## Percentage fast track arithmetic formulae for all competitive examination.

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%

### Suchit prajapati

Suchit Prajapati is the Founder and CEO of Laws Of Nature. He is also an Entrepreneur, motivational speaker, spiritual thinker, affiliate marketer and physics enthusiast.