Percentage- theory and rules

प्रतिशत

प्रतिशत का अर्थ होता है  'प्रत्येक सौ '
"प्रतिशत  एक  भिन्न  होता है जिसका हर 100 है और भिन्न के अंश को रेट प्रतिशत कहा जाता है"| प्रतिशत को  % प्रतीक द्वारा लिखा जाता है| 

प्रतिशत की गणना करने का कांसेप्ट 

यदि हमें x का y% ज्ञात करना है तो, 
x का y% =(x*y)/100

प्रतिशत का भिन्न में परिवर्तन 

• प्रतिशत को दशमलव या साधारण भिन्न में बदलने के लिए उस भिन्न में 100 से  भाग देते है

Expression per cent (x%) into fraction. 
Required fraction=x/100

भिन्न का प्रतिशत में परिवर्तन 

• साधारण भिन्न या दशमलव भिन्न को प्रतिशत भिन्न में बदलने के लिए उस भिन्न में 100 से गुणा करते है

Expressing a fraction (x/y) in per cent.
Required percentage=(x/y)*100)%

अन्य के सम्बन्ध में एक मात्रा की प्रतिशत के रूप में अभिव्यक्ति 

To express a quantity as a per cent with respect to other quantity following formula is used.
(The quantity to be expressed in per cent)/ (2nd quantity (in respect of which the per cent has to be obtained))X100%


महत्वपूर्ण कांसेप्ट और ट्रिक  

1. If x% of A is equal to y% of B, then
z% of A=(yz/x)% of B

2. When a number x is increased or decreased by y%, then the new number will be 
(100+y)*x/100


3. When the value of an object is first changed (increased ) by a% and then changed (increased ) by b%,then
Net effect=[a+b +ab/100]%

4. Suppose in an examination, x% of total number of students failed in subject A and y% of total number of students failed in subject B and z% failed
in both the subject. Then, 
(i) Percentage of students who passed in both the subjects=[100-(x+y-z)]%
(ii) Percentage of students who failed in either subject=(x+y-z)%

5. If due to r% decrease in the price of an item, a person can buy A kg more in Rs.x, then
Actual price of that item= Rs (rx)/((100-r)A) Per kg

Example :If due to 10% decrease in the price of sugar ,Ram can buy 5 kg more sugar in Rs 100 , then find the actual Price of sugar ?

solution : Here r = 10 % ,x = 100 and A = 5 kg 
Actual price of sugar = 10*100/((100-10 )*5) =  Rs. 2(2/9)

6.If the population of a town is P and it increases at the rate of R% per annum, then
(i) Population after n yr= P(1+r/100)^n
(ii) Population, n yr ago=P/(1+r/100)^n


7. If the present population of a city is P and there is a increment  of R1%, R2% ,R3% in first, second and third year respectively, then

Population of city after 3 yr=P(1+R1/100)(1+R2/100)(1+R3/100)

Example : Population of a city in 2010 was 1000000. If in 2011 there is an increment of 15 % , in 2012 there is a decrements of 35 % and in 2013 there is an increment of 45 %, then find the population of city at the end of the year 2013.

solution : Required population = P (1 + R1/100)(1 - R2/100)(1 + R3/100)

= P (1 + 15/100)(1 - 35/100)(1 + 45/100)
= 1083875