Percentage increase and decrease

A newspaper headline says a city's population "rose by 15%," a shop banner shouts "flat 12% off," and a bank statement shows your savings "grew by 8% this year." All three sentences are doing the same piece of arithmetic: comparing a new value with an original value and expressing the gap as a percentage of that original value. Get the base right — always the original, never the new — and every one of these situations collapses into the same short formula. This article builds directly on percentage of a quantity, the skill from earlier in the Fractions in Disguise chapter (CBSE Class 8, NCERT Ganita Prakash Part-II). Here you learn to run that skill in two directions: finding the percentage change between two known values, and finding a new value directly once you know the original value and the percentage change. Both directions turn up again almost immediately, in profit/loss, discount, and simple/compound interest.

The core insight

Percentage increase or decrease always compares the change in a quantity to its original value, never to the new value, and can also be reversed to jump straight from an original value plus a known percentage change to the new value.

The core idea

A percentage change has two ingredients: the original value (the starting point, sometimes called the base) and the new value (what it became). The change itself is New minus Original, and turning that change into a percentage means dividing by the original value and multiplying by 100.

Percentage increase=New value−Original valueOriginal value×100\text{Percentage increase} = \frac{\text{New value} - \text{Original value}}{\text{Original value}} \times 100
Percentage decrease=Original value−New valueOriginal value×100\text{Percentage decrease} = \frac{\text{Original value} - \text{New value}}{\text{Original value}} \times 100

where: New value is the quantity after the change, Original value is the quantity before the change (always the denominator), and the result is a pure percentage with no units. The same relationship can be run backwards. If you already know the original value and the percentage change RR, you can jump straight to the new value without a separate subtraction step:

New value=Original value×(1±R100)\text{New value} = \text{Original value} \times \left(1 \pm \frac{R}{100}\right)

where: the ++ sign is used for an increase and the −- sign for a decrease, and RR is the percentage change expressed as a plain number (12, not 0.12). Reading a word problem correctly is the real skill here. If the problem gives you two actual values and asks "by what percentage," use the first pair of formulas. If it gives you one value and a percentage change and asks for the resulting value, use the direct formula instead. The figure below lines the two directions up side by side.

Figure
Two paths connecting an original value box to a new value box one showing the top path finding the percentage change and the other showing the bottom path finding the new value directly from a percentage
Two directions, one relationship Original value e.g. 4500 New value e.g. 5175 given both values, find percent change (New minus Original) over Original times 100 given Original and percent R, find New directly New equals Original times (1 plus or minus R over 100) Same base, same relationship, read in whichever direction the question asks for

The goal of this derivation is to show, using nothing more than algebra you already know, why the "direct" formula (New value equals Original value times one plus or minus R over 100) is not a separate rule to memorise but is simply the percentage-change formula rearranged to make the new value the subject, so that both formulas describe exactly the same relationship viewed from opposite ends.

Step 1. Start from the percentage increase formula and give the percentage change a name, RR, so that R=New−OriginalOriginal×100R = \dfrac{\text{New} - \text{Original}}{\text{Original}} \times 100.

Step 2. Divide both sides by 100 to isolate the fraction: R100=New−OriginalOriginal\dfrac{R}{100} = \dfrac{\text{New} - \text{Original}}{\text{Original}}.

Step 3. Multiply both sides by Original to clear the denominator: Original×R100=New−Original\text{Original} \times \dfrac{R}{100} = \text{New} - \text{Original}.

Step 4. Add Original to both sides and factor it out on the left: New=Original+Original×R100=Original×(1+R100)\text{New} = \text{Original} + \text{Original} \times \dfrac{R}{100} = \text{Original} \times \left(1 + \dfrac{R}{100}\right).

Step 5. The same four steps, starting from the percentage decrease formula instead, give New=Original×(1−R100)\text{New} = \text{Original} \times \left(1 - \dfrac{R}{100}\right). This confirms that the direct formula is just the percentage-change formula solved for the new value, so remembering one relationship is enough to recover the other by simple rearrangement whenever a problem needs it.

