In “laws of nature” today we are going to talk about the some important arithmetic formulae on PERCENTAGE, which may be very helpful to you , in all types of competitive examination.

## Some important sutras

* If two values are respectively x% and y% more than the third value,then the first is the = (100+x/100+y)×100%

of the second.

* If the two values are respectively x% and y% more than the third value,then second is the=(100+y/100+x)×100% of the first.

* If the two values are respectively x% and y% less than the third value,then the second is the= (100-y/100-x)×100% of the first.

* If the two values are respectively x% and y% less than the third value,then the first is the=(100-x/100-y)×100% of the second.

* If A is x% of C and B is y% of C then A is =x/y×100% of B

* If x% of quantity is taken by the first,y% of remaining is taken by the second and z% of remaining is taken by the third person,now if ₹A is left in the fund, then the fund was in the beginning=

A×100×100×100/(100-x)(100-y)(100-z)

* If A is the initial amount in the fund and x% is taken by the first ,y% is taken by the second and z% is taken by the third person then amount left in the fund is =

A×(100-x)(100-y)(100-z)/100×100×100

* If intial amount is A, then x% of initial amount is added to initial amount,then y% of increased amount is added to the amount,and then z% of increased amount added then initial amount become =

A×(100+x)(100+y)(100+z)/100×100×100

* If the original population of a town is P and annual increase is r% then the population after n years will be=

P(1+r/100)^n

* If the present population of a town is P,and annual increase is r% then the population was n years ago will be=P/(1+r/100)^n

* If the original population of a town is P, and annual decrease is r% then population after n years will be=P(1-r/100)^n

* If the annual decrease in the population of a town is r%,and present population is P,then population of the town in n years ago will be=P/(1-r/100)^n

* The population of a city is P,it increases/decrease by x% in first year,y% in second year and z% in third years.then the population after the three years will be=

P×(100+-x)(100+-y)(100+-z)/100×100×100

* If the population of a town is increase/decrease by x% in first year,y% in second year and z% in third year,and after the three years the population of town is to be noted P then the population of town in the beginning is =

P×100×100×100/(100+-x)(100+-y)(100+-z)

* If the population of town is P1,and in the town male increased by x% and females increased by y% then the population become P2, then the no. Of males and females in the town is given as below=

[P2×100-P1(100+y)]/(x-y) is the no of males

[P2×100-P1(100+x)]/(y-x) is the no of females

* If the value is decreased/decreased successively x% and y% then the net decrease is as follows=[+-x+-y-(+-x)(+-y)/100]

* If one number is decreased/increased by x% and second number is decreased/increased by y% then the effect on the product is =[x+y-+(xy/100)]%

* The passing marks in a examination is x% and a student secures y marks and he fail by z marks,then the maximum marks of examination is =100(y+z)/x

* A student scores x% marks,and fails by a marks,another student who has scored y% marks and he get b marks more than the minimum required marks to be pass, then max marks of the examination is =

100(a+b)/y-x

* In a examination x% fails in physics, y% failed in cosmology and z% students failed in both the subjects, then % of students who has passed the both exams is =100-(x+y-z)

* A man spends x% of his income, his income is increased by y% and his expenditure also increased by z% then the % increased in his savings is =

[100y-xz/100-x]%

* If x% of objects is one type,the remaining y% is of second type, and remaining z% is third type, and the value of remaining objects is A then the total no of objects is =

A[100/(100-x)][100/(100-y)][100/(100-z)]

* The producer of a goods makes a profit/loss of x% , whole seller makes the profit of y%, and retailer makes the profit of z%, if retailer sold it for RsA, then the producing cost of the goods is =

A[100/(100+-x)][100/(100+-y)][100/(100+-z)]

* In L litres of x% sulphuric acid solution,the amount of water to be added/removed to make the y% of acidic solution is =+-L(x-y)/y

Note: here x% is always greater than y% if H2O is added,if H2O is removed then y is greater.

* Certain amount of solvent B is added to the solution of A and B of amount M to change the % amount of A to ∆A ,then amount of B to be added=[(A/∆A)×M]-M

* If the original value of a trust is A and new value is trust is B, then increase or decrease in the consumption such that the expenditure is unaffected.(B-A/B)×100%

* Splitting of a number N into two parts such that one part is x% of other, then the splitted two parts is=[100/(100+x)]×100 and

Nx/(100+x)

* If L litre of water is poured into a tank,but it is still x% empty then amount of water should be added to fill it up to brim= L×x/(100-x) and capacity of tank is=100L/(100-x)

* If x%, y% and z% is successively three discount are given then a net single discount is=[x+y+z-(xy+yz+zx)/100+xyz/100]

*If x% houses contains two or more people and those houses which contains only one male is y% ,then % of all houses which contains exactly one female with no male is = [(100-x)(100-y)/100]

* If the monthly income of A is x% more than that of B, monthly income of B is y% less than that of C, if difference between the monthly income of A and C is M then the monthly income of B and C is =

[100M(100-y)/(100+x)(100-y)-100^2] and

[100^2M/(100+x)(100-y)-100^2] respectively

* Mass of two boys A and B is in the ratio of a:b ,if mass of A is increased by x% then the total mass become M, if mass of B is increased by y% then the total mass become=[(100+y)/100(1+a/b)-{a/b(100+x)/100+1}]×100%

*If a person spend x% of his monthly income on dance bar,and y% of remaining on worship, then after he saves Rs A, then the monthly income of the person is =

A×100^2/(100-x)(100-y)

Monthly amount spent on dance bar is =Ax×100/(100-x)(100-y)

Monthly amount spent on worship is=

Ay/(100-y)

* If r% is decreased in the value of a objects then a person buy A kg more objects in Rs x then the cost of the object is = rx/(100-r)A

* In the election of two candidates, one gets x% votes of total votes but lose by y votes, then the total no of votes is = 100y/(100-2x)

* In a examination between A boyes and B girls, r1% boyes pass and r2% girls get passed, then % of total passed students is =

(Ar1+Br2)/A+B

* If the value of any object is increased by x% and then decreased by x%, then resultant is always get decreased then resultant is given as= x^2/100

* Convert x% in fraction = x/100

* Convert x% in decimal= 0.0x

* Convert X/y in percentage = (x/y)×100%

* If x% of A is equal to the y% of B, then z% of A is = (yz/x)×100%