Understanding Percentage - From Intuition to Problem Solving

Published by Ganit Charcha | Category - Middle School Math Topics | 2025-08-21 12:24:50

Your phone battery jumps from $12$% to $24$%. Did it double? Yes—$24$ is twice $12$—but it’s still only $24$ out of $100$, so don’t start a long video call yet. Percentages tell us “how much out of $100$,” and they let us compare things fairly, even when the original totals are different.

What a Percentage Really Is

A percentage plays three roles. It is a number ($18$% means $0.18$), an operator (“find $18$% of $250$”), and a comparison (“A is $18$% more than B”). Most mistakes come from mixing up the base—the quantity you are taking the percent of.

Mental Math Toolbox

Think in anchors you can combine quickly: $10$% is “move one decimal place,” $1$% is “divide by $100$,” $5$% is “half of $10$%,” $25$% is “one quarter,” $50$% is “one half,” $75$% is “three quarters.”
Compute $15$% of $200$ in your head. Answer: $10$% of $200$ + $5$% of $200$ = $20 + 10 = 30$.

Converting Without Fuss

Percent to fraction: divide by $100$ and simplify ($45% = 45/100 = 9/20$).
Decimal to percent: multiply by $100$ and add the % sign ($0.37 → 37%$).
Fraction to percent: multiply by $100$ ($3/5 → 60$%).
Compute $7/25$ as a percent. Answer: $(7×4)/(25×4)=28/100=28$%.

The Three Classic Unknowns

A jacket marked $₹2400$ is sold with a $25$% discount.
Find the part: $25$% of $₹2400$ is $₹600$; discounted price = $₹1800$.
Find the percentage: If price drops from $₹2400$ to $₹1800$, the drop is $₹600$; $600/2400 = 25$%.
Find the whole: If the discounted price is $₹1800$ at $25$% off, the original is $1800 ÷ 0.75 = ₹2400$.
If $35$% of a number is $70$, what is the number? Answer: $70 ÷ 0.35 = 200$.

Percent Change and Successive Changes

Increase by $p$% multiplies by $1+\frac{p}{100}$; decrease by $p$% multiplies by $1−\frac{p}{100}$. 
Successive changes multiply. “Up $20$% then down $20$%” is $1.20 × 0.80 = 0.96$, so you end $4$% below the start.
“Up $25$% then down $25$%.” Where do you end? 
Answer: $1.25×0.75=0.9375$, i.e., $6.25$% below start.
If A is $25$% more than B, take B=$100$; then A=$125$. B is what percent less than A? The drop is $25$ on a base of $125$, so $25/125 = 20$%. “$+25$%” in one direction is not “$−25$%” in the other.

Percentage vs Percentage Points

A poll moves from $47$% to $49$%. That’s a rise of 2 percentage points. As a relative percentage change, it’s $2/47×100 ≈ 4.26$%. Use “percentage points” when both numbers are already percentages.

Picture It (without drawing)

Imagine a $10×10$ square grid with $100$ small squares. Shade $37$ squares—that’s $37$%. Shade $50$ squares—$50$%. The area you shade is literally “how many out of $100$.”

Modern, Global Uses

Marks: $360$ out of $500$ is $360/500×100 = 72$%.
Shopping: $$80$ with $15$% off gives $80 − 0.15 × 80 = 68$ dollars.

Taxes/Tips: A meal of €$50$ with $10$% tip adds €$5$.
GST/VAT reverse: ₹$1180$ including $18$% GST means base price = $1180/1.18 = ₹1000$.
Sport: A batter scoring $45$ runs off $60$ balls has $45/60×100=75$%. 
Football possession of $62$% means $62$ minutes out of every $100$ minutes of play, with the ball.
Health/Tech: Hydration target at $60$%, storage used $72$%, battery $24$%—all “out of $100$$.”

Note

“$20$% of $50$ equals $50$% of $20$” is true (both equal $10$) because multiplication commutes.
“$20$% off then $20$% on top of that brings you back to the original value” is false (So, $100$ becomes $80$ and then $20$% on top of that makes it $96$).

Worked Examples

1. $15$% of $240$: $10$% of $240$ is $24$ and $5$% of $240$ is $12$. Therefore, $15$% of $240$ is $36$.
2. $18$ out of $24$ as a percent is equal to $18/24=0.75=75%$.
3. Successive discounts of $20$% then $10$% give a single discount of $1−0.8×0.9 = 0.28 = 28$%. Price $₹2000 → ₹1440$.

4. Price rises from $₹800$ to $₹1000$: increase = $200/800 =25$%. If it then falls $25$%, new price is $1000×0.75=₹750$, which is $6.25$% below the original $₹800$.
5. Reverse-percent with tax: “$₹1180$ including $18$% GST” → before-tax = $1180/1.18 = ₹1000$.
6. Base flip: A is $25$% more than B; B is $20$% less than A (If B is $100$, then A is $125$ and hence B is $\frac{125-100}{125}= \frac{25}{100} = 20$% less than A).

Tiny Data Story

A $200$-student survey of favourite fruit records Mango $60$, Apple $50$, Banana $40$, Grapes $30$, Orange $20$. The percentages are $30$%, $25$%, $20$%, $15$%, $10$%—a perfect $100$%. If Mango rises by $10$ students next year (to $70$), its share becomes $70/200=35$%, a $5$ percentage-point rise.

Practice Set 

1) Convert $0.36$ to a percentage.
2) Express $7/25$ as a percentage.
3) Find $12$% of $350$.
4) What poercentage of $60$ is $18$?
5) If $35$% of a number is $70$, find the number?
6) The sale price after a $20$% discount is $₹480$. What was the marked price?
7) A quantity increases from $50$ to $65$. What is the percent increase?
8) A value is increased by $25$% and then decreased by $20$%. What is the overall percent change?
9) A price rises $10$% and then another $10$%. What is the single equivalent percent change?
10) A team won $12$ matches, which is $60$% of its games. How many games did it play in total?

11) A shop offers successive discounts of $20$% and $30$% on a shirt. Find the single equivalent discount $d$%. Also, give the general formula for combining $p$% and $q$% discounts.
12) After a $25$% loss on the original price, a trader increases the new price by $25$%. Overall, compared to the original, is the final price up or down, and by what percent?

Answer Key (check yourself).
1. $36$%

2. $28$%. 
3. $42$. 
4. $30$%. 
5. $200$.
6. $₹600$ (since $480 ÷ 0.8$). 
7. $30% (15/50)$. 
8. Net multiplier $1.25×0.8=1.0$, so 0% change. 
9. $1.1×1.1=1.21\Rightarrow 21$%.
10. $20$
11. $1−0.8×0.7=0.44⇒44$%. General: $1−(1−p/100)(1−q/100)$.
12. Down $6.25$% ($100 → 75 → 93.75$).

Final Thought

Treat every percent question as three decisions: What’s the base? What operation am I doing? Does my answer make sense by estimation? With these habits and the toolbox above, you’ll move smoothly from classroom fluency to solving relatively hard problems appearing in competitive mathematics examinations.




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