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Showing posts with label Quantitative. Show all posts
Showing posts with label Quantitative. Show all posts

## Fully solved Numerical/quantitative aptitude practice questions for bank po and clerk exams

As you know maths or quantitative section is the most crucial section in bank exams.Mostly this section is the most time consuming section in most of the bank and other competitive exams. YOU need thorough preparation of numerical section to clear any bank test.You should understand basic formulas and tricks for solving such questions.After that attend maximum practice questions and mock tests.Here is the complete list of quantitative aptitude practice questions for bank examinations. We have prepared this numerical aptitude questions in pdf format so that you can read and download this practice sets.Each pdf contains section wise  15-20 quantitative practice questions with answers. We have also included questions from previous year question papers.Questions are solved using best shortcuts available.

#### List of Quantitative Aptitude questions PDFs

We are sure that this practice questions will help you in upcoming bank,ssc,upsc,rrb and other competitve exams.We will post more number of practice sets in coming days.
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## Download Quantitative shortcuts and Tricks for IBPS,SBI,RRb,SSC,UPSC and other competitive exams

Here we sharing updated version(V.2) quantitative aptitude shortcuts and formulas pdf ebook for bank and other competitive exams.You can read it in pdf ebook format and you can also download this arithmetic shortcut tricks for various competitive exams for free.

### Contents of the below quant shortcut formula ebook

Chapter 1:Basic Math Shortcuts
1. Division-shortcuts
2. Multiplication-shortcuts
3. Square-Shortcut Tricks
4. Cubes-Shortcut
5. Cube root(for perfect cubes only)

Chapter 2:Ratio and Fractions
1. Ratios-Important rules and shortcuts
2. Comparison of ratios and Fractions

Chapter 3.Simple Interest and Compound Interest
1. Simple Interest
2. Compound Interest

Chapter4:Mixture and Alligation
1. When Two quantities are mixed
2. If more than two different commodities are mixed
3. Removal and replacement

Chapter 5:Work and Time
Chapter 6.Profit and loss formulas and shortcuts
Chapter 7:Pipe And Cistern Shortcuts for Bank Exams

Chapter8:Time distance and speed
1. Ratio of speed
2. Average speed
3. Points to be noted while doing 'train and time' problems
4. Boat and stream problems
Chapter9:Permutation and combination important formulas and shortcuts
1.Permutation Formulas and shortcuts
2.Combination Important formulas

Chapter10:Probability formulas
1.Some random experiments and their outcomes
2.Probability of occurrence of an Event
3.Important Results on probability

### Quantitative shortcuts and tricks ebook V.2

If you find this quantitative aptitude shortcut tricks pdf useful ,then share it with your friend.
Competitive exams pdf.Keep on visiting bankaspire.We will come back with new and improved version of quant shortcuts pdf.Like our facebook page or follow us on google+ to get updates.
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## IBPS PO quantitative practice question paper (Fully solved)

Here we are sharing a new set of quantitative practice questions for ibps po main examination in pdf format.Quantitative is the most crucial section in ibps po prelim and main examinations.Purpose of quantitative section is to test your numerical ability.You need not be a math expert to solve this section.Practicing more questions and attending mock test or solving model papers will improve your speed considerably.Model paper is the best way to assess yourself.By solving this IBPS PO quantitative practice question paper, you can identify your area of strength and weakness and plan your further preparation based on that.This Quantitative Practice Questions for IBPS PO is fully solved using time saving shortcut tricks. The below IBPS PO quantitative aptitude questions pdf ebook contains fully solved 40 questions based on latest IBPS PO syllabus.You can download pdf after reading ,from the link provided at the end.

## ebook:IBPS PO quantitative Aptitude questions for ibps po main exam

Difficulty level of this fully solved IBPS PO Quantitative aptitude is moderate,and it best suit for IBPS PO preliminary and mans examinations.

Other ebooks[pdf] for ibps po quant preparation.

