Percentage applications

A percent is simply a fraction with denominator 100: $P\% = \dfrac{P}{100}$. Three conversions recur constantly, and fluency in all three directions is what Class 8 exams test.

where: $a$ is the part, $b$ is the whole, and multiplying by $100\%$ rescales the fraction onto a denominator of 100.

Two everyday problem types build directly on this. Finding a percentage of a quantity:

where: $N$ is the given quantity and $P$ is the percentage rate; the result is the "part."

Finding what percentage one quantity is of another simply reverses the question — given the part and the whole, the rate is unknown:

A third type, finding the whole when a percentage of it is known, appears whenever the quantity itself is the unknown, as in the mixture problem below. If $P\%$ of an unknown number $N$ equals $x$, then $N = \dfrac{x \times 100}{P}$. It helps to picture a percentage as a multiplying factor rather than a separate step. Increasing a quantity by $P\%$ multiplies it by $\left(1 + \dfrac{P}{100}\right)$; decreasing it by $P\%$ multiplies it by $\left(1 - \dfrac{P}{100}\right)$. This view is faster for mental work, and it becomes essential once two changes are applied one after another, since the factors simply multiply in sequence — which is exactly why successive percentage changes are not additive (see Common traps).

A school had 850 students last year; this year it has 918. Find the percentage increase, then use that same 20%-water idea to size a mixture.

Step 1. Find the increase: $918 - 850 = 68$ students.

Step 2. Apply the percentage-change formula, dividing by the original value 850:

So enrolment increased by 8%.

Step 3. Extend to a "find the whole" problem. A container has 1.5 L of milk; water is added until water is 20% of the new mixture. The milk (1.5 L, unchanged) is then $100\% - 20\% = 80\%$ of the total. Let the total be $N$: $80\%$ of $N = 1.5$, so $\dfrac{80}{100} \times N = 1.5$.

Step 4. Solve: $N = \dfrac{1.5 \times 100}{80} = 1.875$ L. Water added $= 1.875 - 1.5 = 0.375$ L. Total mixture = 1.875 L, water added = 0.375 L (check: $20\%$ of $1.875 = 0.375$ ✓).

Competitive stretch

Suppose the school's enrolment of 918 this year is itself expected to rise by a further 10% next year, then fall back by 10% the year after. Find the enrolment after both changes, and compare it with 918.

Step 1. Apply the 10% increase using the multiplying-factor view: $918 \times \left(1 + \dfrac{10}{100}\right) = 918 \times 1.10 = 1009.8$, which rounds to 1010 students (enrolment must be a whole number).

Step 2. Apply the 10% decrease to this new base of 1010, not to 918: $1010 \times \left(1 - \dfrac{10}{100}\right) = 1010 \times 0.90 = 909$ students.

Step 3. Compare with the original 918. The school ends up at 909 students, which is 9 fewer than the starting 918 — even though the two percentage changes "cancel" in size, they do not cancel in effect, because the second 10% was taken of a larger number than the first. This is the successive-percentage-change trap in action, and the multiplying-factor method ($1.10 \times 0.90 = 0.99$, a net 1% decrease) is the fastest way to see it coming.

• Dividing by the wrong base. Always divide by the original value in percentage change, never the new value.
• Confusing "percentage of" with "percentage more/less than." "20% more than 500" means $500 + 20\%$ of $500 = 600$, not $20\%$ of $500 = 100$.
• Treating successive percentage changes as additive. A 10% increase followed by a 10% decrease does not return to the original, because the second 10% applies to a different base: ₹500 up 10% is ₹550, but ₹550 down 10% is ₹495, not ₹500.
• Forgetting to convert before comparing. $\tfrac{3}{8}$ and $40\%$ cannot be compared directly until one is converted to the other's form ($\tfrac{3}{8} = 37.5\% < 40\%$).

Ask the tutor: if a percentage word problem is tripping you up — especially figuring out what the "original value" or "whole" actually is — ask the AI tutor to work through it with you step by step.