Worked example

A shirt originally priced at ₹800 is marked down by 12%. Find the new price, and separately verify the population example from earlier in the chapter still checks out.

Step 1. Identify the known quantities: Original value = ₹800, percentage decrease R=12R = 12. The question asks for the new value directly, so the direct formula applies with the minus sign.

Step 2. Substitute into New value=Original value×(1−R100)\text{New value} = \text{Original value} \times \left(1 - \dfrac{R}{100}\right):

New value=800×(1−12100)=800×0.88\text{New value} = 800 \times \left(1 - \frac{12}{100}\right) = 800 \times 0.88

Step 3. Compute the product: 800×0.88=704800 \times 0.88 = 704. So the new price of the shirt is ₹704.

Step 4. Verify with the other formula. The actual decrease is 800−704=96800 - 704 = 96. Checking, 96800×100=12\dfrac{96}{800} \times 100 = 12, which matches the given 12%, so the answer is consistent.

Step 5. As a second check on the same core idea, recall the village population that grew from 4,500 to 5,175 (Original = 4500, New = 5175). Using R=5175−45004500×100=6754500×100=15R = \dfrac{5175-4500}{4500}\times100 = \dfrac{675}{4500}\times100 = 15, the increase is 15%. Running this RR back through the direct formula, 4500×(1+15100)=4500×1.15=51754500 \times \left(1+\dfrac{15}{100}\right) = 4500 \times 1.15 = 5175, which lands exactly back on the given new value, confirming both formulas describe the same relationship.

Common traps

  • Dividing the change by the new value instead of the original value — the original value is always the base, even when the problem states the new value first in the sentence.
  • Mixing up the increase formula (1+R/1001 + R/100) with the decrease formula (1−R/1001 - R/100), especially when a problem uses positive-sounding words for a decrease, such as "marked down by" or "fell by."
  • Adding or subtracting successive percentage changes directly (assuming a 10% increase followed by a 10% decrease returns the original value) — each change must be applied to the updated value in turn, not combined arithmetically.
  • Forgetting to convert RR correctly in the direct formula — RR is used as R/100R/100 inside the bracket, not as a decimal typed in twice, so a 12% change means using 12/10012/100, not 0.12/1000.12/100.
  • Losing track of which quantity a percentage in the question is even attached to — "12% off the marked price" and "12% tax added to the cost price" both use the same formula shape but move the price in opposite directions, so read the wording carefully before choosing a sign.

Ask the tutor: confused about why a 10% rise followed by a 10% fall does not bring you back to the original value — ask your AI tutor to walk through the successive-change calculation with you step by step.

Why does the original value always go in the denominator, and never the new value?

Percentage change measures how big the change is relative to where you started. If you divided by the new value instead, the same ₹100 rise on a ₹500 item would give a different percentage than the same ₹100 rise described starting from ₹600, even though intuitively the size of the change relative to the starting point is what matters. Textbooks and exams fix the original value as the base precisely so that everyone measures the change the same way, so always check which number in the problem came first in time — that one is the original.

If a price rises by 10 per cent and then falls by 10 per cent, does it return to the original value?

No, and this is one of the most common traps in this topic. A 10% rise on ₹100 gives ₹110. A 10% fall is now calculated on ₹110, not on the original ₹100, giving 110×0.90=99110 \times 0.90 = 99. The final value is ₹99, slightly less than the original ₹100, because the second percentage is taken on a different (larger) base than the first. Successive percentage changes must always be applied one after another on the updated value, never combined by simply adding or subtracting the percentages.

How do I quickly decide whether a word problem needs the percentage-change formula or the direct new-value formula?

Look at what the question gives you and what it asks for. If it states two actual values (like "population went from 4,500 to 5,175") and asks "by what percentage," you have both values already, so use the percentage-change formula to find RR. If it states one value and a percentage (like "a shirt priced at ₹800 is marked down by 12%") and asks for the resulting price or amount, you already know Original and RR, so use the direct formula to jump straight to the new value without a separate subtraction step.

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Where this leads