### IBPS PO quantitative aptitude ,numerical ability syllabus 2016

• Profit and loss
• Simplifications,Approximation
• Chain rule
• Time and work
• Average
• Simple and compound interest
• Fractions, Square root and cube root
• Decimal fractions
• HCF & LCM
• Ratio and proportions
• Mensuration
• Speed,Time and Distance.
• Partnerships
• Percentage
• Problems based on Ages
• Data Interpretation(Pie charts, Bar graphs, Line graphs, Mixed graphs)
• Permutation and Combination
• Probability
Hope you will find this ebook for IBPS PO quantitative preparation helpful.Share it with your friends on social media.Like our facebook page or follow us on google+ to get more ibps po exam preparation materials like this.
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## Data interpretation for IBPS,SBI PO and Clerk exams

Data interpretation is an important section in bank exams.In IBPS PO and Clerk exams data interpretation section carries 10 marks.In SBI and SBI associate bank Probationary officer and clerical examination data interpretation is a separate section. Many candidates take  this section lightly and doesn't pay required attention to this section.Preparing data interpretation section will help you lot in time management during bank exams.Here we are sharing some data interpretation questions for bank exam preparation.Questions are fully solved with shortest methods.You can also download data interpretation questions for bank po pdf  ebook from the link provided below.To improve your Data interpretation speed solve maximum model questions.By practicing more and more questions, your brain  will eventually adapt to interpret tabular,pie-chart,bar graph and line graph much faster.

### Data interpretation practice questions for bank exams

Data interpretation questions for bank PO and clerk can be categorised into 5 types.Tabular data based data interpretation questions,Pie-chart based data interpretation questions,Line graph data interpretation,Bar graph data interpretation and mixed type data interpretation(Any two of table,line graph,pie-chart,bar graph).Under each section we have provide 2 set of data interpretation questions(5+5 questions each).This Data interpretation questions for ibps po and clerk is fully solved using easiest methods.You can also download this data interpretation questions pdf with solution from the link provided at the end of this page.These fully practice questions on data interpretation is also useful for exams like SBI PO,SBI clerK,RBI assistant,CAT,MAT and other major competitive examinations.

### Data interpretation practice set for bank PO and Clerk

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simon

## Probability questions for bank exams

Probability questions shortcuts for bank exams:In most of the bank PO exams like IBPS PO,SBI PO,IBPS RRB PO and RBI officer , under quantitative section 2 to 4 questions are asked from probability section. Generally candidates take this section lightly and tend to loose 2,3, or 4 valuable marks.If you are ready to spend few hours in preparing probability section during your entire exam preparation period , you can come out of the exam hall with a smile.Here we are sharing important formulas and facts,easy way to solve probability questions for bank PO and Clerical examinations.In clerical examinations only simple questions from probability sections are asked,while questions asked in bank PO examination demand some preparation.So,here we are sharing How to solve probability questions faster for bank examination.We have included all important formulas of probability.At the end you can also find some solved practice questions on probability ,which will be useful in your bank PO exam preparation
Prerequisites For probability section preparation
-Knowledge of basics of set theory
-permutation and combination
Read Shortcuts for permutation and combination for bank exams.

### Probability-Important formulas and facts

Experiment:An operation which can produce some well-defined outcomes is called an experiment.
Random Experiment:An experiment in which all possible outcomes are known and exact output cannot be predicted in advance is called a random experiments.For example, rolling a dice,tossing a coin,drawing a card from a well shuffled pack of cards etc.

### Some random experiments and their outcomes

Tossing a coin:
• When we toss a coin ,either Head(H) or a Tail(T) appears. If two coins are tossed simultaneously then possible outcomes are HH,HT,TH and TT.As number of coins increases possible outcomes also increases.
Rolling a Dice:
• A dice is a solid cube,having 6 faces,marked 1,2,3,4,5 and 6.When we roll a dice possible outcomes are 1,2,3,4,5 and 6.
• If we roll two dice simultaneously possible out comes are combination of two number (1,1)(1,2)(1,3)……………….(6,6)
Card is drawn from a pack of cards:
• A pack of card has 52 cards.
• It has 13 cards of each suit, namely Spades, Clubs, Hearts and Diamonds.
• Cards of Spade and Club are black.
• Cards of Heart and Diamond are red cards.
• There are 4 honours of each suit.These are Aces,Kings,Queens and Jack.These are called face cards.
Sample space: When we perform an experiment, then the set of all possible outcomes is called Sample Space.denoted by ‘S’.
• In tossing a coin S={H,T}
• If two coins are tossed S={HH,HT,TH,TT}
• In rolling a dice S={1,2,3,4,5,6}
Event:Any subset of Sample Space is called an Event

