = 31², 1024= 32², 1089 = 33², 1156 = 34², 1225 = 35².

Example 5: 36, ?, 64, 81, 100, 121
Answer: 36 = 6², 49 = 7², 64 = 8², 81 = 9², 100 = 10², 121 = 11².

Example 6 : 121 , 169 , ? , 289 , 361
Answer : 11² = 121 , 13² = 169 , 15² = 225 , 17² = 289 , 19² = 361.

Example 7 : 121 , 484 , 1089 , 1936 , ? , 4356
Answer : 11² = 121 , 22² = 484 , 33² = 1089 , 44² = 1936 , 55² = 3025 , 66² = 4356.

Example 8 : 961 , 1024 , 1089 , ? 1225
Answer : 31² , 32² , 33² , 34² , 35²

Example 9 : 1849 , ? , 2025 , 2116 , 2209
Answer : 43² , 44² , 45² , 46² , 47²

Example 10 : 2500 , 2401 , 2304 , ? , 2116 , 2025
Answer : 50² , 49² , 48² , 47² , 46² , 45²

Perfect Cube Series

Perfect cube series is a arrangement of numbers in a certain order,where some numbers this Types of Series are based on cube of a number which is in same order and one cube number is missing in that given series.

we need to observe and find the accurate number to the series of numbers. This type of problem are given in Quantitative Aptitude which is a very essential paper in banking exam.Under below given some more example for your better practice.

All numbers are arranged in sequent order. we need to observe and find the accurate number to this type series of numbers. Here we learn the perfect cube series of Example.

This type of problem are given in Quantitative Aptitude which is a very essential in banking exam. Under below given some more example for your better practice.

In perfect cube series number is a combination of cube number are arranged. In example 1) 1331, 1728, 2197, ? where you need to count them in a one step or two step calculation for obtain the difference common result according with the series of ratio numbers .

At first you can calculate missing number in ratio series and that you place the actual missing number in the ? or missing place. Be prepared when you calculate differences because it is either one or two step calculation. So when you calculate and get result of two difference numbers you need follow some step wise.

At first calculate the first number cube value and second number cube value if all number are maintain a sequential order cube value then follow same steps which is carry up to last and after that you get actual missing number by finding the common value when you put the missing number you have noticed that all series numbers are common difference in between them.

This kind of missing series calculation you go thorough some common calculation shortcut tricks using cube and cube shortcut tricks, or you memorize the 1 to 30 cube series number value.

In this type series example questions, it is sounds hard, but it really isn’t. Get it? Once you have done this, by practice with more example then you just easily can do in your way as well competitive and as in bank exam also . So, each of our examples are given below.

Perfect Cube Series:

Example 1 : 1331 , ? , 35937 , 85184 , 166375
Answer : 11³ , 22³ , 33³ , 44³ , 55³

Example 2 : 125, ?, 343, 512, 729, 1000
Answer : 125 = 5³ , 216 = 6³, 343 = 7³, 512 = 8³, 729 = 9³, 1000 = 10³.

Example 3 : 1 , 9 , 125 , 343 , ? , 729
Answer : 1³ , 3³ , 5³ , 7³ , 8³ , 9³

Example 4 : 125, ?, 343, 512, 729, 1000
Answer: 125 = 5³, 216 = 6³, 343 = 7³, 512 = 8³, 729 = 9³, 1000 = 10³.

Example 5 : 8 , 64 , ? , 512 , 1000 , 1728
Answer : 2³ , 4³ , 6³ , 8³ , 10³ , 12³

Example 6 : 4096, 4913, 5832, ?, 8000
Answer: 4096 = 16³, 4913 = 17³, 5832 = 18³, 6859 = 19³, 8000 = 20³.

Example 7 : 1331 , ? , 29791 , 68921 132651
Answer : 11³ , 21³ , 31³ , 41³ , 51³

Example 8 : 1331, 1728, 2197, ?
Answer: 1331 = 11³, 1728 = 12³, 2197 = 13³, 2744 = 14³.

Example 9: 1728, 1331, ?, 729, 512
Answer: 1728 = 12³, 1331 = 11³, 1000 = 10³, 729 = 9³, 512 = 8³.

Example 10 : 1000 , 8000 , ? ,64000 , 125000
Answer : 10³ , 20³ , 30³ , 40³ , 50³

Example 11 : 125000 , 64000 , ? , 8000 , 1000
Answer : 50³ , 40³ , 30³ , 20³ , 10³

Ratio and Proportion Methods shortcut tricks

You all know that math portion is very much important in competitive exams. That doesn’t mean that other sections are not so important. But only math portion can leads you to a good score. A good score comes with practice and practice. All you need to do is to do math problems correctly within time, and only shortcut tricks can give you that success. But it doesn’t mean that without using shortcut tricks you can’t do any math problems. You may have that potential that you may do maths within time without using any shortcut tricks. But so many people can’t do this. Here we prepared ratio and proportion shortcut tricks for those people. Here in this page we try to put all types of shortcut tricks on Ratio and Proportion. But we may miss few of them. If you know anything else rather than this please do share with us. Your little help will help so many needy.

What is Ratio?
A ratio is a relationship between two numbers by division of the same kind. The ration of a to b is written as a : b = a / b, In ratio a : b, we can say that a as the first term or antecedent and b the second term or consequent.

Example : The ratio 4 : 9 we can represent as 4 / 9 after this 4 is a antecedent and, consequent = 9

Rule of ration : In ratio multiplication or division of each an every term of a ratio by the same non- zero number does not affect the ratio.

Different type of ratio problem are given in Quantitative Aptitude which is a very essential topic in banking exam. Under below given some more example for your better practice.

What is Proportion?
The idea of proportions is that two ratios are like equal.
If a : b = c : d, we write a : b : : c : d,
Ex. 3 / 15 = 1 / 5
a and d called extremes, where as b and c called mean terms.
Proportion of quantities
the four quantities like a, b, c, d we can say proportion then we can express it
a : b = c : d
Then a : b : : c : d <–> ( a x d ) = ( b x c )
product of means = product of extremes.

If there is given three quantities like a, d, c of same like then we can say it proportion of continued.
a : d = d : c , d is called mean term. a and c are called extremes.

Geometric Series

Examples 1: 5, 45, 405, 3645, ?
Answer: 5 x 9 = 45, 45 x 9 = 405, 405 x 9 = 3645, 3645 x 9 = 32805.

Examples 2: 73205, 6655, 605, 55, ?
Answer: 5 x 11 = 55, 55 x 11 = 605, 605 x 11 = 6655, 6655 x 11 = 73205.

Examples 3: 25, 100, ?, 1600, 6400
Answer: 25 x 4 = 100, 100 x 4 = 400, 400 x 4 = 1600, 1600 x 4 = 6400.

Examples 4: 9, 54, ?, 1944, 11664
Answer: 9 x 6 = 54, 54 x 6 = 324, 324 x 6 = 1944, 1944 x 6 = 11664.

Mixed Series

Examples 1:

111, 220, 438, ?, 1746
Answer:
from 111 to 220 we get using this 111 x 2 = 222 – 2 = 220,similarly we follow next steps
from 220 to 438 we get using this 220 x 2 = 440 – 2 = 438,
from 438 to ? we get using this 438 x 2 = 876 – 2 = 874,
from 874 to 1746 we get using this 874 x 2 = 1748 – 2 = 1746.

So the missing number is 874

Examples 2:

24, ?, 208, 622, 1864
Answer:
from 24 to ? we get using this 24 x 3 = 72 – 2 = 70, Similarly we follow next steps
from 70 to 208 we get using this 70 x 3 = 210 – 2 = 208,
from 208 to 622 we get using this 208 x 3 = 624 – 2= 622,
from 622 to 1864 we get using this 622 x 3 = 1866 – 2 = 1864.

So the missing number is 70

Examples 3:

11, 24, 50, 102, 206, ?
Answer:
11 x 2 = 22 +2 = 24,
24 x 2 = 48 + 2 = 50,
50 x 2 = 100 + 2 = 102,
102 x 2 = 204 + 2 = 206,
206 x 2 = 412 + 2 = 414.

So the missing number is 414.

Example 4:

0, 6, 24, 60, 120, 210, ?
Answer :
The given series is : 1³ – 1, 2³ – 2, 3³ – 3, 4³ – 4, 5³ – 5, 6³ – 6,
So the missing term = 7³ – 7 = 343 – 7 = 336 .

Example 5:

11, 14, 19, 22, 27, 30, ?
Answer :
The pattern is + 3, + 5, + 3, + 5, …………
So the missing term is = 30 + 5 = 35 .

Example 6:

6, 12, 21, ? , 48
Answer :
The pattern is + 6, + 9, + 12, +15 ………..
So the missing term is = 21 + 12 = 33 .

