Friday, 11 October 2013


  1.    Concept of percentage or Percentage formula
To express x% as a fraction: We have, x%
x/100 Thus, 30%=30/100= 3/10
In simple terms we can conclude that percentage symbol "%" means 1/100
In case we are having a fraction and we have to calculate its percentage then we will multiply it by 100.
that is, a/b=(a/b)×100)%

2.    Increase or decrease in percentage with price
(1) If the price of a commodity increases by R%, then the reduction in consumption so as not to increase the expenditure will be = [R/ (100+R)×100]%
( 2) If the price of a commodity decreases by R%, then the increase in consumption so as not to decrease the expenditure will be = [R/(100−R)×100]

3.    Results on population
Formula's on result of population are very important when we have to calculate the population n years after or n years before,
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

4.    Results on Depreciation
We know that value of a machine depreciate with time, so it will decrease with the time. To calculate the value of machine after n years or before n years, we use these formula's
Let the present value of a machine be P. Suppose it depreciates at the rate of R% per annum.
(1) Population after n years   = P(1−R/100)n
(2) Population before n years =P/(1−R/100)n

5.    Important results for Percentage
The formulas we are going to mention below is same as of increase or decrease in consumption with increase or decrease in the commodity price, just here we are in a bit different context.
(1) If A is R% more than B, then B is less than A by = [R/(100+R)×100]%
(2) If A is R% less than B, then B is more than A by = [R/(100−R)×100]%

1.    Two numbers are less than third number by 30% and 37% respectively. How much percent is the second number less than by the first
(1) 20%                 (2) 30 %
(3) 10%                 (4) 40%

2.    In an examination, 34% of the students failed in mathematics and 42% failed in English. If 20% of the students failed in both the subjects, then find the percentage of students who passed in both the subjects.
(1) 40%                 (2) 41%
(3) 43%                 (4) 44%

3.    In the new budget , the price of kerosene oil rose by 25%. By how much percent must a person reduce his consumption so that his expenditure on it does not increase ?
(1) 20%                 (2) 30%
(3) 22%                 (4) 24%

4.    The value of a machine depreciates at the rate of 10% per annum. If its present is Rs.1,62,000.? What was the value of the machine 2 years ago ?
(1) 200000             (2) 20000
(3) 2000                 (4) None of these

5.    A student multiplied a number by 3/5 instead of 5/3. What is the percentage error in the calculation?
(1) 60%                 (2) 62%
(3) 64%                 (4) 65%

1.    (3) : Let the third number is x.
then first number = (100-30)% of x
= 70% of x = 7x/10
Second number is (63x/100)
Difference = 7x/10 - 63x/100 = 7x/10
So required percentage is, difference is what percent of first number
=> (7x/100 * 10/7x * 100 )% = 10%

2.    (4) : Failed in mathematics, n(A) = 34
Failed in English, n(B) = 42
n(AUB) = n(A) + n (B) – n (A∩ B )
= 34 + 42- 20 = 56
Failed in either or both subjects are 56
Percentage passed = (100−56)%=44%

3.    (1) : Reduction in consumption = [((R/(100+R))*100]%
? [(25/125)*100]%=20%.

4.    (1): Value of the machine 2 years ago
= Rs.[162000/(1-(10/100)2)]=Rs.[162000*(10/9)*(10/9)]= Rs.200000

5.    (3) : Let the number be x
Then, error =  (5/3)x – (3/5)x =(16/15)x
Error% = [(16x/15) /(5x/3)] * 100% = 64%


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weh jo said...

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jo said...

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thank u..

a learner said...

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nayan said...


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Sameer said...

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Vipin Dhoundiyal said...


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