### Probability of occurrence of an Event

Let S be the sample space and E be the Event,then probability of occurrence of E denote by P(E)
P(E)=n(E)/n(S)=Number of favourable outcomes/Number of possible outcomes

Results on probability
• P(S)=1
• 0≤P(E)≤1
• For any events A and B P(A∪B)=P(A)+P(B)-P(A∩B)                                         ∪=Union,∩=Intersection
• P(A)=1-P(not A)

### Probability Practice questions for bank PO and Clerk examination

Question1:Two unbiased coin are tossed .What is the probability of getting at most one head?
Solution: Sample space (All [possible outcomes)S=(HH,HT,TH,TT)
Event(required outcomes)=(TT,HT,TH)
P(E)=n(E)/n(S)=3/4
Question2:Two dice are thrown simultaneously ,what is the probability of getting a total of 7?
Solutions:n(S)=6*6=36
E={(1,6),(6,1),(5,2),(2,5),(4,3),(3,4)}
n(E)=6
P(E)=6/36=1/6

Question3:A bag contains 6 white and 4 black balls. Two balls are drawn at random. Find the probability that they are of same colour.
Solution:n(S)=Number of ways two balls can be drawn from 10 balls(6 white +4 blacks)= 10C2=(10*9)/(2*1)=45
n(E)=Number of ways of drawing 2 balls from 6 white balls or 2 balls from 4 black balls=6C2+4C2=21
P(E)=21/45=7/15

Question4:Two cards are drawn at random from a pack of 52 cards.What is the probability that either both are black or both are queens?
Solution:Total number of possible ways selecting 2 card from 52 cards is n(S)=52C2=(52*51)/(2*1)=1326
Let A be event of getting both black card.
B be event of getting both queens.
Event of getting two black queens A∩B
There are 26 black cards in a pack and 4 queens. Two queens are black
n(A)=Number of ways of selecting 2 blacks from 26 cards=26C2=325
n(B)=Number of ways of selecting 2 queens from 4 cards =4C2=6
n(A∩B)=Chances of getting 2 black queens=2C2=1
P(A)=325/1326
P(B)=6/1326
P(A∩B)=1/1326
P(A∪B)=P(A)+P(B)-P(A∩B)
=(325/1326)+(6/1326)-(1/1326) = 330/1326 =55/221
Tip:Use union '∪'for ‘or’, ‘either’,’neither’.Use intersection '∩^'for ‘and’.
If in the above question ,question if changed like ‘chances of getting black queens’, then solution would be as follows.(ie getting two card that are back and queens)
P(A∩B)= P(A)+P(B)- P(A∪B)=1/1326

Question5:A speaks truth in 75% cases and B in 80% of cases. In what percentage of cases they contradict each other in narrating the same incident?
Solution:
Probability that A is telling the truth P(A)=75/100=3/4
Probability that B is telling the truth P(B)=80/100=4/5
Probability that A is lying P(A’)=1-75/100 =25/100=1/4
Probability that B is lying P(B’)=1-80/100 =20/100=1/5
A telling truth and B lying
Or
B telling truth and A lying
P(Contradiction)= P(A)*P(B’) + P(A’)*P(B)=(3/4) * (1/5) + (1/4)*(4/5)=7/20
7/20 is equal to 35 %.

