Percentage-part 8

Milk and water concept

Problem 1:A mixture of 40 lit of milk and water contains 10% water. How much water should be added so water may 20% in the mixture

Solution: Quantity of mixture  = 40 lit.

Percentage of water = 10%

Quantity of water = \frac{40\times 10}{100}=4 lit

Quantity of milk = 40-4=36

How much water should be added so water may be 20% in new mixture

\frac{4+x}{36}=\frac{20}{80}

x=5

Shortcut :

Water                      Milk                     Water                      Milk

10             :                90          \rightarrow             1               :               9

20             :                80         \rightarrow              1              :                4

To equate the milk multiply (1:9) with 4 and (1:4) with 9

Water                      Milk                                    Water                      Milk

( 1               :               9 )  *4          \rightarrow                      4               :              36

( 1               :               4 )  *9          \rightarrow                       9              :              36

4-9=5

\rightarrow  40

40 \rightarrow  ?     (36+4=40)

\frac{5\times 40}{40}=5lit

Problem 2: Fresh fruit contains 68% water and dry fruit contains 20% water. How many kg of dry fruits can be made from 75kg of fresh fruit?

Solution :

Water                 pulp                               Water                 pulp

Fresh              68%          :         32%            \rightarrow                17           :             8

Dry                    20%          :        80%            \rightarrow                  2          :               8

17+8=25             2+8=10

25%  \rightarrow  75

10%  \rightarrow  ?

\frac{10\times 75}{25}=30

Problem 3:A solution of salt water contains 20% salt by weight. If 20lit of water is evaporated from the solution, the solution contains 25% of salt. What is the original quantity of solution?

Solution:

Salt                     Water                   Salt                     Water

20%       :             80%        \rightarrow             1           :               4

25%      :              75%        \rightarrow              1           :              3

1+4=5          1+3=4

5-4=1

\rightarrow  20lit

\rightarrow  ?

\frac{5\times 20}{1}=100

 

 

 

Percentage part-7

Based on votes and elections

Problem 1: Two persons contested  an election of parliament the winning candidate secured 57% of the total votes polled and won by majority of 42000. The total number of votes polled is?

Solution:Total votes =TV

winner=w     Loser=L

w           L            TV

57%      43%       100%

When you find the word majority in the question use below formula

Majority = w% \sim L%

= 57%  \sim 43%

=14%

14% \rightarrow 42000

100% \rightarrow ?

\frac{42000\times 100}{14}=300000

Problem 2:In an election a candidate secured 40% of the votes but is defeated by the only other candidate by a majority of 298 votes. Find the total number of votes recorded?

Solution:   W                L                 TV

60%            40%           100%

Majority = w% \sim L%

=60%-40%

=20%

20%  \rightarrow 298

100% \rightarrow ?

\frac{2989\times 100}{20}=1490

Problem 3:In an election, three candidates contested. The first candidate got 40% votes and the second candidate got 30% votes. If the total number of votes polled were 36000. Find the number of votes got by the third candidate?

Solution: Total votes 100%=36000

TV                    1st                  2nd                    3rd

100%               40%                 36%                  24%

100% \rightarrow 36000

24% \rightarrow  ?

\frac{3600\times 24}{100}=8640

Problem 4: In a college election between two candidates 10% of votes cast are invalid. The winner got 70% of valid votes and defeats the loser by 1800 votes. How many votes were totally cast?

Solution : Total votes = 100%

votes invalid = 10%

remaining votes = 90%

Winning percentage = 70% of 90%

=(\frac{70}{100}\times 90)%

= 63%

Loser percentage = 90%-63%=36%

36% \rightarrow 1800

100% \rightarrow ?

\frac{1800\times 100}{36}=5000

Percentage part-6

Based on marks

Problem 1: In an examination Ram got 30% of maximum marks and failed by 80 marks. Shyam got 200 marks and failed by 15%. Find the pass percentage marks in the exam?

Solution: Pass mark for Ram = 30%+80

pass mark for shyam = 200+15%

Both are pass mars we can equate these two

30%+80 = 200+15%

30%-15% = 200-80

15% = 120

100% = ?

\frac{120\times 100}{15} = 800

10% of 800 = 80

30%+10%=40%

The passing percentage is 40%

The pass mark is 800

Problem: A got 30% of the maximum marks and get failed by 25 marks where as B get 40% of the total marks in the same exam and set 25% of the passing mark more then the passing mark. Find the passing marks of the exam?

Solution :  Let passing mark=pm

30%+25=pm

40%=\frac{pm\times 125}{100}

pm=32%

30%+25=32%

32%-30%=25

2%=25

Multiply with 16 on both sides so that we can get 32%

16*2%=25*16

32%=400

pm=32%=400

Problem 3: In an examination in which maximum marks are 500, A got 10% less than B. B got 25% more than C. C got 20% less than D. If A got 360 marks,What percentage of maximum marks was obtained by D?

