Maths Tricks for Bank Exams
IBPS Shortcut Tricks –Speed Maths Tricks that every student should know to speed up the calculations in Bank Exams and in competitive Exams 201718.
Before going to start preparation for bank exams.

 At first, one should know the Syllabus and Exam Pattern of various Bank Exams (clerks, POs, Specialist officers of SBI, IBPS and other private banks).
 Get Latest Bank Job Notifications 201718 from here.
 Collect good study material and best books to crack bank exams.
 Have an idea on all the topics of Reasoning, Aptitude, English and concentrate on some important topics and practice more on those topics.
 Learn some Speed Maths Tricks as we are explaining below to save time in solving aptitude.
 Practice previous question papers with time and proper planning while taking the test.
 Analyze the performance and practice more and more. Then, I am sure you can get qualify in bank exams.
First of all, we have to memorize some basic things in Maths.
 We have to memorize 20 multiplication tables completely in such a way that we can able to tell without thinking.
 Learn Squares of numbers up to 30(2^{2},3^{2},…..30^{2).}
 Cubes of numbers up to 20(2^{3},3^{3}………20^{3).}
 Powers of 2 – 22 , 2^{3 }, 2^{4} ……2^{8.}
 Powers of 3 – 3^{2 }, 3^{3 }, 3^{4 }3^{5}
 Memorize some factorial values up to 10!.
 Fractional values 1/2, 1/3, 1/4 ….. up to 1/20.
 Percentage values (useful in solving simplifications, data interpretation, percentage problems.
Quantitative Aptitude Shortcuts: (Speed Maths Tricks)
 Multiplication Tricks
 Addition Tricks
 Squares
 Square Roots
 Cubes
 Cube Roots
 Division
 Subtraction
 Simplification Tricks Using Formulae
 Comparing Fractions: Tips to find Smaller & Larger
 Useful Conversions.
Aptitude Topics: IBPS Shortcut Formulas
 Quadratic Equations
 Percentages
 Time and Work
 Trains, Time and Distance.
 Simple Interest and Compound Interest
 Profit & Loss.
 Probability
 Boats and Streams.
 Permutations and Combinations
 Ages
 Mensuration
Data Sufficiency.
Multiplication Shortcut Methods:
I. Multiplication of 2 two digit numbers Shortcut Tricks
1)24 ×32 = 768
Explanation:
1) The right digit of 8 of the answer is the product obtained by the vertical multiplication of the right digit of the multiplicand and of the multiplier.
2) The left digit 7 of the answer is the product obtained by the vertical multiplication of the left digit of the multiplicand and of the multiplier and by adding carry forward 1 from cross multiplying.
3) The middle digit 6 of the answer is the sum of cross multiplication of the two numbers. ( To obtain the middle digit, one has to multiply “across” and add the two products, in our example 2×2 + 4 ×3).
2) 78 × 86.
II. Mulitplication 2 three digit numbers.
For Example, Let us consider the multiplicand to be ABC and the multiplier to be DEF as shown below:
ABC
DEF
1) The extreme right digit of the answer is obtained (as before) by vertical multiplication as C × F.
2) The extreme left digit is also obtained (as before) by vertical multiplication as A × D.
3) The “middle” digits are obtained (as before) by multiplying across. Progressing one step at a time to the left, middle digits are successive.
B × F + C × E
A × F + C × D + B × E
A × E + B × D
The process is set out in detail in the first two examples below and in condensed form thereafter.
III. Mulitplication 2 Four digit numbers.
The above process of multiplication can be continued indefinitely to numbers of any length.
If the multiplicand and the multiplier are 2 four digit numbers, as in ABCD × EFGH.
Explanation:
1) The extreme right digit is D × H.
2) The extreme left digit is A × E.
3) The middle digits, progressing one step to the left at each stage are successively
C × H + D × G
B × H + D × F + C × G
A × H + D × E + B × G + C × F
A × G + C × E + B × F
A × F + B × E
IV. Mulitplication 2 Five digit numbers.
If the 2 numbers are 5 digit ones, as in ABCDE × FGHIJ
1) The extreme right digit is E ×J
2) The Extreme left digit is A × F
3) The middle digits, progressing to the left one step at a time are successive.
