Chapter 1: Fractions in Disguise

The word per cent means per hundred. To express a fraction as a percentage, we rewrite the fraction with a denominator of $$100$$ (or equivalently, multiply the fraction by $$100$$ and attach the '$$\%$$' sign).

Here the red part is $$\dfrac{3}{4}$$ of the mixture. Multiplying the numerator and the denominator by $$25$$ gives an equivalent fraction with denominator $$100$$:

$$\dfrac{3}{4}=\dfrac{3\times 25}{4\times 25}=\dfrac{75}{100}=75\%.$$

Equivalently, $$\dfrac{3}{4}\times 100=75$$, so red constitutes $$75\%$$ of the mixture.

To express the fraction $$\dfrac{2}{5}$$ as a percentage, we rewrite it with denominator $$100$$. Multiplying the numerator and the denominator by $$20$$:

$$\dfrac{2}{5}=\dfrac{2\times 20}{5\times 20}=\dfrac{40}{100}=40\%.$$

Equivalently, we may compute $$\dfrac{2}{5}\times 100=\dfrac{200}{5}=40$$, so $$\dfrac{2}{5}=40\%$$.

Thus Surya plans to save $$40\%$$ of his prize money.

The symbol '$$\%$$' stands for '$$\div 100$$'. So writing a percentage as a fraction simply means putting it over $$100$$ and then reducing to lowest terms.

$$24\%=\dfrac{24}{100}.$$

The greatest common factor of $$24$$ and $$100$$ is $$4$$. Dividing numerator and denominator by $$4$$:

$$\dfrac{24}{100}=\dfrac{24\div 4}{100\div 4}=\dfrac{6}{25}.$$

Hence $$24\%=\dfrac{6}{25}$$ in lowest terms.

The mixture consists of only red and yellow paint, so the two fractions must add up to $$1$$ (i.e., $$100\%$$ of the mixture).

Red makes up $$\dfrac{3}{4}$$ of the mixture, so the yellow part is

$$1-\dfrac{3}{4}=\dfrac{4-3}{4}=\dfrac{1}{4}.$$

Expressing $$\dfrac{1}{4}$$ as a percentage:

$$\dfrac{1}{4}=\dfrac{1\times 25}{4\times 25}=\dfrac{25}{100}=25\%.$$

Alternatively, since red is $$75\%$$, yellow is $$100\%-75\%=25\%$$.

The whole bar represents $$1$$, which is the same as $$100\%$$. Since the bar is divided into $$5$$ equal parts, each part corresponds to

$$\dfrac{1}{5}=\dfrac{100\%}{5}=20\%.$$

Counting the parts from left to right, the marks are at $$1,2,3,4$$ fifths, so the percentages are $$20\%,40\%,60\%,80\%$$, with $$0\%$$ at the left end and $$100\%$$ at the right end. The completed bar is:

The mark at $$\dfrac{2}{5}$$ therefore reads $$40\%$$, confirming Example 2.

The fraction of marbles that are white is $$\dfrac{15}{25}$$. Converting to a percentage:

$$\dfrac{15}{25}\times 100=\dfrac{15\times 100}{25}=\dfrac{1500}{25}=60.$$

So $$60\%$$ of Nandini's marbles are white, which matches option (iv).

The fraction of students who walk to school is $$\dfrac{15}{80}$$. Converting to a percentage:

$$\dfrac{15}{80}\times 100=\dfrac{1500}{80}=\dfrac{75}{4}=18.75.$$

Therefore $$18.75\%$$ (equivalently $$18\tfrac{3}{4}\%$$) of the students come to school by walking.

Think of the track as a bar from $$0\%$$ (Start) to $$100\%$$ (Finish). Estimating each runner's position by eye:

• Runner A is close to the Start but has moved a bit — roughly one-fifth of the way, i.e. about $$20\%$$.
• Runner B is between one-third and one-half of the way — a value just under $$40\%$$, i.e. about $$38\%$$.
• Runner C is a little past the middle of the track — about $$55\%$$.
• Runner D is very close to the Finish — well past three-quarters, roughly $$93\%$$.

Matching each runner to the closest option from the list:

The two options $$72\%$$ and $$84\%$$ are not used because no runner in the picture is at those positions.

The percentages tell us only what fraction of each biscuit is sugar — not the actual amount of sugar consumed. To find the actual sugar eaten by each person, we would need to know how much biscuit each of them ate.

For example, if Madhu ate a very large biscuit and Madhav ate a very small one, Madhu could still consume more sugar even though her biscuit had a smaller sugar percentage; the reverse is also possible.

So, based on the information given, we cannot decide who ate more sugar. We also need the weights (or amounts) of the biscuits they ate.

Yes. The number $$0.5$$ is just another name for the fraction $$\dfrac{1}{2}$$:

$$0.5=\dfrac{5}{10}=\dfrac{1}{2}=\dfrac{50}{100}=50\%.$$

So for any value $$x$$,

$$50\%\text{ of }x=\dfrac{50}{100}\times x=\dfrac{1}{2}\times x=0.5\times x.$$

All three expressions give exactly the same result. For instance, $$50\%$$ of $$240$$ is $$\dfrac{1}{2}\times 240=120$$, and $$0.5\times 240=120$$ as well.

The minimum marks needed for an A grade are $$80\%$$ of the maximum marks, i.e. $$80\%$$ of $$75$$.

$$80\%\text{ of }75=\dfrac{80}{100}\times 75=\dfrac{80\times 75}{100}=\dfrac{6000}{100}=60.$$

So Zubin must score at least $$60$$ marks out of $$75$$ to get an A grade.

Fraction of millet in the mixture. Out of every $$2+7=9$$ parts of the mixture, $$2$$ parts are millet. So the fraction of millet is

$$\dfrac{2}{9}.$$

Expressing this as a percentage:

$$\dfrac{2}{9}\times 100=\dfrac{200}{9}=22\dfrac{2}{9}\approx 22.22.$$

Therefore millet constitutes about $$22.22\%$$ (exactly $$22\dfrac{2}{9}\%$$) of the mixture.

Amount of millet in $$500\,\mathrm{ml}$$ of the mixture.

$$\dfrac{2}{9}\times 500=\dfrac{1000}{9}=111\dfrac{1}{9}\approx 111.11\,\mathrm{ml}.$$

So about $$111.11\,\mathrm{ml}$$ of millet should be used (with the remaining $$\approx 388.89\,\mathrm{ml}$$ being water).