Example 7:

18, 22, 30, ? ,78, 142
Answer :
The pattern is +4, +8, +16, +32, +64
So the missing term is = 30 + 16 = 46 .

Example 8:

589245773, 89245773, 8924577, 924577, ?
Answer :
The pattern is The digits are removed one by one from the beginning and the end in order alternately, So to obtain the subsequent terms of the missing series is = 92457 .

Example 9:

8, 35, ? , 143, 224, 323
Answer :
The pattern is (3² – 1), (6² – 1),………., (12² – 1), (15² – 1), (18² – 1)
So the missing term is = (9² – 1 ) = 81 – 1 = 80 .

Example 10:

3, 7, 23, 95, ?
Answer :
The pattern is ( x 2 + 1 ),( x 3 + 2) , ( x 4 + 3 ) , ……….
So the missing term is = 95 x 5 + 4 = 479 .

Coding – Decoding Shortcut Tricks

About Coding – Decoding

Friends, today we shall discuss about the Coding and Decoding type questions of Reasoning section. There are different types of Coding and Decoding questions are there, before going into details lets first understand what is meant by coding and decoding.

Coding : A particular code or pattern is used to express a word in English language to express it as a different word. The coded word itself does not make any sense unless we know the code, i.e., unless we know the pattern or code that has been followed.

Decoding : Decoding refers to the process of arriving at the equivalent English word from the code word given. Hence, we can look at two broad categories of questions in coding-decoding.

1st Category

In the 1st category of questions, a particular code is given and on the basis of this given code, we have to find out how another word (in English Language) can be coded. The correct code for the given word has to be selected from the answer choices on the basis of the code given in the question.

Ex : In certain code if the word “VIRTUAL” is coded as “UHQSTZK”, then in this code, how is the word “PAINFUL” coded?

Solution :

By observation, we can find that in coding “VIRTUAL” as “UHQSTZK”, each letter in the given word has been replaced by the letter that comes immediately before it in the English alphabet. Using this code, if we now have to code “PAINFUL”, we need to replace each letter of the word with the letter that comes before it in the English alphabet. Thus the code will be “OZHMETK”.

Note : Here, it can be seen form the example, we treat the English alphabet in a circular fashion. i.e., the letter that comes after Z is A. Similarly, the letter that comes before A is Z.

2nd Category

In the second category of questions, a particular code is given and on the basis of this given code, we have to find out the equivalent word in English language for a word given coded form. The correct word (in English Language) for the given coded word has to be selected from the answer choices on the basis of the code given in the question. Ex : In a particular Code, if the word :SYSTEM” is coded as “UAUVGO”, then in teh same code, what does “HCUJKQP” stand for? Sol : Here, we can observe that each letter in the given word is replaced by the second letter that comes after it in the English alphabet to give us the word in coded form. Hence, to know what “HCUJKOP”, we should replace each letter by a letter which second letter before it in the English alphabet. This gives us the word “FASHION”. The first step to be taken in solving questions in coding – decoding is to crack the code. To do this, it will be helpful to understand the broad types of codes that are used. Ofcourse, the number of codes that can be created and used are infinite but one can always keep the commonly used codes in mind. Secondly, practice will make a student much more comfirtable and conversant with these types of questions.

Coding-Decoding Problems

Some of the more commonly used codes to represent letters of the alphabets in coding decoding problems are

Use the letter that comes one or any fixed number of places before it in the alphabet.
Use a letter as many places from the end of the alphabet as the original letter is from the begining of the alphabet.
Each vowel may be replaced by the next vowel that comes in the alphabet and each consonant by another consonant following certain pattern with reference to the consonant under consideration.
Use one code for all the letters in even places in the given word and a different code for letters n the odd places in the given word to give a new word.
The same letters in the given word may be used in a cyclic or in some other order to give the coded word.
Some numerical values can be attributed to the letters of the word based on the letter positions in the alphabet.
Combination of two or more of the above ways of coding.

Examples of Coding and Decoding

Basic shortcut techniques of Coding – Decoding section of Reasoning. Today we shall discuss some example problems of Coding and Decoding with detailed explanations.

Practice problems of Coding and Decoding

1. If the word “DIAGRAM” is coded as “AGDMIRA”, then the word “PICTURE” can be coded as ?

rpteiuc
ctpeiur
rtpeuic
ctpriuc
rpteicu

Approach : The letters of the given word are written in a jumbled order to give us the word in the coded form. We need to find out the manner in which they have been jumbled. If you keenly observe the given words, the first letter of the word D has been used as the 3rd letter in the code. So, for the word PICTURE, the first letter P will have to be the third letter in the code. This eliminates choices (1) and (5), Then, the second letter I is used as the third from the last and that is not happening in choice (3). So, we now have choice(2) and (4). Since there are two As in the given word, let us leave aside A for the time being and look at some of the other letters. If we take the last letter M, it comes as the fourth letter in the code. SO, E in the word PICTURE should come as the fourth letter in the code. This leaves us only choice (2). So the correct answer is option (2). Note : Here, suppose even this has not given us a unique choice, we would then look at the remaining letters too. 2. If the word “CODING” is represented as DPEJOH, then the word “CURFEW” can be represented as

dvsgfx
dvshfx
dgshfx
dtsgfy
dysgff

Approach : Each letter in the given word “CODING” is replaced by the letter that immediately follows it in the English alphabet. Hence, the correct choice is option (1)3. if “ASHTRAY” is coded as “DVKWUDB”, then what does the word “UDQFLG” in that code mean?

RADISH
MANAGER
RUNNER
RANDOM
RANCID

Approach : If we replace each letter of the word “ASHTRAY” by a letter which is three letters to the right of it in the alphabet, then we get “DVKWUDB“. So, to find out the word whose code is “UDQFLG“, we replace each letter in the code by a letter which comes three places behind in the alphabet. We then get “RANCID” as the word. Option (5). 4. If “RAJESH” is coded as “SZKDTG”, then “PRANESH” should be coded as

QQBMDTG
QSBMDTG
QQBMFRI
QSZMDTG
QQZMFRI

Approach :The first letter is shifted by one alphabet forward, the second by one alphabet backward, the third by one forward and so on… So, the word “PRANESH” will be coded as “QQBMFRI“which is option (3). 5. If “PROMPT” is coded as QSPLOS, then “PLAYER” should be coded as

qmbzfs
qmbxfq
okzxfq
qmbfqx
omzxfq

Approach : The first half of the word has the letters being moved back by one letter in the alphabet and the second half of the word has the letters being moved forward by one letter in the alphabet. So, option (2). 6. If in a certain code, the number 1 is assigned to all letters in the even places in the alphabet and the number 2 is assigned to all letters in the odd places in the alphabet, then the code for the word ALPHABET will be

21112121
21121221
21111221
21112211
21212121

Approach : An easy way to do these sort of problems is to Just remember the place values of the alphabets so that it will be easier for you to solve these type of problems. According to above link, ALPHABET can be written as A — > 1 (Odd Number) so we have to assign value 2L –> 12 (Even Number) so we have to assign value 1P –> 16 (Even Number) so we have to assign value 1H —> 8 (Even Number) so we have to assign value 1A — > 1 (Odd Number) so we have to assign value 2B–> 2 (Even Number) so we have to assign value 1E–> 5 (Odd Number) so we have to assign value 2T–> 20 (Even Number) so we have to assign value 1 So the answer is 21112121. Which is option (1). 7. In a certain code, a is represented by 1, b is by 2, c by 3 and so on; then all multiples of 2 are assigned a code of 2 and non-multiples of 2 are assigned a code of 1. In this scheme of coding, what would the word PAPERS be coded as ?

121212
212121
212122
121211
212112

Approach : According to place values, the given word PAPERS can be written as P –> 16 (multiple of 2) so we can write 2A—> 1 (non multiple of 2) so we can write 1P —> 16 (multiple of 2) so we can write 2E —> 5 (non multiple of 2) so we can write 1R—> 18 (multiple of 2) so we can write 2. So the code will be 21212. Which is option (5). 8. If in a certain code “RANGE” is coded as 12345 and “RANDOM” is coded as 123678, then the code for the word “MANGO” would be

82357
84563
82346
82347
82543

Approach : In these type of questions, we cannot get the code for all the letters of the alphabet but we can get the code for the letters of the word “MANGO” which is of interest to us. Take one by one of these letters of the word “MANGO” and look for the code of that specific letter from the words for which the code is already given to us. Here, in this case, when we observe that R, A and N are the first three letters of both the given words in the same order and the code for both words have 1, 2 and 3 in that order at the begining. Because of lack of any additional information, we then conclude that the code for R is 1, A is 2 and N is 3 and that the letters givne in the code are in the same order as the letters given in the word. Thus, we can make out that the code for MANGO is 82347 on the basis of the code different letters of the word can be given. Choice (3).

