Fractions in Disguise Class 8: Percentages, Profit and Loss, Discounts and Compound Interest

What Fractions in Disguise is about

Key concepts

Percentages as fractions out of 100

NCERT section 1.1 Fractions as Percentages

Per cent means 'per hundred'. So $p\%=\frac{p}{100}$.

• Fraction to percentage: multiply by 100: $\frac{7}{20}=\frac{35}{100}=35\%$; $\frac{3}{8}\times 100=37.5\%$.
• Decimal to percentage: move the point two places right: $0.065=6.5\%$.
• Percentage to fraction: write over 100 and simplify: $24\%=\frac{24}{100}=\frac{6}{25}$.

Why 100? It gives enough detail for everyday use and is easy to picture. Per 10 would be too coarse, per 1000 (per mille) finer than needed for most purposes.

Benchmarks: $50\%=\frac{1}{2}$, $25\%=\frac{1}{4}$, $20\%=\frac{1}{5}$, $10\%=\frac{1}{10}$, $33\frac{1}{3}\%=\frac{1}{3}$, $12.5\%=\frac{1}{8}$. These help you estimate and compare quickly: $\frac{3}{11}$ is a little over $\frac{1}{4}$, so about 27%, clearly less than 61%.

Worked example

Express $\frac{18}{24}$ and $\frac{5}{16}$ as percentages, and 24% as a fraction in lowest terms.

Try it first, then show the solutionSolution

1. $\frac{18}{24}=\frac{3}{4}=75\%$.
2. $\frac{5}{16}\times 100=\frac{500}{16}=31.25\%$.
3. $24\%=\frac{24}{100}=\frac{6}{25}$.

Answer: 75%, 31.25% and $\frac{6}{25}$.

Percentage of a quantity and finding the whole

NCERT section 1.2 Percentage of Some Quantity

Percentage of a quantity: $p\%$ of $Q=\frac{p}{100}\times Q$. So 35% of 240 is $0.35\times 240=84$.

Useful mental routes:

• 10% is one-tenth; 1% is one-hundredth; build others from these (35% = 10% + 10% + 10% + 5%).
• 50% of $x$ = half of $x$; 25% = a quarter.
• Percentages above 100% mean more than the whole: 130% of 70 is 91; a harvest of 650 kg compared with 260 kg is 250%, that is 2.5 times.

What percentage is A of B? $\frac{A}{B}\times 100$: 45 is $\frac{45}{60}\times 100=75\%$ of 60.

Finding the whole from a part: if 80% of a number is 40, then 1% is $\frac{40}{80}=0.5$ and 100% is 50. Bar models (a bar split into 10 or 100 equal parts) show these relationships clearly.

Chains of facts: if 30% of $k$ is 70, then 60% of $k$ is 140 and 90% is 210 (just scale).

Worked example

Find 35% of 240 and 130% of 70. Then find the number of which 45 is 75%.

Try it first, then show the solutionSolution

1. $35\%$ of $240=0.35\times 240=84$ (or 10% is 24, so 30% is 72, 5% is 12; total 84).
2. $130\%$ of $70=1.3\times 70=91$.
3. 75% of $x$ is 45, so 1% is $\frac{45}{75}=0.6$ and $x=60$.

Answer: 84; 91; 60.

Comparing proportions with percentages

NCERT section 1.3 Using Percentages To Compare Proportions

Raw numbers can mislead when the totals differ. Converting to percentages puts everything on the same base of 100.

• Class A: 27 of 36 students passed $=75\%$; class B: 33 of 40 passed $=82.5\%$. Class B did better even though both numbers look similar.
• A score of 42 out of 50 (84%) beats 70 out of 80 (87.5%)? No: 87.5% is higher, so the second is the better performance.

Fraction thinking also works: to compare $\frac{27}{36}$ and $\frac{33}{40}$ you could cross-multiply, but percentages make several comparisons at once easy (a whole table of classes or shops).

Worked example

In a quiz, Meena scored 42 out of 56 and Arjun scored 52 out of 65. Who did better?

Try it first, then show the solutionSolution

1. Meena: $\frac{42}{56}\times 100=75\%$.
2. Arjun: $\frac{52}{65}\times 100=80\%$.
3. Arjun's percentage is higher.

Answer: Arjun (80% against 75%).

Percentage increase and decrease

NCERT section 1.3 Using Percentages To Compare Proportions

Always compare the change with the original value:

• Percentage increase $=\frac{\text{increase}}{\text{original}}\times 100$;
• Percentage decrease $=\frac{\text{decrease}}{\text{original}}\times 100$.

Example: a ticket price rises from ₹45 to ₹54: increase ₹9, so $\frac{9}{45}\times 100=20\%$.

Using multipliers: an increase of $r\%$ multiplies by $(1+\frac{r}{100})$; a decrease multiplies by $(1-\frac{r}{100})$. A 5% rise makes $p$ into $1.05p$.

Watch out: a 20% rise followed by a 20% fall does not bring you back: $1.2\times 0.8=0.96$, a 4% fall overall, because the fall is taken on the bigger number. Also, 'fell by 85%' means the new value is 15% of the old one.

Worked example

A number increased by 25% becomes 150. Find the number. What is the percentage decrease when a price falls from ₹160 to ₹120?

Try it first, then show the solutionSolution

1. An increase of 25% multiplies by 1.25: $1.25x=150$, so $x=120$.
2. Decrease $=160-120=40$.
3. Percentage decrease $=\frac{40}{160}\times 100=25\%$.

Answer: 120; 25% decrease.

Profit, loss and discount

NCERT section 1.3 Using Percentages To Compare Proportions

• Cost price (CP): what the seller paid. Selling price (SP): what the customer pays.
• Profit $=SP-CP$ (when $SP>CP$); loss $=CP-SP$ (when $SP<CP$).
• Profit % (profit margin on cost) $=\frac{\text{profit}}{CP}\times 100$; loss % likewise on CP.
• Marked (list) price and discount: a 15% discount on a marked price of ₹1,200 gives $SP=0.85\times 1200=₹1,020$. The discount is on the marked price; profit is still measured on CP.
• Gross and net profit: gross profit $=$ sales $-$ cost of goods; net profit also subtracts running expenses (rent, salaries). Net profit % is often calculated on the revenue (total sales), for example ₹45,000 net profit on ₹1,50,000 revenue is 30%.

To price for a target margin: $SP=CP\times (1+\frac{m}{100})$. Material worth ₹640 sold at a 35% margin: $640\times 1.35=₹864$.

Worked example

A shopkeeper buys a mixer for ₹850 and marks it at ₹1,200. He gives a 15% discount. Find the selling price and his profit percentage.

Try it first, then show the solutionSolution

1. Discount $=15\%$ of $1200=₹180$, so $SP=1200-180=₹1,020$.
2. Profit $=1020-850=₹170$.
3. Profit $\%=\frac{170}{850}\times 100=20\%$.

Answer: SP ₹1,020; profit 20%.

Taxes such as GST

NCERT section 1.3 Using Percentages To Compare Proportions

Taxes such as GST (Goods and Services Tax) are given as a percentage of the price and are added to the bill; the tax goes to the government, not the shopkeeper.

• Price before tax ₹2,400 with 12% GST: tax $=₹288$, bill $=2400\times 1.12=₹2,688$.
• Working backwards: if a bill of ₹1,180 includes 18% GST, then $1.18\times \text{price}=1180$, so the price is ₹1,000 and the tax ₹180. Do not take 18% of 1,180.

When there is a discount and a tax, apply the discount to the marked price first, then add tax to the discounted price.

Worked example

A bill of ₹1,180 includes 18% GST. Find the price before tax and the tax paid.

Try it first, then show the solutionSolution

1. Price before tax $\times 1.18=1180$.
2. Price before tax $=1180÷1.18=₹1,000$.
3. Tax $=1180-1000=₹180$ (check: 18% of 1000 is 180).

Answer: ₹1,000 before tax; ₹180 GST.

Growth with and without compounding

NCERT section 1.3 Using Percentages To Compare Proportions

When money is deposited at an interest rate of $r$ per annum (p.a.), the principal $p$ earns interest each year.

Without compounding (interest paid out each year): the interest is the same every year, $p\times r$.

• Amount after $t$ years $=p+prt=p(1+rt)$. Growth is linear.

With compounding (interest added back, so next year's interest is on a bigger amount):

• Amount after $t$ years $=p(1+r{)}^t$. Growth is exponential.

Here $r$ is written as a decimal (10% = 0.1). Example: ₹6,000 at 10% p.a. for 3 years: without compounding $6000\times 1.3=₹7,800$; with compounding $6000\times 1.1^3=6000\times 1.331=₹7,986$.

The same multiplier idea gives population growth: 2,00,000 growing 4% a year becomes $200000\times 1.04^2=2,16,320$ after 2 years.

Worked example

₹8,000 is deposited at 5% p.a. for 3 years. Find the amount with and without compounding, and the difference.

Try it first, then show the solutionSolution

1. Without compounding: $8000(1+0.05\times 3)=8000\times 1.15=₹9,200$.
2. With compounding: $8000\times 1.05^3=8000\times 1.157625=₹9,261$.
3. Difference $=9261-9200=₹61$.

Answer: ₹9,200 and ₹9,261; compounding gives ₹61 more.

Decline, depreciation and tricky percentages

NCERT section 1.3 Using Percentages To Compare Proportions

Depreciation: many items lose value with use and age. A fall of $r\%$ each year multiplies the value by $(1-r)$ each year: a ₹80,000 scooter depreciating 10% a year is worth $80000\times 0.9^2=₹64,800$ after 2 years. Falling populations work the same way.

Tricky percentages:

• Successive discounts do not add: '30% + 20% off' means 30% off, then 20% off the reduced price: $0.7\times 0.8=0.56$, so you pay 56% (a 44% discount), not 50%.
• Percentages versus amounts: depositing ₹100 to get ₹300 is a 200% gain; depositing ₹1,000 to get ₹1,500 is only 50%, though the second gives more money. Which is better depends on what you care about.
• Flat versus percentage offers: '20% off or ₹75 off' — 20% is bigger only when the bill exceeds $75÷0.2=₹375$.
• Base matters: 'A is 25% more than B' does not mean 'B is 25% less than A' (it is 20% less).

Worked example

A cake costs ₹200. Bakery P offers '30% + 20%' off and bakery Q offers 50% off. Which is cheaper, and what single discount equals P's offer?

Try it first, then show the solutionSolution

1. P: after 30% off, $200\times 0.7=₹140$; after a further 20% off, $140\times 0.8=₹112$.
2. Q: $200\times 0.5=₹100$.
3. P's price is 56% of the original, a single discount of 44%. Q is cheaper by ₹12.

Answer: Q (₹100); P's offer equals a 44% discount.

Maths in real life: Fractions in Disguise

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Topics in this chapter

• Percentages and Percentage Change
• Profit, Loss, Discount, Tax and Compound Interest

Fractions in Disguise formula sheet

Percentage

$p\%=\frac{p}{100}$

Part as a percentage

$\frac{A}{B}\times 100$

Percentage change

$\frac{\text{change}}{\text{original}}\times 100$

Increase or decrease multiplier

$\times \left(1\pm \frac{r}{100}\right)$

Profit and loss percentage

$\frac{SP-CP}{CP}\times 100$

Discount

$SP=MP\times \left(1-\frac{d}{100}\right)$

Price with tax

$\text{bill}=\text{price}\times \left(1+\frac{t}{100}\right)$

Without compounding

$A=p(1+rt)$

With compounding

$A=p(1+r{)}^t$

r as a decimal, t in years.

Depreciation

$V=V_0(1-r{)}^t$

How the school exam asks Fractions in Disguise

The question patterns to expect, and the method that works for each.

Tricks to remember Fractions in Disguise

Short memory aids. Say each one aloud once, then use it on a question the same day.

• Per cent = per hundred

  Read '%' as '÷ 100' every time.

  Use it when: Any conversion.

• Change over original

  'New minus old, over old' gives percentage change.

  Use it when: Increase and decrease.

• Profit on cost, discount on marked

  Profit looks back at what you paid; discount looks at the tag.

  Use it when: Choosing the correct base.

• Multipliers multiply

  Successive changes multiply ($0.7\times 0.8$); they never simply add.

  Use it when: Successive discounts and compounding.

• Simple is a straight line, compound is a curve

  Without compounding the amount rises by the same step; with compounding the step keeps growing.

  Use it when: Comparing the two options.

Fractions in Disguise: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section E style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 4-mark question split into 1 + 1 + 2 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Fractions in Disguise

The slips examiners see most often, and what to write instead.

• ✗Mistake: Taking percentage change on the new value

  The base is chosen as whichever number is handy.

  Wrong working

  Price rose from ₹45 to ₹54: $\frac{9}{54}\times 100\approx 16.7\%$.

  Right working

  Base is the original ₹45: $\frac{9}{45}\times 100=20\%$.

  ✓Do this: Divide by the original (old) value.

  Usually costs: The answer mark.

• ✗Mistake: Adding successive discounts

  30% + 20% looks like 50%.

  Wrong working

  30% then 20% off ₹200 is 50% off: ₹100.

  Right working

  $200\times 0.7\times 0.8=₹112$: a 44% discount.

  ✓Do this: Apply each discount to the current price, or multiply the multipliers.

  Usually costs: 1-2 marks.

• ✗Mistake: Finding tax backwards by subtracting a percentage of the bill

  The bill is treated as the base.

  Wrong working

  Bill ₹1,180 with 18% GST: price $=1180-18\%$ of $1180=₹967.60$.

  Right working

  Price $\times 1.18=1180$, so price $=₹1,000$.

  ✓Do this: Divide by the multiplier when working backwards.

  Usually costs: Method and answer.

• ✗Mistake: Assuming equal profit and loss percentages cancel

  +20% and −20% look balanced.

  Wrong working

  Two items each sold at ₹60,000, one at 20% profit and one at 20% loss: no profit, no loss.

  Right working

  CPs are ₹50,000 and ₹75,000; total CP ₹1,25,000, SP ₹1,20,000: loss of ₹5,000, that is 4%.

  ✓Do this: Find each cost price before combining.

  Usually costs: Whole answer.

• ✗Mistake: Using the rate as a whole number in the formula

  10% is put in as 10.

  Wrong working

  $A=6000(1+10{)}^3$

  Right working

  $A=6000(1+0.1{)}^3=6000\times 1.331=₹7,986$.

  ✓Do this: Write $r$ as a decimal (or use $\frac{r}{100}$).

  Usually costs: Answer mark.

NCERT exercise guide: Fractions in Disguise

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

Practice questions on Fractions in Disguise

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

1. Q1 · 1 mark · MCQExam level

   140% of 85 is
   (a) 11.9
   (b) 119
   (c) 125
   (d) 140

   Hint

   $1.4\times 85$.

   Answer

   (b) 119

2. Q2 · 1 mark · MCQExam level

   0.065 written as a percentage is
   (a) 0.65%
   (b) 6.5%
   (c) 65%
   (d) 650%

   Hint

   Multiply by 100.

   Answer

   (b) 6.5%

3. Q3 · 1 mark · Very short answerExam level

   26 of the 40 students in a class bring lunch from home. What percentage is that?

   Hint

   $\frac{26}{40}$.

   Answer

   65%

4. Q4 · 2 marks · Very short answerExam level

   The price of a bag falls from ₹250 to ₹200. Find the percentage decrease.

   Hint

   Base is ₹250.

   Answer

   $\frac{50}{250}\times 100=20\%$

5. Q5 · 2 marks · Very short answerHarder exam level

   A fridge was bought for ₹30,800 after a 12% discount. Find its marked price.

   Hint

   $0.88\times MP$.

   Answer

   $30800÷0.88=₹35,000$

6. Q6 · 3 marks · Short answerHarder exam level

   A scooter bought for ₹90,000 loses 15% of its value every year. Find its value after 2 years and the total loss in value.

   Hint

   Multiply by 0.85 each year.

   Answer

   $90000\times 0.85=76500$; $76500\times 0.85=₹65,025$. Loss in value $=₹24,975$.

7. Q7 · 3 marks · Short answerHarder exam level

   A store offers either 20% off or ₹75 off on bills above ₹250. Which is better for a bill of (a) ₹300, (b) ₹500? At what bill amount are they equal?

   Hint

   Compare 20% of the bill with 75.

   Answer

   (a) 20% of 300 = ₹60, so ₹75 off is better. (b) 20% of 500 = ₹100, so 20% is better. Equal when $0.2B=75$, $B=₹375$.

8. Q8 · 2 marks · Very short answerHarder exam level

   A carpenter's materials for a table cost ₹720. He wants a 40% profit on cost. At what price should he sell it?

   Hint

   $CP\times 1.4$.

   Answer

   ₹1,008

9. Q9 · 2 marks · Very short answerHarder exam level

   In two tests, Ravi scored 34 out of 40 and Anu scored 52 out of 64. Who did better? By how many percentage points?

   Hint

   Convert both to percentages.

   Answer

   Ravi 85%, Anu 81.25%; Ravi by 3.75 percentage points.

10. Q10 · 5 marks · Long answerHarder exam level

    Two friends each borrow ₹24,000 for 2 years: Ravi at 9% p.a. without compounding, Sana at 8% p.a. compounded annually. (a) Find Ravi's interest. (b) Find Sana's amount and interest year by year. (c) Who pays more and by how much?

    Hint

    $prt$ and $p(1+r{)}^t$.

    Answer

    (a) $24000\times 0.09\times 2=₹4,320$. (b) Year 1: $24000\times 1.08=₹25,920$; year 2: $25920\times 1.08=₹27,993.60$; interest ₹3,993.60. (c) Ravi pays ₹326.40 more.

11. Q11 · 3 marks · Short answerHOTS / competency

    ₹1,000 is invested at 12% p.a. After how many whole years does it first become at least ₹2,000 (a) without compounding, (b) with annual compounding?

    Hint

    Test year by year.

    Answer

    (a) $1+0.12t\ge 2$ needs $t\ge 8.33$: after 9 years (₹2,080). (b) $1.12^6\approx 1.974$, $1.12^7\approx 2.211$: after 7 years (about ₹2,211).

12. Q12 · 3 marks · Short answerHOTS / competency

    The demand for film cameras fell by 80% over some years. (a) What percentage of the old demand is the new demand? (b) By what percentage must demand now rise to get back to the old level?

    Hint

    Base changes.

    Answer

    (a) 20%. (b) From 20 to 100 is a rise of 80 on a base of 20: $\frac{80}{20}\times 100=400\%$.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1. 1.What does 'per cent' mean?Show answer

   Per hundred: $p\%=\frac{p}{100}$.

2. 2.Write $\frac{3}{8}$ as a percentage.Show answer

   37.5%.

3. 3.How do you find what percentage A is of B?Show answer

   $\frac{A}{B}\times 100$.

4. 4.What is the base for percentage increase?Show answer

   The original value.

5. 5.On what is profit percentage usually calculated?Show answer

   On the cost price.

6. 6.What is the selling price after a $d\%$ discount on marked price MP?Show answer

   $MP\times (1-\frac{d}{100})$.

7. 7.Give the amount formula without compounding.Show answer

   $A=p(1+rt)$.

8. 8.Give the amount formula with annual compounding.Show answer

   $A=p(1+r{)}^t$.

9. 9.What is depreciation?Show answer

   The fall in value of an item because of use and age.

10. 10.What single discount equals 30% followed by 20%?Show answer

    44%.

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Can you convert $\frac{5}{16}$ to a percentage?
• ☐Can you find 62% of 360 mentally using 10% and 1%?
• ☐Can you find a number when 80% of it is given?
• ☐Can you compare two test scores out of different totals?
• ☐Can you find the percentage change from old to new?
• ☐Can you find SP after a discount and the profit % on cost?
• ☐Can you remove GST from a bill that includes it?
• ☐Can you compute an amount with and without compounding?
• ☐Can you find the value of an item after depreciation?
• ☐Can you explain why '30% + 20%' is not 50% off?

Fractions in Disguise: frequently asked questions