Shortcuts & Speed Methods

The Fraction and Percent Table as a Tool

Percentage questions are some of the quickest marks in Tier 1, but only if you stop dividing by 100 every time. You have about 36 seconds a question. Most of the speed comes from two things: knowing the fraction table by heart and writing each rise or fall as a multiplier. Every method below also shows where it goes wrong, which is usually a percentage that only looks like a friendly fraction, or a change taken on the wrong base.

Swap the percentage for a fraction

When to use it: the percentage is a friendly one like 12.5%, $16\frac{2}{3}\%$ or $14\frac{2}{7}\%$, or the number in the question divides neatly by 6, 7, 8, 9 or 11.

Fast way: change the percentage into its fraction, so the whole step becomes one division. The Percentage Basics and Conversions section builds the full table. These are the ones to know cold:

Multiples come straight from these: 37.5% = 3/8, 62.5% = 5/8, 87.5% = 7/8, $28\frac{4}{7}\%$ = 2/7, $42\frac{6}{7}\%$ = 3/7, $66\frac{2}{3}\%$ = 2/3, $83\frac{1}{3}\%$ = 5/6, $27\frac{3}{11}\%$ = 3/11 and $41\frac{2}{3}\%$ = 5/12. Above 100% it works the same way: 125% = 5/4 and 175% = 7/4.

For example: $14\frac{2}{7}\%$ of 2,457 is just 2,457 ÷ 7 = 351. And 87.5% of 1,736 = $\frac{7}{8}$ × 1,736 = 7 × 217 = 1,519.

Where it fails: the percentage has to match exactly. 14% is not 1/7, and 33% is not 1/3. If the question says 33%, it means 33/100, and the options are usually close enough to punish the swap. Watch the place value too: 1/7 is $14\frac{2}{7}\%$, but 1/70 is only $1\frac{3}{7}\%$.

x% of y is the same as y% of x

When to use it: one number is awkward as a percentage but easy as a plain number, like 36% of 25 or 48% of 12.5.

Fast way: swap them. Both are $\frac{xy}{100}$, so pick whichever order is easier.

For example: 36% of 25 = 25% of 36 = 9. And 48% of 12.5 = 12.5% of 48 = 48 ÷ 8 = 6.

Where it fails: it works only for "x% of y", not for "more than". 36% more than 25 is 34, but 25% more than 36 is 45.

[SSC Tip] Examiners pick percentages that land on these fractions. If a percentage looks strange, like $41\frac{2}{3}\%$ or $22\frac{2}{9}\%$, test it against the table before you multiply. Those two are 5/12 and 2/9.

The Multiplier Method

One multiplication for each rise or fall

When to use it: any "increase by r%" or "decrease by r%" step, and above all when several come one after another.

Fast way: a rise of r% multiplies by $\frac{100 + r}{100}$, and a fall of r% multiplies by $\frac{100 - r}{100}$. Write the multiplier as a fraction whenever it is a friendly one.

For example: ₹4,800 raised by 12.5% is 4,800 × $\frac{9}{8}$ = ₹5,400. And 640 cut by 15% is 640 × 0.85 = 544. You never work out the 15% separately and then take it away.

Where it fails: a fall of r% is × (1 − r/100), not ÷ (1 + r/100). A 20% cut is × 0.8, while ÷ 1.2 is a cut of only $16\frac{2}{3}\%$.

Going back to the start: divide by the multiplier

When to use it: you know the value after the change and want the value before it.

Fast way: original = new value ÷ multiplier. With a fraction multiplier, flip it and multiply.

For example: after a 15% rise a bill is ₹2,553. The original is 2,553 ÷ 1.15 = ₹2,220. After a 12.5% fall a price is ₹1,470, so the original is 1,470 × $\frac{8}{7}$ = ₹1,680.

Where it fails: taking 15% of the new value and subtracting it. 2,553 − 15% of 2,553 = 2,170.05, which is wrong, because the 15% was taken on the old value. This trap sits in the options nearly every time.