D×J + E×I
C×J + E×H + D× I
B×J + E × G + C ×I + D ×H
A×J + E ×F + B ×I + D × C + C ×H
A×I + D×F + B ×H +C ×G
A×H + C ×F + B ×G
A × G + B ×F
V. Multiplication of Numbers of Different Lengths
In the examples we saw above, both the multiplicand and the multiplier contained the same number of digits. But what if the two numbers were to contain a different number of digits, For instance, how would we multiply 286 and 78?
Obviously, we could prefix a zero to 78 (so that it becomes 078 as 3 digit numbers. The following examples will clarify the procedure).
1) 286 × 078 =?
2) 998 × 098 =?
Multiplication Tricks
Learn some multiplication Tricks in solving aptitude problems, data interpretation problems and in some simplifications to save time.
Trick 1: Multiplying the two digit numbers which are having the same number in tens place to and ending with 5.
Eg:
1) 25 × 25 = (2×3)25=625.
In this example, 25 × 25. Here the sum of units digit of two numbers 5 + 5 is 10,
the left digit of two numbers is same i.e 2. Then we can apply this trick. (for any numbers as above).
In this case the right digit of the multiplicand and multiplier is 5,
the right part of our answer is always 5 × 5 i.e. 25 and therefore,
we can mechanically set down 25 as right part of the answer without any calculation
all that needed is to find out the left part of the answer and this is done exactly as in the previous section.
2) 35 × 35 = (3×4)25=1225
3) 65 × 65 = (6×7)25=4225.
4) 75 × 75 = (7 × 8)25 = 5625.
5) 95 ×95 =left part (9×10)  25 right part =9025.
6)775 × 775 = (77 × 78)  25 = 600625.
7) 875 × 875 = 765625
8) 995 × 995 = 990025
9) 1005 × 1005 = 1010025
10) 1245 × 1245 = left  right = (124 × 125) 25 = 1550025
11) 8995 ×8995 = (899 ×900)  25 = 80910025.
12) 9995 × 9995 = 999 × 1000  25 = 99900025.
13) 10015 × 10015 = 1001 × 1002  25 = 100300225.
14) 99995 × 99995 = 9999 × 10000  25 = 9999000025.
Trick 2: Multiplication of Fractions using above method for the following examples – the sum of fractional parts is 1 and their integral parts are same.
In simplifications, if you came across the following calculations use this shortcut trick.
1) 6 (1/2) × 6(1/2) = (6 ×7) ¼ = 42 ¼ .
2) 9(1/2) × 9(1/2) = (9 ×10) ¼ = 90 (1/4).
3) 12 (1/2) × 12(1/2) = (12 ×13) ¼ = 156 ¼
4) 6(1/4) ×6 (3/4) = (6 ×7) 3/16= 42 (3/16).
5) 6(1/8) × 6 ( 7/8) = (6×7) 7/64 = 42 ( 7/64).
6) 7 (2/7) × 7 (5/7) = (7 ×8) 10/49 = 56 (10/49)
7) 8(5/11) × 8 (5/11) = (8 ×9) 30/121 = 72(30/121).
Trick 3: When the two numbers, when the sum of unit digits of the two numbers is 5 and other digits are same use this shortcut trick.
1) 24 × 21 = ?
Here the first part is same and the sum of unit digits of two numbers is 5.
(2×2 + ½ ×2 and 4×1)
= ( 4 + 1 and 4 ) = 504.
2) 803 × 802 =?
(80×80 + ½ ×80 and 3×2) = 6400 + 40 and 06 = 644006.
3) 1001 × 1004 =?
(100 ×100 + ½ ×100 and 1×4).
1005004.
4) 74 × 71 =?
= 7 × 7 + ½ ×7  and 4 ×1
= 49 + 3.5  04
= 5254
5) 253 × 251 =?
= (25 × 25 + ½ (25) and 3 ×1)
= 625 + 12.5 and 03
= 63753.
6) 991 × 994 =?
= (99 × 99 + ½ (99) and (1 ×4))
= 9801 + 49.5 and 04
= 985054.
Trick 4: If the sum of the last 2 digits of two numbers is 50 and the other digits are the same. Then, use this shortcut trick.