Let the total distance from Delhi to Agra be $$D$$ km. Then $$40\%$$ of $$D$$ corresponds to $$92$$ km:

$$\dfrac{40}{100}\times D=92\quad\Rightarrow\quad D=\dfrac{92\times 100}{40}=\dfrac{9200}{40}=230.$$

So the total distance is $$230\,\mathrm{km}$$. The distance remaining is

$$230-92=138\,\mathrm{km}.$$

Alternatively: since $$40\%$$ corresponds to $$92$$ km, $$10\%$$ is $$\dfrac{92}{4}=23$$ km, and the remaining $$60\%$$ is $$6\times 23=138$$ km.

The daily target is ₹$$5000$$. For each day we compute (sales/target) $$\times 100$$.

Day 1: Sales $$=$$ ₹$$2000$$.

$$\dfrac{2000}{5000}\times 100=\dfrac{2}{5}\times 100=40\%.$$

So he achieved $$40\%$$ of his target on Day 1.

Day 2: Sales $$=$$ ₹$$3500$$.

$$\dfrac{3500}{5000}\times 100=\dfrac{7}{10}\times 100=70\%.$$

So he achieved $$70\%$$ of his target on Day 2.

Take last year's harvest ($$260\,\mathrm{kg}$$) as the reference $$100\%$$. Then this year's harvest as a percentage of last year's is

$$\dfrac{650}{260}\times 100.$$

Simplify: $$\dfrac{650}{260}=\dfrac{65}{26}=\dfrac{5}{2}$$, so

$$\dfrac{5}{2}\times 100=250.$$

Therefore this year's harvest is $$250\%$$ of last year's harvest — a value greater than $$100\%$$, which correctly signals that the harvest has more than doubled.

Sugar eaten by each person $$=$$ (percentage of sugar) $$\times$$ (weight of biscuits eaten).

Madhu: $$25\%$$ of $$120\,\mathrm{g}$$ $$=$$

$$\dfrac{25}{100}\times 120=\dfrac{1}{4}\times 120=30\,\mathrm{g}.$$

Madhav: $$35\%$$ of $$95\,\mathrm{g}$$ $$=$$

$$\dfrac{35}{100}\times 95=\dfrac{35\times 95}{100}=\dfrac{3325}{100}=33.25\,\mathrm{g}.$$

Since $$33.25>30$$, Madhav consumed more sugar (about $$3.25\,\mathrm{g}$$ more) than Madhu.

Handy shortcuts for mental computation:

• $$25\%$$ of a value is $$\dfrac{1}{4}$$ of the value.
• $$10\%$$ of a value is $$\dfrac{1}{10}$$ of the value (just shift the decimal point one place to the left).
• $$20\%$$ of a value is $$\dfrac{1}{5}$$ of the value, which equals twice the $$10\%$$ value.
• $$5\%$$ of a value is half of $$10\%$$ of the value.

Applying these rules gives:

$$40\%$$ of a value is twice its $$20\%$$ (or four times its $$10\%$$). Using the $$10\%$$ and $$20\%$$ rows already computed:

Relationship among $$20\%$$, $$5\%$$ and $$25\%$$: Since $$20+5=25$$, for any value $$y$$ we get

$$(20\%\text{ of }y)+(5\%\text{ of }y)=25\%\text{ of }y.$$

Verify with $$y=80$$: $$20\%$$ of $$80=16$$, $$5\%$$ of $$80=4$$, and $$16+4=20=25\%$$ of $$80$$. This gives another quick mental route to $$25\%$$: add the $$20\%$$ and $$5\%$$ values.

Similarly, $$15\%$$ of $$y$$ can be split as $$10\%$$ of $$y$$ + $$5\%$$ of $$y$$ (since $$10+5=15$$). Using the $$10\%$$ and $$5\%$$ values we already know:

Break each percentage into easy pieces we already know how to compute mentally — namely $$50\%,25\%,10\%,5\%$$ — and then add or subtract.

• $$75\%$$ of $$y=50\%$$ of $$y+25\%$$ of $$y$$. Equivalently, $$75\%$$ of $$y=y-25\%$$ of $$y$$ (take three-quarters).
• $$90\%$$ of $$y=y-10\%$$ of $$y$$ (i.e. total minus one-tenth). Or add $$50\%+25\%+10\%+5\%$$.
• $$70\%$$ of $$y=50\%$$ of $$y+20\%$$ of $$y$$, or equivalently $$y-30\%$$ of $$y=y-(20\%+10\%)$$ of $$y$$.
• $$55\%$$ of $$y=50\%$$ of $$y+5\%$$ of $$y$$.

As a quick check with $$y=200$$: $$75\%$$ of $$200=150$$ ($$=100+50$$); $$90\%$$ of $$200=180$$ ($$=200-20$$); $$70\%$$ of $$200=140$$ ($$=100+40$$); $$55\%$$ of $$200=110$$ ($$=100+10$$).

$$10\%$$ means $$\dfrac{10}{100}=\dfrac{1}{10}=0.1$$. So $$10\%$$ of a quantity $$x$$ is

$$10\%\text{ of }x=\dfrac{10}{100}\times x=0.1\times x.$$

Multiplying by $$0.1$$ is the same as shifting the decimal point of $$x$$ one place to the left — a very quick mental operation.

For any percentage $$p\%$$, we have $$p\%=\dfrac{p}{100}$$ (which we then simplify) and the corresponding decimal is just $$p\div 100$$.

Use $$\dfrac{\text{sales}}{\text{target}}\times 100$$ each day, with target $$=$$ ₹$$5000$$.

Day 3 sales $$=$$ ₹$$5000$$:

$$\dfrac{5000}{5000}\times 100=1\times 100=100\%.$$

He exactly met his target.

Day 4 sales $$=$$ ₹$$6000$$:

$$\dfrac{6000}{5000}\times 100=\dfrac{6}{5}\times 100=120\%.$$

He exceeded his target by $$20\%$$.

Percentage of target achieved on Day 4 $$=$$

$$\dfrac{\text{sales on Day 4}}{\text{target}}\times 100=\dfrac{6000}{5000}\times 100=\dfrac{6}{5}\times 100=120\%.$$

So Kishanlal achieved $$120\%$$ of his target on Day 4, i.e. he exceeded his target by $$20\%$$.

Use $$\dfrac{\text{sales}}{\text{target}}\times 100$$, target $$=$$ ₹$$5000$$.

Day 5 sales $$=$$ ₹$$7800$$:

$$\dfrac{7800}{5000}\times 100=\dfrac{7800}{50}=156\%.$$

Day 6 sales $$=$$ ₹$$9550$$:

$$\dfrac{9550}{5000}\times 100=\dfrac{9550}{50}=191\%.$$

So Kishanlal achieved $$156\%$$ of his target on Day 5 and $$191\%$$ on Day 6.

Sales $$=$$ percentage achieved $$\times$$ target $$=$$ percentage $$\times$$ ₹$$5000$$.

Day 7: $$150\%$$ of $$5000=$$

$$\dfrac{150}{100}\times 5000=1.5\times 5000=7500.$$

So sales $$=$$ ₹$$7500$$.

Day 8: $$210\%$$ of $$5000=$$

$$\dfrac{210}{100}\times 5000=2.1\times 5000=10500.$$

So sales $$=$$ ₹$$10{,}500$$.

For every $$p\%$$, the corresponding fraction is $$\dfrac{p}{100}$$ and the decimal is $$p\div 100$$.

On the bar from $$0$$ to $$4$$: mark $$15\%\ (0.15)$$ and $$28.9\%\ (0.289)$$ close to $$0$$; $$90\%\ (0.9)$$ just before $$1$$; $$110\%\ (1.1)$$ just after $$1$$; $$173\%\ (1.73)$$ between $$1$$ and $$2$$; $$200\%\ (2)$$ at $$2$$; $$250\%\ (2.5)$$ midway between $$2$$ and $$3$$; $$305\%\ (3.05)$$ just after $$3$$; and $$358\%\ (3.58)$$ between $$3$$ and $$4$$ (closer to $$4$$).

Red paint is $$\dfrac{3}{4}$$ (i.e. $$75\%$$) of the mixture. So the amount of red paint used is

$$\dfrac{3}{4}\times 60\,\mathrm{ml}=\dfrac{180}{4}\,\mathrm{ml}=45\,\mathrm{ml}.$$

Therefore Surya used $$45\,\mathrm{ml}$$ of red paint (and $$60-45=15\,\mathrm{ml}$$ of yellow paint).

Convert both quantities to the same unit — say, hours.

$$10\%\text{ of a day}=\dfrac{10}{100}\times 24\,\text{h}=2.4\,\text{h}=2\,\text{h}\;24\,\text{min}.$$

$$1\%\text{ of a week}=\dfrac{1}{100}\times (7\times 24)\,\text{h}=\dfrac{168}{100}\,\text{h}=1.68\,\text{h}=1\,\text{h}\;40.8\,\text{min}.$$

Since $$2.4>1.68$$, $$10\%$$ of a day is longer than $$1\%$$ of a week (by $$0.72\,\text{h}\approx 43\,\text{min}$$).

Similar questions students could try: Is $$25\%$$ of a month longer than $$5\%$$ of a year? Is $$50\%$$ of a minute longer than $$1\%$$ of an hour?

On the $$n$$th day the bull is given $$(n+1)$$ units of fodder and eats $$n$$ units, so the percentage of the fodder the bull consumed on day $$n$$ is

$$\dfrac{n}{n+1}\times 100\%.$$

A few sample days:

Observation: The percentages $$50\%,66.67\%,75\%,80\%,\ldots,99\%$$ keep increasing as the days go by, but they never reach $$100\%$$. The bull always leaves exactly $$1$$ unit uneaten, and this $$1$$ unit is a smaller and smaller share of the total each day, so the percentage eaten approaches (but never touches) $$100\%$$.

$$100\%$$ of the plantation is $$5$$ times $$20\%$$ of the plantation. If the rate of work is constant, five times as much area needs five times as much time.

$$\text{Total days}=5\times 18=90\text{ days}.$$

Equivalently, if $$20\%$$ takes $$18$$ days, then $$1\%$$ takes $$\dfrac{18}{20}=0.9$$ day, and $$100\%$$ takes $$100\times 0.9=90$$ days.

Why the assumption matters. If the workers work faster/slower on different portions of the plantation (e.g. because some patches have thicker foliage, or because more/fewer workers are available, or because the terrain varies), then the relationship 'time is proportional to area' breaks down, and we cannot simply multiply by $$5$$. The assumption of a constant rate is what lets us use direct proportion.

The percentages $$10\%,80\%,10\%$$ add up to $$100\%$$, so they represent shares of the total session of $$90$$ min.

Warm up: $$10\%$$ of $$90=\dfrac{10}{100}\times 90=9$$ min.

Play: $$80\%$$ of $$90=\dfrac{80}{100}\times 90=72$$ min.

Cool down: $$10\%$$ of $$90=9$$ min.

Check: $$9+72+9=90$$ min. ✓

The world population in 2025 is approximately $$8.2$$ billion (i.e. $$8{,}20{,}00{,}00{,}000$$ people). Then

$$90\%\text{ of }8.2\text{ billion}=\dfrac{90}{100}\times 8.2\text{ billion}=0.9\times 8.2\text{ billion}=7.38\text{ billion}.$$

So approximately $$7.38$$ billion people (about $$7{,}38{,}00{,}00{,}000$$) live in the Northern Hemisphere.

Neither raw marks scored (Reema) nor raw marks lost (Eesha) can be compared directly because the two tests have different maximum marks. To compare fairly, convert each score to a percentage of its maximum.

English:

$$\dfrac{42}{50}\times 100=\dfrac{4200}{50}=84\%.$$

Science:

$$\dfrac{70}{80}\times 100=\dfrac{7000}{80}=87.5\%.$$

So Eesha actually scored $$84\%$$ in English and $$87.5\%$$ in Science. Her performance in Science ($$87.5\%$$) is slightly better than in English ($$84\%$$).

Vishu is right that the raw scores cannot be compared directly (both Eesha and Reema were using an unfair comparison). Once we convert them to percentages, Reema's conclusion — that Eesha did better at Science — turns out to be correct as well.

The two packets have different total weights ($$150\,\mathrm{g}$$ and $$400\,\mathrm{g}$$), so we can't compare the grams directly. Convert each ingredient's weight into a percentage of the packet's total.

DEF (total $$150\,\mathrm{g}$$):

• Badam powder: $$\dfrac{12}{150}\times 100=\dfrac{1200}{150}=8\%$$.
• Food chemicals: $$\dfrac{9}{150}\times 100=\dfrac{900}{150}=6\%$$.

Zacni (total $$400\,\mathrm{g}$$):

• Badam powder: $$\dfrac{40}{400}\times 100=10\%$$.
• Food chemicals: $$\dfrac{24}{400}\times 100=6\%$$.

Comparison. Zacni has a larger share of badam ($$10\%$$ vs. $$8\%$$). Both products have the same share of food chemicals ($$6\%$$ each), so neither uses a smaller proportion of food chemicals — they are equal.

Let the population in 1961 be $$P$$ (i.e. take $$P$$ as $$100\%$$).

Statement (i): Population in 1991 $$=165\%\text{ of }P=\dfrac{165}{100}P=1.65\,P$$.

Statement (ii): Population in 1991 $$=P+65\%\text{ of }P=P+\dfrac{65}{100}P=P\left(1+\dfrac{65}{100}\right)=\dfrac{165}{100}P=1.65\,P$$.

Both statements give exactly the same 1991 population, namely $$1.65\,P$$. So yes, the two statements say the same thing. In general, saying '$$X$$ is $$100\%+k\%$$ of $$Y$$' is the same as saying '$$X$$ has increased by $$k\%$$ from $$Y$$'.

Cost Price (CP) $$=$$ ₹$$300$$; Selling Price (SP) $$=$$ ₹$$430$$. The marked price is not needed for the profit calculation.

Profit:

$$\text{Profit}=\text{SP}-\text{CP}=430-300=130.$$

Profit percentage (always computed on the cost price):

$$\text{Profit }\%=\dfrac{\text{Profit}}{\text{CP}}\times 100=\dfrac{130}{300}\times 100=\dfrac{13000}{300}=\dfrac{130}{3}=43\dfrac{1}{3}\%\approx 43.33\%.$$

Cost Price of $$10\,\mathrm{kg}$$ of rice $$=$$ ₹$$35\times 10=$$ ₹$$350$$.
Selling Price of the same $$10\,\mathrm{kg}$$ $$=$$ ₹$$300$$.

Loss: $$\text{CP}-\text{SP}=350-300=$$ ₹$$50$$.

Loss percentage (on cost price):

$$\text{Loss }\%=\dfrac{\text{Loss}}{\text{CP}}\times 100=\dfrac{50}{350}\times 100=\dfrac{5000}{350}=\dfrac{100}{7}=14\dfrac{2}{7}\%\approx 14.29\%.$$

Cost Price (CP) $$=$$ ₹$$2650$$. A loss of $$18\%$$ means the selling price is $$100\%-18\%=82\%$$ of the cost price.

$$\text{SP}=82\%\text{ of }2650=\dfrac{82}{100}\times 2650=\dfrac{82\times 2650}{100}.$$

Compute the numerator: $$82\times 2650=82\times 2000+82\times 650=164000+53300=217300$$. So

$$\text{SP}=\dfrac{217300}{100}=2173.$$

Therefore Shyamala will get ₹$$2173$$ for the damaged vase (the loss on it is ₹$$2650-2173=$$ ₹$$477=18\%$$ of ₹$$2650$$).

We work out both cases: without compounding (simple interest) and with annual compounding.

Option 1: Without compounding (simple interest). Each year, interest is $$10\%$$ of the original principal ₹$$6000$$:

$$\text{Annual interest}=10\%\text{ of }6000=\dfrac{10}{100}\times 6000=600.$$

Interest for $$3$$ years $$=3\times 600=1800$$. Total amount after $$3$$ years $$=6000+1800=$$ ₹$$7800$$.

Option 2: With annual compounding. Each year, the amount is multiplied by $$1+\dfrac{10}{100}=1.1$$.

Total amount after $$3$$ years with compounding $$=$$ ₹$$7986$$.

Compounding gives ₹$$7986-$$ ₹$$7800=$$ ₹$$186$$ extra over simple interest.

For each option, we compute $$\dfrac{\text{Amount received}}{\text{Amount deposited}}\times 100$$.

Option 1 (no compounding):

$$\dfrac{7800}{6000}\times 100=\dfrac{7800}{60}=130\%.$$

So he receives $$130\%$$ of the deposit, an increase of $$30\%$$ over 3 years — consistent with $$10\%\times 3$$ years of simple interest.

Option 2 (with compounding):

$$\dfrac{7986}{6000}\times 100=\dfrac{7986}{60}=133.1\%.$$

So he receives $$133.1\%$$ of the deposit, an increase of $$33.1\%$$. Compounding gives $$3.1\%$$ extra compared to simple interest.

Case 1: Without compounding (simple interest). Each year adds $$10\%$$ of the original principal, i.e. ₹$$600$$. Over $$t$$ years, the interest is $$600t$$ and the total amount is

$$A_1=6000+6000\times \dfrac{10}{100}\times t=6000+600\,t=6000\,(1+0.10\,t).$$

Case 2: With annual compounding. Every year, the principal is multiplied by $$1+\dfrac{10}{100}=1.1$$. After $$t$$ years the amount is

$$A_2=6000\times \Bigl(1+\dfrac{10}{100}\Bigr)^{t}=6000\times (1.1)^{t}.$$

Quick check for $$t=3$$: $$A_1=6000(1+0.3)=7800$$ and $$A_2=6000\times 1.1^{3}=6000\times 1.331=7986$$, matching Example 7.

Depreciation of $$5\%$$ means the value after one year is $$100\%-5\%=95\%$$ of the original.

$$\text{Value after 1 year}=95\%\text{ of }21000=\dfrac{95}{100}\times 21000=0.95\times 21000=19950.$$

Alternative calculation: depreciation amount $$=5\%$$ of $$21000=\dfrac{5}{100}\times 21000=1050$$, so value $$=21000-1050=$$ ₹$$19{,}950$$.

A $$10\%$$ decrease each decade means the population is multiplied by $$100\%-10\%=90\%=0.9$$ each decade.

After $$3$$ decades the population becomes

$$1250\times (0.9)^{3}.$$

Compute step by step:

• End of decade 1: $$0.9\times 1250=1125$$.
• End of decade 2: $$0.9\times 1125=1012.5$$.
• End of decade 3: $$0.9\times 1012.5=911.25$$.

So the expected population after 3 decades is about $$911$$ people (or $$911.25$$ if fractional).

Let the marked price of a cake be $$P$$.

Cakely — successive discounts of $$30\%$$ then $$20\%$$. After the first $$30\%$$ discount, the price is $$70\%$$ of $$P=0.7P$$. The second $$20\%$$ discount is applied on this reduced price, leaving $$80\%$$ of it:

$$\text{Cakely price}=0.8\times 0.7\,P=0.56\,P.$$

So Cakely's total effective discount is $$100\%-56\%=44\%$$ of $$P$$.

Cakify — flat $$50\%$$ discount.

$$\text{Cakify price}=0.5\,P.$$

Comparison. $$0.5\,P<0.56\,P$$, so Cakify is cheaper. In percentage terms, Cakify gives a full $$50\%$$ discount whereas Cakely's combined discount is only $$44\%$$ — successive percentage discounts do not simply add up.

For each ingredient, compute (weight of that ingredient / total weight) $$\times 100$$.

DEF (total $$150\,\mathrm{g}$$):

• Sugar: $$\dfrac{99}{150}\times 100=66\%$$ (already given).
• Milk solids: $$\dfrac{30}{150}\times 100=20\%$$.
• Badam powder: $$\dfrac{12}{150}\times 100=8\%$$.
• Food chemicals: $$\dfrac{9}{150}\times 100=6\%$$.

Zacni (total $$400\,\mathrm{g}$$):

• Sugar: $$\dfrac{272}{400}\times 100=68\%$$.
• Milk solids: $$\dfrac{64}{400}\times 100=16\%$$.
• Badam powder: $$\dfrac{40}{400}\times 100=10\%$$.
• Food chemicals: $$\dfrac{24}{400}\times 100=6\%$$.

Completed table:

Using the percentages computed in the previous question:

DEF: $$66\%+20\%+8\%+6\%=100\%$$. ✓

Zacni: $$68\%+16\%+10\%+6\%=100\%$$. ✓

This is expected: since the four ingredients together make up the whole packet, their percentages must sum to $$100\%$$. Any total other than $$100\%$$ would signal an arithmetic error.

Profit percentage is always calculated on the cost price: $$\dfrac{\text{SP}-\text{CP}}{\text{CP}}\times 100$$.

Manufacturer: CP $$=230$$, SP $$=253$$. Profit $$=253-230=23$$.

$$\text{Profit }\%=\dfrac{23}{230}\times 100=\dfrac{2300}{230}=10\%.$$

Wholesaler: CP $$=253$$, SP $$=300$$. Profit $$=300-253=47$$.

$$\text{Profit }\%=\dfrac{47}{253}\times 100=\dfrac{4700}{253}\approx 18.58\%.$$

So the manufacturer made a profit of $$10\%$$ and the wholesaler made a profit of approximately $$18.58\%$$ per sweater.

Cost price per book (CP) $$=$$ ₹$$36$$. A $$20\%$$ profit means the selling price is $$100\%+20\%=120\%$$ of the cost price.

$$\text{SP}=120\%\text{ of }36=\dfrac{120}{100}\times 36=1.2\times 36=43.2.$$

Alternatively, the profit is $$20\%$$ of $$36=$$ ₹$$7.20$$, and SP $$=36+7.20=$$ ₹$$43.20$$.

A $$25\%$$ profit means the selling price is $$125\%$$ of the cost price. Let $$C$$ be the cost price. Then

$$125\%\text{ of }C=50\ \Rightarrow\ \dfrac{125}{100}\,C=50\ \Rightarrow\ C=\dfrac{50\times 100}{125}=\dfrac{5000}{125}=40.$$

So Shambhavi bought each crayon box from the wholesaler for ₹$$40$$. (Check: $$25\%$$ of $$40=10$$, and $$40+10=50=$$ SP. ✓)

Yes. Working per kg:

CP per kg $$=$$ ₹$$35$$. SP per kg $$=\dfrac{300}{10}=$$ ₹$$30$$. Loss per kg $$=35-30=$$ ₹$$5$$.

$$\text{Loss }\%=\dfrac{5}{35}\times 100=\dfrac{500}{35}=\dfrac{100}{7}=14\dfrac{2}{7}\%\approx 14.29\%.$$

This is the same value obtained using the total for $$10\,\mathrm{kg}$$ in Example 5 (loss ₹$$50$$ on CP ₹$$350$$ gives $$\dfrac{50}{350}\times 100=\dfrac{100}{7}\%$$).

Why they agree: The loss percentage is a ratio. Dividing both the loss and the CP by the same factor ($$10$$ in this case) does not change the ratio. Algebraically, if the loss on $$n$$ kg is $$nL$$ on a cost of $$nC$$, then

$$\dfrac{nL}{nC}\times 100=\dfrac{L}{C}\times 100.$$

A loss of $$12\%$$ means the selling price is $$100\%-12\%=88\%$$ of the cost price. Let CP $$=C$$ (per kg). Then

$$88\%\text{ of }C=80\ \Rightarrow\ \dfrac{88}{100}\,C=80\ \Rightarrow\ C=\dfrac{80\times 100}{88}=\dfrac{8000}{88}=\dfrac{1000}{11}\approx 90.91.$$

So the cost price is $$\dfrac{1000}{11}\approx$$ ₹$$90.91$$ per kg.

Check: $$12\%$$ of $$90.91\approx 10.91$$, and $$90.91-10.91=80$$. ✓

Selling price after discount. A discount of $$35\%$$ on MRP leaves $$65\%$$ of MRP.

$$\text{SP}=65\%\text{ of }1800=\dfrac{65}{100}\times 1800=\dfrac{65\times 1800}{100}=\dfrac{117000}{100}=1170.$$

So the selling price is ₹$$1170$$.

Profit and profit percentage. With CP $$=$$ ₹$$900$$,

$$\text{Profit}=1170-900=270.$$

$$\text{Profit }\%=\dfrac{270}{900}\times 100=\dfrac{27000}{900}=30\%.$$

Verify each line:

• Amount: $$3\times 150=450$$. ✓
• Sub Total: $$450$$ (only one line item). ✓
• CGST $$9\%$$: $$9\%$$ of $$450=\dfrac{9}{100}\times 450=40.50$$. ✓
• SGST $$9\%$$: Similarly $$40.50$$. ✓
• Total: $$450+40.50+40.50=531$$. ✓

All the figures in the bill are correct.

Let the principal be $$P$$, the annual interest rate be $$r\%$$ (so as a decimal the rate is $$\dfrac{r}{100}$$), and the time period be $$t$$ years.

Without compounding (simple interest). Interest each year is $$\dfrac{r}{100}P$$, and over $$t$$ years:

$$\text{Interest (simple)}=P\times \dfrac{r}{100}\times t=\dfrac{Prt}{100}.$$

Total amount received: $$A=P+\dfrac{Prt}{100}=P\Bigl(1+\dfrac{rt}{100}\Bigr)$$.

With annual compounding. Each year the amount is multiplied by $$1+\dfrac{r}{100}$$, so after $$t$$ years the amount is

$$A=P\Bigl(1+\dfrac{r}{100}\Bigr)^{t}.$$

The interest earned is the amount minus the principal:

$$\text{Interest (compound)}=P\Bigl(1+\dfrac{r}{100}\Bigr)^{t}-P=P\left[\Bigl(1+\dfrac{r}{100}\Bigr)^{t}-1\right].$$

Percentage gain $$=\dfrac{\text{Gain}}{\text{Deposit}}\times 100$$.

Option A: Deposit $$100$$, receive $$300$$. Gain $$=300-100=200$$.

$$\%\text{ gain}=\dfrac{200}{100}\times 100=200\%.$$

Option B: Deposit $$1000$$, receive $$1500$$. Gain $$=1500-1000=500$$.

$$\%\text{ gain}=\dfrac{500}{1000}\times 100=50\%.$$

Which is better? It depends on what you care about:

• Percentage return is higher for A ($$200\%>50\%$$) — every rupee invested grows more.
• Absolute rupee gain is higher for B ($$500>200$$) — you walk away with more money.

Since the contest lets you choose only one option and you actually take home the entire returned amount, Option B is the better real-world choice because you receive ₹$$1500$$ (a profit of ₹$$500$$) versus only ₹$$300$$ (a profit of ₹$$200$$) in Option A.

Let Arun have $$x$$ marbles. Then Ariba has

$$A=120\%\text{ of }x=\dfrac{120}{100}x=\dfrac{6}{5}x.$$

So Arun's marbles as a fraction of Ariba's are

$$\dfrac{x}{A}=\dfrac{x}{(6/5)x}=\dfrac{5}{6}.$$

Expressed as a percentage:

$$\dfrac{5}{6}\times 100=\dfrac{500}{6}=83\dfrac{1}{3}\approx 83.33.$$

So Arun could say: "The number of marbles with me is $$83\dfrac{1}{3}\%$$ (about $$83.33\%$$) of the marbles Ariba has." Equivalently, he has $$16\dfrac{2}{3}\%$$ fewer marbles than Ariba.

CP $$=$$ ₹$$75$$, SP $$=$$ ₹$$110$$. Profit $$=$$ SP $$-$$ CP $$=110-75=35$$.

Profit percentage on cost:

$$\dfrac{35}{75}\times 100=\dfrac{3500}{75}=\dfrac{140}{3}=46\dfrac{2}{3}\%\approx 46.67\%.$$

A profit margin of $$50\%$$ (on cost) means the selling price is $$100\%+50\%=150\%$$ of the cost price.

$$\text{SP}=150\%\text{ of }475=\dfrac{150}{100}\times 475=1.5\times 475.$$

Compute: $$1.5\times 475=475+237.5=712.5$$. So the chair should be sold at ₹$$712.50$$.

Profit margin (with respect to cost/expenditure) of $$25\%$$ means Revenue = Expenditure + $$25\%$$ of Expenditure = $$125\%$$ of Expenditure. Let expenditure $$=E$$ (in crore rupees). Then

$$\dfrac{125}{100}\,E=2.5\ \Rightarrow\ E=\dfrac{2.5\times 100}{125}=\dfrac{250}{125}=2.$$

So the total expenditure was ₹$$2$$ crore. (Check: profit $$=2.5-2=0.5$$ crore $$=25\%$$ of ₹$$2$$ crore. ✓)

A discount of $$25\%$$ means Anwar pays $$100\%-25\%=75\%$$ of the marked price.

$$\text{Price to pay}=75\%\text{ of }300=\dfrac{75}{100}\times 300=\dfrac{3}{4}\times 300=225.$$

Alternatively, discount amount $$=25\%$$ of $$300=75$$, so amount payable $$=300-75=225$$.

Percentage increase is computed on the original value (₹$$60$$).

Increase in price $$=100-60=40$$.

$$\%\text{ increase}=\dfrac{40}{60}\times 100=\dfrac{4000}{60}=\dfrac{200}{3}=66\dfrac{2}{3}\%\approx 66.67\%.$$

The closest match in the options is (iv) $$66.66\%$$.

A discount of $$15\%$$ means Samson paid $$100\%-15\%=85\%$$ of the original marked price. Let the original price be $$M$$. Then

$$85\%\text{ of }M=440000\ \Rightarrow\ \dfrac{85}{100}\,M=440000\ \Rightarrow\ M=\dfrac{440000\times 100}{85}=\dfrac{44000000}{85}.$$

Compute: $$\dfrac{44000000}{85}=517647.058\ldots$$ Rounding to the nearest rupee, $$M\approx$$ ₹$$5{,}17{,}647$$ (exactly $$\dfrac{88\,00\,000}{17}$$).

Check: $$15\%$$ of $$517647.06\approx 77647.06$$; $$517647.06-77647.06=440000$$. ✓

Percentage of votes the winner got:

$$\dfrac{500}{1600}\times 100=\dfrac{50000}{1600}=\dfrac{125}{4}=31.25\%.$$

Minimum number of candidates. The remaining votes are $$1600-500=1100$$. For the winner to actually have won, no other candidate could have received $$500$$ or more votes. So every other candidate got at most $$499$$ votes.

If there are $$c$$ other candidates, they can share at most $$499c$$ votes and must share at least all $$1100$$ remaining votes. So we need

$$499\,c\ge 1100\ \Rightarrow\ c\ge \dfrac{1100}{499}\approx 2.20.$$

Thus $$c\ge 3$$, i.e. there must be at least $$3$$ other candidates. Including the winner, there were at least $$4$$ candidates in the election. (With $$3$$ other candidates the remaining $$1100$$ votes could split as, say, $$499+499+102$$ — all less than $$500$$ — which is possible.)

Increase in price $$=42-38=4$$ rupees. Inflation is the percentage increase computed on the older price (₹$$38$$).

$$\text{Inflation}=\dfrac{4}{38}\times 100=\dfrac{400}{38}=\dfrac{200}{19}\approx 10.53\%.$$

So the rate of inflation for rice from 2024 to 2025 is about $$10.53\%$$.

Let the original number be $$x$$. Increasing $$x$$ by $$20\%$$ means multiplying by $$1+\dfrac{20}{100}=1.2$$:

$$1.2\,x=90\ \Rightarrow\ x=\dfrac{90}{1.2}=\dfrac{900}{12}=75.$$

Check: $$20\%$$ of $$75=15$$; $$75+15=90$$. ✓

Find the cost price of each buffalo from its own SP and profit/loss percentage.

Buffalo 1 (sold at $$5\%$$ profit). SP $$=$$ ₹$$80{,}000$$ $$=105\%$$ of CP$$_{1}$$.

$$\text{CP}_{1}=\dfrac{80000\times 100}{105}=\dfrac{8000000}{105}=\dfrac{1600000}{21}\approx 76190.48.$$

Buffalo 2 (sold at $$10\%$$ loss). SP $$=$$ ₹$$80{,}000$$ $$=90\%$$ of CP$$_{2}$$.

$$\text{CP}_{2}=\dfrac{80000\times 100}{90}=\dfrac{8000000}{90}=\dfrac{800000}{9}\approx 88888.89.$$

Total CP: $$\dfrac{1600000}{21}+\dfrac{800000}{9}$$. Common denominator $$63$$:

$$\dfrac{1600000\times 3}{63}+\dfrac{800000\times 7}{63}=\dfrac{4800000+5600000}{63}=\dfrac{10400000}{63}\approx 165079.37.$$

Total SP: $$2\times 80000=160000$$.

Overall: SP $$-$$ CP $$\approx 160000-165079.37=-5079.37$$. So the milkman incurs a loss of about ₹$$5079.37$$.

Overall loss percentage: $$\dfrac{5079.37}{165079.37}\times 100\approx 3.08\%$$.

Increasing $$p$$ by $$5\%$$ gives $$p+\dfrac{5}{100}\,p=\Bigl(1+\dfrac{5}{100}\Bigr)\,p=1.05\,p$$.

Comparing with the options:

• (i) $$0.5\,p$$ — halves the population (a $$50\%$$ decrease). Wrong.
• (ii) $$0.05\,p$$ — leaves only $$5\%$$ of $$p$$. Wrong.
• (iii) $$1.5\,p$$ — a $$50\%$$ increase. Wrong.
• (iv) $$1.05\,p$$ — a $$5\%$$ increase. ✓
• (v) $$p+1.50$$ — this adds a constant $$1.5$$ (not a percentage of $$p$$). Wrong.

Let the demand a decade ago be $$D$$. A fall of $$85\%$$ means the demand now is $$D-85\%\text{ of }D=D(1-0.85)=0.15\,D$$, i.e. $$15\%$$ of the demand a decade ago.

Check each option:

• (i) Now $$=85\%$$ of past $$=0.85\,D$$. This would be only a $$15\%$$ fall. Wrong.
• (ii) Past $$=85\%$$ of now, so now $$=\dfrac{D}{0.85}\approx 1.176\,D$$, an increase. Wrong.
• (iii) Now $$=15\%$$ of past $$=0.15\,D$$. Matches. ✓
• (iv) Past $$=15\%$$ of now, so now $$=\dfrac{D}{0.15}\approx 6.67\,D$$, a large increase. Wrong.
• (v) Past $$=185\%$$ of now, so now $$=\dfrac{D}{1.85}\approx 0.54\,D$$, a fall of about $$46\%$$. Wrong.
• (vi) Now $$=185\%$$ of past, an increase of $$85\%$$. Wrong.

Principal $$P=20000$$, rate $$r=10\%$$ p.a., time $$t=2$$ years.

Without compounding (simple interest). Interest each year $$=10\%$$ of $$20000=2000$$.

Interest over $$2$$ years $$=2\times 2000=4000$$; Total $$=20000+4000=24000$$.

With annual compounding. Each year the amount is multiplied by $$1.1$$.

Total with compounding $$=24200$$.

Difference: $$24200-24000=200$$. Compounding gives ₹$$200$$ extra over $$2$$ years, because the second year's interest is computed on the higher amount ₹$$22{,}000$$ instead of the original ₹$$20{,}000$$.

Principal $$P=20000$$, rate $$r=5\%$$ p.a., time $$t=4$$ years.

Without compounding. Interest each year $$=5\%$$ of $$20000=1000$$. Over $$4$$ years: interest $$=4\times 1000=4000$$, total $$=20000+4000=24000$$.

With annual compounding. Each year the amount is multiplied by $$1.05$$.

Total with compounding $$\approx$$ ₹$$24{,}310.13$$ (using the formula, $$20000\times 1.05^{4}=20000\times 1.21550625=24310.125$$).

Difference: $$24310.13-24000=310.13$$. Compounding gives about ₹$$310.13$$ more over $$4$$ years.

Comparing the two problems side-by-side:

• Yahapur: $$10\%$$ for $$2$$ years. Without compounding $$=$$ ₹$$24000$$. With compounding $$=$$ ₹$$24200$$.
• Wahapur: $$5\%$$ for $$4$$ years. Without compounding $$=$$ ₹$$24000$$. With compounding $$\approx$$ ₹$$24310.13$$.

Observations.

1. The simple interest is exactly the same in both cases (₹$$24000$$). This makes sense because in both scenarios the product 'rate $$\times$$ time' is $$10\%\times 2=20\%=5\%\times 4$$, and simple interest depends only on this product.
2. The compound amounts are different: Wahapur ($$5\%$$ for $$4$$ years) actually gives more money (₹$$24310.13$$) than Yahapur ($$10\%$$ for $$2$$ years) (₹$$24200$$). Splitting the same total 'rate $$\times$$ time' into more, smaller compounding periods gives the interest more opportunities to be added back to the principal, so it grows a little faster.
3. Compounding always earns more than simple interest for the same principal, rate and time period (whenever $$t>1$$).

Without compounding, the amount after $$t$$ years at rate $$r\%$$ is

$$A=p+p\times \dfrac{r}{100}\times t.$$

Here $$r=6$$ and $$t=4$$, so $$\dfrac{r}{100}=0.06$$ and

$$A=p+p\times 0.06\times 4=p\,(1+0.24)=1.24\,p.$$

Test each option (with $$p=100$$; the correct answer must give $$100\times 1.24=124$$):

• (i) $$100\times 6\times 4=2400$$. Wrong.
• (ii) $$100\times 0.6\times 4=240$$. Wrong.
• (iii) $$100\times \dfrac{0.6}{100}\times 4=2.4$$. Wrong.
• (iv) $$100\times \dfrac{0.06}{100}\times 4=0.24$$. Wrong.
• (v) $$100\times 1.6\times 4=640$$. Wrong.
• (vi) $$100\times 1.06\times 4=424$$. Wrong.
• (vii) $$100+(100\times 0.06\times 4)=100+24=124$$. ✓

So only expression (vii) $$p+(p\times 0.06\times 4)$$ correctly gives the total amount.

Principal $$P=50000$$, rate $$r=7\%$$ p.a., time $$t=3$$ years.

Simple interest (no compounding).

$$\text{SI}=\dfrac{P\,r\,t}{100}=\dfrac{50000\times 7\times 3}{100}=\dfrac{1050000}{100}=10500.$$

Total amount $$=50000+10500=60500$$.

Compound interest (annual compounding). Each year the amount is multiplied by $$1.07$$.

Total with compounding $$=$$ ₹$$61{,}252.15$$; compound interest $$=61252.15-50000=11252.15$$.

Extra earned by compounding: $$11252.15-10500=752.15$$. So one gets about ₹$$752.15$$ more with compounding.

Principal $$P=12500$$, time $$t=3$$ years for both.

Giridhar (simple interest at $$12\%$$):

$$\text{Interest}=\dfrac{P\,r\,t}{100}=\dfrac{12500\times 12\times 3}{100}=\dfrac{450000}{100}=4500.$$

Raghava (compound interest at $$10\%$$, compounded annually): Amount $$=P(1.1)^{3}=12500\times 1.331$$.

Compute year by year:

• End of year 1: $$12500\times 1.1=13750$$.
• End of year 2: $$13750\times 1.1=15125$$.
• End of year 3: $$15125\times 1.1=16637.5$$.

Interest $$=16637.5-12500=4137.5$$.

Compare: Giridhar's interest ₹$$4500$$ vs. Raghava's interest ₹$$4137.50$$. Giridhar pays more by $$4500-4137.5=$$ ₹$$362.50$$.

We want the amount to reach $$2\times 1000=2000$$.

Without compounding. Each year adds ₹$$100$$ (i.e. $$10\%$$ of ₹$$1000$$). After $$t$$ years the amount is $$1000+100\,t$$. Setting this equal to $$2000$$:

$$1000+100\,t=2000\ \Rightarrow\ t=\dfrac{1000}{100}=10\text{ years}.$$

With annual compounding. Amount after $$t$$ years is $$1000\times (1.1)^{t}$$. We need $$(1.1)^{t}\ge 2$$.

At $$t=7$$ the amount is ₹$$1948.72$$ (just short of $$2000$$), and at $$t=8$$ it is ₹$$2143.59$$ (well past $$2000$$). So with annual compounding the money doubles between years $$7$$ and $$8$$ — under the usual once-a-year rule, at the end of year $$8$$.

Linear vs. exponential.

• Without compounding the amount grows by a fixed rupee amount each year (₹$$100$$). The formula $$1000+100t$$ is linear in $$t$$, so this is linear growth.
• With compounding the amount is multiplied by the same factor $$1.1$$ each year. The formula $$1000\times (1.1)^{t}$$ has $$t$$ in the exponent, so this is exponential growth — it grows slowly at first, then faster and faster.

A $$3\%$$ annual increase means the population is multiplied by $$1+\dfrac{3}{100}=1.03$$ each year. After $$3$$ years the population is

$$1.5\text{ crore}\times (1.03)^{3}.$$

Compute $$(1.03)^{3}$$:

• $$(1.03)^{2}=1.0609$$.
• $$(1.03)^{3}=1.0609\times 1.03=1.092727$$.

So the population $$=1.5\times 1.0927270\approx 1.63909$$ crore, i.e. about $$1.64$$ crore (approximately $$1{,}63{,}90{,}905$$ people).

Year-by-year check:

• Year 1: $$1.5\times 1.03=1.545$$ crore.
• Year 2: $$1.545\times 1.03=1.59135$$ crore.
• Year 3: $$1.59135\times 1.03\approx 1.63909$$ crore.

A $$2.5\%$$ hourly increase means the count is multiplied by $$1+\dfrac{2.5}{100}=1.025$$ every hour. After $$2$$ hours the count is

$$506000\times (1.025)^{2}=506000\times 1.050625.$$

Compute step by step:

• End of hour 1: $$506000\times 1.025=506000+2.5\%\text{ of }506000=506000+12650=518650$$.
• End of hour 2: $$518650\times 1.025=518650+2.5\%\text{ of }518650=518650+12966.25=531616.25$$.

So the expected count at the end of $$2$$ hours is $$531616.25$$, i.e. about $$5{,}31{,}616$$ bacteria (rounding to a whole number since bacteria are discrete).

$$250\%$$ of $$50$$ lakhs $$=\dfrac{250}{100}\times 50$$ lakhs $$=2.5\times 50=125$$ lakhs.

$$125$$ lakhs $$=12.5$$ million $$=1.25$$ crore $$=1{,}25{,}00{,}000$$ people.

So Bengaluru's population in 2025 is approximately $$1.25$$ crore (or $$125$$ lakhs). This represents a $$150\%$$ increase (since $$250\%=100\%+150\%$$) over the 2000 population.

World population $$\approx 8.2$$ billion $$=8200$$ million $$=8.2\times 10^{9}$$. For each country compute $$\dfrac{\text{country population}}{8.2\text{ billion}}\times 100$$.

• (i) Germany, $$83$$ million: $$\dfrac{83}{8200}\times 100=\dfrac{8300}{8200}\approx 1.01\%$$. Closest option: (e) $$1\%$$.
• (ii) India, $$1.46$$ billion $$=1460$$ million: $$\dfrac{1460}{8200}\times 100=\dfrac{146000}{8200}\approx 17.8\%$$. Closest option: (c) $$18\%$$.
• (iii) Bangladesh, $$175$$ million: $$\dfrac{175}{8200}\times 100=\dfrac{17500}{8200}\approx 2.13\%$$. Closest option: (g) or (h) $$2\%$$.
• (iv) USA, $$347$$ million: $$\dfrac{347}{8200}\times 100=\dfrac{34700}{8200}\approx 4.23\%$$. None of the given options ($$13, 8, 18, 10, 1, 35, 2, 2, 0.1$$) is close to $$4\%$$; the nearest is (b) $$8\%$$ though it is not a good match.

Matching each country to the closest available option:

Adding an $$18\%$$ GST to a price $$P$$ gives

$$P+18\%\text{ of }P=P+0.18\,P=1.18\,P.$$

Here $$P=8250$$, so the correct final price is

$$1.18\times 8250=9735,$$

i.e. ₹$$9{,}735$$. This can be written equivalently as $$8250+8250\times 0.18$$.

Check each option:

• (i) $$8250+18=8268$$. Wrong.
• (ii) $$8250+1800=10050$$. Wrong (GST would be $$1800$$ only if $$P=10000$$).
• (iii) $$8250+\dfrac{18}{100}=8250.18$$. Wrong.
• (iv) $$8250\times 18=148500$$. Wrong.
• (v) $$8250\times 1.18=9735$$. ✓
• (vi) $$8250+8250\times 0.18=8250+1485=9735$$. ✓
• (vii) $$1.8\times 8250=14850$$. Wrong ($$1.8$$ corresponds to $$180\%$$, not $$118\%$$).

A change of $$+r\%$$ multiplies by $$1+\dfrac{r}{100}$$ and a change of $$-r\%$$ multiplies by $$1-\dfrac{r}{100}$$. Successive percentage changes are multiplied, not added.

• Month 1 ($$+5\%$$): multiply by $$1.05$$.
• Month 2 ($$-2\%$$): multiply by $$0.98$$.
• Month 3 ($$-3\%$$): multiply by $$0.97$$.

Population after $$3$$ months $$=p\times 1.05\times 0.98\times 0.97$$.

Compute the product: $$1.05\times 0.98=1.029$$; $$1.029\times 0.97\approx 0.99813$$. So the final population is about $$0.99813\,p$$, i.e. slightly less than $$p$$ (a net decrease of about $$0.19\%$$).

Check each statement:

• (i) $$p\times 0.05\times 0.02\times 0.03$$: uses the decimals for the changes themselves rather than $$1\pm$$ them; gives about $$0.00003\,p$$. False.
• (ii) $$p\times 1.05\times 0.98\times 0.97$$: exactly matches our derivation. True.
• (iii) $$p+0.05-0.02-0.03=p$$: adds percentages as if they were absolute numbers. False.
• (iv) Final population $$=p$$: would need the product to be exactly $$1$$, but it is $$0.99813$$. False.
• (v) Final $$>p$$: not so, since the product is $$<1$$. False.
• (vi) Final $$

  True.