Percentage — Applications

The change in a product when both factors change by different percents.

The idea

A field’s length rises 25% and its breadth falls 20%. Adding the percents ($25-20=5\%$) to get the effect on the area is wrong. The right way: take a “unit number” for each factor. 25% $=\frac{1}{4}$: length 4 to 5. 20% $=\frac{1}{5}$: breadth 5 to 4. Original area $4\times 5=20$, new $5\times 4=20$: no change at all!

This is the same successive-change rule: $x+y+\frac{xy}{100}$ (falls negative). Here the factors multiply each other; there the changes came one after another — the arithmetic is identical.

1. Turn each factor’s percent into a fraction $\frac{a}{b}$.
2. Take that factor as $b$; the new value is $b\pm a$.
3. Original product = $b_1\times b_2$, new = $(b_1\pm a_1)(b_2\pm a_2)$.
4. Change ÷ original product $\times 100$.

The change in consumption needed to keep expenditure the same, and consumption from a given change in expenditure.

The idea

Expenditure = price × consumption. If the price rises 25% and the expenditure must stay the same, consumption must fall — but not by 25%. Let the price go from 4 to 5; to keep the product 20, consumption goes from 5 to 4, i.e. in inverse proportion: price $4:5$, consumption $5:4$. The fall is 1 part and the base is the old consumption 5: $\frac{1}{5}=20\%$.

This is the same “getting back” rule: if the price factor is $\frac{b\pm a}{b}$ the consumption factor is its reciprocal $\frac{b}{b\pm a}$. If the expenditure also changes, the consumption factor $=\frac{\text{expenditure factor}}{\text{price factor}}$.

Here: $x$ = the change in price in %; $a$, $b$ = numerator and denominator of its fraction.

1. Write the price change as $\frac{a}{b}$: price $b\to b\pm a$.
2. Product fixed: consumption reversed, $b\pm a\to b$.
3. Change in consumption $=\frac{a}{b\pm a}$ (base: the old consumption).
4. If the expenditure changes too, turn all factors into numbers (original 100) and divide.

The original consumption and prices from the extra kg bought with the same money, and the “₹ per kg reduction” questions.

The idea

The money is the same, so the quantity is inversely proportional to the price. The price rises 25%: price $4\to 5$ means quantity $5\to 4$. The gap is 1 part. The question says that 1 part is 6 kg, so the original quantity is $5\times 6=30$ kg and the new one $4\times 6=24$ kg. Price = money ÷ quantity gives both prices.

1. From the price percent write the inverse-ratio parts (old $b\pm a$, new $b$).
2. The gap in quantity is $a$ parts; equate it to the given kg and find 1 part.
3. Old / new quantity = parts × 1 part.
4. Price = money ÷ quantity (units matched).

The change in area and volume from a percent change in length, the reverse, and measuring error.

The idea

Area is the square of the length and volume its cube. So if the length rises 10% the area does not rise 10% but ${\left(\frac{11}{10}\right)}^2=\frac{121}{100}$, that is 21%; the volume goes to ${\left(\frac{11}{10}\right)}^3=\frac{1331}{1000}$, that is 33.1%. This is a special case of the double change, where both factors change together and equally.

The reverse question: if the area changes by $y\%$, the length factor is $\sqrt{1+\frac{y}{100}}$. The $y$ are chosen so that $1+\frac{y}{100}$ is a perfect-square fraction: 21%, 44%, 69%, 125%, −19%, −36%.

1. Write the length factor: $\frac{b\pm a}{b}$.
2. Square it for area, cube it for volume; the same factor for perimeter.
3. The factor $-1$ is the change; $\times 100$ gives the percent.
4. In the reverse case take the square root of the area factor (cube root for volume).

The percent change in saving when income and expenditure change, and the condition to keep saving fixed.

The idea

Saving is the difference of income and expenditure, so it moves faster than either. A person who spends 60% of his income saves only 40%. If income and expenditure both rise 10%, saving also rises 10%; but if income rises 20% and expenditure only 10%, saving rises 35%. So always take the percent change of saving on the old saving.

1. Take income $=100$ (or the given amount); find expenditure and saving.
2. Get the new income and new expenditure with factors.
3. New saving $=$ new income $-$ new expenditure.
4. $\frac{\text{new saving}-\text{old saving}}{\text{old saving}}\times 100$.