1) 421 × 429 =?
Here, the sum of 21 and 29 is 50 and other digits are the same.
(4 × 4 + ½ × 4 and 21 ×29)
= 16 + 2 and 0609
= 180609.
2) 1823 × 1827 =?
In this example, 23 + 27 = 50 and other digits 18 is same.
= (18 × 18 + ½ ×18 and 23 × 27)
=324+9 and 0621
= 3330621.
Trick 5: Multiplication of 2 digit numbers ending with 5 and also their difference should be 10.
Eg:
1) 15 × 25 = (1×3)75 = 375.
2) 35 × 45 = (3×5)75 = 1575.
3) 25 × 35 = (2×4)75 = 875.
Trick 6: Multiplication of the two digit numbers having the same number in the tens place and the addition of unit digits should be 10.
Eg:
1) 23 × 27= (2×3) 21= 621.
2) 24 × 26 = (2×3) 24= 624.
3) 36 × 34 = (3×4)24 = 1224.
4) 94 × 96
5) 982 × 988
In all of them, the sum of unit digits of two numbers is 10.
1)3+7 = 10 , 2) 4+6 = 10, 3) 6+4 = 10, 4) 4+6 = 10, 5) 2+8 = 10.
In such cases, this shortcut trick can be used.
(2×3) 21= 621. [ the first digit in two numbers is 2, 2 is multiplied with next digit i.e. 3 and then next digits of the answer is numbers .multiplication of unit digits (3× 7) = 21].
To get the right part of the answer, multiply the unit digits i.e. the extreme right digit numbers.
To obtain the left part of the answer, multiply the other (i.e. the left digits by one more than itself /themselves.
For example, if the left digits of the 2 numbers is/are
3, then multiply 3 with 4
4, then multiply 4 with 5.
Note: The only thing you have to be careful about is to ensure that the right part of the answer always has 2 digits.
For example, in 29 × 21, the right part will be written as 09 and not as 9 because the right part has to contain 2 digits.
Therefore, 29 × 21 = 609.
19 × 11 = 209
18 × 12 = 216
17 × 13 = 221
16 × 14 = 224
15 × 15 = 225
29 × 21 = 609
28 × 22 = 616
27 × 23 = 621
26 × 24 = 624
25 × 25 = 6 25
52 × 58 = 3016
87 ×83 = 7221
86 × 84 = 72 (24)
104 × 106 = 11024
111 × 119 = 13209
199 × 191 = 38009
892 × 898 = 801016
992 × 998 = 990016
1002 × 1008 = 1010016
1009 × 1001 = 1010009
10008 × 10002= 100100016
999991 × 999999 = 999990000009
Trick 7: Multiplying the two digit numbers having 9 in the tens place and different numbers in unit place.
Eg:
1)92×93= (927) (8×7) = 8556
10093=7, 10092=8. Smallest of 8 and 7 can be deducted from the smaller number of 92 and 93.Here we take 927.
2)93×97= (933) (7×3) = 9021.
Trick 8: If the unit digit or ten’s digit of the two numbers are same, then use this trick to solve in less time.
1) 83 × 93 = ?
= (8×9) + (17×3)  (3×3)
= 72 + 51  9
= 7 71 9
The middle term is obtained by multiplying 3 by 17, 17 being the sum of 8 and 9.
2) 28 × 23 = 644
In this case, the middle term is 2 multiplied by 11 and 11 being the sum of 8 and 3.
SQUARES:
How to Find the square of the number in Using Shortcuts:
I. Squaring of 2 digit numbers which begin with 5.
Example:
 (51)^{2} = (5^{2 }+ 1), (01)^{ 2} è [1^{st} two digits, 2^{nd} two digits].
= 2601.
 (56)^{2} = (5^{2} + 6), (6^{2})
= 3136.
 (58)^{2} = 5^{2} + 8 , 8^{2}
= 3364.
 (59)^{2} = 5^{2} + 9 , 9^{2}
= 3481.
II. Squaring numbers which are near to 100. ( i.e. 80 to 130)
Examples:
1. (86)^{2 }= 1^{st} part /2^{nd} part
= (86 14) / 14^{2}
= 72 _{1} 96 = 7396.
