View More


MATHS: CHAPTER 8 : PERCENTAGE

Introduction

There is a saying from all Quant faculties that you can clear quant section of any competitive examination which also includes data interpretation, if you have good command over two topics, which are
1. Percentage
2. Ratios and Proportion
Concept of percentage is used in all others chapters as well for e.g. in Profit and Loss, Simple Interest and Compound Interest, Mixture and Alligation, Data Interpretation etc. and each and every government exam consist at least 2-3 questions from this section.

What is the percentage?
Percentage is generally denoted by the sign ‘%’
Percentage is generally a number or ratio expressed as a fraction of 100
                                Per       cent
                                 Per       100

For e.g. If your results shows you have scored 85% overall. Then it means for every 100 marks you have scored 85 marks.
Let’s understand the concept and importance of percentage by one example 

e.g.1. There are the two students A and B. A has scored 150 marks out of 200 in exam ad B has scored 200 marks from 250 now find whose performance is better?
Solution:
In these kind of sums when common ground is not same 
It is difficult to compare their performances.
Here percentage comes into the picture.
Marks scored by A is 150 out of 200
Marks scored by B is 200 out of 250
Here we will find for every ‘100’ marks how much both of them have scored.
For A                       150 from 200
                                ?   From 100
                                ? = 150/200× 100
                                   = 75
Similarly for B          200 from 250 
                                ?   from 100
                                ? = 200/250×100
                                  = 80
    A’s marks per cent (per 100) = 75
                    A’s percentage = 75%        
    B’s marks per cent (per 100) = 80
                    B’s percentage = 80%        
            B% > A%    hence performance of B is better

e.g.2. In class of 60 students 30% students passed in mathematics and 70% students passed in English find the number of who passed in English and mathematics.
Solution:
Total number of students = 60
Students who passed in mathematics = 30% of total students 
                                                      = 30/100×60 
                                                      = 9
Students who passed in mathematics = 70% of total students 
                                                       = 70/100×60         
                                                       = 42
NOTE:
           30% means 30 from every 100 -> 30     100
                                                              ?        6
                                                              ?  = 30/100×60 
            X % of Y can be written as = X/100 × Y

 

 TYPES OF PROBLEMS IN PERCENTAGE

  1. Conversion from fraction value to percentage and vice a versa

  2. X is what % of Y or X is what % more/less than Y

  3. Application based problems

Type – 1 Conversion from fraction value to percentage and vice a versa

Any percentage can be written in the form of fraction (a/b) for e.g.
50% = 50/100 = 1/2
Similarly 25% of A = 25/100 of A = 1/4×A = A/4
By converting percentage into fraction value it makes calculation easier and time saving in quant section as well as in DI section. Similarly we can convert fraction value to percentage for e.g. to convert 2/5 to percentage multiply the fraction by 100.
2/5 = 2/5×100 = 40%,  3/7 = 3/7×100 = 42%

Solved Problems
Q.1. Find the fraction value of the following percentages
   A. 10%                          F. 83 1/3 %
   B. 33 1/3 %                   G. 57 1/7 %
   C. 75%                          H. 37 1/2 %       
   D. 16 2/3 %                   I. 2 %
   E. 11 1/9 %                   J. 15%

Solutions:

   A. 10% 
       10% = 10/100 = 1/10   
   B. 33 1/3 %
       (33 1/3)/100    = (100/3) x 1/100  =  1/3  
   C. 75%
        75/100 =  3/4
   D. 16 2/3 %
        16 2/3 % = (16 2/3)/100  = (50/3)/100 = 50/(100×3) = 1/(2×3) = 1/6
   E. 11 1/9 %
        11 1/9 % = (100/9) x 1/100 = 1/9
   F. 83 1/3 %
       83 1/3 % = ((83×3)+1)/(3×100) = 250/300 = 5/6
   G. 57  1/7 %
        57  1/7 % = ((57×7)+1)/(7×100) = 400/700 = 4/7
   H. 37 1/2 %       
        37 1/2 % = (37×2+1)/(2×100) = 75/200 = 3/8     
   I.  2%
       2% = 2/100   = 1/50
   J. 15%
       15% = 15/100 = 3/20

