Percentage of a quantity

Percentages are just fractions out of 100 wearing a "%" sign, so finding a percentage of a quantity is nothing more than fraction-of-a-quantity multiplication in disguise. Convert the percentage to a fraction (or decimal), then multiply it by the quantity:

where: Percentage is the given rate (a plain number, without the % sign, once written as a fraction over 100), Whole is the total quantity the percentage is taken of, and Part is the resulting value — always smaller than the Whole when the percentage is less than 100%. Very often the question is flipped: you are told a part and the percentage it represents, and asked to find the whole. Rearranging the same relationship gives the "find the whole" formula:

where: the symbols mean exactly the same things as above, just solved for a different unknown. Both formulas are two faces of one relationship, $\text{Percentage} = \dfrac{\text{Part}}{\text{Whole}} \times 100$, so it helps to think of Part, Whole and Percentage as three corners of a triangle: cover the one you want, and what remains tells you whether to multiply or divide. Reading a word problem correctly means first deciding which two of the three corners are given. If the Whole and the Percentage are known, multiply to get the Part. If the Part and the Percentage are known, divide (using the second formula) to get the Whole. Getting this identification right is usually the harder half of the problem — the arithmetic that follows is routine once the correct formula is chosen. A practical shortcut worth knowing is benchmark splitting: break an awkward percentage into easy pieces built from 10%, 25% and 50%. For example, 35% splits neatly into $30\% + 5\%$, and each piece is quick to find from the 10% benchmark alone — useful both for fast mental estimates and for double-checking a calculator answer.

A student scored 72 marks in a test, and this was known to be 60% of the maximum marks. Find the maximum marks for the test.

Step 1. Identify which two corners of the triangle are given. The Part is 72 marks, and the Percentage is 60%. The Whole — the maximum marks — is the unknown, so this is a "find the whole" problem, meaning the second formula applies. Call the maximum marks $M$.

Step 2. Set up the equation from the basic relationship. Since 60% of $M$ equals 72, $\dfrac{60}{100} \times M = 72$.

Step 3. Solve for $M$ using the rearranged formula. $M = \dfrac{72 \times 100}{60} = \dfrac{7200}{60} = 120$. So the maximum marks for the test were 120.

Step 4. Verify by substituting back into the original relationship: 60% of 120 is $\dfrac{60}{100} \times 120 = 72$, which matches the given part exactly, confirming the answer.

Competitive stretch

A shop advertises a discount of ₹450 on an item, and this discount is 15% of the marked price. Using the same "find the whole" formula, the marked price is $\dfrac{450 \times 100}{15} = \dfrac{45000}{15} = ₹3000$. The same rearranged formula that recovers a maximum-marks total from a test score works identically to recover an original price from a discount amount — the triangle of Part, Whole and Percentage does not care whether the context is marks or money.

• Using the "Part × 100 ÷ Percentage" formula upside down — sanity-check by asking whether the Whole you found is bigger than the Part, since it always must be when the percentage is under 100%.
• Multiplying by the raw percentage number instead of the fraction, writing $35 \times 260$ instead of $\dfrac{35}{100} \times 260$ — forgetting the division by 100 changes the answer by a factor of 100.
• Confusing "percentage of a quantity" with "percentage increase or decrease" — this topic asks only for a straight percentage value, with no addition or subtraction of the result to the original quantity.
• Rounding a fraction like $\dfrac{35}{100}$ before multiplying instead of simplifying exactly at the end, which can quietly shift the final answer away from the correct whole-number result.

Ask the tutor: stuck on whether a word problem wants you to find the part or the whole — ask your AI tutor to help identify which quantity is missing before setting up the equation.