Here: $I$ = income, $E$ = expenditure, $S$ = saving; $s$ = the % of income spent.

Net income falling when tax rises, the tax-free part, and slab-wise tax.

The idea

Income $=$ tax $+$ net income (what is left in hand after tax). With the income unchanged, whatever the tax gains the net income loses — in rupees. In percent the two have different bases (one is tax, the other is net income), so: tax $\times$ (% change in tax) $=$ net income $\times$ (% change in net income). That gives tax : net income.

The tax-free part: if the share of income that is not taxed is the same in both situations, it cancels from the equation. Slabs: each rate applies only to its own slice of income; the total tax is the sum over the slices.

Here: $r$ = the tax rate (% of income); $a$, $b$ = the percent changes of tax and of net income.

1. Treat tax and net income as separate parts; with income $=100$, tax $=r$, net income $=100-r$.
2. Find the change in tax in rupees/units; it is the opposite change in net income.
3. Take the percent on the old net income.
4. In slab questions first see which slab the income falls in.

Total sales from a salesman’s earning or deposit, when a commission comes with a bonus or in two slabs.

The idea

Commission is a percent of sales. The difficulty comes when the bonus applies only to sales above a limit, because the earning is then $\frac{c}{100}S+\frac{b}{100}(S-L)$. A clever move: pretend the bonus is on the whole sales; the rate becomes $c+b$ and the equation is direct. But then a bonus of $b\%$ on $L$ was counted extra; correct for it by adding it back (or subtracting).

When the salesman deposits the sales after deducting his earning, the deposit is $D=S-E$. The same move: if the bonus were on all sales, the deposit would be $(100-c-b)\%$ of sales. But no bonus was actually paid on $L$, so the real deposit is $\frac{bL}{100}$ more than that. So the imagined deposit $=D-\frac{bL}{100}$.

1. What is asked — earning, deposit, or the gap between two plans? Choose the equation/move accordingly.
2. Imagine one rate on all the sales.
3. Find in rupees the part counted too much or too little (the bonus on $L$ or the rate gap) and correct it.
4. Solve the corrected equation and check.

Here: $c_1$, $c_2$ = the commission % of the two slabs; $D$ = the sum deposited with the company; $F$ = the fixed salary, $d$ = the commission % of plan 2.

The size of each group when two groups change at different rates, the % married, and births–deaths–migration.

The idea

A town has 10,000 people. Males rise 6% and females 10%, and the total becomes 10,840. The total gain is 840. If everyone had grown at 6% (the lower rate) the gain would be 600. The extra 240 came because females grew 10% instead of 6%, that is $(10-6)=4\%$ extra: females $=\frac{240\times 100}{4}=6,000$, males $=4,000$.

Married people: the number of married men = the number of married women (each married man has one wife; the question has to state this assumption). So married people $=2\times$ married men, but the % of married women is on the total women. Births and deaths: both rates act on the starting population, so net growth $=b-d$; a fixed number of migrants is added or subtracted at the end.

Here: $x_1$, $x_2$ = the starting sizes of the two groups; $b$, $d$ = the birth and death rates (%).

1. Apply the lower rate to everyone and find the total gain.
2. Subtract it from the actual gain: the leftover came from the faster group’s extra rate.
3. Faster group $=\frac{\text{leftover}\times 100}{a-b}$.
4. The other group $=$ total $-$ the faster group.

Finding the maximum and the passing percentage by keeping the pass mark fixed, and the overall percent of papers with different maxima.

The idea

The marks needed to pass are a fixed number. One student fails by 18 marks at 30%, another passes with 30 marks to spare at 42%. Between them there is a gap of $(42-30)=12\%$ of the marks, and the same gap in marks is $18+30=48$. So $12\%=48$, 1% $=4$ and the maximum $=400$.

If in both cases the student passes (or fails in both) take the gap $d_2-d_1$. And with two papers of different maxima, the overall percent = (total scored ÷ total maximum) $\times 100$; the plain average of the two percents is wrong.

Here: $d$ = shortfall/surplus in marks; $M_1$, $M_2$ = the maxima of the two papers; $s_1$, $s_2$ = the % scored in them.

1. Find the gap between the percents of the two cases.
2. Find the marks of that gap: the sum of the failing and passing margins, or their difference if both are on one side.
3. Get the marks of 1% and then the maximum ($100\%$).
4. Pass marks $=p_1\%$ of the maximum $+d_1$; passing % $=$ pass marks ÷ maximum $\times 100$.

