# Competitive Exam Notes for Percentage (प्रतिशत)

*प्रतिशत का अर्थ होता है ‘प्रत्येक सो ‘*

*प्रतिशत का अर्थ होता है ‘प्रत्येक सो ‘*

*“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 **

**Expression per cent (x%) into fraction. **

**Required fraction=x/100**

**भिन्न का प्रतिशत में परिवर्तन **

**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 suject. 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% increase 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 20004 was 1000000. If in 2005 there is an increment of 15 % , in 2006 there is a decrement of 35 % and in 2007 there is an increment of 45 %, then find the population of city at the end of the year 2007.**

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

**P (1 + 15/100)**

**(1 – 35/100)**

**(1 + 45/100)**

**= 1083875**