Tricks to solve Percentage Problems

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Tricks to solve Percentage Problems

Here we shall discuss about Tricks to solve Percentage Problems of the Quantitative aptitude section.  Now a days this topic  have become an important part of the Quant test in all Entrance Exams.So Here we will help you in this.We will  provide short tricks on Percentage problems.

Percentage is a fraction whose denominator is always 100. x percentage is represented by x%.

Percentage word which is a combination of a word ” per cent “ that means for every hundred. The number, which is in a fraction form whose denominator is “100” is called as a percentage and the numerator of the same number is called as a rate per cent .



How to calculate Percentage?

Percentage = ( Value / Total Value ) X 100

3D man pushing percentage sign isolated over white

Percentage to Decimal Conversion or Fraction Conversion

1 = 100%

1/2 = 50%

1/3 = 33 1/3%

1/4 = 25%

1/5 = 20%

1/6 = 162/3%

1/7 = 142/7%

1/8 = 121/2%

1/9 = 111/3%

1/10 = 10%

1/11 = 91/11%

1/10 = 81/3%

1/13 = 79/13%

1/16 = 6.25%

5/4 = 125%

3/2 = 150%

2 = 200%

7/2 = 350%





To express x% as a fraction :

We know, x% = x/100

Thus 10% = 10/100 (means 10 parts out of 100 parts) = 1/10 (means 1 part out of 10 parts)

To express x/y as a percentage :

We know that x/y = (x/y× 100 )

Thus 1/4 = ( 1/4 ×100 )% = 25% and 0.8 = ( 8/10 ×100 )% = 80%

If the price of a commodity increases by R%, then reduction in consumption as not to increase the expenditure is = [ R/(100+R)×100 ] %

If the price of a commodity decreases by R%, then the increase in consumption as not to decrease the expenditure is = [ R/(100-R)×100 ] %

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

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

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




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

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

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

For two successive changes of x% and y%, net change = {x + y +xy/100}%




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  • ◀benzene▶

    one problem i have studied hindi till class 8 but dont have any proof to show becoz report card of that time i have lost so now what should i do , i know hindi very well reading writing everything

  • Yes you can apply. Yoy need to givr a address from Bihar

  • ◀benzene▶

    ibps rrb i can apply from bihar ,being from jharkhand???