Question6:A bag contains 2 red, 3 green and 2 blue balls.Two balls are drawn at random.What is the probability that none of the drawn ball is blue?
Solution:Possible out comes(E) or Favourable outcomes are
E1.2 red ball or
E2.2 green ball or
E3.1 red and 1 green
Number of selecting 2 balls from two red balls n(E)=2C2=1
Number of ways of selecting 2 balls from 3 green balls n(E2)=  3C2=3
Number of ways of selecting 1 red ball and 1 green ball n(E3)= 2C1*3C1=2*3=6
Sample space S=Number of ways of selecting 2 balls from 7(2+3+2) balls n(S)=7C2=21
n(E)=n(E1)+ n(E2)+ n(E3)=1+3+6=10
Required probability P(E)=n(E)/n(S)=10/21

Hope you will find this page on "Probability questions for bank PO and Clerk examination" useful.Share it with your friends.
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## Permutation and combination shortcut formula for bank exams

Here we are sharing important formulas and shortcuts of permutation and combination questions.In IBPS PO,SBI PO,RBI assistant/officer exams permutation and combination questions are asked repeatedly. In IBPS clerk examinations also some simple permutation and combination questions are asked.Here you can find permutation and combination easy tricks to solve these questions.We are sure that this article will help you solve permutation and combination for bank po much faster.Only important formulas from permutation and combination sections are given.We have also provided few permutation and combination problems with solutions.This will help you prepare for various bank's probationary officer exams.

### Permutation and combination important formulas and facts

Permuation formulas and shortcuts
Factorial:Let n be a positive integer.Then factorial of n denoted by n!  is defined as
n!=1*2*3*……………..(n-2)*(n-1)*n
Note:0!=1
Permutation: Different arrangement of a given number of things by taking some or all at a time.
Example:All arrangements made with letters a,b,c by taking two at a time are ab,ba,bc,cb,ca,ac.
All arrangements made with letters a,b,c by taking all at a time are abc,bca,cab,cba,acb,bac.
Number of permutation:Total number of possible arrangements(permutation) of n things, taken r at a time, is given by:
nPr =n!/(n-r)!=n(n-1)(n-2)………….(n-r+1)
Example 1.Arrangement of 3 items taken 2 at atime
3P2=3!/(3-1)!=3*2=6
Example2. Arrangement of 4 items taken all at a time
4P4=4!/(4-4)!=4!/0!=4*3*2*1=24

### Permutation :Important point to note

• Number of all permutation of n things all at a time is n!
• If there are n objects ,m numbers are alike Then number of permutations of these objects is :n!/(m!)
• If there are n objects ,p1 numbers are alike of one kind,p2 objects are alike of another kind,p3 are alike of third kind and so on and pr are alike of rth kind,such that p1+p2+p3+……pr=n Then number of permutations of these objects is : n!/(p1!.p2!…..pr!)
Combination shortcuts and formulas

Each of different groups or selection which can be formed by taking some or all of a number of object, is called a combination.
Suppose we want to select two students from a group of three students namely A,B and C.Then, possible selections are AB,BC and CA.
Note AB and BA represents same selection. But in permutation/arrangement AB and BA represents two different arrangements.
If we want to select ‘all at a time ‘, then there is only one possibility ABC.
Number of combinations:The number of all combination of n things, taken r at a time is:
nCr=n!/((r!)(n-r)!)=[n(n-1)(n-2)….upto r factors]/r!
nCr= nC(n-r)
nCn=1
nC0=1
Example: 10C3=10!/(3!)(10-3)!=(10*9*8)/(1*2*3)=120

### Permutation and Combination Practice Questions

Question1.How many arrangements are possible using all the letters of the word BIHAR?
Solution:The word BIHAR contains 5 different letters.
Required number of arrangements = 5P5=5!=120
Question2.How many word can be formed using all the letters of the word DAUGHTER so that vowels always come together?
Solution:There are 8 different letters in the given word,but it is given that vowels should come together.Treat vowels as a single entity ‘AUE’
So total number of letters is 5 different letters+ vowel entity=6
Total arrangement possible using 6 letters=6P6=6!=720
Vowels can be arranged in 3P3 ways. 3P3=6
Total number of arrangement=720*6=4320
Question3.In how many ways a football team of 11 players can be selected from a group of 15 candidates?
Solution:Number of ways of selection=15C11=15C15-11=15C4=(15*14*13*12)/(1*2*3*4)=1365
Question4.In how many ways, a committee of 5 members can be selected from a group of 6 men and 5 ladies ,consisting of 3 men and 2 ladies?
Solution :3 men out of 6 and 2 ladies out of 5 are to be chosen.
So number of ways selections can be made is = 6C3*5C2=200
If you find this permutation and combination shortcuts for bank po exam useful ,share it with your friends.
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## IBPS Clerk quantitative aptitude practice set pdf