Input – Output Arrangement Shortcuts

Important Info

In today’s post we shall discuss one of the most promising, easiest and time taking model of the Reasoning paper Input – Output arrangements. People often leave this area as this is little time taking. But with little practice and concentration you can easily get 5 out of 5 marks in this area within less time.

First of all lets see how the problem will look like. In these type of problems they will give you an input line of words and numbers and rearranges them in a few steps (normally 5 to 7 steps). and you should find out the logic behind the rearrangement and work with the problems given by them. Lets see an example.

Directions : Study the following information carefully and answer the following questions : A word and number arrangement machine when given an input line of words and numbers rearranges them following a particular rule in each step. The following is an illustration of input and rearrangement :

Input : Now 41 28 Credit Join 37 Go 61

Step 1 : 61 Now 41 28 Credit Join 37 Go

Step 2 : 61 Credit Now 41 28 Join 37 Go

Step 3 : 31 Credit 41 Now 28 Join 37 Go

Step 4 : 61 Credit 41 Go 37 Now 28 Join

Step 5 : 61 Credit 41 Go 37 Now 28 Join

Step 6 : 61 Credit 41 Go 37 Now 28 Join

Step 7 : 61 Credit 41 Go 37 Join 28 Now

Step 7 is the last step for this input.
As per the rules followed in the above steps, find out in each of the following Questions the appropriate step for the given input. (followed by some questions).

Now first lets have a look at the given problem. The logic in the arrangement is : The input is the combination of words and numbers. Firstly, the numbers got arranged in descending order. Whereas the words get arranged in alphabetical order. Numbers occupy the odd places and words occupy even places in the final step. When any element gets arranged, the previous element occupying that place shifts one place towards right.

And one more basic rule here we have to remember is, we can make only one change in one step. In step One, 61 occupies the first place from the left end and the other elements are pushed one place rightward.

Similarly, in the step 2 now occupies the second place from the left end and the other elements are pushed one place rightward.

So, alternate arranging of numbers and words finally gives the last step in which the odd places from the left are occupied by numbers and the even places are occupied by the words.

Shortcuts

When ever you see this type of problems just see the last step first. So that you can understand the logic without wasting your valuable time (generally these arrangements will be of assenting and descending orders). Then just check from bottom to top for the arrangement of words. And then only check the questions.
If you cant get answers mentally, you should write them on paper, but its waste of time. In those type of situations just write the first letters of the words. So, for the above example you can write
- N 41 28 C J 37 G 61 and work with this example. Suppose you encounter two words with the same starting letter then you should write two letters instead of one.
  - Ex : If you encounter Gun and Goat as two words you will be confused if you write two Gs. so just write Go and Gu to avoid confusion.
Without giving you the raw data if they give you some third or fourth step and ask you to find out the Sixth step there is no need to solve the problem completely. If the given step is the third one atleast three words will be already arranged in order. So just check the given arrangement and check how many words are arranged in order and just strike those words with pencil. And just work with the remaining words (as there is no need to change the already arranged words again). It will save your time and effort.
In some cases the required word or number which is to be arranged will be the first letter in the first letter of the resulting arrangement, in this case we will cut tha word or number but we will not increase the step counter as we do not have to shift it anywhere, it was already at its place
Keep in mind that If they give you an arrangement and asks you to guess the prior steps (I mean giving you the 5th or 6th step and asking you the prior steps 2nd or 4th), then the answer would be Cant be Determined. Because there wont be any rule for guessing backwards as the word may come from anywhere.

Input – Output Arrangements – Examples

Note : To save the space, We are not posting the previous example problem here. So please refer to the example problem here for reference. As per the rules followed in the above steps, find out in each of the following questions the appropriate step for the given input. 1 ) Input : Chair Wood 21 42 59 Height Bench 78. How many steps will be required to complete the rearrangement?
1. Three
2. Four
3. Five
4. Six
5. None of the above.
Solution : write the first letters of the given problem.

Step 1 : 78 C W 21 42 59 H B
Step 2 : 78 B C W 21 42 59 H
Step 3 : 78 B 59 C W 21 42 H
Step 4 : 78 B 59 C 42 W 21 H
Step 5 : 78 B 59 C 42 H 21 W
Step 6 : 78 B 59 C 42 H 21 W

So, according to the logic, step 6 is the last step for this input (Numbers are in descending order and words are in alphabetical order in this step 6). So 6 Steps will be required to complete the rearrangement. So the answer is option 4.

2) Input : When You 22 Special 31 16 47 Town
Which of the following steps will be the last but one?

1. 4th
2. 6th
3. 5th
4. 7th
5. Non of the above

Solution :

Step 1 : 47 W Y 22 S 31 16 T
Step 2 : 47 S W Y 22 31 16 T
Step 3 : 47 S 31 W Y 22 16 T
Step 4 : 47 S 31 T W Y 22 16
Step 5 : 47 S 31 T 22 W Y 16
Step 6 : 47 S 31 T 22 W 16 Y

So, the answer is Step 5. So option 3 is correct.

3) Step 4 of an input is : 74 Again 69 Call 17 32 Horse Desk
Which of the following is definitely the input?

1. Again Call 74 69 17 32 Horse Desk
2. 74 Call Again 17 69 Horse 32 Desk
3. Call 74 Again 69 17 32 Desk Horse
4. Cannot be Determined
5. None of these

Sol : As we have already discussed, this is the case of arrangement. So, the previous steps cannot be determined with certainty. We cannot estimate the exact input. 4) Step 3 of an input is :82 Brown 74 Sugar Hobby Lady 32 49Which of the following will be step 6?

1. 82 Brown 74 Hobby 49 Sugar Lady 32
2. 82 Brown 74 Hobby 49 Lady Sugar 32
3. 82 Brown Hobby 74 Hobby 49 Lady 32 Sugar
4. Cannot be determined
5. None of these.

Sol : Step 3 is 82 Brown 74 Sugar Hobby Lady 32 49, And we can find that first 3 words are in a proper order. So just strike them off with pencil and work with remaining words.

Step 4 : H S L 32 49
Step 5 : H 49 S L 32
Step 6 : H 49 L S 32

So, option 2 is correct. Note: If you don’t have enough time and you want to take risk then just see the answers. They’ve given the step 3, so there are chances that the first 3 letters wont change (in this case). So obviously the answer will be either option 1 or option 2. But this method is not suggestible. 5) Input : Goal Team Ask 12 92 85 42 Sound.Which of the following will be the step 4 ?

- 1. 92 Ask 85 Goal 42 Sound 12 Team
- 2. 92 Ask 85 Goal 42 Sound Team 12
- 3. 92 Ask 85 Goal 42 Team 12 Sound
- 4. 92 Ask 85 Goal 12 42 Sound
- 5. None of These

Sol : Input : G T A 12 92 85 42 S

Step 1 : 92 G T A 12 85 42 S
Step 2 : 92 A G T 12 85 42 S
Step 3 : 92 A 85 G T 12 42 S
Step 4 : 92 A 85 G 42 T 12 S

So the answer is option 3.

Percentage Shortcut Tricks

Important Info

Percentage shortcut tricks are very important thing to know for your exams. Competitive exams are all about time. If you know time management then everything will be easier for you. Most of us skip that part. Few examples on percentage shortcuts is given in this page below. All tricks on percentage are provided here. Visitors please read carefully all shortcut examples. These examples will help you to understand shortcut tricks on Percentage.

Before doing anything we recommend you to do a math practice set. Choose any twenty math problems and write it down on a page. Do first ten maths using basic formula of this math topic. You also need to keep track of Timing. After solving all ten math questions write down total time taken by you to solve those questions. Now practice our shortcut tricks on percentage and read examples carefully. After this do remaining ten questions and apply shortcut formula on those math problems. Again keep track of timing. You will surely see the improvement in your timing this time. But this is not all you need. You need more practice to improve your timing more.

You all know that math portion is very much important in competitive exams. That doesn’t mean that other sections are not so important. But if you need a good score in exam then you have to score good in maths. You can get good score only by practicing more and more. You should do your math problems within time with correctness, and this can be achieved only by using shortcut tricks. Again it does not mean that you can’t do maths without using shortcut tricks. You may have that potential to do maths within time without using any shortcut tricks. But so many people can’t do this. For those we prepared this percentage shortcut tricks. We always try to put all shortcut methods of the given topic. But it possible we miss any. We appreciate if you share that with us. Your little help will help so many needy.

