| 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. |
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 \((2)\), $$ = \frac{70}{\left[1 + \frac{10}{100}\right]^5} $$ $$ = \frac{70}{\left[\frac{11}{10} \times \frac{11}{10} \times \frac{11}{10} \times \frac{11}{10} \times \frac{11}{10}\right]} $$ $$ = \frac{70 \times 100,000}{161051} $$ $$ = 43.464 \ kg \ (Answer)$$