6. Percentages Introduction: The Term “percent” Means Per Hundred Or

6. PERCENTAGES INTRODUCTION: The term “percent” means per hundred or for every hundred. H is the abbreviation of the latin phrase percentum. The term percent is sometimes abbreviated as P.C. the symbol % is often used for the term percent. SOME BASIC FORMULAE 1) To convert a fraction into a percent: To convert any fraction l/m into rate percent, multiply it by 100 and put % sign i.e. l/mx100% Ex: 6.2 can be expressed in terms of percentage as Sol: 6.2 × 100 =620% 2) To convert a percent into a fraction: To convert a percent into a fraction, drop the percent sign and divide the number by 100. 3)To find a percentage of a given number X% of given number(N) = x/100 xN Ex: What is 15 percentage of Rs34. Sol:

15 × 34 = Rs 5.10 100

1) Fractional equivalents of Important percents:

1 100 1 5%= 20 1%=

1 16 1 8 13 % = 12 6 14 % =

1 50 1 10%= 10

1 25 1 20%= 5

2% =

2 25 2 40%= 5

4%=

1 8 1 16 23 % = 6 12 12 % =

25%=

1 4

33 13 % =

16 25 3 60%= 5

8%=

16%=

3 8 2 66 23 % = 3

37 12 % = 1 3

50%=

1 2

83 13 % =

5 6

16 25 2 80%= 4 5 5 7 87 12 % = 8 4 133 13 % = 3 64%=

24 25 6 120%= 5 96%=

100%=1



 x × 100 %  (100 + x ) 

5) a) if A is x% more than that of B, then B is less than A by 

Ex: If A’s salary is 25% more that B, then how much percentage is B’s salary less that of A ? Sol: B’s salary is less by =

100 × 100 %=20% 100 + 25 

 x × 100  (100 − x ) 

b) If A is x% less than that of B, then B is more than that of A by 

Ex: If A’s salary is 30% less that B, then how much percentage is B’s salary more that of A ? Sol: B’s salary is less by =

100 × 100 = 42 67 % 100 − 30

6) If A is x% of C and B is y% of C, then A is

x × 100 % of B. y

Ex: Two number are respectively 20% and 25% of third number . What percentage is the first of the Second ? Sol: first is of second by

20 × 100 =80% 25

7)a) If two numbers are respectively x% and y% more than a third number, then the first number is

100 + x  100 + y × 100 % of the second and the second is  

100 + y  100 + x × 100 % of the first.

Ex: A Ram got 20% marks and Rahul got 50% marks more that Raj then by what percentage is Ram is of the Rahul and Rahul is of what percentage of Ram?

100 + 20 × 100 =80% of Rahul and 100 + 50 100 + 50 × 100 =125% of Ram. Rahul is 100 + 20

Sol: Ram is

b) If two numbers are respectively x% and y% less than a third number, then the first number is

100 − x  100 − y  100 − y × 100 % of the second and the second is 100 − x × 100 % of the first.    

Ex: Two worker get 30% and 40% less than the third person, by what percentage is first person less by second and second is less of first by what percentage?

100 − 40 × 100 = 85 75 % less by second and 100 − 30 100 − 30 40 × 100 = 11 60 second person is less of first by what percentage 100 − 40

Sol: first person is

8) x% of a Quantity is taken by the first y% of the remaining is taken by the second and z% of the remaining is taken by third person now if A is left in the quantity, then there was

A × 100 × 100 × 100 (100 − x )(100 − y )(100 − z ) the

beginning. Ex: From a tank 10% of wine is taken out, 20% of remaining wine is taken out next time volume of wine remaining is 144liters. What was the amount of wine in the beginning. Sol: Amount of wine in the starting was

144 × 100 × 100 (100 − 10)(100 − 20) =200liters

9) x% of a Quantity is added, again, y% of the increased Quantity is added, again z% of the increased Quantity is added, now it becomes A, then the initial amount is given by

A × 100 × 100 × 100 (100 + x )(100 + y )(100 + z )

Ex: A man deposited 20% of the amount in his locker in the first year. In the second year, he deposited 25% of increased amount in his locker, he raise it to Rs4880. What was initial amount deposited by the man. Sol: The man has Rs4800 at the end of second year He deposit’s

4800 × 100 × 100 (100 + 20)(100 + 25) =Rs 3200 initially in his bank.

10) Population formula: 1) If the present population of a town (or value of an item) be P and the population (or value of item) changes at r% per annum then

r   a) Population (or value of item) n years after = P1 +   100  P b) Population (or value of item) n years ago =

r   1 −   100 

n

n

Ex: If the annual increase in the population of a 4% and the present number of people is 15,625, what will the population in 3years and 3 years ago?

 

Sol: The population after 3 year will be 15625 1 +

3

26 25 25 4  =17576  = 15625 × × × 25 26 26 100 

15625

25 25 25 3 =17660 4  = 15625 × × ×  24 24 24 1 −   100 

The population before 3 year is

where r is +ve or –ve according as the population (or values of item) increases or decreases. 2) When the rate of growth different for different yrs: a) The population of a town [or value of an item] is P. If increases by x% during the first year, increases by y% during the second year and again increases by Z% during the third year. Population(or value of item) after 3 yrs will be

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

Ex: The population of the town is 8000. It increased by 10% during the first year and 20% during second year. What is the population at the end of second year? Sol: Population after 2years would be

8000(100 + 10 )(100 + 20 ) =10,560 100 × 100

b) The population of a town (or value of an item) is P, if increases x% during the first year, decreases by y% during the second year, and again decreases by z% during the third year the population (or value of item) after 3 yrs.

