Find the latest bookmaker offers available across all uk gambling sites www.bets.zone Read the reviews and compare sites to quickly discover the perfect account for you.
Home / Aptitude / Quantitative Aptitude Shortcut Tips and Tricks For IBPS Bank Exam – Part I

Quantitative Aptitude Shortcut Tips and Tricks For IBPS Bank Exam – Part I

Quantitative aptitude is one of the most important and must-do part of your IBPS banking exam. You could never avoid general or common aptitude preparation, particularly when you are determined to appear for any type of banking exams. But, to get the taste of success, you must need to expert on solving the aptitude questions within the specific amount of time during the examination. If you don’t know the tricks or shortcut methods of solving the problems, then you will simply be undone after the exam. Your hard preparation throughout the year will go in absolute vain. But, some exquisite tricks or methods can do magic for you. Some of the tricks and methods are given below part-by-part.

The whole aptitude covers chapters from different sections like Problems on Trains, Time and Work, Profit and Loss, Problems on Ages, Average, Permutation and Combination, Problems on H.C.F and L.C.M, Square Root and Cube Root Problems, Alligation or Mixture, Probability, Banker’s Discount, Time and Distance, Simple Interest and Compound Interest, Partnership, Calendar, Problems on Numbers – Numbers, Decimal Fraction, Logarithm, Odd Man Out and Series, Height and Distance, Percentage, Simplification, Ratio and Proportion, Boats and Streams, True Discount, Chain Rule, Partnership, Area, Surds and Indices, Pipes and cistern, Volume and Surface, Race and Games.

Aptitude Shortcut Tips and Tricks for Banking Exam

Source

It’s a hard job to cover all the topics in a single article. So all the readers are requested to read every part of this article, which will be released in a sequential order (like Part I, Part II, Part III………..) in coming days. Don’t loose your heart so soon, keep yourself motivated and updated with the brilliant application and methods to succeed in career.

Part I consists of three sections: Problems on Trains, Time and Work, Profit and Loss.

Shortcut methods and tricks to solve questions from ‘Problems on Trains’ are given below:

Conversion of kilo-meter (km) per hour (hr) to meter per second (m/s):

To convert km/hr to m/s:

u (km/hr) = u*5/18 (ms/s)

Conversion of meter per second kilo to meter per hour :

To convert m/s to km/hr:

v (m/s) = v*18/5 (km/hr).

speed = distance / time or length /time.

——————————————————————————————————–

Shortcut methods and tricks to solve questions from ‘Time and Work’ are given below:

Basic formula = 1/totalwork = (1/workrate1 + 1/workrate2 + …………+ 1/workrateN)

If X can do a certain work in y days, then work done by X in 1 day = 1/y

If P can do W1 works in D1 days working H1 hours per day and Q can do W2 works in D2 days working H2 hours per day, then

P( D1H1 / W1) = Q (D2H2 / W2)

If P can do a work in x days and Q can do in y days, then they together can do a work in: [xy / (x + y)] days.
Some important rules and their applications:

Suppose P = person(s), D = days, W = works, T = time and E = efficiency;

Then Universal Rule:

P1D1W2 = P2D2W1
P1D1T1W2 = P2D2T2W1 (if the time is said)
P1D1T1E1W2 = P2D2T2E2W1 (if efficiency is added)

Rule 2: If X can do a piece of work in n days.

In one day work done by X is 1/n.

Rule 3: X can do a work in D1 days, Y can do same work in D2 days. X and Y together do the same work in how many days ?

Formula= (D1*D2)/(D1+D2)

Rule 4: If X is twice as good as Y, then X will take 1/2 time taken by Y to complete a work.

Y alone can finish in 2p days ( p is work done by A )

Rule 5: If X is thrice as good a as B, then A will take 1/3rd of the time taken by B to complete a work.

B alone can finish in 3x days

If P is x times as good a workman as Q, then:

1. Ratio of work done by P and Q = x:1
2. Ratio of times taken by P and Q to finish a work = 1:x ie; P will take (1/x)th of the time taken by Q to do the same work.

P and Q can do a piece of work in ‘x’ days and ‘y’ days respectively, then working together:
1. They will complete the work in xy/(x+y)  days
2. In one day, they will finish [(x+y)/xy]th part of work.

If P can do a piece of work in x days, Q can do in y days and R can do in z days then,
P, Q and R together can finish the same work in: [pqr/(pq+qr+rp)] days

If P can do a work in x days and P and Q together can do the same work in y days then,
Number of days required to complete the work if Q works alone= [(xy/x−y)]days
If P and Q together can do a piece of work in x days, Q and R together can do it in y days and R and P together can do it in z days, then number of days required to do the same work:
If P, Q, and R working together = [2xyz/(xy+yz+zx)] If P working alone = [2xyz/(xy+yz−zx)] If Q working alone = [2xyz/(−xy+yz+zx)] If R working alone = [2xyz/(xy−yz+zx)]

If A and B can together complete a job in x days.
If A alone does the work and takes a days more than A and B working together.
If B alone does the work and takes b days more than A and B working together.

Then,x=√ab days

If m1 men or b1 boys can complete a work in D days, then m2 men and b2 boys can complete the same work in

[Dm1b1/(m2b1+m1b2)] days.

If m men or w women or b boys can do work in D days, then 1 man, 1 woman and 1 boy together can together do the same work in

[Dmwb/(mw+wb+bm)] days

If the number of men to do a job is changed in the ratio x:y, then the time required to do the work will be changed in the inverse ratio. ie; y:x

If people work for same number of days, ratio in which the total money earned has to be shared is the ratio of work done per day by each one of them.

A, B, C can do a piece of work in x, y, z days respectively. The ratio in which the amount earned should be shared is (1/x:1/y:1/z=yz:zx:xy)

A certain no of men can do the work in D days. If there were m more men, the work can be done in d days less. How many men were there initially?

Let the initial number of men be M
Number of man days to complete work = MD

If there are M+m men, days taken = D−d
So, man days = (M+m)(D−d)
ie; MD=(M+m)(D−d)
M(D–(D−d))=m(D−d)
M=m(D−d)/d

A certain no of men can do the work in D days. If there were m less men, the work can be done in d days more. How many men were initially?

Let the initial number of men be M
Number of man days to complete work = MD
If there are M−m men, days taken = D+d
So, man days = (M−m)(D+d)
ie; MD=(M−m)(D+d)
M(D+d–D)=m(D+d)
M=m(D+d)/d

——————————————————————————————————–

Shortcut methods and tricks to solve questions from ‘Profit and Loss’ are given below:

C.P = Cost Price; S.P = Selling Price
Profit = S.P – C.P; Loss = C.P – S.P
Gain%= (Gain *100 / C.P);
Loss%=(Loss*100 / S.P)
S.P= (100 + Gain%) / 100 * (C.P);
S.P=(100 – Loss%) / 100 * (C.P)
C.P=(100/(100 + gain%)* (S.P));
C.P=(100/(100 – Loss%) * (S.P))

——————————————————————————————————–

                                                                                                                           (…….to be continued in Part II)

About Puzzleduniya

I am a quiz, puzzle lover, who always intended to post different type of study materials, which will help the all the competitive job seekers all around the world.

Check Also

Spherical Triangle

Quantitative Aptitude Shortcut Tricks For IBPS Exam – Part X

(Part IX) – common aptitude for competitive exam like IBPS consists of shortcut methods, tips ...

One comment

  1. Thank you for any other informative web site. Where else may just I get that kind of information written in such an ideal way? I have a mission that I am just now working on, and I’ve been on the look out for such info.

Leave a Reply

Visit blogadda.com to discover Indian blogs