Quantitative Practice Questions for IBPS Clerk:Here we are sharing a set of 40 fully solved quantitative aptitude practice questions for your IBPS clerk main exam preparation.We have also include answer key with this question paper.After reading you can download the IBPS clerk quantitative practice set from the link given at the end. We are sure that this IBPS Clerk quantitative questions will help you in your upcoming  IBPS Clerk exam preparation.We have prepared this practice set based on the IBPS Clerk previous year exam patterns.You may also like to check IBPS Clerk quantitative syllabus from here.

### IBPS Clerk  2016 Quantitative Aptitude syllabus

• Profit and loss
• Basic maths/Simplifications,Approximation
• Chain rule
• Time and work
• Average
• Simple and compound interest
• Fractions, Square root and cube root
• HCF & LCM
• Ratio and proportions
• Mensuration
• Speed,Time and Distance.
• Partnerships
• Percentage
• Problems based on Ages
• Data Interpretation(charts/Tables)

## ebook:IBPS Clerk quantitative questions

Other Practice Papers you might interested in
We are sure that you will find this quantitative practice questions useful for your clerk main examination preparation.Don't forget to share this. Like us on facebook for more practice papers.

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## Saturday, 14 November 2015

### Quantitative Aptitude Shortcuts and tricks Free eBook Download(for Bank, SSC,UPSC and other competitive exams)

Quantitative aptitude is one of the most important section in every competitive exam.Usually candidates find this section time consuming.Whether it is  IBPS PO/Clerk, ssc, upsc or any other competitive exam ,quantitative section is little time consuming in comparison with reasoning, general awareness and English section. To crack any competitive exam you need to excel in all sections.As you  know,  there is sectional cutoff for most of the competitive exam.You need to score minimum in each section.This is applicable to bank exams also.Many candidates fail to clear sectional cutoff in  quantitative section. You need thorough preparation in quantitative aptitude.Most of the people are unaware of numerous shortcuts and tricks that are helpful in  solving quantitative aptitude questions faster.You can save your precious minutes during exams by practicing this Quantitative Aptitude Shortcuts.Here we are sharing Quantitative Aptitude Shortcuts for bank exam pdf with you.
These are the topics covered in this Quantitative Aptitude Shortcuts Free eBook for bank exams.

### Chapter 1:Basic Math Shortcuts

1. Division-shortcuts
2. Multiplication-shortcuts
3. Square-Shortcut Tricks
4. Cubes-Shortcut
5. Cube root(for perfect cubes only)

### Chapter 2:Ratio and Fractions

1. Ratios-Important rules and shortcuts
2. Comparison of ratios and Fractions

### Chapter 3.Simple Interest and Compound Interest

1. Simple Interest
2. Compound Interest

### Chapter4:Mixture and Alligation

1. When two quantities are mixed
2. If more than two different commodities are mixed
3. Removal and replacement

### Chapter8:Time distance and speed

1. Ratio of speed
2. Average Speed
3. Train and time problems
4. Boat and stream problems
Chapter9:Permutation and combination important formulas and shortcuts

1. Permutation Formulas and shortcuts
2. Combination Important formulas
Chapter10:Probability formulas

1. Some random experiments and their outcomes
2. Probability of occurrence of an Event
3. Important Results on probability

simon

## IBPS PO preliminary Quantitative model paper

IBPS PO Preliminary Quantitative Practice Questions PDF download:IBPS PO preliminary is a 100 mark test for the duration of one hour.This online preliminary exam will have 3 sections:Quantitative aptitude, Reasoning and English language. Quantitative and Reasoning carry 35 marks each and English language carries 30 marks.Here we are sharing a set of 35 questions from Quantitative section for PO/MT preliminary exam.Along with this sample questions ,we also provide answer key.Hope this Quantitative sample questions will help you in IBPS PO preliminary exam preparation.After reading you can also download PDF of this questions with answer key from the link provided at the end of this post.
Before going to the questions you may like to check IBPS PO Quantitative Detailed syllabus
IBPS PO preliminary 2015 syllabus and exam pattern

### Quantitative Practice Questions

This IBPS PO quantitative practice set contains 35 question.The pattern of this practice question set is same as that of IBPS PO preliminary online common written examination.

You can download above IBPS PO/MT preliminary Quantitative sample questions  in PDF from here
If you have any doubt  regarding above questions ,you can comment below .Those who need explanation for any question , drop your request in comment box below.Don't forget to like and share this IBPS PO Preliminary Quantitative practice questions pdf with your friends on facebook,twitter and google plus.

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## Age Related Problems

Generally problems based on ages are asked in all competitive examinations.In bank exams this section is comparatively  easy.This questions comes under quantitative section. You can score full mark in this section if you are familiar with  Algebra.These questions are solved using algebraic equations, either linear or quadratic equations.

Points to note:
• If present age of a person is x,then age of that person n years ago is x-y and age after n years will be x+n.
• The present ages of A and B are x and y respectively.If A is n times older than B,then x=ny .If A is n years older than B,then x=y+n.
Now let us practice some age related problems .
Example1:Age of Raghav who is 30 years old is 3/5 times of Ram and Ram is older than Mohan by 35 years.If vijay's age is just between the age of Ramesh and and Mohan and Ramesh is 21 years old.What is age of vijay?
1. 20
2. 25
3. 18
4. 27
Solution:
Raghav's age=3/5 Ram's age
=>30=3/5 Ram's age
=>Ram's age=30/3 * 5=50
Ram's age=Mohan's age +35
=>Mohan's age=Ram's age -35
=>Mohan's age=50-35
=>Mohan's age =15
Vijay's age=(Ramesh's age +Mohan's age)/2
=>Vijay's age=(21+15)/2
=>Vijay's age=18 years.

Example2:The sum of the ages 4 members of a family 5years ago was 94.Today,when the daughter has been married off and replaced by a daughter-in-law,the sum of their ages is 100.Assuming that there has been no other change in the family structure and all members are alive,what is the difference in the ages of daughter and daughter-in-law ?
1. 22
2. 11
3. 25
4. 14
Solution:sum of ages of 4 members 5 years ago=94
=>sum of present ages of 4 members=94+ 4*5=114
Difference in the sum of the ages=Difference in the ages of daughter and daughter-in-law
Difference in the sum of the ages=114-100=14
=>Difference in the ages of daughter and daughter-in-law=14

Exampl3:The sum of ages of Aswin ,Sachin and Sumesh is 93 years.10 years ago ratio of their age was 2:3:4.what is the present age of sachin.
1. 19
2. 21
3. 31
4. 41
Solution:Sum of the present ages of Aswin ,Sachin and Sumesh=93
Sum of the ages of Aswin ,Sachin and Sumesh 10 years ago=93- 10*3=63
Age of sachin 10 years ago=(3/(2+3+4)) * 63 = 21.
Present age of Sachin=21+10=31

Example4:Ratio of Rani's and Kamal's age is 3:5. Ratio of Kamal's and Pooja's age is 2:3.If Rani's age is two-fifth of Pooja's age, what is Rani's age?
1.  10 years
2.  15 years
3.  24 years
4.  Cannot be determined
5. None of these

Solution:Here in this question only ratio is given.No other information like sum ,difference or product is given.So given data is inadequate to find a solution.