Percentage is a very important chapter and it uses and it help for calculation most chapters for solution. Percentage Type1 Shortcut Tricks is based on, to find the missing number and find the percentage error or find the number that is given in this type of questions. Some difference percentage and rational number are given so At first we need to follow some traditional rule then we go through the shortcut tricks.

This type of problem are given in Quantitative Aptitude which is a very essential paper in banking exam. Under below given some more example for your better practice.

when we say 100 percent in mathematical notation we write 100% so 35 % means 35 per 100 and 65% means 65 per 100 it is a proportion per hundred and it is used to find marks, profit percent or loss percent of a particular product. it is also used in find out the depreciation at the rate percent per annul.

we can expressed ration number as percent, Here is some different percent related problems are given with shortcut tricks below.

Percentage Type1 Shortcut Tricks

Example 1:

If the difference between 45% of a number and 3/5th of that number is 18. What is the number?
Solution:
Let the number be a.
Then a x 45% – a x 3/5 = 18
a x 45% – a x 60% = 18 ∴ ( 60% = 60 / 100 = 3 / 5 )
a x 15% =18
a = 100 x 18 = 1800/15 = 120.
Tricks:
we know that 3 / 5 = 60%
45% – 60% = 15 %
15% = 18
= 120.
The number is 120.

Example 2:

Find the missing term.
? % of 35 = 672
Answer :
Let x % of 35 = 672
Then, ( x / 100 ) of 35 = 672
672 x 100 = 67200 / 35 = 1920
The missing term is 1920.

Example 3:

If 65 percent of a number is 51 less than 4 / 5 of that number. Find the number is ?
Solution:
Let the number be Y.
Then, 4 / 5 Y – ( 65% of Y ) = 51
4 / 5 Y – 65 / 100 Y = 51
15 Y = 5100
Y = 340.
The number is 340.

Example 4 :

If declaring a length 71.472 Km as almost as possible with three important digits, What would be the percentage error ?
Solution :
Error = ( 71.5 – 71.472 )Km = 0.028.
So Required percentage = ( 0.028 / 71.472 X 100 ) % = 0.039%.
The percentage error is 0.039%.

Example 5 :

The ratio 6 : 4 expressed as a percent equals to as.
Solution :
6 : 4 = 6 / 4
6 / 4 = (6 / 4 x 100 )%
= 150%

Example 6 :

Evaluate the Example ?
25% x 280 + 24% x 750
Solution: The percentage error is 0.039%.
25 / 100 x 280 + 24 /100 x 750
70 + 180 = 250

Example 7 :

A Nokia dealer sold a smart phone at 15% discount on print rate. If the customer paid Rs, 8500 and the printed price is 25% more than its cost price, What is the cost price of smart phone ?
Answer :
Cost price of smart phone = 8500 x 100 / 85 x 100 / 125
=85000000 / 10625 = 8000 .
The cost price of smart phone is 8000.

Example 8 :

The difference between 60% of a number and 30% of the same number is 24500. What is 75% of that number ?
Answer :
difference between 60% of a number and 30% of the same number is 24500
60% – 30% = 24500
30% = 24500
24500 x 100 / 30
So, 75% of that number is = 24500 x 75 / 30
= 2450 x 25 = 61250 .
The 75% of that number is 61250.

Example 9 :

Evaluate the Example ?
58% of 30102 – 32% of 17344
Answer :
17459.16 – 5550.08 = ?
? = 11909.08

Example 10 :

Evaluate the Example ?
52% of 8540 + 15% of 7860 = ? + 2540
Answer :
4440.8 + 1179 = ? + 2450
5619.8 – 2540 = ?
? = 3079.8

Percentage Type2 Shortcut Tricks

Example 1:

Bibhas work in shop and his 30% of income is Rs.1800 . Now Find the 65% of his income and also find the 1 / 5 of Bibhas income .

Solution: 30% = 1800 So, 65% = 1800 x 65 / 30 = 3900
Shortcut Trick: 1800 x 65 / 30 = 3900.
Solution: we know that 1 / 5 = 20% so, 1800 x 20 / 30 = 1200.

Example 2:

Bibhas ‘s salary was decreased by 20% and later increased by 20%, How much percent does he lose?
Solution:
Let Bibhas’s original salary = Rs.100
Now salary is = 120% of ( 20% of 100 ) = 120 x 80 x 100 / 100 x 100 = 96.
So, Bibhas Decrease salary is = 4%

Example 3:

If the sum of 40% & 20% of a number is 520. Find the 75% of that number ?
Answer :
40% + 20% = 520
60% = 520
1% = 520 / 60
75% = 520 x 75 / 60 = 650.

Example 4:

Karan’s monthly income is Rs. 14000. If his monthly income is increased by 10%. Find the total income after 2 months ?
Solution :
= 14000 x 110 x 110 / 100 x 100
= 16940

Example 5:

Jamal’s salary is increased by 10% & he got now Rs. 6000 . Find the salary before the increase ?
Answer :
6000 x 100 / 110 = 5454.54

Example 6:

If the product of 15% & 20% of a number is 360 . Find the sum of the 15% and 20% of that number ?
Answer :
15% x 20% = 360
300% = 360
1% = 360 / 300
Sum of ( 15 + 20 )% = 35%
35% = 360 x 35 / 300 = 42.

Example 7:

If Rahul’s income is 20% more than Rohan then how much percentages Rohan’s income is less than that of Rahul’s income ?
Answer :
We applied formula = ( increment% / 100 + increment% ) x 100
20 / 100 + 20 = 20 x 100 / 120 = 50 / 3.

Example 8:

Falguni’s salary is 40% of Dipika salary which is 25% of Rima’s salary . What percentage of Rima’s salary is Falguni’s salary ?

Answer :
Falguni : Dipika : Rima = 40 : 100 : 400
= 2 : 5 : 20
So , Required percent = 2 x 100 / 20 = 10%

Example 9:

Nilam who invest 25% of her income and she is able to save Rs.1500 per month. How much money her monthly expenses :
Solution :
Let her monthly income be Rs. X
Then ( 100 – 25)% of X = 1500
(75 % of X) = 1500
(75 /100) x X = 1500
X = 1500 x 100 / 75
X = 2000
Monthly expenses is = Rs.(2000 – 1500) = Rs. 500

Example 10 :

Raju has a Electronic shop, he marked his goods price always 25% more than the cost price. Find the cost price of a iPod, If a customer paid for iPod Rs.2890.
Answer :
2890 x 100 / 125 = 2312

Percentage Type 3 Shortcut Tricks

We all know that the most important thing in competitive exams is Mathematics. That doesn’t mean that other topics are less important. You can get a good score only if you get a good score in math section. A good score comes with practice and practice. All you need to do is to do math problems correctly within time, and you can do this only by using shortcut tricks. But it doesn’t mean that you can’t do math problems without using any shortcut tricks. You may have that potential to do maths within time without using any shortcut tricks. But so many other people may not do the same. For those we prepared this percentage shortcut tricks. We try our level best to put together all types of shortcut methods here. But if you see any tricks are missing from the list then please inform us. Your little help will help others.

Example 1:

Bijoy invest his money from his saving in different way that is he invest 30% on buying books, 10% on nutrition, 15% on wages and 22% on purchase a bike and after that all expenditure he saved 4600. Find the how much he spent on bike.
Answer :
Let the total income of Bijoy x. then total expenditure from income X x ( 30% + 10% + 15% + 22% ) = X x 77% = Total savings = X x 23%
X = 4600 x 100 / 23 = 20000 and expenditure on Bike = 22% so, 20000 x 22 / 100 = 4400.

Shortcut Tricks
The total income as 100% so, ( 100% – 30% + 10% + 15% + 22% ) = 77% and now ( 100% – 77% ) = 23%
Bike Expenditure is now 4600 x 22 / 23 = 4400.

Example 2:

If X is 80% of Y, then What percent of X is Y ?
Solution :
X = 80 / 100 x Y
X = 4 / 5 x Y
Y / X = 5 / 4.

Example 3 :

54.5% of 600 + 30.5% of 1800 = (?) + 147
Answer :
327 + 549 = (?) + 147
(?) = 729
? = 27.

Example 4 :

What is 25% of 25% equal to ?
Answer :
25% of 25% = 25 / 100 x 25 / 100 = 1 / 16 = 0.0625.

Example 5:

Subtracting 60% of a number from the number, We get the result as 20. The number is :
Solution :
Let the number is x, then ,
x – 60 / 100x = 20
x – 3 / 5 = 20
2x / 5 = 20
x = 20 x 5 / 2 = 50.

Example 6 :

In Mumbai due to reduction by 10% in the price of sugar, a person is able to buy 2 kg more for Rs. 280 . Find the reduced rate/kg of sugar ?
Answer :
10% of 280 is due to reduction of price in sugar
Rs. 28 = 2 kg
Rs. 14 = 1 kg.

