** Percentage: Concept, Short Trick, and Example with Solution **

**Percentage:** Percentage refers to “Per hundred” i.e. 1

**Conversion:** (i) To convert a fraction/Decimal into percentage multiply it by 100.

Example: 0.35 = $\frac{35}{100}$= 3

(ii) To convert a percentage to fraction/Decimal divide it by 100.

Example: 2

**Rule1**:

**Example:** 5

**Solution: **5

⇒ $\frac{50}{100}\times \mathrm{x}=60$

⇒ x = $\frac{60\times 100}{50}=120$

**Rule2**: P is a number. After

∴ Q = P + $\frac{\mathrm{x}}{100}\times \mathrm{P}$ = $\mathrm{P}\left( 1+\frac{\mathrm{x}}{100} \right) $

= $\mathrm{P}\left( \frac{100+\mathrm{x}}{100} \right) $

∴ Q = $\mathrm{P}\left( \frac{100+\mathrm{x}}{100} \right) $ or P = $\frac{100\mathrm{Q}}{100+\mathrm{x}}$

**Example:** New fare of bus Rs. 12 after 2

**Solution:** Using above relation, Original fare = $\frac{100\times 12}{100+20}=10$

Practice: All Arithmetic or Quantitative Aptitude MCQ Questions with Solution

**Rule3:** P is a number. After

∴ Q = P – $\frac{\mathrm{x}}{100}\times \mathrm{P}$ = $\mathrm{P}\left( 1-\frac{\mathrm{x}}{100} \right) $

= $\mathrm{P}\left( \frac{100-\mathrm{x}}{100} \right) $

∴ Q = $\mathrm{P}\left( \frac{100-\mathrm{x}}{100} \right) $ or P = $\frac{100\mathrm{Q}}{100-\mathrm{x}}$

**Example:** As production increased, commodity prices fell by 2

**Rule4:** If the cost of an article is increased by

Or

If A’s salary is

$\frac{\mathrm{x}}{100+\mathrm{x}}\times 100$

**Example1:** If the price of sugar goes up by 2

**Solution:** Required percentage = $\frac{25}{100+25}\times 100=20$

**Example2:** If A’s salary is 2

**Solution:** B’s salary is less than A’s =$\frac{20}{100+20}\times 100=16\frac{2}{3}$

**Rule5:** If the cost of an article is decreased by

Or

If A’s salary is

$\frac{\mathrm{x}}{100-\mathrm{x}}\times 100$

**Example1:** If the price of sugar decreased by 2

**Solution:** Required percentage = $\frac{20}{100-20}\times 100=25$

**Example2:** If A’s salary is 1

**Solution:** B’s salary is more than A’s salary =$\frac{10}{100-10}\times 100=11\frac{1}{9}$

**Rule6:** If a number is first increased by

$\mathrm{x}+\mathrm{y}+\frac{\mathrm{xy}}{100}$

**N.B:**– Remember to use the ‘+’ sign in case of increase and the ‘–’ sign in case of decrease. After the calculation, if ‘+’ then gain or increase and if ‘–’ then loss or decrease

**Example:** The salary of one person is increased by 1

**Solution: **Total percentage increase = $10+20+\frac{10\times 20}{100}=32$

**Rule7:** If a number is increased by

$\mathrm{x}-\mathrm{y}-\frac{\mathrm{xy}}{100}$

N.B:- ‘+’ then gain or increase and if ‘–’ then loss or decrease

**Example:** The price of an article 4

Solution: Gain or loss = $40-10-\frac{40\times 10}{100}=26$

**Rule8:** If a number is first decreased by

$-\mathrm{x}-\mathrm{y}+\frac{\mathrm{xy}}{100}$

‘–’ for decrease

**Example:** Total decrease of an article after 1

**Solution:** Total change = $-10-20+\frac{10\times 20}{100}=-28$

∴ Total decrease = 2

**Rule9:** If an amount is increased by

$\frac{\mathrm{x}^2}{100}$

**Example:** If one side of a rectangular increase by 1

**Solution:** Area decrease percentage = $\frac{10^2}{100}=1$

**Rule10:** If an amount is increased by

Initial amount (P) = $\frac{100\times 100\times 100\times \mathrm{Q}}{\left( 100+\mathrm{x} \right) \left( 100+\mathrm{y} \right) \left( 100+\mathrm{z} \right)}$

**Example:** Population of a city after three year 91080. Every year population increase by 1

**Solution:** initial population of the city = $\frac{\left( 100+10 \right) \left( 100+15 \right) \left( 100+20 \right) \times 60000}{100\times 100\times 100}=91080$

**Rule11:** If

Initial amount = $\frac{100\times 100\times 100\times \mathrm{Q}}{\left( 100-\mathrm{x} \right) \left( 100-\mathrm{y} \right) \left( 100-\mathrm{z} \right)}$

