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Quantitative Aptitude
hardQ. No. 5483-the-population-of-a-town-becomes-16212-after-an-increase-of-12-percent-what-was-the-original

The population of a town becomes 16212 after an increase of 12 percent. What was the original population?

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The percentage expression “The population of a town becomes 16212 after an increase of 12 percent. What was the original population” is evaluated with Original population = final population ÷ (1 + rate)., producing 14475.0 as the option-specific result.

A14472.0
B14475.0
C14456.0
D14431.0
Correct Answer: 14475.0Your Answer: B

Explanation

Solving “The population of a town becomes 16212 after an increase of 12 percent. What was the original population” requires the relationship Original population = final population ÷ (1 + rate).; substituting the stated quantities gives 16212 ÷ 1.12 = 14475.; hence 14475.0 is the calculated result and option B is correct.

Important Notes

  • Method for “The population of a town becomes 16212 after an increase of 12 percent. What was the original population”: use Original population = final population ÷ (1 + rate)..
  • Substitution for “The population of a town becomes 16212 after an increase of 12 percent. What was the original population” produces 16212 ÷ 1.12 = 14475..
  • Calculated answer for “The population of a town becomes 16212 after an increase of 12 percent. What was the original population” is 14475.0, matching option B.
  • Verification for “The population of a town becomes 16212 after an increase of 12 percent. What was the original population”: retain the stated base quantity throughout the percentage calculation.

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Common questions and clear answers for this topic.

The population of a town becomes 16212 after an increase of 12%. What was the original population?

The answer is 14475.0. Using Original population = final population ÷ (1 + rate), 16212 ÷ 1.12 = 14475.

Which formula is used?

Original population = final population ÷ (1 + rate)

What is Percentage and how is it used in competitive exam problems?

Percentage means per hundred and is used to express a number as a fraction of 100. Formula: Percentage = (Value/Total Value) x 100. In competitive exams like SSC CGL, IBPS Bank PO, and CAT, percentage questions appear as: finding percentage of a number, percentage increase/decrease, percentage change, successive percentage change, and percentage in Data Interpretation.

What are the important percentage-fraction equivalents to memorize?

Important percentage-fraction equivalents: 10% = 1/10, 12.5% = 1/8, 16.67% = 1/6, 20% = 1/5, 25% = 1/4, 33.33% = 1/3, 37.5% = 3/8, 50% = 1/2, 62.5% = 5/8, 66.67% = 2/3, 75% = 3/4. Converting percentages to fractions makes calculations faster. For example: 25% of 480 = 480/4 = 120. Memorizing these fractions is essential for quick Quantitative Aptitude calculations.

How to calculate Percentage Increase and Decrease?

Percentage Increase = (Increase/Original Value) x 100. Percentage Decrease = (Decrease/Original Value) x 100. If price increases by x% then decreases by x%, net effect is a decrease of x squared/100 percent. For Successive Percentage Changes: if first change is a% and second is b%, overall change = a + b + (ab/100) percent. This formula is crucial for profit/loss and interest problems in competitive exams.

How to solve population growth problems using percentage?

Population problems use percentage increase formula. If population is P and annual growth rate is r%, then after n years: Population = P x (1 + r/100)^n. Example: Town population 50,000 grows at 10% per year. After 2 years = 50000 x 1.1 x 1.1 = 60,500. If growth rate decreases, use (1-r/100). These problems appear in SSC CGL, IBPS, and banking exams.

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