# Successive Percentage Change Important Formulas, Definitions, & Examples:

#### Overview:

 Topic Included: Formulas, Definitions & Exmaples. Main Topic: Quantitative Aptitude. Quantitative Aptitude Sub-topic: Percentage Aptitude Notes & Questions. Questions for practice: 10 Questions & Answers with Solutions.

#### Successive Percentage Change:

If the present value of an item is k and the value of item changes $$n \ \%$$ per year then.

Successive Percentage Change Formula after few years: Value of the item after $$x$$ years $$= k \ \left[1 + \frac{n}{100}\right]^x .......(1)$$

Successive Percentage Change Formula few years ago: Value of the item $$x$$ years ago $$= \frac{k}{\left[1 + \frac{n}{100}\right]^x} .......(2)$$

Example (1): If the present population of the country is $$100,000$$ and population of country increases $$5 \ \%$$ yearly, then find the population of country after $$3$$ years as well as population of country $$3$$ years ago ?

Solution: Given values are, $$k = 100,000$$, $$n = 5$$, $$x = 3$$, then according to equation $$(1)$$,

Population of the country after $$3$$ years, $$= 100,000 \left[1 + \frac{5}{100}\right]^3$$ $$= 100,000 \left[\frac{21}{20}\right]^3$$ $$= 100,000 \times \frac{21}{20} \times \frac{21}{20} \times \frac{21}{20}$$ $$= 100,000 \times \frac{9261}{8000}$$ $$= 115,762.5 \ (Answer)$$

Population of the country $$3$$ years ago, according to equation $$(2)$$, $$= \frac{100,000}{\left[1 + \frac{5}{100}\right]^3}$$ $$= \frac{100,000}{\left[\frac{21}{20} \times \frac{21}{20} \times \frac{21}{20}\right]}$$ $$= \frac{100,000 \times 8,000}{9261}$$ $$= 86,383.76 \ (Answer)$$

Example (2): If the present weight of a person is $$70 \ kg$$ and the weight of the person increases $$10 \ \%$$ per year, then find the weight of the person after $$5$$ years as well as $$5$$ years ago?

Solution: Given values are, $$k = 70 \ kg$$, $$n = 10$$, $$x = 5$$, then according to equation $$(1)$$,

weight of the person after $$5$$ years, $$= 70 \left[1 + \frac{10}{100}\right]^5$$ $$= 70 \left[\frac{11}{10}\right]^5$$ $$= 70 \times \frac{11}{10} \times \frac{11}{10} \times \frac{11}{10} \times \frac{11}{10} \times \frac{11}{10}$$ $$= 70 \times \frac{161051}{100000}$$ $$= 112.7357 \ kg \ (Answer)$$

weight of the person $$5$$ years ago, according to equation $$(2)$$, $$= \frac{70}{\left[1 + \frac{10}{100}\right]^5}$$ $$= \frac{70}{\left[\frac{11}{10} \times \frac{11}{10} \times \frac{11}{10} \times \frac{11}{10} \times \frac{11}{10}\right]}$$ $$= \frac{70 \times 100,000}{161051}$$ $$= 43.464 \ kg \ (Answer)$$