P(100 + x )(100 − y )(100 − z ) 100 × 100 × 100

Ex: The population of a town is 10,000. It is increased by 10% during the first year. During the second year, it decrease by 20% and increase by 30% during third year. What is the population ? Sol: The required population is

10000(100 + 10 )(100 − 20 )(100 + 30) =11440 100 × 100 × 100

11) Reduction in consumption : expenditure = consumption X rate of item for keeping the expenditure fixed, it is essential that if the rate of the item increases, the consumption will be reduced or if the rate of the item decreases, then consumption will increase. 1) If price of a commodity increases by r%, then the reduction in consumption so as not to increase the

expenditure, is

 100  100 + r × 100 %

Ex: If the price petrol is increased by 30%, by how much percentage a car owner must reduce his consumption in order to maintain the same budget? Sol: In order to maintain same budget the consumer have to reduce his consumption by

100 × 100 = 23 131 % 100 + 30 2) If the price of a commodity decreases by r% then increase in consumption, so as not to decrease expenditure on this item is

 100  100 − r × 100 %

Ex: The price of wheat falls by 16%. By what percentage a person can increase, the consumption of wheat so that his overall budget doesn’t change? Sol: Consumption of wheat = 12) x as a percentage of y =

100 × 100 =19% 100 − 16

x × 100 % y

Ex: What rate per cent is 6P of Re1?

Solo:

6P 6 = × 100 =6% Re 1 100

13) If the value of a number is first increased by x% and later decreased by x% the net change is always decrease which is equal to x% of x or

x2 100

Ex: A shop keeper marks the price of his goods 12% higher than its orginal price. After that, he allows a discount of 12%. What is his percentage profit are loss ? In this case, there is always a loss. And the % value of loss is

(12) 2 100

=1.44% loss

14) If a number is changed(increased/decreased) successively by x% and y% then net% change is given by

xy    x + y + 100  % which represents increase or decrease in value according as the sign is +ve or –ve.

If x or y indicates decrease in percentage then put –ve sign before x or y otherwise +ve sign. Ex: The salary of a person is increased by 10% and then decreased by 20%. What is the net change in the salary? Sol: increased % is taken as +10% and decreased % as -20% from above formula

10 − 20 +

10 × −20 = -12% , -ve sine indicate decrease in salary 100

15) If two parameters A and B are multiplied to get a product and if A is changed (increased/decreased) by x% and another parameter B is changed(increased/decreased) by y% then the net% change in the product(AxB) is given by

xy    x + y + 100  which represents increase or decrease in value according as the sign is

+ve or –ve, if x or y indicates decrease in percentage, then put –ve sign before x or y , otherwise +ve sign. 16) Effect on revenue : (i) If the price of commodity is diminished by x% and its consumption is increased by y%

(ii) Or if the price of a commodity is increased by x% and consumption is decreased by y% then the effect m revenue = inc% value-Dec% value – Inc% valuexDec% of value/100. Ex: The number of seats in the cinema hall is increased by 25% and price on a ticket is also increased by 10%. What is the effect on the revenue collected? Sol: Since there is an increase in the seats as well as in the price, we use Decrease= -(increase) Thus formula becomes %effect=25-(-10) −

25 × ( − 10) =35+2.5=37.5 100

Thus there is 37.5% increase in the revenue.

17) The pass marks in an examination is x%. If a candidate who secures y marks fails by Z marks then the maximum marks. M=

100( y + z ) x

Ex: Passing marks in an examination is 55%. If a candidate who secures 100 marks fails by 10 marks then the maximum marks. Sol: Maximum of the exam is

100(100 + 10 ) = 200 marks 55

18) A candidate scoring x% in an examination fails by ‘a’ marks, while another candidate who scores y% marks get ‘b’ marks more than the minimum marks then the maximum marks for that examination are M=

100( a + b )   y−x   

Ex: A candidate scores 25% and fails by 30 marks, while another candidate who scores 50% marks, more than minimum required marks to pass the examination. Find the maximum marks for the examination. Sol: Maximum marks=

100( 30 + 20 ) =200 50 − 25

19) In measuring the sides of a rectangle or side is taken x% is excess and the other y% in deficit. The error percent in area calculated form

xy    x − y − 100  the measurement is in excess or deficit according to the +ver or

–ve sign. Ex: If the length of a rectangle is decresed by 40% and the breadth is increased by 30%, then find the % change in the area of the rectangle. Sol: Here, both the length and breadth is changed. Where length=-40, breadth=+30 Net change % in areas of rectangle is -40+30 − 22%

30 × 40 =-22, -ve sine signifies decrese in area by 100

20) If the sides of a triangle, rectangle, square, circle, rhombus (or any 2 dimensional figure) are increased by x%. Its area is increased by

x2  x( x + 200 )  % or 2 x +  100  100

Ex: The radius of the circle is so increased that its circumference increased by 5%. The area of the circle is increased by? Sol: Here x=5, from above formula We get

2×5 +

52 =10.25% 100

21) If the radices of a circle increased or decreases by x% circumference in order x%. 22) In an examination, x% failed in English and y% is failed in maths. If z% of students failed in both the subjects the percentage of students who passed in both the subjects is 100-(x+y-z). Ex: In an examination, 40% of the students failed in Maths, 30% failed in English and 10% in both. Find the percentage of student who passed in the both the subjects. Sol: The required %of student passed in both subject = 100-(40+30-10) = 40%

 x+ y  × 100 or 23) P1% of a number is x and p2% of a number is y then the number is   p1 + p 2   x− y  × 100   p1 − p 2  Ex: