## 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 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 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

$\rightarrow$  ?

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