2. (89)^{2} = (8911) / (11)^{2}
= 78 _{1 }21 = 7921
3. (97)^{2 }= 973 / 3^{2}
= 9409
4. (99)^{2} = (991) / 1^{2}
=9801
5. (106)^{2 }=1^{st} part / 2^{nd} part
= (106 +6) / 6^{2}
= 11236
6. (107)^{2 }= (107 + 7) / 7^{2}
= 11449.
7. (112)^{2 }= 112+12 / 12^{2}
= (124) / _{1} 44
= 12544.
8. (119)^{2} = (119 +19) / 19^{2}
= 138 / _{3}61
= 14161.
III. Squaring the numbers which are near to 1000.
Examples:
1. (990)^{2} = 1^{st} 3 digits/ 2^{nd} 3 digits
= (990 10) /10^{2}
= 980100.
2. (992)^{2} = (990 8) / 8^{2}
= 982064.
3. (1006)^2 = 1006 +6 /6^{2}
= 1012036.
IV. Already we remember squares of two digit numbers up to 30. Then we follow the below methods to find the squares of numbers from 30 to 80.
 Here we take the base number as 50 and then split the given number.
( if we want to find square of 39 it can split to 5011 ) 50^{2}=2500, 11^{2}=121 
Eg: 39^{2}= (5011)


 2500

1100 ()
1400
121 (+)
1521
If we practice then these calculations we can do mentally without using paper.I will explain with one more example.
Eg 2: 63^{2}= 50+13
 2500
1300 (+)
3800
169 (+)
3969
Squaring numbers from 90 to 99.
Here, 10094 = 6 
Eg: 1) 94^{2}=(946)6^{2}
= 8836
Here, 10096 = 4 
2)96^{2}=(964)4^{2}
= 9216.
Addition :
The method of adding is very simple and everyone can do this. But, the thing is adding numbers in shortest possible time is important.
I.Single Column Addition:
For Example: Add below numbers from bottom to top by following simple technique.
 9
 7
 8
 5
 4
 7
 6
while adding the above numbers from bottom to top. Don’t say 6 plus 7 is equal to 13 and 13 plus 4 is equal to 17 and 17 plus 5 is equal to 22 and so on. It is too timeconsuming. Here, we have to use strategy saying like this. 6 then 13, 17,22, 30 and so on and do mind calculations and don’t require to say out each and everything its very time taking.
II. Double Column Addition:
Double column addition is absolutely useful for quick calculations in competitive exams.
Example :
89
56
23
78
45
12
The addition of numbers starting from bottom starts as 12 +40(of 45) and then giving 57, 57+70( of 78) and then giving 135 and so on.
III. The addition of Multiple column Numbers:
Example :
58964
45896
52639
78954
Here, First start adding double columns each time. It takes less time compared to the normal method. Once the 1st set of double column(units and tens place digit numbers ) is mastered. It becomes easy to add the 2nd set of double column i.e. 100’s place and 1000’s place digits.
Subtraction:
Solving questions without using pen and getting answers mentally saves much time in bank exams. Here, we are providing some subtraction shortcut methods which are useful in solving DI problems and also simplifications in bank exams.
Suppose, if we need to subtract 66 from 92. Mentally increase the number to the nearest multiple of 10 that means to increase 66 to 70 by adding 4 and at the same time mentally increase the number to the nearest multiple of 10 i.e. increase 66 to 70 by adding 4 and at the same time, increase the other quantity by the same amount i.e. by 4 then 92 becomes 96.
Therefore, the problem now is 96 minus 70 with this we can get instant answers mentally without using pen i.e. 26.
Eg: 3 5 8
2 6 7 5
+6 5 8 8
3 5 6 9
2 7 0 2
Generally, we find questions as below or these type of calculations are much useful in solving data interpretation problems in bank exams.
Q.1) 358 – 2675 +6588 3569 =?
If we try to solve in double column wise. Sometimes, we no need to solve entire answer by knowing the answer of last two digits from the options we can able to checkout correct answer.