Error questions: wrong answer over right answer

When to use it: a student divided by a fraction when he should have multiplied by it (or used $\frac{b}{a}$ for $\frac{a}{b}$), and the question asks for the percentage error.

Fast way: for the same number, write wrong answer ÷ right answer. The error is how far that ratio is from 1, written as a percentage. The base is always the right answer.

For example: he had to multiply by $\frac{4}{5}$ but divided by $\frac{4}{5}$. Wrong ÷ right = $\frac{5}{4} \div \frac{4}{5} = \frac{25}{16}$. That is $\frac{9}{16}$ above 1, so the error is 56.25%.

Where it fails: the same mix-up the other way round gives a different answer. If he had to divide by $\frac{4}{5}$ but multiplied instead, wrong ÷ right = $\frac{16}{25}$, so the error is $\frac{9}{25}$ = 36%. Both numbers will be in the options. If the question gives the actual gap ("his answer was 36 more"), use number × (difference of the two multipliers) = gap. The Percentage Change and Successive Change section has the other error types, like a digit read wrongly.

"More Than" Turned Round, and Getting Back

x% more means $\frac{x}{100 + x}$ less

When to use it: "A is x% more than B. By what percent is B less than A?", or the same with "more" and "less" swapped.

Fast way:

$$\text{more to less: } \frac{x}{100 + x} \times 100\% \qquad \text{less to more: } \frac{x}{100 - x} \times 100\%$$

The fraction form is quicker. If A is $\frac{1}{n}$ more than B, then B is $\frac{1}{n + 1}$ less than A. If A is $\frac{1}{n}$ less than B, then B is $\frac{1}{n - 1}$ more. For a fraction like $\frac{3}{8}$, add the top to the bottom to go from "more" to "less" ($\frac{3}{11}$), and take it away to go from "less" to "more" ($\frac{3}{5}$).

For example: A earns 37.5% more than B. 37.5% = $\frac{3}{8}$, so B earns $\frac{3}{11}$ less than A, which is $27\frac{3}{11}\%$. And if a bus fare is $16\frac{2}{3}\%$ = $\frac{1}{6}$ less than a train fare, the train fare is $\frac{1}{5}$ = 20% more than the bus fare.

Where it fails: the base changes. "A is 37.5% more than B" is measured on B, while "B is less than A" is measured on A. That is why the two numbers differ, and the trap option is the same 37.5% again.

The percentage that undoes a change

When to use it: "A price is cut by r%. By what percent must it now rise to get back to where it was?"

Fast way: it is the same formula. Undoing an r% fall needs a rise of $\frac{r}{100 - r} \times 100\%$, and undoing an r% rise needs a fall of $\frac{r}{100 + r} \times 100\%$.

For example: a share falls 40% = $\frac{2}{5}$. To get back it must rise $\frac{2}{3}$ = $66\frac{2}{3}\%$. A price that rose 60% = $\frac{3}{5}$ must fall $\frac{3}{8}$ = 37.5% to come back.

Where it fails: after two changes, combine them first. A 20% cut and then a 25% cut leave $\frac{4}{5} \times \frac{3}{4} = \frac{3}{5}$ of the start, a total cut of $\frac{2}{5}$. Getting back needs a rise of $\frac{2}{3}$ = $66\frac{2}{3}\%$. Adding the two separate undos, 25% + $33\frac{1}{3}\%$ = $58\frac{1}{3}\%$, is wrong.

[SSC Tip] The undo is always bigger than the cut and smaller than the rise. A 40% cut needs more than 40% to recover, and a 60% rise needs less than 60% to undo. Use that to strike out options in a second.

Successive Changes

Two changes: $a + b + \frac{ab}{100}$

When to use it: two percentage changes, one after the other, on the same quantity.

Fast way: the net change is

$$a + b + \frac{ab}{100}$$

with a minus sign for any fall. A positive answer is a net rise and a negative one a net fall.

