Tips & Tricks for Quantitative Aptitude Section | All Bank Exams 2019

Quantitative Aptitude is hard in most cases especially in exams like Banks and Insurance. Many Banks exam has a two-tier examination pattern i.e., Prelims and Mains. Most of them have changed their exam patterns and set a sectional timing of 20 minutes on each section. Quantitative aptitude is important for every exam because proper strategy and enough practice can help you score full marks in this section. There may not be assured in the language section and you may be stuck while solving reasoning questions but quants is a scoring subject and assure full marks if the calculation is correct.

So to help you ace the quants and to save your precious time during exam hours Adda247 providing some quant tricks to help aspirants.

Recapitulation of some important hacks that we shared on the previous posts. Let’s revise what we learned till date. 

Some solved questions based on-

Concept 1– Multiplication by 11

Example 1- 

24*11= 2_addition of (2+4) 4

hundred and ones digit is fixed, the tens digit is the sum of digits.

a) 2 always in hundred place

b) tens place- addition of digits

c) ones place- 4

Concept 2– Addition of interchange number 

Addition of interchange number= 11*(addition of digits)

Example-

1) 26+62

Now addition of 26+62 be done

 =11*(2+6)= 88.

Concept 3– Subtraction of Interchange Number

Subtraction of interchange Number= 9*( difference of digits)

Note – a sign of a number is according to

For 26-62= 9*(4)=-36 … the sign is according to sign with a bigger number.

Some Questions based on these rules-

Example1– A number consists of two digits. The sum of the digits is 9. If 63 is subtracted from the number, its digits are interchanged. Find the number?

Solution- Two digit number is 10x+y

given sum of digits(x+y)=9 –equ(1)

digits are interchanged as-

10x+y-63=10y+x—-equn(2)

basic method-solve equn (2)

9x-9y=63 or we can skip this equation if we know interchanging of number concept. 

interchange Number= 9*( difference of digits).

now 

x-y=7—equ(3)

solve equ 1 and 3

solve here 

for x=9+7/2=8

for y=9-7/2=1

the required number is 10x+y=81

Example2-Qty1 and Qty 2

Qty1. A certain number of two digits is three times the sum of its digits and if 45 be added to it, the digits are reversed. The number is:

Solution- Given 10x+y=3(x+y)

7x-2y=0 OR x/y=2/7 equn1

10x+y+45= 10y+x

9(x-y)=45 

x-y=5 equn2

solving equn 1 and 2

x=2 and y=7

the required number is 27.

Qty2– A two digit number is such that the product of the digits is 8. When 18 is added to the number, then the digits are reversed. The number is.

Solution-given xy=8  eun1

10x+y+18=10y+x

9x=9y-18 equn 2

solving equn 2

y-x=2, put y=2+x

x(2+x)=8

x^2+2x-8=0— equn3

solving quadratic equn3 

when equation has (+,-) roots are in (- bigger root,+ smaller root)

so x=2 and -4

x=2, y=4

the number is 24.

Qty 1> Qty2

Example4– Combination of interchange number-

44+89-64+78+44-98+46-87

Rearrage numbers(44+89-64+78+44-98+46-87) and apply interchange concept of addition and subtraction.

11*8-9*(1+2+1) =88-36=52.

Concept4– Multiplication of a number that has 5 in unit place.

We Know multiplication of two number ending with 5 gives 25 always

Example- 65*35 {Here the difference between number is 30}

= 3*(6+1) for 1st part 25 comes after it + (6-3)*50

=2125+150=2275

Concept5– (a³+b³)= (a+b)*(a²-ab+b²)

(a³-b³)= (a-b)*(a²+ab+b²)

(a²-b²)=(a+b)(a-b)

Example1–68% of 720+41% of 390

1/10(68*72)+(41*39)

1/10(70-2)(70+2)+(40-1)(40+1)

(70^2-4)+(40^2-1)

(489.6+159.9)= 649.5

Example2- 30.6% of 2940

1/100(306*294)

1/100(300+6)(300-6)

=899.64.

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