Simplification Methods:
In some simplification problems may sometimes involves application of algebraic formulae listed below:
 (a+b)^{2} = a^{2} + 2ab + b^{2}
 (a +b)^{2 }= a^{2 }– 2ab + b^{2}
 (a+b)^{3} = a^{3} +3ab (a+b) +b^{3}
 (ab)^{3} = a^{3} – 3ab (a+b) – b^{3}
 (a^{2} –b^{2}) = (a+b)(ab)
 (a^{3}+ b^{3}) = (a+b)(a^{2}ab+b^{2})
 (a^{3}– b^{3}) = (ab)(a^{2}+ab+b^{2})
Tips to Solve Fractions Ascending/Descending Order :
In some cases, we find comparision of fractions in bank test papers.
It’s timeconsuming if we do by calculating the decimal value and compare. But the simple technique to do in seconds is comparing by doing cross multiplication of two fractions .whichever is the big number after cross multiplying that fraction is big as viceversa.
For example:
Compare, 3/4 and 1/4.
By seeing itself, we can say 3/4 is big. Implement here our logic of cross multiplication of two fractions.
¾ ¼ we get, 12 and 4 we know 12 > 4 so we can say ¾ >1/4
Apply the same logic big fractions it’s very simple and takes less time.
Problem 1: Arrange the following fractions in ascending order.
5/6, 3/4, 2/9, 1/5
Sol: Ascending>Arise ( go up)( small to big)
Descending> down( big to small)
Here compare the cross multiplication values whichever is smaller that fraction is the small one.
We have compared remaining fractions 3/4 and 5/6 as same as above process.
Finally, we get, 1/5 <2/9 <3/4 <5/6.
If we understand the logic behind this and practice 3 problems it will take less time without paper.
Some of the useful conversion factors:
 1foot = 12 inches =30.48 milimetres
 1 inch = 2.54 centimetres = 25.4 millimetres
 1 mile = 1.609 kilometres
 1 acre = 4000 sq.metres = 43560 sq. fts
 1 quintal = 100 kgs
 1 hectare = 2.5 acres = 10000 sq. metres
 1 Metric Ton = 1000 kgs
 1kg = 2.204 pounds
 1 gallon = 4.5 litres
 1 degree = 12 inches = 60minutes
Divisibility:
When we do division of big numbers. we can’t know whether it is divisible or not. For this to find before, Here are some rules for testing divisibility of a number by 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12.
A Number is divisible by  IF 
2  The last digit is 0,2,4 6 or 8 ( any even number) 
3  The sum of all digits is divisible by 3 
4  The last 2 digits form a number divisible by 4 or 00. 
5  The last digit is either 0 or 5. 
6  It is divisible by both 2 and 3. 
8  The last 3 digits form a number divisible by 8 or 000. 
9  The sum of all digits is divisible by 9. 
10  The last digit is 0. 
11  The difference between the sum of digits at odd and even places is either 0 or in multiples of 11. 
12  It is divisible by both 3 and 4. 
SQUARES  CUBES  


Factorial Values:
Important Factorial Values  
2!  2 
3!  6 
4!  24 
5!  120 
6!  726 
7!  5040 
8!  40320 
9!  362880 
10!  3628800 
Basic Percentage Values to Remember
 1/2 = 50%
 1/3 = 33.33%
 ¼ = 25%
 1/5 = 20%
 1/6 =16.66%
 1/7 = 14.28%
 1/8 = 12.5%
 1/9 = 11.11%
 1/10 = 10%
 1/11 =9.09 %
 1/12 = 8.33%
 1/13 = 7.69%
 1/14 = 7%
 1/15 = 6.66%
 1/16 = 6.25%
 1/17 = 5.88%
 1/18 =5.55%
 1/19 = 5.26%
 1/20= 5%
 5%= 1/20 = 0.05
 10% =1/10 =0.1
 15% =3/20
 20% = 1/5
 25% = ¼
 30% =3/10
 40% =2/5
 50% =1/2
 55%=11/20
 60% = 3/5
 70% = 7/10
 75% = 3/4
 80% = 4/5
 90% = 9/10
 100%=1
 6 ¼ % = 1/6
 12 ½ % = 1/8
 16 2/3 % = 1/6
 33 1/3 % = 1/3
 66 2/3 % =2/3
 125% =5/4
 150% = 3/2