For example: up 20% then down 15%: $20 - 15 + \frac{20 \times (-15)}{100}$ = 20 − 15 − 3 = 2% up. Up 10% then up 30%: 10 + 30 + 3 = 43% up.

Where it fails: the sign of the last term. With one rise and one fall, $\frac{ab}{100}$ is negative. Writing +3 in the first example gives 8%, a ready-made trap.

Three or more changes: multiply the fractions

When to use it: three changes, or any changes whose multipliers are friendly fractions.

Fast way: multiply the multipliers as fractions, and most of them cancel. You can also use $a + b + \frac{ab}{100}$ twice: first on two changes, then on that result with the third.

For example: up 25%, down 20%, down 10%: $\frac{5}{4} \times \frac{4}{5} \times \frac{9}{10} = \frac{9}{10}$, so the net effect is 10% down. The formula agrees: the first two give 25 − 20 − 5 = 0, and 0 followed by −10 is −10.

Where it fails: just adding the rates. 25 − 20 − 10 = −5 is wrong. The order of the changes doesn't matter, because you can multiply in any order, so don't waste time on it.

Same x up and down: always a loss of $\frac{x^2}{100}\%$

When to use it: a rise of x% and a fall of the same x%, in either order.

Fast way: put b = −x into the formula: $x - x - \frac{x^2}{100}$. So the result is always a loss of $\frac{x^2}{100}\%$.

For example: up 30% then down 30% is a loss of $\frac{900}{100}$ = 9%. If the final price is ₹2,730, the start was 2,730 ÷ 0.91 = ₹3,000.

Where it fails: it is a loss even when the fall comes first. And it needs the same rate both ways: up 30% then down 20% is 30 − 20 − 6 = 4% up, not a loss.

Population and Depreciation

A steady rate: raise the multiplier to a power

When to use it: a population, a price or a machine's value changes by the same percentage every year.

Fast way: after n years, value = start × $m^n$, where m is the yearly multiplier. Keep these powers ready:

For example: a town of 60,000 grows 5% a year. After 2 years it has 60,000 × 1.1025 = 66,150 people. A car bought for ₹5,00,000 loses 20% of its value each year, so after 2 years it is worth 5,00,000 × 0.64 = ₹3,20,000.

Different rates in different years need one multiplier per year. Up 10%, then down 5%, then up 20% is 1.1 × 0.95 × 1.2 = 1.254, a rise of 25.4% over the three years.

For births and deaths given as yearly percentages of the population, the net rate is births minus deaths. Births of 3.5% and deaths of 1.5% make a growth of 2% a year. People moving in or out are added or taken away in the same way.

Going back in time: divide

When to use it: "The population now is … What was it two years ago?"

Fast way: divide by each year's multiplier, or flip the fractions and multiply.

For example: a town grows 10% a year and now has 84,700 people. Two years ago it had 84,700 ÷ 1.21 = 70,000.

Where it fails:

• Adding the rates. 10% a year for 3 years is a 33.1% rise, not 30%.
• The wrong step back. After a 10% fall, the step back is ÷ 0.9, which is a rise of $11\frac{1}{9}\%$, not × 1.1.
• Compound interest uses exactly these multipliers (see Chapter 8), but here you need only the idea.

The Product Rule and the 1/n Trick

When the product stays fixed

When to use it: two quantities multiply to something that doesn't change. Price × consumption = spending, length × breadth = area, and speed × time = distance.

Fast way: if one factor goes up by $\frac{1}{n}$, the other must go down by $\frac{1}{n + 1}$. If one goes down by $\frac{1}{n}$, the other must go up by $\frac{1}{n - 1}$. It is the "more than turned round" rule again.

For example: sugar costs 25% more. To keep the monthly bill the same, a family must use $\frac{1}{5}$ = 20% less sugar. A train runs $33\frac{1}{3}\%$ faster over the same route, so the journey takes $\frac{1}{4}$ = 25% less time.