Example 7:

In an party election between two candidates, one got 45% of the total valid votes, 20% of the votes were invalid. If the total number of votes was 7500, the number of valid votes that the other candidate got, was
Answer :
Number of valid votes
= 80% of 7500
= ( 80 / 100 ) x 7500 = 6000.
one got 45% of the votes
So other Valid votes polled by other candidate = 55% of 6000
= ( 55 / 100 ) × 6000 = 3300.
The number of valid votes that the other candidates got is 3300.

Example 8 :

If the price of diesel is increased by 25%, by how much percent a car owner must reduce his consumption in order to maintain the same budget ?
Answer :
25% = 25 x 100 / 125 = 20% less
20% a car owner must reduce his consumption in order to maintain the same budget.

Example 9 :

If the sugar price is increased by 8%, then by how much percent should a housewife reduce her consumption of sugar, to have no extra expenditure ?
Answer :
Less% = 8 x 100 / 108
= 800 / 108
= 7.40%

Example 10 :

In an general election a candidate who gets 64% of the total votes and wins by 476 votes. What is the total number of votes polled ?
Answer :
64% – ( 100% – 64% )
= 64% – 36%
= 28% = 476
= 476 x 100 / 28 = 1700.

Percentage Type 4 Shortcut Tricks

At first we need to follow some traditional rule then we go through the shortcut tricks. Percentage Type 4 Shortcut Tricks is based on, to find the number of valid votes that the other candidate got, Percentage is a very important chapter and it uses most chapters for calculation. some problems are based on Relation between Percentage or x and y.

This type of problem are given in Quantitative Aptitude which is a very essential paper in banking exam. Under below given some more example for your better practice.

we can expressed ration number as percent, Here is some different percent related problems are given with shortcut tricks below.

Example 1 :

Sanjay got 88 marks in Hindi, 81 marks in Science, 74 marks in Maths, 68 marks in Social Science and 57 marks in History. The maximum marks of each subject is 100. How much overall percentage of marks did he get ?
Answer :
Percentage = 368 x 100 / 500 = 73.6.

Example 2 :

In a class test, it is required to get 45% marks to pass. Joy got 618 marks and failed by 57 marks. What is the maximum marks in class test ?
Answer :
Joy got 618 marks and failed by 57 marks
X x 45% = 618 + 57
X x 45% = 675
X = 675 x 100 / 45 = 1500.

Example 3 :

A engineering student has to secure 60% marks to pass. He gets 70 and fails by 50 marks. Find the maximum marks.
Answer :
He gets 70 and fails by 50
So , 60% = 70 + 50
60% = 120
100% = 120 x 100 / 60 = 200 marks
So , the maximum marks is 200.

Example 4 :

The average marks of Rahim in 9 subject is 68. His average marks in 8 subjects except Math is 65. How many marks did he get in Math ?
Answer :
( 68 x 9 ) – ( 65 x 8 ) = ?
612 – 520 = 92.

Example 5 :

Ranjan got 82 marks in Math, 78 marks in Physics, 65 marks in Computer, 68 marks in English, The maximum marks of each subject are 85. How much overall percetage of marks did ranjan get ?
Answer :
Total marks get in all subject ( 82 + 78 + 65 + 68 ) = 293
Maximum marks ( 85 x 4 ) = 340
Percentage = 293 x 100 / 340 = 86.17.

Example 6 :

A student scores 25% & failed by 35 marks while another student who scores 65% get 45 marks more than minimum required marks to pass. Find the maximum marks in the exam ?
Answer :
25% – 35 = 65% + 45
65% – 25% = 45 + 35
40% = 80
100% = 80 x 100 / 40 = 200 marks.

Example 7 :

If N is equals to 20% of M and P is equals to 30% of N, then which one of the following equals to 40% of P ?
Answer :
N = 20% of M = ( 20 / 100 ) x M = 0.2 M.
P = 30% of N = ( 30 / 100 ) x N = 0.3 N = 0.3 x 0.2 M.
So here 40% of P = ( 40 / 100 ) x P = ( 0.4 )( 0.3 )( 0.2 M )
= 0.024 M.

Example 8 :

In an examination it is required to get 57% of the aggregate marks to pass. A student gets 237 marks and is declared failed by 7% marks. What are the maximum aggregate marks a student can get ?
Answer :
57% = 237 + 7%
57% – 7% = 237
50% = 237
100% = 237 x 100 / 50 = 474.

Example 9 :

A student has obtain 34% of the total marks to pass in paper. He got 113 and failed by 40 marks. The maximum marks are :
Answer :
Let the maximum number is X
Then, 34% of X = 113 + 40
34 / 100 x X = 153
X =153 x 100 / 34
X = 15300 / 34 = 450.

Example 10 :

Rajah has to score 60% to pass exam. He scores 225 marks & failed by 15%. Find the maximum marks of exam.
Answer :
60% – 15% = 225
45% = 225
100% = 225 x 100 / 45
= 500 marks.

Percentage calculation Shortcut Tricks

We all know that the most important thing in competitive exams is Mathematics. That doesn’t mean that other sections are not so important. But if you need a good score in exam then you have to score good in maths. A good score comes with practice and practice. All you need to do is to do math problems correctly within time, and you can do this only by using shortcut tricks. Again it does not mean that you can’t do maths without using shortcut tricks. You may have that potential that you may do maths within time without using any shortcut tricks. But so many other people may not do the same. So Percentage calculation shortcut tricks here for those people. We try our level best to put together all types of shortcut methods here. But it possible we miss any. We appreciate if you share that with us. Your little help will help so many needy.

Percentage is a very important chapter and it uses most chapters for calculation.Shortcut Tricks is based on, to find the number that is obtain in this type of questions. we go through the shortcut tricks.

This type of problem are given in Quantitative Aptitude which is a very essential paper in banking exam. Under below given some more example for your better practice.

we can expressed ration number as percent, Here is some different percent related problems are given with shortcut tricks below.

Example 1 :

Find what percent is 4% of 5% ?
Answer :
( 4 x 100 x 100 / 100 x 5 ) % = 80%.

Example 2 :

1 / 3 of 1206 is what % of 134 ?
Answer :
Let 1 x 1206 / 3 = y% of 134 . Then, y x 134 / 100= 402
x = ( 402 x 100 / 134 ) = 300.

Example 3 :

M is what % of N ?
Answer :
100 M / N %.

Example 4 :

How is 3 / 2% expressed as a percentage ?
Answer :
3 x 100 / 2 = 150%.

Example 5 :

35% of 250 + 25% of 350
Solution:
35 X 2.50 + 25 X 3.50 = 87.5 + 87.5 = 175
( We can write 250 / 100 = 2.50 and 350 / 100 = 3.50 when any number divide by 100 then put the two decimal point above the number which help in quick calculation ).

Example 6:

20% of 400 + 60% of 800
Solution:
( 20 x 400 / 100 ) + ( 60 x 800 / 100 )
= ( 80 + 48 )
= 128.

Example 7 :

48 % of 200 gm – 18% of 400 gm
Solution:
48 x 2 = 96, 18 x 4 = 72, ( 96 – 72 ) = 24.

Example 8 :

5 is what percent of 50 ?
Solution:
(5 / 50) = 10 and (100 / 10 )% = 10%.

Example 9 :

How is 3 / 2 % expressed as a decimal fraction ?
Answer :
3 / 2 % = ( 3 / 2 x 1 / 100 ) = 0.015.

Example 10:

A is six times as large as B. The percent that B less than A, is:
Solution :
A = 6B So, B is less than A by 5B
Required percentage is :
(5B / 100)% = (5B / 6B x 100)% = 250 / 3 % .

Percentage calculation on population Shortcut Tricks

We all know that the most important thing in competitive exams is Mathematics. It doesn’t mean that other topics are not so important. But only math portion can leads you to a good score. Only practice and practice can give you a good score. You should do your math problems within time with correctness, and only shortcut tricks can give you that success. But it doesn’t mean that without using shortcut tricks you can’t do any math problems. You may do math problems within time without using any shortcut tricks. You may have that potential. But so many other people may not do the same. Here we prepared percentage calculation on population shortcut tricks for those people. Here in this page we try to put all types of shortcut tricks on Percentage calculation on population. But we may miss few of them. If you know anything else rather than this please do share with us. Your help will help others.

Percentage is a very important chapter and it uses most chapters for calculation.Shortcut Tricks is based on, to find the population of a particular town,city in increased and decreased order that is obtain in this type of questions.

we go through the shortcut tricks. This type of problem are given in Quantitative Aptitude which is a very essential paper in banking exam.Under below given some more example for your better practice.

when we say 100 percent in mathematical notation we write 100% so, 35 % means 35 per 100 and 65% means 65 per 100 it is a proportion per hundred and it is used to find marks percentage of a student, profit percent or loss percent of a particular product.

