Percentage
PERCENTAGE
The Percentage problems are the basic methodology in the Numerical Ability section. These percentage problems are frequently asked in any levels of the difficulty in every competitive exam. Especially these questions are focused mainly on every bank examinations. The percentage problems are asked as separate questions are in the combination with the DI. Percentage problems are the easily understood with few concepts. We provide you with those concepts and shortcut methodologies to crack percentage questions in a few seconds. We also provide you with problems based on percentage quiz on a daily basis to improve your performance in the exam. The candidates preparing for banking and other competitive exams can validate your ability here.
| Fraction | Percentage |
| 1/100 | 1% |
| 1/20 | 5% |
| 1/10 | 10% |
| 1/8 | 121/2Â % |
| 1/5 | 20% |
| 1/4 | 25% |
| 1/3 | 331/3% |
| 1/2 | 50% |
| 3/4 | 75% |
| 4/5 | 80% |
| 9/10 | 90% |
| 99/100 | 99% |
Check Below the Practice Questions
What is the Percentage: A fraction with its denominator as ‘100’ is called a percentage. Percentage means per hundred. So it is a fraction of the form 6/100, 37/100, 151/100 and these fractions can be expressed as 6%, 37% and 151% respectively. By a certain percent, we mean that many hundredths.
Thus x percent means x hundredths, written as x%.
To express x% as a fraction: We have, x% = x/100.
Thus, 20% =20/100 =1/5; 48% =48/100 =12/25, etc.
To express a/b as a percent: We have, a/b = ((a/b)*100)%
Thus, ¼ =[(1/4)*100] = 25%; 0.6 =6/10 =3/5 =[(3/5)*100]% =60%
Why Percentage: Percentage is a concept evolved so that there can be a uniform platform for comparison of various things. (Since each value is taken to a common platform of 100)
Example: To compare three different students depending on the marks they scored we cannot directly compare their marks until we know the maximum marks for which they took the test. But by calculating percentages they can directly be compared with one another.
Important Points to Remember:
a)Â If the price of a commodity increases by R%, then the reduction in consumption so as not to increase the expenditure is
[R / (100+R))*100] %
b)Â If the price of the commodity decreases by R%, then the increase in consumption so as to decrease the expenditure is
[(R / (100-R)*100] %
c) If A is R% more than B, then B is less than A by
[(R/ (100+R))*100]%
d) If A is R% less than B, then B is more than A by
[(R/ (100-R))*100]%
Results on Population: Let the population of the town be P now and suppose it increases at the rate of R% per annum, then:
1.Population after n years = P [1+(R/100)]n
2.Population n years ago = P / [1+(R/100)]n
Results on Depreciation: Let the present value of a machine be P. Suppose it depreciates at the rate R% per annum. Then,
1.Value of the machine after n years = P [1-(R/100)]n
2.Value of the machine n years ago = P / [1-(R/100)]n
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