Where it fails: the product must really stay fixed. If the price rises 25% and the bill is allowed to rise 10%, consumption becomes 1.10 ÷ 1.25 = 0.88 of what it was. That is a 12% cut, not 20%.

When both factors change

When to use it: both factors change and you want the change in the product, like an area, a bill or a shop's takings.

Fast way: $a + b + \frac{ab}{100}$ again. A square's side up 10% means its area goes up 10 + 10 + 1 = 21%. Length up 30% and breadth down 10% means the area goes up 30 − 10 − 3 = 17%. For a cube, cube the multiplier: every edge up 10% gives $1.1^3 = 1.331$, so the volume rises 33.1%.

Where it fails: a circle's area goes with $r^2$, so a 10% longer radius gives 21% more area, but the circumference grows only 10%. A triangle behaves like a rectangle, since its area is $\frac{1}{2}$ × base × height.

The price changes, so the same money buys more or less

When to use it: "A fall of r% in the price lets a person buy k kg more for ₹M", or k kg less after a rise.

Fast way: new price = (r% of M) ÷ k. It works for a rise too, with k the kilograms lost. Either way it gives the new price.

For example: sugar falls 20%, and ₹720 now buys 4 kg more. 20% of 720 = 144, and 144 ÷ 4 = ₹36 a kg, the new price. The old price was 36 ÷ 0.8 = ₹45. Check: ₹720 bought 16 kg before and buys 20 kg now.

Where it fails: the formula gives the new price, and many students report it as the old one. Read which price the question wants.

Income, Expenditure and Savings

This type comes up often in recent CGL Tier 1 papers. It all starts from one line: income = expenditure + savings, and the Product Rule section has every form of the question in full.

Take the income as 100

When to use it: only percentages are given, and the answer is a percentage.

Fast way: let the income be 100. Split it into expenditure and savings, apply each change to its own part, and subtract at the end.

For example: a man saves 30% of his income. His income goes up 20% and his expenditure goes up 10%. Expenditure: 70 becomes 77. Income: 100 becomes 120. Savings: 120 − 77 = 43, up from 30. So his savings rise by $\frac{13}{30}$ = $43\frac{1}{3}\%$.

The weighted-average shortcut

When to use it: you know the percentage changes in two of income, expenditure and savings, and need the third.

Fast way: the change in income is the average of the other two changes, weighted by the old expenditure E and the old savings S.

$$E \times (\text{change in } E) + S \times (\text{change in } S) = (E + S) \times (\text{change in income})$$

For example: the same man has E : S = 70 : 30 = 7 : 3, income +20% and expenditure +10%. So 7 × 10 + 3 × s = 10 × 20, which gives 3s = 130 and s = $43\frac{1}{3}\%$. No subtraction needed.

Where it fails: use the old E and S as the weights, not the new ones. Savings can also fall while income rises, so let s come out negative if it has to.

Two people: ratios of incomes and expenditures

When to use it: "Their incomes are in the ratio a : b, their expenditures in the ratio c : d, and each saves ₹k."

Fast way: write the incomes as ax and bx and the expenditures as cy and dy. Equal savings means ax − cy = bx − dy, so (a − b)x = (c − d)y. That links x and y in one line. Put it back into either saving.

For example: incomes 4 : 3, expenditures 3 : 2, and each saves ₹5,000. Then 4x − 3y = 3x − 2y, so x = y. Now 4x − 3x = x = 5,000, and the incomes are ₹20,000 and ₹15,000. Check: they spend ₹15,000 and ₹10,000.

Where it fails: this needs equal savings. If the savings are in a ratio like 2 : 1, write (ax − cy) : (bx − dy) = 2 : 1 instead of setting them equal.

Elections: Work on the Valid Votes

The margin as a percentage of valid votes

When to use it: two candidates, the winner's share is a percentage, and the winning margin is given in votes.

Fast way: if the winner gets w% of the valid votes, the loser gets (100 − w)%. So the margin is (2w − 100)% of the valid votes, and one division finishes the question.