Solution: Assume D=100

C got 20% less than D =  100\times \frac{80}{100}=80

B got 25% more than C = 80\times \frac{125}{100}=100

A got 10% less than B = 100\times \frac{90}{100}=90

90% \rightarrow 360

100%  \rightarrow ?

\frac{100\times 360}{90}=400

% of maximum marks obtained by D is = \frac{400}{500}\times 100= 80%

Maximum marks obtained by D is = 400

 

 

Percentage-part5

More or less concept

Problem 1; A reduction of 25% in the price of the sugar enables a person to purchase 4kg more for 800 RS. Find the original and current price per kg?

Solution ; Let the original price = x

Reduced price = \frac{3x}{4}

\frac{800}{\frac{3x}{4}}-\frac{800}{x}=4

\frac{800}{3x}=4

Original price x=66.66

Reduced price=\frac{3x}{4}=\frac{3\times 66.66}{4}=50

Shortcut: 25% \rightarrow \frac{1}{4}

4-3=1

1 \rightarrow 4

3 \rightarrow 3*4=12

4 \rightarrow 4*4=16

\frac{800}{12}=66.66         \frac{800}{16}=50

Problem 2:Due to 30% increase in price of apples, 6 apples are less available for RS 520. Find the new price and old price of a apple?

Solution :30%  \rightarrow \frac{3}{10}

3 \rightarrow 6

1 \rightarrow 2

13*2=26

10*2=20

\frac{520}{26}=20            \frac{520}{20}=26

old price=20               new price=26

Percentage-part4

Income and expenditure concept

Problem 1:A person spends 40% of his salary on house rent, on remaining 10%  spends on travels, on remaining 16\tfrac{2}{3}% spends on food and remaining is saved. If he saved RS 6750 what amount he spent on food?

Solution : Assume total income=100

40% spends on salary=100-40=60

on remaining 10% spends on travels=60\times \frac{10}{100}=6

=60-6=54

On remaining 16\tfrac{2}{3}% spends on food=54\times \frac{1}{6}=9

=54-9=45

saved=45

Amount he spent on food=9

45  \rightarrow  6750

9    \rightarrow  ?

\frac{9\times 6750}{45}=1350

The amount spent on food is 1350

Observe the below figure

Assume income =100

Problem 2: A man  spends 75% of his income. His income is increased by 20% and he increased his expenditure by 10%. His savings are increased by?

Solution : A man sends his income  is 75%

The fractional value of 75%=\frac{3}{4}

Here 4 is the actual income and 3 is the expenditure

Savings are 4-3=1

He increases his income by 20%

4\times \frac{20}{100}=0.8

4+0.8=4.8

He increases his expenditure by 10%

3\times \frac{10}{100}=0.3

3+0.3=3.3

Savings = 4.8-3.3=1.5

Percentage change = \frac{Final volume-initial volume}{initial volume}\times 100

= \frac{1.5-1}{1}\times 100

=50%

Observe the below figure

Problem 3:The price of  sugar is increased by 16\tfrac{2}{3}% and the consumption of a family is decreased by 20%. Find the percentage change in expenditure?

Solution :16\tfrac{2}{3}%=\frac{1}{6}       20%=\frac{1}{5}

Actual price is 6 and increased price is 7

Actual consumption is 5 and decreased consumption is 4

Price * Consumption = Expenditure

6*5=30        7*4=28

Percentage change=\frac{Final volume-initial volume}{initial volume}\times 100

=\frac{30-28}{30}\times 100=6.66%

Observe the below figure

Problem 4 : The sale of a cinema ticket is increased by 57\tfrac{1}{7}% and the price of a ticket is increased by 16\tfrac{2}{3}%. Find the change in the return?

Solution:   57\tfrac{1}{7}%=\frac{4}{7}         16\tfrac{2}{3}%=\frac{1}{6}

Percentage change = \frac{Final volume-initial volume}{initial volume}\times 100

= \frac{77-42}{42}\times 100

=83\tfrac{1}{3}%

Percentages Part 3

Increase and decrease concept

Problem 1: If the radius of the right circular cylinder is increased by 40% and its height is reduced by 37.5% then find the percentage change in its volume?

Solution: Radius increased to 40% = \frac{2}{5}

Height decreased to 37.5%=\frac{3}{8}

Volume of a cylinder  =  \pi r^{2}h

Actual volume  =  \pi \times 5\times 5\times 8= 200\pi

Radius increased=5+2=7

Height decreased  =  8-3  =5

Changed volume=\pi \times 7\times 7\times 5=245\pi

% change in its volume = \frac{Final volume-initial volume}{initial volume}\times 100

=\frac{245-200}{200}\times 100

=22.5

Problem 2: If the length of the rectangle is increased by 20% and the breadth of the rectangle is decreased by  20% find the percentage change in its area?

