Important Percentage Notes For SSC CGL 2018

Percentage Concepts for SSC And Railway Exam:

Quantitative Aptitude is an equally important section for SSC CGL, CHSL, MTS exams and has an even more abundant importance in some other exams conducted by SSC. Generally, there are questions asked related to basic concepts and formulas of Time & Work. 

To let you make the most of QUANT section, we are providing important facts related to Percentage. Also, Railway Exam is nearby with bunches of posts for the interested candidates in which quantitative aptitude is a major part. We have covered important notes and questions focusing on these prestigious exams. We wish you all the best of luck to come over the fear of Mathematics section.

Percentage Tricks

Percentage: A fraction whose denominator is 100 is called percentage. and The numerator of the fraction is called the rate percent.

1. To express x% as a fraction:

We have, x% = x/100

      Thus, 30% = 30/100 =3/10

2. To express fraction as percentage,

we have ,

 a/b =  {a/b x 100}%

3. Percentage : Ratio Equivalence

1/3 ✖ 100 = 33.33%	1/10 ✖ 100 = 10%
1/4 ✖ 100 = 25%	1/11 ✖ 100 = 9.09%
1/5 ✖ 100 = 20%	1/12 ✖ 100 = 8.33%
1/6 ✖ 100 = 16.66%	1/13 ✖ 100 = 7.69%
1/7 ✖ 100 = 14.28%	1/14 ✖ 100 = 7.14%
1/8 ✖ 100 12.5%	1/15 ✖ 100 = 6.66%
1/9 ✖ 100 11.11%	1/16 ✖ 100 = 6.25%

                           

4. If A is R% more than B, then B is less than A by

  {(R/(100+R)x100}% 

5. If A is R% less than B, then B is more than A by

   {(R/(100-R)x100}% 

6. If the price of a commodity increases by R%, then the reduction in consumption so as not to increase the expenditure is:

  {(R/(100+R)x100}% 

7. If the price of a commodity decreases by R%, then the increase in consumption so as not to decrease the expenditure is:

  {(R/(100-R)x100}% 

8. Let the population of a town be P now and suppose it increases at the rate of R% per annum, then

   1. Population after n years = P(1+R/100)^n

   2.Population before n years =P/(1 + R/100)^n

9. Let the present value of a machine be P. Suppose it depreciates at the rate of R% per annum.

   1.Value of the machine after n years= P(1-R/100)^n

                                                        

   2.Value of the machine n years ago=P/(1-R/100)^n 

                         

10. For two successive changes of x% and y%, net change

   {x + y +xy/100}%

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