Percentage
30 Questions
Practice Problems
Percentage Interview Questions
30 Questions
Percentage — Concept Notes
A quick refresher on the core formulas for percentages. Read through this once, then head into practice.
- The term percent means 'per hundred' or 'for every hundred'. It is the abbreviation of the Latin phrase 'percentum'.
- Scoring 60 percent marks means that out of every 100 marks, the candidate scored 60 marks.
- The term percent is sometimes abbreviated as p.c.
- The symbol % is often used for the term percent. Thus, 40 percent is written as 40%.
- A fraction whose denominator is 100 is called a percentage, and the numerator of that fraction is called the rate percent.
- For example, 5/100 and 5 percent mean the same thing, i.e., 5 parts out of every hundred parts.
- 1. To convert a fraction into a percent: multiply the fraction l/m by 100 and put the % sign, i.e., (l/m) × 100%.
- 2. To convert a percent into a fraction: drop the percent sign and divide the number by 100.
- 3. To find a percentage of a given number: x% of N = (x/100) × N.
- (a) If A is x% more than B, then B is less than A by [x / (100 + x) × 100]%.
- (b) If A is x% less than B, then B is more than A by [x / (100 − x) × 100]%.
Example
Explanation for (a): Given A = B + (x/100)B, so A = [(100 + x)/100] B. Therefore, A − B = [(100 + x)/100] B − B = (x/100) B. So, (A − B)/A = x / (100 + x). Hence, B is less than A by [x / (100 + x) × 100]%. (The result for (b) is proved similarly.)
- If A is x% of C and B is y% of C, then:
- A = (x/y) × 100% of B
- B = (y/x) × 100% of A
Example
Explanation: Given A = (x/100)C and B = (y/100)C. Therefore, A/B = (x/100)/(y/100) = x/y. Hence, A = (x/y) × 100% of B. Similarly, B = (y/x) × 100% of A.
- (a) If two numbers are respectively x% and y% more than a third number, then the first number is [(100 + x)/(100 + y) × 100]% of the second, and the second number is [(100 + y)/(100 + x) × 100]% of the first.
- (b) If two numbers are respectively x% and y% less than a third number, then the first number is [(100 − x)/(100 − y) × 100]% of the second, and the second number is [(100 − y)/(100 − x) × 100]% of the first.
Example
Explanation: Let A, B and C be the three numbers. If A is x% more than C: A = C + (x/100)C = [(100 + x)/100] C. If B is y% more than C: B = C + (y/100)C = [(100 + y)/100] C. Therefore, A = [(100 + x)/(100 + y)] B, i.e., A = [(100 + x)/(100 + y) × 100]% of B. The formula for the second number is obtained similarly. For decreases, replace x and y with −x and −y.
- (a) If the price of a commodity increases by P%, the reduction in consumption required so as not to increase expenditure is [P / (100 + P) × 100]%.
- (b) If the price of a commodity decreases by P%, the increase in consumption required so as not to decrease expenditure is [P / (100 − P) × 100]%.
Example
Explanation: Let the original price of the commodity be ₹100. If the price increases by P%, the new price becomes (100 + P). To keep expenditure unchanged, consumption must be reduced by an amount equivalent to ₹P out of the new price. So the reduction in consumption as a percentage of the new price is [P / (100 + P)] × 100. Hence, Reduction in consumption = [P / (100 + P) × 100]%. Similarly, if the price decreases by P%, the required increase in consumption is [P / (100 − P) × 100]%.
- If a number is changed (increased/decreased) successively by x% and y%, the net percentage change is (x + y + xy/100)%.
- The result represents an increase or decrease according to the sign. If x or y is a decrease, use a negative sign before that value.
Example
Explanation: Let the given number be N. After increasing by x%, it becomes N(100 + x)/100. After further changing by y%, the value becomes [N(100 + x)/100] × (100 + y)/100 = N(100 + x)(100 + y)/100². So the net change is N(100 + x)(100 + y)/100² − N = N(100x + 100y + xy)/100². Hence, % change = (x + y + xy/100)%.
- If two parameters A and B are multiplied to get a product, and A is changed by x% while B is changed by y%, the net percentage change in the product A×B is (x + y + xy/100)%.
- The result represents an increase or decrease according to the sign. If either parameter decreases, use a negative sign for that percentage.
Example
Explanation: Let P = A × B. If A changes by x%, it becomes A(1 + x/100); if B changes by y%, it becomes B(1 + y/100). So the new product is AB(1 + x/100)(1 + y/100) = AB[1 + x/100 + y/100 + xy/10000]. Therefore, the net percentage change is (x + y + xy/100)%.
- If the present population of a town (or value of an item) is P, and it changes at r% per annum:
- (a) Population or value after n years = P(1 + r/100)ⁿ
- (b) Population or value n years ago = P / (1 + r/100)ⁿ
- Here, r is positive or negative according as the population or value increases or decreases.
Example
Explanation: Population at the end of the first year = P + (r/100)P = P(1 + r/100). At the end of the second year = P(1 + r/100)². Continuing this way, population at the end of n years = P(1 + r/100)ⁿ. Therefore, population n years ago = P / (1 + r/100)ⁿ.
- If a number A is increased successively by x%, then y%, then z%, the final value is A(1 + x/100)(1 + y/100)(1 + z/100).
- If the value decreases by any of these percentages, use a negative sign before that percentage.
- In an examination, the minimum pass percentage is x%. If a student secures y marks and fails by z marks, then the maximum marks in the examination are m = 100(y + z) / x.
Example
Explanation: Let the maximum marks be m. The student needs (y + z) marks to pass. So, (x/100) × m = y + z. Hence, m = 100(y + z) / x.
- If x% of students fail in one subject, y% fail in another subject, and z% fail in both subjects, then the percentage of students who pass in both subjects is (100 − x − y + z)%, i.e., 100 − (x + y − z)%.
Example
Explanation: Percentage who failed in one subject only = (x − z)%. Percentage who failed in the other subject only = (y − z)%. Percentage who failed in both subjects = z%. Total percentage who failed in at least one subject = (x − z) + (y − z) + z = x + y − z. Hence, percentage who passed in both subjects = 100 − (x + y − z), i.e., (100 − x − y + z)%.