## Percentages 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%

The quantity of water = $\frac{40\times&space;10}{100}=4$ lit

The quantity of milk = 40-4=36

How much water should be added so water maybe 20% in the 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)

$\frac{5\times&space;40}{40}=5lit$

Problem 2: Fresh fruit contains 68% water and dry fruit contains 20% water. How many kgs 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%  $\rightarrow$  ?

$\frac{10\times&space;75}{25}=30$

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

$\rightarrow$  ?

$\frac{5\times&space;20}{1}=100$

## Percentages Part 7

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

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% $\rightarrow$ ?

$\frac{42000\times&space;100}{14}=300000$

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% $\rightarrow$ ?

$\frac{2989\times&space;100}{20}=1490$

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?

TV                    1st                  2nd                    3rd

100%               40%                 36%                  24%

100% $\rightarrow$ 36000

24% $\rightarrow$  ?

$\frac{3600\times&space;24}{100}=8640$

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%

Winning percentage = 70% of 90% =$(\frac{70}{100}\times&space;90)$% = 63%

Loser percentage = 90%-63%=36%

36% $\rightarrow$ 1800

100% $\rightarrow$ ?

$\frac{1800\times&space;100}{36}=5000$

## Percentages Part 6

Based on marks

Problem 1: In an examination, Ram got 30% of the 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

The 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% = ?

$\frac{120\times&space;100}{15}$ = 800

10% of 800 = 80

30%+10%=40%

The passing percentage is 40%

The pass mark is 800

Problem 2: A got 30% of the maximum marks and get failed by 25 marks whereas B gets 40% of the total marks in the same exam and set 25% of the passing mark more than the passing mark. Find the passing marks of the exam?

Solution: Let passing mark=pm

30%+25=pm

40%=$\frac{pm\times&space;125}{100}$

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 =  $100\times&space;\frac{80}{100}=80$

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

A got 10% less than B = $100\times&space;\frac{90}{100}=90$

90% $\rightarrow$ 360

100%  $\rightarrow$ ?

$\frac{100\times&space;360}{90}=400$

% of the maximum marks obtained by D is = $\frac{400}{500}\times&space;100=&space;80$%

Maximum marks obtained by D is = 400

## Percentages Part 5

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=$\frac{3x}{4}=\frac{3\times&space;66.66}{4}=50$

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 the price of apples, 6 apples are less available for RS 520. Find the new price and the old price of an 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

## Percentages Part 4

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=$60\times&space;\frac{10}{100}=6$ =60-6=54

On remaining $16\tfrac{2}{3}$% spends on food=$54\times&space;\frac{1}{6}=9$ =54-9=45

Saved=45

Amount he spent on food=9

45  $\rightarrow$  6750

9    $\rightarrow$  ?

$\frac{9\times&space;6750}{45}=1350$

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%. Are his savings 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%

$4\times&space;\frac{20}{100}=0.8$

4+0.8=4.8

He increases his expenditure by 10%

$3\times&space;\frac{10}{100}=0.3$

3+0.3=3.3

Savings = 4.8-3.3=1.5

Percentage change = $\frac{Final&space;volume-initial&space;volume}{initial&space;volume}\times&space;100$

= $\frac{1.5-1}{1}\times&space;100$

=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=$\frac{Final&space;volume-initial&space;volume}{initial&space;volume}\times&space;100$

=$\frac{30-28}{30}\times&space;100=6.66$%

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 = $\frac{Final&space;volume-initial&space;volume}{initial&space;volume}\times&space;100$

= $\frac{77-42}{42}\times&space;100$

=$83\tfrac{1}{3}$%