Percentage Problems And Formulas | Learn To Solve

Percentage Problems And Formulas
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Percentage Problems And FormulasPercentage problems are generally asked in all major exams like SSC, IBPS,CAT, GATE etc.These problems become complex if we do not have basic concept of problems.We will deal with concept of percentage problems and formulas are given at the end of this chapter to help students revise important formulas and concepts.

Concept Of Percentage

By certain percentage we mean many hundredths.It means X % means X hundredths.
There are two things which should be clear to us.

1.Conversion into percentage
In order to convert some quantity(say X) into percentage.We simply multiply that quantity with 100.
For example:
Convert 5/2 into percentage.
We simply multiply it with 100=(5/20)*100=250%
2.Conversion into Fraction from given percentage
If we have given X %.In order to change it into fraction we simply divide it with hundred i.e X/100.
For example:
Convert 20% into fraction
20/100 = 1/5

In percentage, a term ‘OF’ is used.Like, 20% “OF” all students play cricket and rest play football. If there are 50 students in the class, 20% “OF” 50 students play cricket.
It means (20/100) * 50 =10 students play cricket.Replace “of” with a multiplication sign.

Intext Question:-
  1. Convert 0.6% into fraction
    Solution- Divide 0.6 with hundred i.e 0.6/100 = 6/1000

  2. Convert 30/20 into rate percent
    Solution-Multiply fraction with 100 = (30/20) * 100 = 150%

  3. Evaluate 30% of 200 + 50% of 500
    Solution- (30/100) * 200 + (50/100) * 500 = 60 + 250 = 310

  4. What percent of 60 is 15
    Solution- Let that percent be X.
    X percent of 60 = 15
    X * (60/100) = 15
    or, X = (15/60) * 100 =25%


Percentage Important Formulas
  • Percentage Increase/Decrease:

    If the price of a commodity increases by R%, then the reduction in consumption so as not to increase the expenditure is: \left [ \frac{R}{(100+R)}\times 100 \right ] %

    If the price of a commodity decreases by R%, then the increase in consumption so as not to decrease the expenditure is:\left [ \frac{R}{(100-R)}\times 100 \right ] %

  • Results on Population:

    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\left ( 1+\frac{R}{100} \right )^{^{n}}

    2. Population n years ago = \frac{p}{\left ( 1+\frac{R}{100} \right )^{^{n}}}

  • Results on Depreciation:

    Let the present value of a machine be P. Suppose it depreciates at the rate of R% per annum. Then:
    1. Value of the machine after n years = p\left ( 1-\frac{R}{100} \right )^{^{n}}

    2. Value of the machine n years ago = \frac{p}{\left ( 1-\frac{R}{100} \right )^{^{n}}}

    3. If A is R% more than B, then B is less than A \left [ \frac{R}{(100+R)}\times 100 \right ] %

    4. If A is R% less than B, then B is more than A by \left [ \frac{R}{(100-R)}\times 100 \right ] %

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