Important Formulas and Tricks for Quantitative Aptitude (Bank Exams)

Quantitative aptitude is a crucial topic that appears in all bank exams. You may do well in this topic if you understand the fundamental ideas, but you must be aware of several Quantitative Aptitude Tricks if you want to do well. Data Interpretation, Arithmetic problems, Number series, Quadratic quantity comparison, Simplification/Approximation, and other topics are covered in the quantitative aptitude portion. In this article, we will share with you Important Formulas and Tricks for Quantitative Aptitude (Bank Exam)

If you know how to make a quick calculation and use strategies, you can answer several topics like Simplification/Approximation, Number Series, and Quadratic Equations Comparison extremely successfully and in less time. In bank (IBPS/SBI and other) tests, these areas account for almost half of the weight age in the quantitative ability part.

We’ll discuss several Quantitative Aptitude Tricks in this post to assist you to improve your calculating speed so you can do well in the exam.

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Important Formulas and Tricks for Quantitative Exams

Operation on Numbers

• An even number will always be divisible by 2.
• If the total of the digits is divisible by three, the number is divisible by three.
• If the number generated by the final two digits is divisible by four, the number is divisible by four.
• If the units digit is either 5 or 0, the figure is divisible by 5.
• If a number is divisible by two and three, it is divisible by six.
• If the number obtained by the last three digits is divisible by 8, the number is divisible by 8.
• If the total of the digits is divisible by nine, the number is divisible by nine.
• If the units digit is 0, a number is divisible by 10.
• If the difference between the sum of its digits in odd places and the sum of its digits in even places is divisible by 11, the number is divisible by 11.
• If any number is divisible by 3 and 4, then it will automatically be divisible by 12.

Tips to follow for special cases: –

• For Divisibility of 7, we multiply the unit digit by 2 and then subtract.
• For 13 Divisibility, multiply the unit digit by 4 and then add.
• For the Divisibility of 17, we multiply the unit digit by 5 and then subtract.
• For the Divisibility of 19, we multiply the unit digit by 2 and then add.

Example

Is 2961 divisible by 7, for example?

296 – 1X2 = 294 (Step I)

29 – 4X2 = 21 (Step II)

Because 21 is divisible by 7, we may conclude that 2961 is likewise divisible by 7.

Note: The same procedure will be used for larger numbers.

If a number follows the rules below, it is divisible by 17.

Is 1904 divisible by 17?

190 – 4X5 = 170.

We can see that 170 is divisible by 17. Then we may conclude that 1904 is divisible by 17 as well.

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LCM AND HCF

H.C.F. stands for Highest Common Factor/ G.C.M. for Greatest Common Measure/ G.C.D. for Greatest Common Divisor.

L.C.M: Lowest Common Factor / Lowest Common Divisor (L.C.D) and Lowest Common Measure (L.C.M)

• The biggest number that divides each of two or more numbers completely is the H.C.F.
• Their L.C.M. is the smallest number that is perfectly divisible by each of the provided numbers.
• If the H.C.F. of any two numbers will be 1, then they are said to be co-prime.

H.C.F. of fractions = L.C.M. of denominators/H.C.F. of numerators.

L.C.M. of fractions = H.C.F of denominators/ G.C.D. of numerators.

The product of two numbers will be equal to the product of their H.C.F. and L.C.M.

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Simplification by using BODMAS rule

 BODMAS Rule:

This Rule demonstrates the proper order in which calculations should be performed in order to determine the value of a given expression.

B – Bracket

O – Of

D – Division

M – Multiplications

A – Addition

S – Subtractions

• (a +b)² = a ² + 2ab + b²
• (a + b) (a – b) = (a² – b²)
• (a + b)² = (a² + b² + 2ab)
• (a – b)² = (a² + b² – 2ab)
• (a + b + c)² = a² + b² + c² + 2(ab + bc + ca)
• (a³ + b³) = (a + b) (a² – ab + b²)
• (a³ – b ³) = (a – b) (a² + ab + b²)
• (a³ + b³ + c³ – 3abc) = (a + b + c) (a² + b² + c² – ab – bc – ac)

When a + b + c = 0, then a 3 + b 3 + c 3 = 3abc.