Q.2. Convert the following fractions to percentages 
  1. 2/5              2. 1/7            3. 3/8              4. 11/13
Solutions:
   1. 2/5
       2/5×100 = 2 x 20 = 40%
   2. 1/7            
       1/7× 100 = 14.28 OR 14 2/7 %                      
   3. 3/8          
       3/8×100 = 37.5 OR 37 1/2 %                               
   4. 11/13
       11/13×100 = 84.61%

Q.3. In class of 60 students (2/5)th of total students gave the exam of mathematics and (1/5) th of total students gave the exam of English, overall how much per cent students gave either mathematics or English exam?
Solution:
Students who gave mathematics exam = 2/5×60
                                                         = 24
Students who gave English exam = 1/5×60
                                                = 12
Total students who gave exam = 12+24
                                             = 36
Per cent of students who gave either mathematics or English exam = 36/60× 100
                                                                                                 = 60%

PERCENTAGE -> FRACTION VALUE TABLE


Per cent to fraction                             Fraction to per cent
1% =1/100           2% = 1/50                 1/2 = 50%
3% = 3/100          4% = 1/25                 1/3 =  331/3%
5% = 5/100          61/4% = 1/16            1/4 = 25%
81/3% = 1/12       10% = 1/10               1/5 = 20%
121/2% = 1/8       20% = 1/5                 1/6 = 162/3%
25% = 1/5            30% = 1/3                 1/7 = 142/7%
331/3% = 1/3       40% = 2/5                 1/8 = 121/2%
50% = 1/2            60% = 3/5                 1/9 = 111/9%
66  2/3 % = 2/3    70% = 7/10               1/10 = 10%
75% = 3/4            80% = 4/5                 1/11 = 11/11%
90% = 9/10          100% = 1                  1/12 = 81/3%
150% = 3/2          200% = 2    
                                                              1/13 = 71/13%
                                                              1/14 = 71/7%
                                                              1/15 = 62/3%
                                                              1/16 = 61/4%
                                                              1/20 = 5%
                                                              1/25 = 4%
                                                              1/30 = 31/3%
                                                              1/40 = 21/2%

   
Type – 2 X is what % of Y or X is what % more/less than Y
                           
In this category of sums we will be given two numbers and we will have to find one number is what per cent of another number or how much per cent more or less than another number.
(1) X is what per cent of Y.
e.g.(i) find 5 is what percent of 20
Solution:
            = 5/20×100
            = 20%
Hence 5 is 20% of the number 20
Now in the above sum what if 20 is what percent of 5 had been asked then,

e.g.(ii) 20 is what per cent of 5
Solution:
             = 20/5×100
             = 400%
So, it is clear that 20 is 400% of 5

(2) X is what percent more/less than Y
e.g. (i) Find 25 is how much percent more than 20?
Solution:
          = ((20-25)/20)×100
          = (5/20)×100
          = 25%
Hence 25 is 25% more than 20.

e.g.(ii) in above sum in two numbers 20 and 25, 20 is how much per cent less than 25?
Solution:
            = ((25-20)/25)×100
            = (5/25)×100
            = 20
Hence 20 is 20% less than 25.
Above four examples clearly explains the concept of percentage now we will solve some sums to understand if in detail.

Q.1. Find weather X% of Y and Y% of X are similar or not, if not then which one is greater?
Solution:
X% of Y = (X/100) x Y = XY/100                   …(1)
Y% of X = (Y/100) x X = YX/100 = XY/100         …(2)
From (1) and (2) X% of Y and Y% of X are always similar.