Solution: Length increased to 20%

Breadth decreased to 20%

Area of the rectangle  =  Length * Breadth

Actual area  =  10*10  =  100

Changed area  =  12*8  =96

% change in the area  =  \frac{Final volume-initial volume}{initial volume}\times 100

=\frac{96-100}{100}\times 100

=-4%

Here ‘-‘ indicates area decreased to 4%

Shortcut method: Direct formula \frac{a\times a}{100} = \frac{20\times 20}{100}  =  4

Note:

  • The above method is applicable when there is increase or decrease in the value of 10%,20%,30,%40%,50%,60%, etc.
  • If such values occur follow first problem method. It is not applicable when the fractional values occur.
  • The fractional values like  37.5%,57\tfrac{1}{7}%, etc., If such values occur follow first problem method.

Percentages Part 2

Based on the concept in part-1, let’s try some problems which are asked in many bank exams.

problem 1: 65% of a number is 21 less than that  \frac{3}{4} of the number. What is the number?

Solution: Convert \frac{3}{4}  into percentage

\frac{3}{4}=75%

65% \sim 75%   \rightarrow  21

10%  \rightarrow  21

100%  \rightarrow  ?

\frac{21\times 100}{10}=210

The number is 210.

Problem 2: If of 66\tfrac{2}{3} a number is added with it self then result becomes 3900. Find the original number?

Solution: Convert 66\tfrac{2}{3}% into fraction

Split 66\tfrac{2}{3}% into 50%+16\tfrac{2}{3}%

50%=\frac{1}{2}    ,     16\tfrac{2}{3}%=\frac{1}{6}

\frac{1}{2}+\frac{1}{6} = \frac{3+1}{6}=\frac{4}{6}=\frac{2}{3}

66\tfrac{2}{3}% = \frac{2}{3}

Here 3 is the original number

3+2  \rightarrow  3900

\rightarrow  3900

\rightarrow  ?

\frac{3900\times 3}{5}=2340

The original number is 2340.

Problem 3: If 96 is added in the number then number becomes 157\tfrac{1}{7}% of itself. Find the number?

Solution : Convert 157\tfrac{1}{7}% into fraction

Split 157\tfrac{1}{7}% into 100% and 57\tfrac{1}{7}%

100%=1    ,     57\tfrac{1}{7}%=\frac{4}{7}

1+\frac{4}{7}=\frac{11}{7}

Here 7 is the original number and 11 occur when we add 96 to the original number

11-7=4

\rightarrow  96

\rightarrow  ?

\frac{96\times 7}{4}=168

The number is 168.

Percentages Part 1

How to convert fractions into percentage

To convert a fraction into percentage just multiply with 100

  • Example1: Convert  \frac{1}{2}  into the percentage

solution:\frac{1}{2}\times 100=50%

  • Example 2: Convert  \frac{1}{6}  into the percentage

Solution:\frac{1}{6}\times 100=\frac{100}{6}= 16\tfrac{2}{3}% or 16.66%

 

How to convert percentages into fractions

To convert a percentage into fraction just divide by 100

  • Example1: Convert 20% into the fraction

Solution: \frac{20}{100}=\frac{1}{5}

  • Example 2: Convert 14\tfrac{2}{7}% into the fraction

Solution:14\frac{2}{7}= \frac{100}{7\times 100}=\frac{1}{7}

 

Remember this chart shown below which is used to solve problems in a  shortcut way

1=100%                                        \frac{1}{11}=9\tfrac{1}{11}% or 9.09%

\frac{1}{2}=50%                                         \frac{1}{12}=8\tfrac{1}{3}% or 8.33%

\frac{1}{3}=33\tfrac{1}{3}%                                      \frac{1}{13}=7\tfrac{9}{13}% or7.69%

\frac{1}{4}=25%                                        \frac{1}{14}=7\tfrac{1}{7}% or 7.142%

\frac{1}{5}=20%                                        \frac{1}{15}=6\tfrac{2}{3}% or 6.66%

\frac{1}{6}=16\tfrac{2}{3}% or 16.66%                \frac{1}{16}=6\tfrac{1}{4}%% or 6.25%

\frac{1}{7}=14\tfrac{2}{7}% or 14.28%               \frac{1}{20}=5%

\frac{1}{8}=12\tfrac{1}{2}% or 12.5%                  \frac{1}{25}=4%

\frac{1}{9}=11\tfrac{1}{9}% or 11.11%                \frac{1}{30}=3\tfrac{1}{3}% or 3.33%

\frac{1}{10}=10%                                    \frac{1}{50}=2%