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Algebra

Sum of first z natural numbers = z(z+1)/2

Sum of the squares of first z natural numbers = z(z+1) (2z+1)/6

Sum of the cubes of first z natural numbers = [z(z+1)/2]2

Sum of first z natural odd numbers = z2

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Arithmetic Progression

Average = (Sum of items)/Number of items

An A.P. has the form a, a+d, a+2d, a+3d…, where a is the ‘first term’ and d is the ‘common difference.’ The nth term of an A.P. is tn = a + (n-1)d.

Sn = n/2[2a+(n-1)d] or Sn = n/2(first term + final term) is the sum of the first n terms of an A.P.

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Geometrical Progression (G.P.)

A G.P. has the form a, ar, ar2, ar3…, with a being the ‘initial term’ and r being the ‘common ratio.’

the nth term of a G.P. tn = arn-1

Sn = a|1-rn|/|1-r| is the Sum of the first n terms in a G.P.

 

Probability

The probability formula states that the ratio of the number of favorable outcomes to the total number of outcomes equals the probability of an event occurring.

Probability of an event occurring P(E) = Number of favorable outcomes/Total Number of favorable outcomes

Students frequently confuse “favorable outcome” with “desired outcome.” This is the fundamental formula. There are, however, other formulas for certain situations or events.

Examples: –

1) There are 6 flowers in a garden, 3 are red, 2 are yellow and 1 is blue. What is the probability of a child picking a yellow flower?

Ans: The probability is equal to the number of yellow flowers in the garden divided by the total number of flowers, i.e. 2/6 = 1/3.

2) There is a garage full of cars – red, blue, green, and orange. Some of the cars are selected for shifting to another garage. Aman did this 1000 times and got the following results:

• No. of blue cars shifted: 300
• No. of red cars shifted: 200
• No. of green cars shifted: 450
• No. of orange cars shifted: 50

 

1. a) What is the probability that Ama will pick a green car?

Ans: For every 1000 cars shifted, 450 are green.

Therefore, P(green) = 450/1000 = 0.45

1. b) If there are 100 cars in the old garage, how many of them are likely to be green?

Ans: The experiment implies that 450 out of 1000 cars are green.

Therefore, out of 100 cars, 45 are green.

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Probability of Occurrence of an Event

Assume that an event E can happen in r ways out of a total of n equally probable ways. The probability of the event occurring or its success is therefore stated as –

P(E) = r/n.

The probability of non-occurrence of an event is called its failure and is expressed as:

P(E’) = (n-r)/n = 1-(r/n)

Here, E indicates the non-happening of an event.

Now, it can be concluded that –

P(E) + P(E’) = 1.

This indicates that in each random test or experiment, the sum of all probability = 1.

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Average

The formula that we follow in order to calculate Average is (Sum of Observations / Number of Observations)

Suppose a man travels a particular distance at A kmph and another at B kmph. The average speed for the whole route is thus [2AB / (A+B)].

 

Ratio and Proportion

A fraction a/b is represented by the ratio a: b. Here, a is an antecedent and b is a consequent.

Proportion is defined as the equivalent of two distinct ratios.

If a: b = c: d, then a, b, c, and d are all in proportion. This is denoted by the letters a: b: c: d.

In the case of a: b = c: d, we obtain a* d = b * c.

If a/b = c/d then ( a + b ) / ( a – b ) = ( d + c ) / ( d – c ).

 

So these were some important formulas and tricks for or quantitative Aptitude (bank exams). Use it and improve your quantitative aptitude game. For more such content, surf our website: icdkollam.in.

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