Example5:The ratio of ages of A and B 1 year ago was 3:4.After 1 year it will be 5:6.what is the present age of A.
1. 1
2. 2
3. 3
4. 4
5. none of the above
Solution:Let x be the age of A and y be the age of B
hen ratio of their ages one year ago (x-1):(y-1)=3:4

=>(x-1)/(y-1)=3/4............(1)
Ratio of their ages after one year (x+1):(y+1)=5:6
=>(x+1)/(y+1)=5/6..........(2)
Solving (1) and  (2)
(1)=>4(x-1)=3(y-1)
=>4x-4=3y-3
=>4x-3y=1.......................(3)
(2)=>6(x+1)=5(y+1)
=>6x-5y=-1......................(4)

3*(3) - 2*(4)

12x - 9y=3   -
12x - 10y=-2
___________
y   = 5
___________

(3)=>4x-15=1
=>4x=16
=>x=16/4=4

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## Number Series questions for bank po and clerk

Number series questions for bank po and clerk:Number Series is an important part of quantitative aptitude section in almost all bank exams.Actually there is no 'specific method' available for solving this kind of questions.Because there are numerous types of number series are available.By giving you these kind of questions ,examiner want to test your ability to analyse a problem from your own perspective and find a solution.But by practicing different kind of series you can easily identify the given series and solve it faster.  Usually two types of number series questions are asked in bank exams.

Number series Type1: Spot the odd one
In this type of questions,you will be given a number series in which one term doesn't follow the common relation that connects other terms.You have to spot that odd one.
Number series Type2: Find the missing one
In this type of questions, you will be given  a  number series in  which one term is missing.You have to find out that missing term.

Number series can also be classified based on the relation between successive terms.In other words 'how succeeding term is obtained from previous term'.

In this type of series,succeeding term is obtained by adding/subtracting a particular number to the previous term.To solve this type of question,find the difference between successive terms.These 'differences' will form another series.solve this 'differences' series first,then you can easily solve the original series.

Example1: Find the next term in the series 6   7    9   13    21    37     ?
Take the difference between successive terms.
• 7-6=1
• 9-7=2
• 13-9=4
• 21-13=8
• 37-21=16
• ?-37
Series formed by 'difference' terms is as follows.
1  2  4   8   16   ?
What comes in place of "?" ?.....Yes it is 32.

Example2: Spot the odd one in the series 5     6     10    18      35     60
• 6-5=1=1^2
• 10-6=4=2^2
• 18-10=8
• 35-18=17
• 60-35=25=5^2
Here you can see,no specific relation exist between 10&18 and 18&35.
So 18 is the odd one
Correct series is as follows

5     6     10    19      35     60
• 6-5=1=1^2
• 10-6=4=2^2
• 19-10=9=3^3
• 35-19=16=4^2
• 60-35=25=5^2
2.Multiplication/Division number series
In this  type of series succeeding term is obtained by multiplying/dividing previous term by a particular number.To solve this type of question ,find the multiplication factor between terms and write it as a series.Solve these 'multiplication factors' series first.Then you can you can easily solve the original series.
Example1:Find the next term in the series. 3   6   15   45   157.5    ?

• 6=3*2
• 15=6*2.5
• 45=15*3
• 157.5=45*3.5
Series formed by multiplication factors is 2  2.5   3   3.5   ?
What comes in place of "?" ? It is 4.
So ?=157.5*4=630.

Example2:Find the odd term in the series.1   2   8    72     1250
• 2=1*1^2
• 8=2*2^2
• 72=8*3^2
Here you can see  multiplication terms form a series 1^2    2^2   3^2   ?.
What comes in place of "?" ?.It is 4^2. So, 72*4^2=1152 comes in place of 1250.Thus odd one is 1250.

3.Combination of Addition/Subtraction and Multiplication /division series
In this type of series succeeding term is obtained by multiplying previous term by a number and adding/subtracting some other number.Like in above cases this multiplying/adding/subtracting number will form a series or a particular pattern.Try to identify the patterns that multiplying/adding/subtracting numbers are following.Then you can easily figure out pattern behind
original series.

Example: 4     3     4     9    32
At first,by looking at the series you may not be able to figure out relation between successive terms.As already said,actually there exist  no 'specific logic ' to solve number series questions.You need to practice more and more questions,so that your brain will get adapted to these kind of problems.Now lets look into the pattern that above series is following.
• 3=4*1-1
• 4=3*2-2
• 9=4*3-3
• 32=9*4-4
You can see that multiplying terms and subtracting terms follows a certain pattern.

