Simplification Questions for competitive exams: An introduction
In every competitive exam ,under quantitative aptitude section you can find simplification questions. You can expect simplification and approximation questions in exams like Common Aptitude Test (CAT), MAT, GMAT, IBPS PO,Clerk Exams, CSAT, CLAT, SBI Bank Competitive Exams, ICET, UPSC Competitive Exams, CLAT, SSC CGL, state PSCs, XAT, GRE, Defense service , L.I.C,General insurance and many other competitive examinations. Many candidates take this section very lightly and fail to score well in this section. To avoid minor errors you must thoroughly understand basic maths rules and formulas.Understanding some tricks and shortcuts will improve your calculation speed considerably. Here we have provided basic formulas and shortcuts for solving simplification questions much faster.You can also find three sets of practice questions for simplification below.
BODMAS rule
BODMAS stands for Brackets, Orders, Division, Multiplication, Addition and subtraction. To simplify a mathematical expression you must do brackets first, then division, then multiplication, then addition and subtraction. When all your operations are the same level (for example,division or multiplication, addition or subtraction, powers or roots ), simply work from left to right.
B=Brackets
O=Orders (powers and roots)
D=Division
M=Multiplication
A=Addition
S=Subtraction
Some useful formulas
Formulas and shortcuts for basic arithmetic operations and simplification
In division instead of direct division, use factoring method
Example1:1848/264=(2*3*4*7*11)/(2*3*4*11)=7
Example2:=594/54=(11*9*3*2)/(9*3*2)
i)Sum-10 method:Example: 78 and 72. These two numbers, if we add the numbers in the unit's place, the resultant is 10 and the numbers in the ten's place are both the same. In such cases, we can have a simple solution.
Step1: multiply the numbers in the unit's place and write down the result.
(8*2 = 16)
Step2: say, the number in the ten's digit is a, then multiply a*(a+1) and write down the result.
=> (7*(7+1) = 56)
Step3: write the final result: 5616
Example:118*112 follow above steps
8*2 = 16; and 11*(11+1) = 11*12 = 132. And hence the result is: 13216.
In short: ab*ac = (a*(a+1))(b*c)
ii)Base Method:Base numbers, in general, are nothing but multiples of 10. If the given numbers are nearer to base numbers, then you can follow this method to multiply them.
Example: 98*95 =?
Here 98 is ,2 less than the base number 100 and 95 is ,5 less than 100. We can write them like this:
98 -2
95 -5
The first step will be deducting/subtracting the resultant of the diff between the base number and the given number with the given number in a crossway! That is, you need to subtract 98 and 5 (which is the resultant of difference between the base number and 95) or you can also cross-subtract 95 and 2, the result will be same. This result forms the 1st part of the resultant at the start. The last part of the resultant will be multiplication of the differences from base numbers (i.e., 2 * 5 = 10)
The first step will be deducting/subtracting the resultant of the diff between the base number and the given number with the given number in a crossway! That is, you need to subtract 98 and 5 (which is the resultant of difference between the base number and 95) or you can also cross-subtract 95 and 2, the result will be same. This result forms the 1st part of the resultant at the start. The last part of the resultant will be multiplication of the differences from base numbers (i.e., 2 * 5 = 10)
98 -2
95 -5
(98 – 5) (-2 * -5)
(98 – 5) (-2 * -5)
Hence, the answer will be: 9310
Example: 998*997 =?
998 -2
997 -3
Observe carefully, in the second part, the multiplication of difference yield in a single digit number, but no. of zeroes in the base number, here 1000, is three. Hence add two zeroes before the result. Therefore, the answer will be:
(998-3) | (-2 * -3) = 995006Observe carefully, in the second part, the multiplication of difference yield in a single digit number, but no. of zeroes in the base number, here 1000, is three. Hence add two zeroes before the result. Therefore, the answer will be:
What if the numbers we get are like this? I mean, the base is 50 here. We will follow the same procedure as above but a small difference that the resultant in the first part will be halved. And if the base is 200, then the number will be doubled and so on based on the base number.
iii)Multiplication with 5,25,50 etc...
Substitute 5 by 10/2,25 by 100/4 and 50 by 100/2.
Substitute 5 by 10/2,25 by 100/4 and 50 by 100/2.
Examples:
1. 5*18=18*10/2=180/2=90
2. 24*25=24*100/4=2400/4=6001. 5*18=18*10/2=180/2=90
3. 73*50=73*100/2=7300/2=3650
iv)Multiplication with 9, 99,999 etc..
Examples:
1. 13*9=13*(10-1)=130-13=117 Examples:
2. 26*99=26*(100-1)=2600-26=2574
3. 350*999=350*(1000-1)=350000-350=349650
Simplification and approximation practice questions
SBI PO 2017 prelim model question paper pdf download