For example: the winner gets 57% and wins by 2,100 votes. The margin is 57 − 43 = 14% of the valid votes, so there were 2,100 ÷ 0.14 = 15,000 valid votes.

Where it fails: with three or more candidates, the margin is only the gap between the top two. If A gets 41%, B 34% and C 25%, A beats B by 7% of the valid votes, and C doesn't come into it.

Voter list, votes cast, valid votes: one chain

When to use it: some people on the voter list did not vote, and some of the votes cast were invalid.

Fast way: valid votes = voters on the list × (fraction who voted) × (fraction of cast votes that were valid). To go back, divide one step at a time.

For example: 25% of the voters on the list did not vote, 4% of the votes cast were invalid, and the winner of a two-way contest got 60% of the valid votes, winning by 2,880. The margin is 20% of the valid votes, so valid = 2,880 ÷ 0.2 = 14,400. Cast = 14,400 ÷ 0.96 = 15,000. On the list = 15,000 ÷ 0.75 = 20,000.

Where it fails: ask "per cent of what?" at every step. Invalid votes that are "4% of the votes cast" are not the same as "4% of the voters on the list". And if a share is a percent of the total votes cast, the margin is a percent of that total, not of the valid votes.

[SSC Tip] Votes are whole numbers. If an option would make a 37.5% share a fraction of a vote, strike it out.

Exams: Pass Marks from the Difference of Percentages

Two students, one either side of the pass mark

When to use it: one student gets a% and fails by p marks, and another gets b% and scores q marks more than the pass mark.

Fast way: the gap between their percentages covers both gaps: (b − a)% of the maximum marks = p + q. Then the pass mark is a% of the maximum + p.

For example: a student with 32% fails by 26 marks, and another with 45% gets 39 marks more than the pass mark. So 13% of the maximum = 26 + 39 = 65, and the maximum is 500. The pass mark is 160 + 26 = 186.

Where it fails: when both students are on the same side. If one with 30% fails by 50 and one with 36% fails by 20, the gap of 6% is 50 − 20 = 30 marks, so subtract instead of adding (the maximum is again 500). Draw a line, mark the pass mark on it, and the sign is never in doubt.

Failing in one subject, the other, or both

When to use it: "x% failed in Maths, y% in English and z% in both."

Fast way: failed in at least one = x + y − z. Passed in both = 100 − (x + y − z).

For example: 30% failed in Maths, 22% in English and 9% in both. So 30 + 22 − 9 = 43% failed in at least one subject, and 57% passed both. If 912 students passed both, then 912 ÷ 0.57 = 1,600 students took the exam.

Where it fails: "failed in Maths" already includes those who failed in both. Failed only in Maths is 30 − 9 = 21%. If the question gives "only" figures, just add them, with nothing to subtract.

Spending Chains: Multiply What Is Left

Each step takes a part of what remains

When to use it: a person spends part of a salary, then part of what is left, and so on, and the last amount (often the savings) is given.

Fast way: for each step, write the fraction that is left, not the part spent. Multiply these fractions. Then salary = last amount ÷ that product.

For example: rent takes 30% of a salary, food takes $\frac{1}{7}$ of the rest, travel takes $\frac{1}{3}$ of what is left after that, and ₹10,400 is saved. The fractions left are $\frac{7}{10}$, $\frac{6}{7}$ and $\frac{2}{3}$, and

$$\frac{7}{10} \times \frac{6}{7} \times \frac{2}{3} = \frac{2}{5}$$

So the salary is 10,400 × $\frac{5}{2}$ = ₹26,000.

Where it fails: mixing "of the salary" with "of the rest". If rent is 20% of the salary and school fees are 10% of the salary, add those two first (70% is left), and only then start multiplying for the "of the rest" steps.

One item as a share of the salary

When to use it: the question asks how much went on one item in the middle of the chain.

Fast way: the item's share of the salary = its fraction × the share that was left just before it.

