# Percentages: Concepts and Formulas for Placement Aptitude Tests

Percentage is one of the simplest tool for comparison of data, who will know that better than students 😛

##### 1. Any percentage can be expressed as a decimal fraction by dividing the percentage by 100

And conversely, any decimal fraction can be converted to percentage by multiplying it by 100.

Q: How will you express 3% in decimal?

Sol: 3% = 3/100 = 0.03

Q: How will you convert 0.8 in percentage?

Sol: 0.8 = 0.8*100% = 80%

**2.** Percentage change $= \frac{Absolute \ Value \ Change}{Original \ Quantity}*100$

Q: Amara scored 80 marks in second aptitude test while she scored 50 in first one. What was the percentage change in Amara’s score?

Sol: Percentage change $= \frac{(80-50)}{50}*100 = 30*100/50 = 60%$

**Note – **The base used for the sake of Percentage Change calculations is always the original quantity unless otherwise stated. For instance in above example denominator in calculation is 80 (original quantity) and not 50 (final quantity). This is the most common mistake students generally make. So be careful! 😐

**3.** If the percentage increase is p%, then the new value $ = Original \ Value*{\LARGE [} \frac{P}{100} + 1 {\LARGE ]}$

Q: A politician received 12% more votes this year compared to last year. If he received 50,000 votes last year, how many votes did he receive this year?

Sol: New vote count is $ = 50000* [ \frac{12}{100} + 1 ] = 56000$

**Note – **You may also just calculate the increase in votes which is 12% of 50000 = 6000, and add it to original vote count to arrive at 56,000. That’s the beauty of aptitude. As long as you are conceptually clear, you will always get the right answer irrepective of what method you choose. In fact it is better to develop your own method over time 😐

**4.** If the new value is “k” times the old value, then the percentage increase $ = [k – 1]*100$

Q: A country’s population has tripled in last few decades. What is the percentage increase in the country’s population?

Sol: Percentage increase in country’s population $ = [3 – 1]*100 = 200\% $

**Note –** You can also solve it by taking initial population as x, and then basics by dividing the change in population by original population $= (3x – x)/x *100 = 200\% $

##### 5. If an item’s value goes up/down by x%, the percentage change to bring it back to the original level is = ${\LARGE [} \frac{100x}{100 \pm x}{\LARGE ]} {\%}$

Q: Price of a newly launched phone is 10000, which increased by 10%. How much percentage discount coupon is needed so that customer can buy it at original price

Sol: discount coupon needed ${\LARGE [} \frac{100*10}{100 + 10}{\LARGE ]} {\%} = \frac{100}{11} \% $

Q: Population of tiger decreased by 10% in last 1 year. By how much percentage should it increase in next year so that original population is restored.

Sol: Increases needed to restore tiger population = ${\LARGE [} \frac{100*10}{100 – 10}{\LARGE ]} {\%} = \frac{100}{9} \% $

**Note –** If you solve these problems by basics, you will get the same answer. No shortcut or formula is needed to be remembered. We have included various scenarios to help you think and understand. Practice and conceptual clarity are the only two things, that will help you in aptitude improvement journey.

**6.** Two more shortcuts can be derived from above exercise:

If A is x% more/less then B, then B is what percent less/more than A is calculated as – ${\LARGE [} \frac{100x}{100 \pm x}{\LARGE ]} {\%}$

If the price of an item goes up/down by x%, then the quantity consumed should be reduced/increased by what percentage so that the total expenditure remains the same – ${\LARGE [} \frac{100x}{100 \pm x}{\LARGE ]} {\%}$

Q: If student A 15% more marks than student B. Then by what percentage student B’s marks are less than student A’s marks?

Sol: Percentage by which student B’s marks are less than student A’s marks = ${\LARGE [} \frac{100*15}{100 + 15}{\LARGE ]} {\%} = \frac{300}{23} \% $

Alternate way to solve using basics: Let’s assume student B has 100 marks, then student A’s marks = 100 + 15 = 115. Now B’s marks are 15 less than students A’s marks which is 115. So B’s marks are less by $ = \frac{15*100}{115} \% $

Q: Petrol prices were at Rs 100/ litre. Prices have been announced to increase by 25%. By how much percentage will a student need to decrease petrol consumption for his bike, if he doesn’t wish to increase his fuel expense?

Sol: Decrease in petrol consumption = ${\LARGE [} \frac{100*25}{100 + 25}{\LARGE ]} {\%} =20\% $

**7. Successive Change**

For successive increases / decrease of p%, q% and r% in three stages, the effective percentage increase = ${\LARGE [} {\LARGE (} 1 + \frac{p}{100} {\LARGE )} {\LARGE (} 1 + \frac{q}{100} {\LARGE )} {\LARGE (} 1 + \frac{r}{100}{\LARGE )} – 1 {\LARGE ]} * 100\% $

Note:- If one or more of p, q, and r are decrease percentage figures, then it will be taken as a negative figure and not as a positive figure.

Q: What is the percentage change in a city’s population at the end of 3 years if it has changed as follow year on year:

Year 1: increased by 10%; Year 2: decreased by 20%; Year 3: increased by 30%

Sol: Percentage change = ${\LARGE [} {\LARGE (} 1 + \frac{10}{100} {\LARGE )} {\LARGE (} 1 – \frac{20}{100} {\LARGE )} {\LARGE (} 1 + \frac{30}{100}{\LARGE )} – 1 {\LARGE ]} * 100\%$

**8. **Some easy to remember percentage – fraction conversion (not expected but good to know)

Pattern 1 | Pattern 2 | Pattern 3 |

20 % = \( \frac{1}{5} \) | 25 % = \( \frac{1}{4} \) | 11.11 % = \( \frac{1}{9} \) |

40 % = \( \frac{2}{5} \) | 50 % = \( \frac{2}{4} \) | 22.22 % = \( \frac{2}{9} \) |

60 % = \( \frac{3}{5} \) | 75 % = \( \frac{3}{4} \) | 33.33 % = \( \frac{3}{9} \) |

80 % = \( \frac{4}{5} \) | 100 % = 1 | and so on |

You can also refer following videos to grasp basics of percentage in more details

Refer Topic: Ratio and Proportion: https://www.learntheta.com/placement-aptitude-ratio-proportion/

Read more about LearnTheta’s AI Practice Platform: https://www.learntheta.com/placement-aptitude/