**Example:** Decrease population of a city after three month 36720 due to COVID-19. Decrease in 1^{st} month 1

**Solution:** initial population of the city = $\frac{100\times 100\times 100\times 36720}{\left( 100-10 \right) \left( 100-15 \right) \left( 100-20 \right)}=60000$

**Rule12:** On increasing/decreasing the cost of a certain article by

Increased/decreased cost of the article = $\frac{\mathrm{xy}}{100\mathrm{a}}$ and initial cost = $\frac{\mathrm{xy}}{\left( 100\pm \mathrm{x} \right) \mathrm{a}}$(Negative sign when decreasing and Positive sign when increasing)

**Example:** The Government reduced the price of sugar by 10 per cent. By this a consumer can buy 6.2 kg more sugar for Rs. 837. The reduced price and initial price per kg of sugar is

Solution: The reduced price = $\frac{10\times 837}{100\times 6.2}=13.5$ and initial price = $\frac{10\times 837}{\left( 100-10 \right) \times 6.2}=15$(Here Negative sign for reduce)

**Example:** Due to an increase of 5

**Solution:** present rate of eggs = $\frac{50\times 24}{100\times 4}=3$ Initial rate of eggs = $\frac{50\times 24}{\left( 100+50 \right) \times 4}=2$

**Rule13:** If a number is first increased by

$\mathrm{x}+\mathrm{y}+\mathrm{z}+\frac{\mathrm{xy}+\mathrm{yz}+\mathrm{zx}}{100}+\frac{\mathrm{xyz}}{\left( 100 \right) ^2}$

**Example:** Side increase of a cuboid by 1

**Solution:** Volume change = $10+20+30+\frac{10\times 20+20\times 30+30\times 10}{100}+\frac{10\times 20\times 30}{100^2}=71.6$

**Rule14:** If a person saves ‘S’ rupees after spending

Monthly income = $\frac{100}{100-\left( \mathrm{x}+\mathrm{y}+\mathrm{z} \right)}\times \mathrm{S}$

**Example:** Radha spends 4

**Solution:** Salary of Radha = $\frac{100}{100-\left( 40+20+10 \right)}\times 1500$ = 5000

**Rule15:** The amount of acid/milk is

Amount of water added = $\frac{\mathrm{L}\left( \mathrm{x}-\mathrm{y} \right)}{\mathrm{y}}$

**Example:** In one litre of a mixture of alcohol and water, water is 3

**Solution:** Amount of alcohol added = $\frac{1\left( 30-15 \right)}{15}=1$ litre

**Rule16:** In a certain examination passing marks is

Total marks = $\frac{100\left( \mathrm{b}+\mathrm{c} \right)}{\mathrm{a}}$

**Example:** A student has to secure 4

**Solution: **Total marks = $\frac{100\left( 90+10 \right)}{40}=250$

Also Practice:Mathematics and Pedagogy MCQ Questions with Answer

**Rule17:** One candidate gets

Total Mark of exam = $\frac{100\left( \mathrm{a}+\mathrm{b} \right)}{\left( \mathrm{y}-\mathrm{x} \right)}$

**Example:** One candidate gets 3

**Solution:** Total Mark = $\frac{100\left( 30+18 \right)}{\left( 43-35 \right)}=600$ and Pass mark = $\frac{35}{100}\times 600+35=245$

**Rule18:** In a certain examination, ‘B’ boys and ‘G’ girls participated.

Percentage of passed students of the total students = $\frac{\mathrm{B}.\mathrm{b}+\mathrm{G}.\mathrm{g}}{\mathrm{B}+\mathrm{G}}$

**Example: **In an examination, there were 1000 boys and 800 girls. 6

**Solution: **Percentage of passed students = $\frac{1000\times 60+800\times 50}{1000+800}=55.5556$

The percent of the candidates failed = 100 – 55.55 = 44.45

**Rule19:** If a candidate got

∴ Total no. of votes = $\frac{50\mathrm{x}}{50\sim \mathrm{A}}$

**Example:** Two candidates contested in an election. One got 6

**Solution:** Total no. of votes = $\frac{50\times 1600}{60-50}=8000$

**Rule20:** **F _{AB} = 100 – (P_{A} + P_{B} – P_{AB}) and P_{AB} = 100 – (F_{A} + F_{B} – F_{AB})**

Where P_{A} = A- subject / percentage pass

P_{B} = B- subject read / percentage pass

P_{AB} = AB- read / percentage pass

F_{A} = don’t read about A / drop percent

F_{B} = B- don’t read about / drop percent

F_{AB} = don’t read about AB / Percentage

**Example: **In one college 5

**Solution:** F_{AB} = 100 – (P_{A} + P_{B} – P_{AB})

= 100 – (50 + 40 – 20 )

= 3

∴ percentage of students do not read any of Bengali and English = 3

**Example: **In any exam, 8

**Solution: **F_{E} = 100 – 85 = 1

P_{EM} = 100 – (F_{E} + F_{M} – F_{EM}) = 100 – (15 + 20 – 10) = 7

∴ percentage passes in both subjects = 7

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