For example: in the chain above, travel took $\frac{1}{3}$ of $\frac{7}{10} \times \frac{6}{7} = \frac{3}{5}$. That is $\frac{1}{5}$ of the salary, so ₹5,200. Food took $\frac{1}{7}$ of $\frac{7}{10}$, which is $\frac{1}{10}$ of the salary, so ₹2,600.

Where it fails: reading "$\frac{1}{3}$ on travel" as a third of the salary when the question says a third of what is left.

Budget lists are simpler. When every part is a percentage of the same total (food 35%, rent 20% and so on), the percentage left over is the savings, and one known amount gives the total. "How much more on food than on rent" is just (35 − 20)% of the total.

Mixtures: Track the Part That Doesn't Change

Adding water, or letting it evaporate

When to use it: water is added to a solution, or water evaporates or dries off, and the percentage of something changes.

Fast way: find the part whose amount stays the same (the milk, the salt, the dry pulp). New total = that amount ÷ its new share.

For example: 40 litres of milk and water is 15% water. How much water must be added to make it 32% water? The milk, 34 litres, doesn't change, and it must be 68% of the new mixture. New total = 34 ÷ 0.68 = 50 litres, so add 10 litres of water.

Drying works the same way. Fresh grapes are 90% water and raisins are 20% water. 50 kg of grapes have 5 kg of dry part, and that is 80% of the raisins, so you get 5 ÷ 0.8 = 6.25 kg of raisins.

Where it fails:

• Tracking the part that changes. When water is added, the milk stays fixed, not the water.
• Using the wrong share. Raisins with 20% water are 80% dry part, so divide by 0.8, not 0.2.

Mixing two solutions

When to use it: two solutions of different strengths are poured together.

Fast way: add the actual amounts of the substance, then divide by the total volume.

For example: 20 litres of 30% acid and 30 litres of 20% acid hold 6 + 6 = 12 litres of acid in 50 litres. That is 24%.

Where it fails: averaging the two percentages gives 25%, which is right only when the volumes are equal. The answer always leans towards the bigger volume.

Ratios, Assumed Values and Option Elimination

a% of A = b% of B means A : B = b : a

When to use it: "35% of A equals 21% of B. Find A : B", or "A is what percent of B?"

Fast way: cross-swap the percentages. A : B = b : a = 21 : 35 = 3 : 5, so A is 60% of B.

If the two sides are in different forms, make both into fractions first. $12\frac{1}{2}\%$ of A = $\frac{1}{6}$ of B means $\frac{A}{8} = \frac{B}{6}$, so A : B = 8 : 6 = 4 : 3.

The same idea works on sums and differences. If 20% of (A + B) = 50% of (A − B), then (A + B) : (A − B) = 50 : 20 = 5 : 2. Componendo and dividendo (Chapter 2) turns that into A : B = (5 + 2) : (5 − 2) = 7 : 3.

Where it fails: the swap is the whole trick. A goes with b, not with a, so the quantity with the bigger percentage is the smaller one. Writing 35 : 21 = 5 : 3 is the trap.

Assume 100, or assume the LCM

When to use it: no actual amount is given, or fractions like $\frac{1}{6}$ and $\frac{1}{4}$ sit next to percentages.

Fast way: choose a starting value that clears every fraction: 100 for plain percentages, the LCM of the bottoms for fractions.

For example: a shop cuts its price by $16\frac{2}{3}\%$ and sells 25% more items. What happens to its takings? $16\frac{2}{3}\%$ is $\frac{1}{6}$ and 25% is $\frac{1}{4}$, so take the price as 6 and the number sold as 4. Takings go from 6 × 4 = 24 to 5 × 5 = 25, a rise of $\frac{1}{24}$ = $4\frac{1}{6}\%$.

Where it fails: assume a value only once. If the question gives a real amount later, scale your answer at the end instead of starting again.

Option elimination

When to use it: before, or instead of, the full working, especially in long word problems.

Fast way: three quick checks.