Non-voters, invalid votes, the winner’s share and the winning margin: solved with a tree.

The idea

An election question is a ladder. Some voters on the list did not vote; of those who voted some votes were invalid; the valid votes were split between two candidates; and the winning margin is the difference of their shares. The base changes at every step, so take the list as 100 units and climb down.

1. Take the list as $100$ units; votes cast $=(100-a)$ units.
2. Valid votes $=(100-i)\%$ of votes cast.
3. Winner $w\%$, loser $(100-w)\%$: margin $=(2w-100)\%$ of valid.
4. Match the given number with its units to get 1 unit and multiply by the units of the thing asked. If a count is given, climb the ladder backwards.

Here: $a$, $i$, $w$ = the percents of the figure above; $L$ = total voters.

From the percents of two groups and their common part: “neither”, “only one”, “exactly one” and “both”.

The idea

If 58% of students passed in maths and 45% in science, adding them gives $103\%$, more than 100% — some students were counted in both lists, that is twice. Those who passed both (30%) must be removed once. So passed in at least one subject $=58+45-30=73\%$.

Here: $A$, $B$ = the percents of the two groups; “both” = their common part (%); total $=100\%$.

1. Spot the % in both: that is the middle region.
2. Find only A and only B: take “both” out of each.
3. Find “neither” as the rest of 100, or check that the four regions add to 100.
4. Read off the region that is asked (only one, exactly one, both, neither).

The salt when water evaporates or is added, the pulp when fruit dries, and the old group’s count when new people join — one principle.

The idea

A solution has 5% salt. Water evaporates, so the solution shrinks and the percent of salt rises, but the amount of salt is unchanged. Hence: first percent $\times$ first solution $=$ second percent $\times$ second solution. The same idea holds for drying fruit (pulp fixed), adding water (salt fixed), adding salt (water fixed) and “boys join a class” (girls fixed).

The two totals come out in the inverse ratio: the smaller percent goes with the bigger total. The parts in the gap of the ratio are the water evaporated or added. For dried fruit: the fixed part is the pulp, so $(100-w_1)\times F=(100-w_2)\times D$, where $w$ is the water %.

Here: $F$, $D$ = the weights of fresh and dried fruit; $w_1$, $w_2$ = their water %.

1. Identify the fixed substance and write its % before and after.
2. Invert the ratio of the percents to get the ratio of the totals; simplify.
3. The gap in the ratio of totals $=$ the substance removed/added ⟹ 1 part.
4. Read the asked total or substance off the parts.

The overall percent from two groups’ different percents, the ratio, and a missing percent.

The idea

A school has 60 boys of whom 80% passed and 40 girls of whom 65% passed. What percent passed overall? $\frac{80+65}{2}=72.5\%$ is wrong, because there are more boys and their percent is higher. The right way: add the real counts: $48+26=74$ passed out of 100, that is 74%. This is a weighted average: each group’s percent is weighted by its size.

Reverse question: given the overall % and the two parts’ percents, the ratio of the parts comes from the cross — the part whose percent is farther from the overall is the smaller part. For a missing percent, subtract the known part from the total count and divide by the remaining size.

Here: $\bar{p}$ = the overall average %.

1. Find the real count in each group ($n\times \frac{p}{100}$).
2. Add the counts, divide by the total size, multiply by 100.
3. For the ratio use the cross: (upper end − overall) : (overall − lower end).
4. For a missing %, total count − known count, then divide by the remaining size.

Swapped quantities in a bill, exams with negative marking, and tickets at relative prices: turning a tangled sentence into one equation.

The idea

Some questions are not solved by parts and factors but by one unknown $x$ and one equation. The art is to turn the sentence into the right equation: words like “times”, “% more”, “interchanged” become multiplication, addition and swaps. Then solve it and test the answer in the original statement.

Here: $x$ = the unknown number/price; $c$, $w$ = marks gained for a right and lost for a wrong answer; $Q$ = total questions; $n_a$, $n_c$ = numbers of adults and children; $r$ = the child ticket’s %.

1. Name the unknown and fix its unit.
2. Write every phrase as a small equation/expression.
3. The sentence that joins two expressions (“equals”, “more than”) makes the equation.
4. Solve it; test the answer in the original statement.