Q.2. Find 20% of 30% of 10% of 300
Solution:
20% of 30% of 10% of 300 = 20/100×30/100×10/100×300
                                         = 180 Ans.
Q.3. If Suhag earns 20% more than Vishal then what % less does Vishal earn than Suhag?
Solution:
Let the earning of Vishal be ‘X’
Now, earning of Suhag = X+20% of X
                                  = X+  20/100 X
                                  = X+0.2X
                                  = 1.2X
Per cent less earned by Vishal and Suhag = (1.2X/1.2X)×100
                                                            = (0.2X/1.2X)×100
                                                            = 1/6×100
                                                            = 100/6
                                                            = 16 2/3%
Shortcut Method
20% = 20/100 = 1/5
Now, by ratio method
1/5 = 1+5 = Suhag
Vishal = 5
Answer = ((Suhag-vishal)/suhag) × 100
             = ((6-5)/6) × 100
             = 100/6  = 162/3%

Q.4. If 20 per cent of a number is less than (2/5)th of the number by 60 then find the number be X
Solution:
Let the number be X
20% of number = 20/100X = X/5
And (2/5)th of number = 2/5× X = 2X/5
Now as per the condition,
X/5 = 2X/5 – 60
60 = (2X-X)/5
X = 300
Required number = 300

Q.5. The sides of rectangle are 12cm and 15cm, 12cm being length and 15cm being breadth if length is increased by 12% and breadth is decreased by 10% then find percentage of change in area?
Solution:
L = 12cm
B = 15cm
Now L1 = 1.2 × 12 = 14.4
B1 = 0.9 × 15 = 13.5
Now A = L × B
          = 180
And new area A1 = 194.4
Percentage change in area = ((194.4-180)/5) × 100
                                       = 14.4/180×100
% change = 8%

Q.6. Saurav spends 80% of his income. His income gets increased by 25% and he increases his expenditure by 5% his Savings are increased by what percentage?
Solution:
Here no exact value is asked. So we can use ‘method 100’
Let his income = 100
Expenditure = 80
Saving = income – expenditure
           = 100 – 80
           = 20
Now
New income = 125
New expenditure = 80 + 5% of 80                      new savings = 125 – 84
                                = 80 + 4                                             = 41
                                = 84
% increase in savings = ((41-20)/20) × 100
                                = 21/20 × 100
                                = 21 × 5
                                =105%
Hence his savings got increased by 105%

Q.7. In an examination, 30% of students failed in English and 40% students failed in history if 20% students failed in both the subjects then what is the percentage of students who passed in both the subjects?
Solution:
As no exact value is asked
Let the total number of students = 100
Students failed in English = 30
Students failed in history = 40
Students failed in both subjects = 20
Total no. of students passed in = 30 + 40 – 20
History and English = 70 – 20
                            = 50
Students passed in both the subjects = 100 – 50
                                                      = 50
% of students passed in both subjects = 50/100×100
                                                        =50%

Type – 3 Application based problems 

            
(1) If population of village is P and it increases every year at the rate of R% then it’s population   after ‘n’ years and before ‘n’ years can be calculated by following formulas.
      Population after ‘n’ years = P (1 + R/100)n
      Population before ‘n’ years = P/((1 + R/100)n)
(2) Depreciation of machine:
      If the present value of machine is ‘m’ and it depreciation at R % per annum then the value of machine after ‘X’ years and value of machine ‘n’ years ago can be calculated by,
      Value of machine ‘n’ years after = M (1 – R/100)n
      Value of machine ‘n’ years ago = M/((1- R/100)n)
(3) If the price of a commodity is increased by R% then the reduction in consumption so as not to increase expenditure can be calculated by the formula:-
     Reduction in consumption = R/((100+R))×100%
     While in other case if the price of the commodity decrease by R% then the increase in consumption so as not to decrease expenditure is,
     Increase in consumption = (R/(100-R))×100%

Posted By:

Hardik Kaneriya

COMMENTS

View 0 more comments

The great mentor Leaderboard

Copyright @ 2018 | Star Universal Mentors LLP | All Rights Reserved.