• Whole numbers. People, votes and marks are whole. If 37.5% = 3/8 of a class are girls, the class size is a multiple of 8.
• Size. A "same x up and down" answer is below the start. When income rises faster than expenditure, savings rise even faster than income.
• Direction. An original price found after a rise must be smaller than the new price.

For example: 37.5% of a class are girls, and 20% of the girls play chess. The class could have (a) 96 (b) 140 (c) 160 (d) 180 students. 140 and 180 are not multiples of 8. With 96 there are 36 girls, and 20% of 36 is 7.2, which can't be. Only (c) 160 survives, with 60 girls and 12 chess players.

Where it fails: when all four options pass the checks, you still need the real working.

Solved Examples

Each answer below uses the quick method, and the time check compares it with the long way.

Question 1

Find $16\frac{2}{3}\%$ of 1,524 + $14\frac{2}{7}\%$ of 1,617 − $11\frac{1}{9}\%$ of 2,961.

Answer. The three percentages are $\frac{1}{6}$, $\frac{1}{7}$ and $\frac{1}{9}$. So it's 1,524 ÷ 6 + 1,617 ÷ 7 − 2,961 ÷ 9 = 254 + 231 − 329 = 156.

Time check: about 20 seconds. Multiplying by 16.67, 14.29 and 11.11 takes over a minute and still comes out slightly off.

Question 2

After an 8% rise, a scooter costs ₹82,620. What did it cost before the rise?

Answer. An 8% rise is × $\frac{108}{100}$, which is $\frac{27}{25}$. To go back I flip it: 82,620 × $\frac{25}{27}$. 82,620 ÷ 27 = 3,060, and 3,060 × 25 = ₹76,500. ₹76,010.40, from taking 8% off 82,620, is the answer to avoid.

Time check: about 20 seconds.

Question 3

A student was asked to multiply a number by $\frac{5}{6}$, but he divided it by $\frac{5}{6}$ instead. What is the percentage error in his answer?

Answer. Wrong ÷ right = $\frac{6}{5} \div \frac{5}{6} = \frac{36}{25}$. That is $\frac{11}{25}$ more than 1, so the error is $\frac{11}{25}$ of the right answer, which is 44%. I never needed the number itself.

Time check: about 15 seconds. Picking a number like 30 and working out both answers (36 and 25) also works, but takes longer.

Question 4

45% of A is equal to 27% of B, and A + B = 1,280. Find B.

Answer. Cross-swap: A : B = 27 : 45 = 3 : 5. So B is $\frac{5}{8}$ of the total, and $\frac{5}{8}$ × 1,280 = 800. Check: A = 480, and 45% of 480 = 216 = 27% of 800.

Time check: about 15 seconds.

Question 5

The price of a phone is raised by 30%, then cut by 20%, and then cut again by 10%. What is the net change in its price?

Answer. Multipliers: 1.3 × 0.8 × 0.9. First 1.3 × 0.8 = 1.04, then 1.04 × 0.9 = 0.936. So the price ends at 93.6% of the start, a 6.4% fall. Just adding the rates gives 0%, which is wrong.

Time check: about 15 seconds.

Question 6

Meera's salary was cut by 12% and later raised by 12%. Overall she now earns ₹360 a month less than at first. What was her salary at first?

Answer. Same rate down and up, so the loss is $\frac{12^2}{100}$ = 1.44% of the first salary. So 1.44% of the salary is ₹360, and the salary is 360 ÷ 0.0144 = ₹25,000.

Time check: about 15 seconds. Writing x, then 0.88x, then 0.88x × 1.12 and solving takes about a minute.

Question 7

A town's population grew 10% in the first year, fell 10% in the second and grew 20% in the third. It is now 47,520. What was it three years ago?

Answer. The multiplier for all three years is 1.1 × 0.9 × 1.2 = 0.99 × 1.2 = 1.188. So the start was 47,520 ÷ 1.188 = 40,000. Adding the rates (10 − 10 + 20 = 20%) would give 39,600, which is wrong.