4:Apart from above categories ,series can be formed by adding/subtracting multiple numbers with previous term to get succeeding term.
Example:1   2     7     50     2507
• 2=1^2 +1
• 7=2^2 + 3
• 50=7^2 +5
• 2507=50^2 +7
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## Time distance and speed formulas and shortcuts for bank exams

Time distance and speed formulas and shortcuts for bank exams:On an average you can found 2-3 question on Time ,distance and speed in every bank exams.By understanding few formulas and shortcuts you can score full mark from this section.We have provided main formulas and shortcuts of time,distance and speed for IBPS Clerk and PO exams.

Relation between time distance and speed is given by the equation
speed=distance/time
Unit of speed is km/hr or m/s
If speed is given in km/hr,then inorder to convert it  in to m/s multiply by 5/18
1km/hr=5/18m/s
If speed is given in m/s,then inorder to convert it  in to km/hr multiply by 18/5

### Ratio

• If ratio of speed of two moving object is a:b, then ratio between times taken for covering same distance is b:a.
• If two objects A and B moving in opposite direction from two different places  reach at common point in t1 and t2 hrs respectively                 Then Speed of A:Speed of B=sqrt(t2/t1)
Average speed
• average speed=total distance/total time taken.
• If a moving object covers a certain distance with a speed of x km/hr and again covers same distance with a speed of ykm/hr, then average speed is 2xy/(x+y).
• If a moving object covers a certain distance with a speed of x km/hr and again covers same distance with a speed of ykm/hr and again with zkm/hr,then average speed is=3xyz/(xy+yz+xz)

### Points to be noted while doing 'train and time' problems

• If two trains are travelling in same direction ,then their relative speed is equal to difference of their speeds.Then Time taken by the fast train to cross the slower train is                                                                            =Sum of       lengths of both trains/difference of their speed
• If two trains are travelling in opposite direction ,then their relative speed is equal to sum of their speeds.Then time taken to pass one another is   =Sum of lengths of both trains/sum of their speed.
• when a train is clearing a pole or a point, then distance covered by train is equal to its length
• When a train is covering a platform or bridge or tunnel ,then distance covered by train is equal to sum of the length of train and the length of platform/tunnel/bridge.
• When a moving train crosses another train, then distance covered is equal to sum of lengths of both trains.
Example:A 480-metre-long train crosses a platform in 140 seconds. What is the speed of the train?
Ans:Cannot be determined,since length of platform is not given

Example:A train 100m long is running at 21km/hr and another train 150m is running at 36km/hr in the same direciton.how long will the faster train take to pass the other train?
Sum of length of both train=100+150=250m
difference of their speed is=36-21 km/hr=15km/hr=15*(5/18) m/s=25/6 m/s
Time taken=250/(25/6)=60 seconds.

### Boat and stream problems

• If speed of stream=xkm/hr and speed of boat in still water is ykm/hrthen speed of boat in downstream=x+y km/hr                                           speed of boat in upstream=y-x km/hr
• If speed of boat in upstream and speed of boat in down stream is given then,                                                                                        speed of boat in still water=1/2(speed in upstream+speed in downstream)                                                                                                      speed of stream=1/2(Speed in downstream - speed  in upstream)
Example: A boat is moving at 30 km/hr upstream, when it travels down stream its speed is 36km/hr.What is the speed of boat in still water and what is the speed of stream?
Speed of boat in still water=1/2 (30+36)=66/2=33 km/hr.
speed of stream=speed of boat downstream-speed of boat in still water
=36-33=3km/hr
Note:
A person walks at x kh/hr he reaches destination t1 hrs late,if he walks at y km/hr,then reaches t2 hrs early
then distance to the destination =(xy/(y-x)) * (t1+t2)

Example:A person walking at 2km/hr reaches his office 6 minutes late .If he walks at 3km/hr he reaches there 6 minute early. How far is the  office from his house?
Distance=(2*3)/(3-2) *((6+6)/60)=6*12/60=1.2 km .

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