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Home / Aptitude / Quantitative Aptitude Shortcut Tricks For IBPS Exam – Part IV

Quantitative Aptitude Shortcut Tricks For IBPS Exam – Part IV

In the previous part (Part III), I have already discussed the shortcut methods, tips and tricks to solve question from Problems on H.C.F and L.C.M”, Square Root and Cube Root Problems, and Alligation and Mixture. This section (Part IV) includes another three very important section from quantitative aptitudes including “Probability”, “Banker’s Discount”, and “Time and Distance”.

Aptitude Tricks for Banking Exam

 

Source

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

Probability is another important section of common aptitude questions and answers you will find in different type of banking examinations. Here some basic tricks from ‘events’, ‘sample space’, ‘probability of occurrence of an event’ are discussed below:

Experiment:

To gain well-defined outcomes by operations is called experiment.

Probability or Chance

Probability is defined as the quantitative measure of the chance of occurrence of a particular event.

Random Experiment

It’s a type of an experiment, where all the possible outcomes of an experiment are known but the exact output cannot be predicted in advance, called random experiment.

Example:

1. Tossing of a fair coin,

2. From a pack of shuffled cards, drawing a particular card,

3. Throwing an unbiased die

4. From a set of balls with different color, you need to take a particular ball randomly.

Now the actual explanation is given below:

1. During tossing a coin, the outcome will be either Head (H) or Tail (T)

2. A card pack has 52 cards in 4 different sets of

Spades (♠)

Clubs (♣)

Hearts (♥)

Diamonds (♦)

Each of the above mentioned categories has 13 cards, 9 cards numbered from 2 to 10, an Ace, a King, a Queen and a jack.

Hearts and Diamonds are red faced cards whereas Spades and Clubs are black faced cards.

Kings, Queens and Jacks are called face cards

3. A die is a cube shaped item with 6 faces and 6 unique number (1,2,3,4,5,6) denoted on every face with a dot mark. When it’s rolled on a plain surface, the outcome is always the number appears on it’s upper face and it is a random integer from 1 to 6, each value being equally likely.

Sample Space

While performing an experiment, then the set S of all possible outcomes is called the sample space.

Example

When a coin is tossed, S = {H, T} where H = Head and T = Tail

When a dice is thrown, S = {1, 2 , 3, 4, 5, 6}

When two coins are tossed, S = {HH, HT, TH, TT} where H = Head and T = Tail

Event

Any subset of a Sample Space is an event. Events are generally denoted by capital letters A, B , C, D etc.

Example

When a coin is tossed, outcome of getting head or tail is an event

When a die is rolled, outcome of getting 1 or 2 or 3 or 4 or 5 or 6 is an event

Equally Likely Events

Events are said to be equally likely if there is no preference for a particular event over the other.

Example

When a coin is tossed, Head (H) or Tail is equally likely to occur.

When a dice is thrown, all the six faces (1, 2, 3, 4, 5, 6) are equally likely to occur.

Probability of en Event

Let E be an event and S be the sample space. Then probability of the event E can be defined as

P(E) =n(E) / n(S)

where P(E) = Probability of the event E, n(E) = number of ways in which the event can occur and n(S) = Total number of outcomes possible

Important formulas

P(S) = 1

0≤P (E)≤1

P(ϕ) = 0 (∵ Probability of occurrence of an impossible event = 0)

Addition Theorem

Let A and B be two events associated with a random experiment. Then

P(A U B) = P(A) + P(B) – P(A∩B)

If A and B are mutually exclusive events, then P(A U B) = P(A) + P(B) because for mutually exclusive events,

P(A∩B) = 0

If A and B are two independents events, then P(A∩B) = P(A).P(B)

Odds on an event

Let E be an event associated with a random experiment. Let x outcomes are favorable to E and y outcomes are not favorable to E, then

Odds in favor of E are x:y, i.e., x / y

and Odds against E are y:x, i.e., y / x

P(E) = x / (x+y)

P(Ē) =y / (x+y)

Binomial Probability distribution

A binomial experiment is a probability experiment which satisfies the following requirements.

Each trial can have only two outcomes. These outcomes can be considered as either success or failure.

There must be a fixed number of trials.

The outcomes of each trial must be independent of each other.

The probability of a success must remain the same for each trial.

The probability of achieving exactly r successes in n trials in a binomial experiment, can be given by:

P (r successes in n trials) = (n r)prqn−r

where p = probability of success in one trial

q = 1 – p = probability of failure in one trial

(n r) = nCr = n! / [(r!)(n−r)!] = [n(n−1)(n−2)⋯(n−r+1)] / r!

More Shortcut Formulas

Suppose n fair coins are tossed:

Total number of outcomes in the sample space = 2n

The probability of getting exactly r-number of heads when n coins are tossed = nCr / 2n

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Shortcut methods and tricks to solve questions from ‘Banker’s Discount’ are given below:

Banker’s Gain (B.G.) = Banker’s Discount (B.D.) – True Discount (T.D.) for the unexpired time.

If the bill date is not given, grace days will not be added.

B.D. = S.I. on bill for the unexpired time.

B.G. = (B.D.) – (T.D.) = S.I. on T.D. = (T.D. * T.D.) / P.W.

T.D. = √ (P.W. x B.G.)

B.D. = [(Amount x Rate x Time) / 100]

T.D. = [(Amount x Rate x Time) / 100 + (Rate x Time)]

Amount = (B.D. x T.D.) / (B.D. – T.D.)

T.D. = (B.G. x 100) / (Rate x Time)

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

Basic tricks and formulas are given below:

Speed = Distance / Time

Time = Distance / Speed

Distance = Speed * Time

Meter/Second convert into Km/hr = A*(18/5) km/hr

Km/hr convert into Meter/Second = A*(5/18) km/hr

Speed of whole journey = [2pq / (p + q)] km/hr

1. If A covers a distance x1 km at p1 km/hr and then x2 at p2 km/hr, then the average speed during the whole journey is given by:

Average speed =

[p1p2(x1+x2) / (p1x2+p2x1)] km/hr

2. If A goes from a to b at p1 km/hr and comes back from b to a at p2 km/hr. Then the average speed during the whole journey is given by:

[2(p1p2) / (p1 + p2)]

3. Somebody goes certain distance at a speed of p1 km/hr and returns back to the same place at a speed of p2 km/hr. If he take T hours in all, then the distance between first place to the second place is:

T [(p1p2) / (p1 + p2)]

4. If two persons X and Y start at the same time from two points A and B towards each other and after crossing they take T1 and T2 hours in reaching B and A respectively, then:

(X Speed / Y Speed) = (√T2 / √T1)

5. If the new speed is a/b of the original speed, then the change in time to cover the same distance is given by:

Change in time = (y/x – 1) x Original Time

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                                                                                                                      (…….to be continued in Part V)

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.

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