Time check: about 25 seconds. If the division looks heavy, try a round guess: 40,000 × 1.188 = 47,520, so 40,000 it is.

Question 8

The price of petrol goes up by $14\frac{2}{7}\%$. By what percent must a driver cut his use to keep his spending the same? If he used 56 litres a month before, how much can he use now?

Answer. $14\frac{2}{7}\%$ is $\frac{1}{7}$ up, so his use must go down by $\frac{1}{8}$, which is 12.5%. And 56 × $\frac{7}{8}$ = 49 litres.

Time check: about 10 seconds.

Question 9

The price of rice falls by 12.5%, so ₹1,120 now buys 4 kg more. What was the original price per kg?

Answer. New price = 12.5% of 1,120 ÷ 4 = 140 ÷ 4 = ₹35 a kg. A 12.5% fall is × $\frac{7}{8}$, so the original price was 35 × $\frac{8}{7}$ = ₹40 a kg. Check: ₹1,120 bought 28 kg at ₹40 and buys 32 kg at ₹35.

Time check: about 20 seconds.

Question 10

A man saves 40% of his income. His income rises by 20% and his expenditure by 12%. By what percent do his savings rise?

Answer. E : S = 60 : 40 = 3 : 2. Weighted average: 3 × 12 + 2 × s = 5 × 20, so 36 + 2s = 100 and s = 32%. Check with 100: income 120, expenditure 67.2, savings 52.8 instead of 40. That is 12.8 more on 40, which is 32%.

Time check: about 15 seconds.

Question 11

The incomes of P and Q are in the ratio 8 : 5 and their expenditures in the ratio 2 : 1. Each of them saves ₹9,000. Find P's income.

Answer. Incomes 8x and 5x, expenditures 2y and y. Equal savings: 8x − 2y = 5x − y, so y = 3x. P's saving is then 8x − 6x = 2x, so 2x = 9,000 and x = 4,500. P's income is 8 × 4,500 = ₹36,000. (Q earns ₹22,500, and they spend ₹27,000 and ₹13,500.)

Time check: about 20 seconds.

Question 12

In an election between two candidates, 15% of the voters on the list did not vote and 8% of the votes cast were invalid. The winner got 56% of the valid votes and won by 4,692 votes. How many voters were on the list?

Answer. The margin is 56 − 44 = 12% of the valid votes, so valid votes = 4,692 ÷ 0.12 = 39,100. Votes cast = 39,100 ÷ 0.92 = 42,500. Voters on the list = 42,500 ÷ 0.85 = 50,000.

Time check: about 30 seconds. Every division comes out exact, which is a good sign you are on track.

Question 13

A student who scored 28% failed by 42 marks. Another who scored 45% got 26 marks more than the pass mark. Find the maximum marks and the pass percentage.

Answer. The gap of 45 − 28 = 17% covers 42 + 26 = 68 marks. So 1% is 4 marks and the maximum is 400. The pass mark is 28% of 400 + 42 = 112 + 42 = 154, which is $\frac{154}{400}$ = 38.5%.

Time check: about 20 seconds.

Question 14

Arjun spends 30% of his salary on rent, $\frac{1}{4}$ of the rest on food and 20% of what is left after that on travel. He saves the remaining ₹12,600. What is his salary?

Answer. The fractions left are 0.7, then $\frac{3}{4}$, then 0.8. Their product is 0.7 × 0.75 × 0.8 = 0.42. So the salary is 12,600 ÷ 0.42 = ₹30,000.

Time check: about 20 seconds.

Question 15

45 kg of sugar syrup is 12% sugar by weight. How much water must boil off to make it 18% sugar?

Answer. The sugar doesn't change: 12% of 45 = 5.4 kg. At 18% the whole syrup must be 5.4 ÷ 0.18 = 30 kg. So 15 kg of water must boil off.

Time check: about 15 seconds. Even quicker: the sugar share goes from 12% to 18%, which is × $\frac{3}{2}$, so the weight becomes $\frac{2}{3}$ of 45 = 30 kg.