Percentage with solved examples using shortcut methods

Percentage

The percentage is a fraction whose denominator is always 100. X percentage is represented by x%.

i. To express x% as a Fraction:

We know that x% = x/100

ii.To express x/y as a percentage:

x/y = x/y × 100

iii.  To find the percentage of a given number or to find how much percentage is of another.

% of given number n = x/100 × n

 % = the quantity to be expressed in percent / 2^nd quantity * 100

vi.

v. Comparison between two values x and y.

a.       If x is compared to y then we always assume that y is always equal to 100%

b.       When any question asks y is what percentage of x then x is always written in the denominator.

       

y/x × 100

vi. When a number x is increased or decreased by y% then the new number is will be

For increase = (x+y)/100 × x

For decrease = (x-y)/ 100 × x

vii. If A is R% more than B, then B is less than A by

R/(100+R) × 10

viii. If A is R% less than B, Then B is more than A by

R/(100-R) × 100

Ex. If the income of A is 40% less than that of B, then how much percent B’s income more than that of A?Sol. 40/60 × 100 = 66.66%

viii. If the price of the commodity increases by R% then % reduction in the consumption so as not to increase the expenditure is

R/(100+R) × 100

If the price of a commodity decreases by R% then % increase in consumption so as not to decrease the expenditure is

R/(100-R) × 100

ix.  Results in population:

For population increase R% per annum.

1.       Population after n years = P (1+ R/100) ⁿ

2.       Population n years ago = P/ (1+ R/100) ⁿ

For population decrease at the rate of R% per annum,

1.       Population after n years = P(1-R/100) ⁿ

2.       Population n years ago = P(1-R/100) ⁿ

If the Population is increased by x%, y%, z%. successively then the final value will be

P(1+x/100)(1+y/100)(1+z/100)

x. If the value of an object is first increased or decreased by x % then increased and decreased by y% Then

Net percentage change

 ± x± y ± x y/100

xi. If the value of an object is first increased by a% then later decreased by a%, then the net effect is always decreased which is equal to

(a/10)²

xii. The passing marks in an examination is p% if a candidate score R marks and fails by f marks, then the maximum marks

M = 100(R+F)/P

A candidate score x% marks in an examination and fails by marks, while another candidate who scores y% marks gets b marks more than the minimum required marks, then the maximum marks for the examination are given as

M = 100(a+b)/ (y-x)

Or 100(sum of marks )/ difference in marks percentage

Solved examples based on Percentage:

Ex.1 Find the value of  3600 if it is increased by 16 ⅔ %.

Sol. Equivalent fraction of  16 ⅔ % is 1/6 

To increase the value by 1/6 

we add denominator with numerator i.e

  (6+1)/6 =  7/6 

and then multiply 

3600* 7/6 = 4200 Ans.

Ex.2 Find the value of 1400 if it is decreased by 14 ⅔ % ?.

Sol. The equivalent fraction of 14 ⅔ % is 1/7

To decrease the value we subtract the denominator with the numerator

i.e (7-1)/7 = 6/7

And then multiply

1400*6/7 = 1200  Ans

Ex.3 If the income of A is 20,000 Rs. It increases by 20% and then decreases by 40%. Find income.

Sol. If income is increased by 20% then it becomes 120%. 

and then decrease by 40% then it becomes 60%

Total income is 

20,000 * 120/100 * 60/100

14400 Ans

Ex.4 The value of an article decreases 10% annually if it was purchased 3 years ago and its present value is 5832 Rs. then find its initial cost.

Sol.   A = P(1-R/100) ⁿ

Let value of article 3 years ago is x

x * (1 - 10/100)³

x * 90/100 * 90/100 * 90/100 = 5832

x = (5832*1000)/(9*9*9)

x = 8000 Rs. Ans

Ex.5 If the length of a rectangle is increased by 20% and breadth is decreased by 10% then find the percentage change in its area.

Sol. The area of the rectangle is length * breath

Method 1.

let the length and breadth of the rectangle are 10.

now increase 10 by 20% and decrease 10 by 20%

10 * 10 = 100

12 * 8 = 108 

108 is 8% of 100 which is the answer

Method 2.

Use the formula of net effect.

+20 - 10 + (20 * -10)/100

10-2 = 8% Ans

Ex.6 If the radius of a circle is increased by 20% then find the % change in diameter circumference and area.

Sol. Now radius r is increased by 20%.

diameter = 2r        Circumference = 2πr        Area = πr²

 Radius r remains constant in diameter and circumference.

thus diameter and circumference is increased by 20%

The area is increased by 44%. By using the formula of net effect. Ans

Ex.7 If the length of the rectangle is increased by 25% then what percent of breath will be reduced for no change in the area.

Sol. Using formula of net effect

0 = 25- y + (25y)/100

y = 20% Ans

Ex.8 If the income of A is 10% more than B then how much percent B's income is less than A.

Sol. Let B = 100, Then A = 110

[(110 - 100)* 100]/110 

(10/110) * 100 = 100/11  Ans

Ex.9 In an exam 35% is the passing mark a student got 135 marks and failed by 40 marks then find the max marks.

Sol. passing marks = passing percentage

135 + 40 = 35%

175 = 35/100

Max marks = (35/100 )*175

 = 500 Ans.

Method 2.

By using formula M = 100(R+F)/P

M = [100(135+40)]/ 35 = 500 Ans

Ex.10 If a student score 25% marks he is failed by 210 marks but if he scores 55% then he scores 240 marks more than the passing marks then finds the maximum and passing marks.

Sol. Passing marks = passing marks

25% + 210 = 55% - 240

30% = 450

1 = 450 * 100/30 = 1500

maximum marks = 1500

Passing marks 25*1500/100 = 375+210 = 585

(585/1500) *100 = 39% Ans

Ex.11 Due to a reduction of 20% in the price of Apples enables a person to purchase 16 more for 320 rs. To find the percent and old price of 1 Apple.

Sol. Method 1.

Now 20% of 320 is

(20/100 )*320 = 64 Rs.

This means 16 Apple for 64 Rs.

16 A = 64 Rs

1 A = 4 Rs.

80% = 4 Rs.

100% = 4 * (100/8) =5 Rs. Ans