It is also used in find out the depreciation at the rate percent per annul or population of a particular city or town where due to some reason rate of population increases or decreases.

we can expressed ration number as percent, Here is some different percent related problems are given with shortcut tricks below.

Example 1:

The population of a town has rapidly increases by 20% every year. If the population of the town in 2011 was 6,00,000 than, What would be its population in 2014 ?

Answer: Population in that town in 2014
600000 x 120 x 120 x 120 / 100 x 100 x 100 = 1036800.

Example 2:

In a town has present population 52500 and If it is decreased by 20% per annul. What would be the population its 2 years hence.

Solution:
We know
Population after n year = P(1+R/100)n
So, Here we
P = 52500
R=20% ( decreased )
n = 2 years.
Population after 2 years

Short Tricks :
= 52500 x 80/100 x 80/100
= 33600.

Example 3:

The population of a city was 2 years ago 24000, population decreases every year at the rate of 5 %. Find the present population of the city.
Solution :
population was decreases at a particular rate so

Present population of that city is
=24000 x ( 1 – 5 %)²
=24000 x ( 1 – 5 / 100)²
=24000 x ( 19 / 20 x 19 / 20 )
=21660.

Example 4:

The population in pune city was 1,50,000 three years ago. If it increased by 4%, 5% and 6% respectively in the last three years , then Find the present population of the pune city.

Solution :
Here population increased by continuous three years respectively, So Present population is the city is
=150000 x (1 + 4 / 100 )( 1 + 5 / 100 )( 1 + 6 / 100 )
= 150000 x 104 / 100 x 105 / 100 x 106 / 100
= 150000 x 52 x 21 x 106 / 50 x 20 x 100
=173628.

Example 5 :

60% of the population of a city are men and 15% are women. If the number of children are 20000, then the number of men will be of that city ?
Answer :
25% = 20000
60% = 20000 x 60 / 25 = 48000 .

Example 6 :

The total population of a village is 6000. The number of males and females increases by 10% and 20% respectively and consequently the population of the village becomes 7500. What was the number of males in the village ?
Answer :
The total % increased population
= 7500 – 6000 x 100 / 6000
= 1500 x 100 / 6000 = 25%
By allegation method
Male Female
+10 +20

. +25

5 15

Male population before increment
= 6000 x 5 / 20 = 1500

Example 7:

The population of a town is 56600. If in a 1st year it is decreased by 10% and 2nd year it is increased by 20%. What would be the population of that city after 2 years?

Answer :
Shortcut tricks :
56600 x 90 x 120 / 100 x 100 = 61128
So, after 2 years the population of that city is 61128.

Example 8 :

The population of a city increases by 10% annually. If its population in 2010 was 12000 then, What it was in 2007 ?

Solution :
We need to find the population of city 3 years back, and also the population of a city increases by 10% annually.
So, the population of city was in 2007
=12000 / (1 + 10%)³
=12000 x ( 22 / 20 x 22 / 20 x 22 / 20 )
= 15972.

Example 9:

If in a city every year population is decreased by 6%, What would be after 2 years of this towns population while its present population has 65000.
Answer: 65000 x 94 x 94 / 100 x 100 = 57434.

Shortcuts for finding Percentages

Percentage is nothing but a fraction, whose denominator is 100. The actual meaning of Per-Cent is per every 100.

It is usually denoted by the sign % and sometimes shortened as P.C.

Percentage is the the standard way to compare two quantities. Suppose we want to compare to fractions (4/50) and (9/25), make the denominator 100 for both the fractions.

Eg. 5/100 is called 5 per cent (or 5%)

Here 5, which is the numerator is called the rate per cent.

In the same way 18% means 18/100 (18 out of 100)

So,4/50 = 8/100 => means 8 out of 1009/25 = 36/100 => means 36 out of 100 so, (4/50) < (9/25)

Have a look at some important points about Percentages.

5% of Rs. 200 means (5/100)X(200/1) = 10 (remember guys, ‘Of’ means multiplication).
75% is equivalent to 75/100 = 3/4
Increase % = (increase/original) X (100)

Have a look at some short forms of percentages before doing some problems (better try to remember these)

5 % = 5/100 = 1/20 = 0.05
6 1/4 % = 25/400 = 1/16 = 0.0625
10 % = 10/100 = 1/10 = 0.1
12 1/2 % = 25/200 = 1/8 = 0.125
16 2/3 % = 50/300 = 1/6 = 0.166
20% = 20/100 = 1/5 = 0.2
25% = 25/100 = 1/4 = 0.25
33 1/3 % = 100 / 300 = 1/3 = 0.33
40 % = 40 / 100 = 2 / 5 = 0.4
50 % = 50 / 100 = 1/2 = 0.5
60 % = 60 / 100 = 3/5 = 0.6
66 2/3 % = 200/300 = 2/3 = 0.66
75 % = 75 / 100 = 3/4 = 0.75
80 % = 80 / 100 = 4 / 5 = 0.8
90 % = 90 / 100 = 9/10 = 0.9
100% = 100 / 100 = 1
125 % = 125 / 100 = 5 / 4 = 1.25
150 % = 150 / 100 = 3/2 = 1.5

Problems on Percentages

1. A’s income is 20% more than that of B while B’s income is 20% less than that of C. Whose income is the highest amongst all?

Sol : Let the income of C be Rs. 100/- So, The income of B is Rs. 80/- A’s income is 20% more than B = (120/100) X 80 = Rs. 96/-

So, obviously C’s income is Highest

2. A’s income exceeds that of B’s by Rs.600 while B’s income is 20% less than that of C’s. If total income of all of them put together is Rs. 3.850, what is the income of C ?

Sol : Let the income of C be Rs. X

B’s income is 20% less than C’s
So, B’s income is 80% of X = 4 X / 5

So, A’s income is (4 X /5) + 600  => ((4 X /5)+600) + (4 X /5) +  X  = 3,850
=> (8 X  / 5) +  X  = 3,250
=> 13  X  = 5 x 3,250
   X  = (5 x 3,250) / 13 = Rs. 1,250

3. Shivani spends 15% of her salary on Shopping, 20% on House – Rent, 50 % on Food and the remaining on Education of Her children. If education costs her Rs 3000, how much does she spend on House – Rent ?

Sol : Let the salary of Shivani is Rs. 100

So, amount spent on Education = 100 – ( 15 + 20 + 50) = Rs. 15

But, her expences towards education is Rs. 3000

So, 15% of salary = (3000 x 100) / 15 = Rs. 20,000

She spent 20% of her salary on house rent

So, 20% of 20,000 = Rs. 4000/-

4. In a class, 30% students passed in Maths, 50% students passed in English and 10% students passed in Both. What per cent of students failed in both of these subjects?

Sol : Let the number of total students = 100

So, Number of students passed in Maths = 30

Number of students passed in English = 50

Number of students passed in Maths and English = 10

So, total number of students passed = (30 + 50 – 10) = 70

Number of students failed = 30

So, Percentage of Students failed = 30 %

5. A trader announced 10% reduction in the unit price of an article. As a result, the sales volume went up by 10%. What was the net effect on the sales revenue?

Sol : Let the unit price be Rs. 10/-

Let the number of articles sold = 10

So, the sales revenue = 10 x 10 = Rs. 100

The trader has reduced the unit price by 10%

So, Reduced price = 90% of Rs 10 = Rs. 9 /-

and, increase in the sales volume = 10% = 110% of 10 = 11

So, The net effect = 9 x 11 = Rs. 99

The sales revenue has come down by Re. 1/-

So, the sales revenue has come down by 1%.

Note : There is a short cut for these type of problems.

If price is reduced by X% and sales increase by Y%,

the total effect on sales revenue : = increased % value – Decreased % value

= (Increased % value x Decreased % value) / 100

The effect on revenue is increased or decreased according to the positive or negative sign otained

= 10 – 10 – ((10 x 10) / 10)) = 0 – 1 = -1

The sign obtained is negative. Thus, the total revenue decreases by 1%

Simple Interest & Compound Interest Shortcut Tricks

Important Info

Shortcut tricks on simple interest and compound interest are one of the most important topics in exams. Competitive exams are all about time. If you manage your time then you can do well in those exams. Most of us skip that part. Here in this page we give few examples on Simple Interest and Compound Interest shortcut tricks. We try to provide all types of shortcut tricks on simple interest and compound interest here. We request all visitors to read all examples carefully. These examples will help you to understand shortcut tricks on Simple Interest and Compound Interest.

Before starting anything just do a math practice set. Write down twenty math problems related to this topic on a page. Do first ten maths using basic formula of this math topic. You also need to keep track of Timing. Write down the time taken by you to solve those questions. Now practice our shortcut tricks on simple interest and compound interest and read examples carefully. After finishing this do remaining questions using Simple Interest and Compound Interest shortcut tricks. Again keep track of the time. You will surely see the improvement in your timing this time. But this is not enough. You need to practice more to improve your timing more.

You all know that math portion is very much important in competitive exams. It doesn’t mean that other topics are not so important. But only math portion can leads you to a good score. A good score comes with practice and practice. You should do your math problems within time with correctness, and this can be achieved only by using shortcut tricks. But it doesn’t mean that without using shortcut tricks you can’t do any math problems. You may have that potential that you may do maths within time without using any shortcut tricks. But other peoples may not do the same. So Simple Interest and Compound Interest shortcut tricks here for those people. We always try to put all shortcut methods of the given topic. But it possible we miss any. We appreciate if you share that with us. Your little help will help others.

What is Interest ?

When some one take up some money from other for the personal or commercial purpose we pay some additional money to him after a certain period of time is called Interest. So we can also called this Interest as Simple Interest. This type of problem are given in Quantitative Aptitude which is a very essential paper in banking exam. Under below given some more example for your better practice.

What is Principle ?

When money borrow for a certain time period called Principle or Sum.

What is Amount ?

The Addition of Simple Interest and Principle is called the Amount.
A = S.I + P ( Principle ).
S.I = A ( Amount ) – P ( Principle ).

What is Per annul means ?

Per annul means For a year.

P = Principle
R = Rate of per annul
T = Number of years

When we Add Simple Interest into Principle. It becomes into Amount.

Formulas Need to Remember
S.I=( P X R X T / 100 )
Here, P = Principle.
R = Rate per annul.
T = Number of years.

Formula:
In case S.I ( Simple Interest ), T ( Number of years ) and R (Rate per annul ) are given in Question then we can easily find the Principle or Sum.
P = ( S.I X 100 / R X T ).

Formula:
In case S.I ( Simple Interest ), T ( Number of years ) and P ( Principle ) are given in question then we can easily find the R (Rate per annul ).
R = ( S.I X 100 / P X T ).

Example 1:

Find the simple interest on Rs 500 for 5 years at 5 per cent ?

Answer :
SI = 500 x 5 x 5 / 100
Simple interest in 5 years is Rs 125.

Compound Interest Shortcut Tricks

Some important formula of Compound Interest

A = Amount.
P = Principal.
R = Rate of Interest.
N = Number of Years.
Type I : Interest compounded yearly :
A = P ( 1 + r / 100 )ⁿ
Type II : Interest compounded half – yearly :
Amount = P [ 1 + r / 2 / 100 ]⁴ⁿ or = P = [ 1 + r / 200 ] ²ⁿ
Type III : Interest compounded quarterly :
Amount = P [ 1 + r / 4 / 100 ] or = P [ 1 + r / 400 ] ⁴ⁿ

In Compound Interest problems asked in exams up to the period of 3 years.

In case we apply basic formula: Amount = Principle ( 1 + r / 100 )ⁿ here r = Rate and n = Time

As consider if Principle is Rs. 1, then the it will be in first year and second and third years.

( 1 + r / 100 )¹

( 1 + r / 100 )²

( 1 + r / 100 )³

If the rate of interest is 3%, then the value will be …….

In first year = (23 / 21 ) = 23 / 21.

In second year = ( 23 / 21 )² = 529 / 441.

In Third year = ( 23 / 21 )³ = 12167 / 9261.

Compound Interest Examples

Example 1. = A. Given an investment of $3,000 at 5% compounded quarterly for 6 years, find the interest earned and the future value. Prepare a table showing the growth of the account balance and illustrate that growth with a chart.

r = 0.05
ppy = 4
i = r/ppy = 0.05/4 = 0.0125
t = 6
n = (t)(ppy) = (6)(4) = 24
P = 3,000
A = ?
I = ?

Calculator Solution

Compare the $1,042.05 interest earned to the $900 that would have been earned with simple interest.

Notice that both the future value and the interest given by the formulas is off by one cent.

B. For this same investment, suppose the interest is compounded monthly instead of quarterly. Find the interest earned and the future value.

r = 0.05
ppy = 12
i = r/ppy = 0.05/12
t = 6
n = (t)(ppy) = (6)(12) = 72
P = 3,000
A = ?
I = ?

Calculator Solution

Notice that almost five dollars more interest will be earned if the interest is compounded monthly instead of quarterly.

Example 2

A. Find the present value of an investment if the future value is $1,000. The investment pays 4.5% compounded semiannually for seven years.

r = 0.045
ppy = 2
i = r/ppy = 0.045/2 = 0.0225
t = 7
n = (t)(ppy) = (7)(2) = 14
P = ?
A = 1,000
I

Calculator Solution

The present value for the corresponding simple interest problem was $760.46. Remember that with compound interest more interest is earned because the interest is periodically added to the balance. Consequently, the interest itself earns interest. Since more interest is being earned, it requires less of an investment to achieve the same future value.

B. Suppose the interest is compounded daily instead of semiannually. Find the present value.

r = 0.045
ppy = 365
i = r/ppy = 0.045/365
t = 7
n = (t)(ppy) = (7)(365) = 2555
P = ?
A = 1,000
I

Calculator Solution

Notice that the present value is somewhat lower than in the example above. Since the interest is paid more frequently (daily instead of semiannually) the total interest paid is greater which lowers the present value even more. The change, however, is much less dramatic than going from simple interest to interest compounded semiannually.

Example 3

The interest on a 4.5 year investment paying 3.6% compounded monthly was $245. How much was invested and what was the future value?

r = 0.036
ppy = 12
i = r/ppy = 0.003
t = 4.5
n = (t)(ppy) = (4.5)(12) = 54
P = ?
A = ?
I = $245.00

Since we know neither P nor A, we cannot use either the future value formula nor the present value formula directly to answer this question. However, with a little algebra we can derive a formula that will give us the present value. The only thing we know is that the interest is $245 so let’s start with the interest formula:

I = A -P

Substitute the future value formula for A:

Factor P out of the two terms on the right hand side of the equation:

Divide both sides by :

Substituting in the known values for I, i, and n, we obtain the following:

Example 4

What is the future value of an investment of $600 at 2.3% compounded daily for 10 years?

r = 0.023
ppy = 365
i = r/ppy = 0.023/365
t = 10
n = (t)(ppy) = (10)(365) = 3650
P = 600
A = ?
I = ?

Compare the future value of $755.15 with the $738.00 that would have resulted from simple interest.

Example 5

What is the purchase price of a $500 savings bond that earns 6% compounded monthly and matures in 5.5 years? How much interest is earned?

r = 0.06
ppy = 12
i = r/ppy = 0.06/12 = 0.005
t = 5.5
n = (t)(ppy) = (5.5)(12) = 66
P = ?
A = 500
I = ?

Age Problems: Concepts & Tricks

About Age Problems

Age problems are one of the most common topics in IBPS, CAT, GMAT and other banks exams. Students waste lots of time in this question as it look very simple but when they start solving with triditional methods, it takes a lot of time. Today I am solving few solving with shortcut trick. In case of any problem, please comment below.

Best way to solve Age questions is to assume fixed period with which further conditions will be compared. For example taking 2000 as fixed year.

Application of this rule

Example 1

Raman’s age after 15 years will be 5 times his age 5 years back. What is his present age ?Solution – Let’s assume right now it is year 2000 Age of Raman in 1995 = xAge of Raman in 2015 = 5xPresent age of Raman (in 2000) = x+5 or 5x-15 we will solve these two equation to find x. X= 5. Then Raman’s present age becomes = x +5 = 10

Example 2

Rahul was 4 times old as his son 8 years back and he will be 2 times old as his son after 8 years. Calculate Rahul and his son’s age. Assume that currently it is year 2000.In 1992 Rahul’s age = 4x, Age of Rahul’s son = xIn 2008 Rahul’s age = 2y and Age of Rahul’s son = y Now we get two equations 2y – 4x = 16 and y – x = 16By solving this equation x = 8, so Rahul’ son’s current age = 16 years and Rahul’s age = 40 years.

Time, Speed & Distance Concepts Tricks

Concepts

1) There is a relationship between speed, distance and time:

Speed = Distance / Time OR

Distance = Speed* Time

2) Average Speed = 2xy / x+y

where x km/hr is a speed for certain distance and y km/hr is a speed at for same distance covered.

**** Remember that average speed is not just an average of two speeds i.e. x+y/2. It is equal to 2xy / x+y

3) Always remember that during solving questions units must be same. Units can be km/hr, m/sec etc.

**** Conversion of km/ hr to m/ sec and m/ sec to km/ hr

x km/ hr = (x* 5/18) m/sec i.e. u just need to multiply 5/18

Similarly, x m/sec = (x*18/5) km/sec

4) As we know, Speed = Distance/ Time.

Now, if in questions Distance is constant then speed will be inversely proportional to time i.e. if speed increases ,time taken will decrease and vice versa.

Time and Distance Problems

Problem 1: A man covers a distance of 600m in 2min 30sec. What will be the speed in km/hr?

Solution:

Speed =Distance / Time
⇒ Distance covered = 600m, Time taken = 2min 30sec = 150sec
Therefore, Speed= 600 / 150 = 4 m/sec
⇒ 4m/sec = (4*18/5) km/hr = 14.4 km/ hr.

Problem 2: A boy travelling from his home to school at 25 km/hr and came back at 4 km/hr. If whole journey took 5 hours 48 min. Find the distance of home and school.

Solution:

In this question, distance for both speed is constant.
⇒ Average speed = (2xy/ x+y) km/hr, where x and y are speeds
⇒ Average speed = (2*25*4)/ 25+4 =200/29 km/hr
Time = 5hours 48min= 29/5 hours
Now, Distance travelled = Average speed * Time
⇒ Distance Travelled = (200/29)*(29/5) = 40 km
Therefore distance of school from home = 40/2 = 20km.

Problem 3: Two men start from opposite ends A and B of a linear track respectively and meet at point 60m from A. If AB= 100m. What will be the ratio of speed of both men?

Solution:

According to this question, time is constant. Therefore, speed is directly proportional to distance.
Speed∝Distance

⇒ Ratio of distance covered by both men = 60:40 = 3:2
⇒ Therefore, Ratio of speeds of both men = 3:2

Problem 4: A car travels along four sides of a square at speeds of 200, 400, 600 and 800 km/hr. Find average speed.

Solution:

Let x km be the side of square and y km/hr be average speed
Using basic formula, Time = Total Distance / Average Speed

x/200 + x/400 + x/600 + x/800 = 4x/y ⇒ 25x/ 2400 = 4x/ y⇒ y= 384
⇒ Average speed = 384 km/hr

Syllogism Questions

About Syllogism

Syllogism is an important chapter in almost every competitive exam. In IBPS PO 2013, there were 5 questions from syllogism. Many candidates face problem in solving these simple questions so I decided to write a detailed tutorial on this chapter.

TYPES OF CONDITIONS:

1) All pigs are animals. Some animals are mammals.

In this example, there are two sub-conditions.

Condition 1 – All pigs are animals. So circle Pig will be covered by circle Animals.
Conditions 2 – Some animals are mammals. This sub-conditions give two situations :-

1) Some pigs are mammals
2) No Pig is mammal.

2) All pigs are animals. All animals are mammals.

This example is bit simple. All pigs are part of category animal and all animals are part of category mammals.

3) Some pigs are animals. All animals are mammals.

In this example there are two conditions. There is no proper order to apply these conditions. Some pigs are animals. All animals are mammals so Animals circle is covered Mammals circle. Conclusion is some but not all pigs are mammals.

In exam, try to make circles for every syllogism question. These questions take hardly 20-30 seconds to solve with this method. This year these type of questions replaced mirror image questions, so never skip these marks rich questions.

Squaring Technique to speed-up

About Squaring Technique

Many readers have requested me to teach them a simple method to square two and three digit numbers. In bank exams calculation speed is very crucial. So today I have decided to explain my Squaring technique.

Technique

In case of two digit number deduct last digit and add it to another number and then add square of same.

In this technique we simplify the squaring method by making one unit’s digit zero. It is far easy to multiply 50*24 than 54*24. So I used this technique. Try practice more to become expert in this technique.

Let’s take some examples

Find square of 53.=(53*53)= (53+3) * (53-3) + (3*3)=(56*50) + 9= (560*5) + 9= 2800 + 9 = 2809 Let’s take another example Find square of 69= (69*69)= (69+1) * (69-1) + (1*1)= (70*68) + 1= (680*7) + 1= 4761 Let’s take one more exampleFind square of 45= (45*45)= (45-5)*(45+5) + (5*5)= (40*50) + 25= 2000+25= 2025

Blood Relation Tricks

Important Info

Cousin : Mother’s or Father’s Brother’s or Sister’s Son or Daughter (or) Parent’s siblings’ son or daughter (or) Uncle’s or Aunt’s son or daughter.

Nephew : Brother’s or Sister’s son

Niece : Brother’s or Sister’s daughter.

Uncle : Father’s or Mother’s brother.

Aunt : Father’s or Mother’s sister

Father-in-Law : Spouse’s father (or) Wife’s or Husband’s Father

Mother-in-Law : Spouse’s mother (or) Wife’s or Husband’s Mother

Son-in-Law : Daughter’s Husband

Daughter-in-Law : Son’s wife

Brother-in-Law : Spouse’s brother (or) Sister’s husband

Sister-in-Law : Spouse’s sister (or) Brother’s Wife

Maternal : of or related to Mother

Paternal : of or related to Father

Siblings : brothers or sisters

Spouse : Husband or Wife

Note : Cousin is a COMMON GENDER, There are NO such words Cousin Brother / Cousin Sister

Generations :

-2 Generation : Grand father and Grand Mother (Maternal & Paternal)

-1 Generation : Mother, Father, Brother, Sister, Brother-in-Law, Sister-in-Law, Cousin

+1 Generation : Son, Daughter, Son-in-Law, Daughter-in-Law, Nephew, Niece

+2 Generation : Grand Sons and Grand Daughters

Points, to Remember

If the question is “How is A related to B”. Then you must know the gender of A to answer the question.

Without knowing A’s gender, you cannot determine the relation from A to B.

If a person is Uncle or Aunt to “A”, A is that person’s nephew (if A is Male) or niece (if A is female)

Cousin is a common Gender (I mean, you can use this word for both Male and Female)

Ex : A says to B, “you are the son of my grand father’s only son”.
- - How is B related to A?
  - How is A related to B
  - Ans :
- - B is Brother of A.
  - My grand father’s only son = A’s father (grand father’s only son means only ONE son.. No other son and no daughter also)
  - A is either brother or sister to B
    - A and B are siblings. B is brother to A as we know the Gender of B. But we dont know the Gender of A. So we cant say whether A is Brother or Sister to B.

Lets have a look at some practice problems

The mother of Ranbir is the only daughter of Neetu’s father. How Neetu is related to Ranbir?
- Sister
- Mother
- Aunt
- Cant determine
- None of These

Answer : Mother Explnation : Only daughter of Neetu’s father = Neetu only. The mother of Ranbir is Neetu, so Neetu is the mother of Ranbir 😀

Amit is the son of Ajit’s grand father’s only daughter. How is Ajit’s father related to Amit?
- Grand Father
- Uncle
- Father
- Data Inadequate
- None of these

Answer : Father Explanation : Ajit’s grand father’s only daughter means Ajit’s mother. Amit is the son of Ajit’s mother. So, Amit and Ajit are Siblings. Ajit’s father is Amith’s father too

Pointing to a boy Rekha said, “He is the son of my mother-in-law’s only Child”. How is the boy related to Rekha?
- Son
- Grand Son
- Nephew
- Brother
- Cant determined

Ans : Son Explnation : My mother-in-law’s only child = Rekha’s mother-in-law’s only child = Rekha’s husband. The boy is the son of the Rekha’s Husband. So obviuosly Rekha is the mother to that boy and that boy is the Son of Rekha.

B is the husband of C. A is the sister of B. D is the sister of C. How is D related to B?
Son
- Uncle
- Sister-in-Law
- Cant be determined

Ans : Sister-in-Law Explnation :

- B is Husband and C is his Wife.
- A is the sister of B, so A is the Sister-in-Law of C ( Husband’s sister)
- D is the sister of C so, D is the Sister-in-Law of B (Wife’s Sister)
Pointing towards a girl, a Person said, “She is the only daugher of the only son of the wife of the father-in-law of my wife”. How is the girl related to the Person?
- Niece
- Daughter
- Sister
- Daughter-in-Law
- Cant be Determined

Ans : Daughter Explnation :

- Father-in-Law of my wife = Father in Law of the Person’s Wife = Person’s Father
- Wife of the Father-in-law of my wife = Wife of the Person’s Father = Person’s Mother.
- Only son of Person’s Mother = That Person only (because Person’s parents dont have any other child)
- So, she is the only daughter of Person as the Person does not have any other Children.