Quantitative Aptitude: Percentages, Ratios, Averages, and Mixtures

Welcome back. In the previous lesson, you built a decision rule for attempt now, defer, or skip. Quantitative Aptitude can become a reliable source of early marks only if the setup becomes almost automatic; otherwise, a seemingly simple percentage or average question can consume several minutes through avoidable interpretation errors.

This lesson develops a compact toolkit for percentages, ratios, averages, and mixtures. These are not isolated chapters: each concerns a quantity relative to a meaningful total. By the end, you should be able to choose the right representation quickly, solve standard GATE-level forms, and check whether an answer is even plausible before committing to it.

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One underlying principle: preserve the quantity that matters

Many aptitude errors occur because a percentage, ratio, or average is treated as a decorative number rather than a statement about a whole.

• A percentage states a part out of a particular whole.
• A ratio compares quantities in common units.
• An average is total quantity divided by number of observations.
• A mixture preserves the amount of each component, even when the total amount changes.

Before calculating, write a short label beside every number:

Number given	Ask yourself
$40\%$	Forty percent of what total?
$4:3$	Four and three units of what comparable quantity?
Average $=72$	What is the total: average times how many?
$10\%$ sugar solution	Ten percent of the final solution is sugar mass

That small discipline directly addresses a common GATE mistake: applying a percentage to the wrong base.

For a timed question, use this setup routine:

1. Identify the requested quantity: count, percentage, ratio, mean, cost, or concentration.
2. Identify the total or base against which each percentage is measured.
3. Translate ratios into variables with a common multiplier, such as $4k$ and $3k$.
4. Prefer totals over averages, and component amounts over mixture labels.
5. Perform a range check before finalizing.

If the representation is clear within the first read, these are strong Pass 1 questions under the strategy from the previous lesson.

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Percentages: the base is the denominator

A percentage is a fraction with denominator $100$:

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

The important phrase is not “percent”; it is “of the original quantity” or “of the new quantity.”

Fast percentage anchors

Build calculations from simple fraction equivalents rather than multiplying mechanically.

Percentage	Fraction
$50\%$	$\frac{1}{2}$
$25\%$	$\frac{1}{4}$
$20\%$	$\frac{1}{5}$
$12.5\%$	$\frac{1}{8}$
$10\%$	$\frac{1}{10}$
$5\%$	$\frac{1}{20}$
$33.33\dots \%$	$\frac{1}{3}$
$66.66\dots \%$	$\frac{2}{3}$

For example, to calculate $48\%$ of $1200$, use $50\%-2\%$:

$$50\%\text{ of }1200=600$$ $$2\%\text{ of }1200=24$$ $$48\%\text{ of }1200=600-24=576$$

This is not a trick to memorize. It is flexible decomposition: choose a nearby percentage whose fraction is easy.

Percentage increase: compare with the old value

If a value changes from $A$ to $B$, the percentage increase is:

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

The denominator is the old value, because the question asks how large the change is relative to where you started.

Suppose a fiscal deficit is $4\%$ of GDP in one year. GDP is indexed as $100$, so the actual deficit is $4$. In the next year, GDP rises by $10\%$, becoming $110$, and the deficit is $5\%$ of this new GDP:

$$\text{New deficit}=5\%\text{ of }110=5.5$$

The actual deficit changed from $4$ to $5.5$, so:

$$\frac{5.5-4}{4}\times 100=\frac{1.5}{4}\times 100=37.5\%$$

Notice the distinction:

• The deficit rate changed from $4\%$ to $5\%$: an increase of 1 percentage point.
• The actual deficit changed from $4$ to $5.5$: an increase of $37.5\%$.

These are different statements. “Percentage points” compare rates directly; “percentage increase” compares a change to an original value.

Successive changes multiply, not add

An increase of $20\%$ followed by a decrease of $20\%$ does not restore the original value.

Take a convenient base of $100$:

$$100\times 1.20=120$$ $$120\times 0.80=96$$

The final value is $96$, so the net result is a $4\%$ decrease.

In general, if changes are $a\%$ and $b\%$, with a decrease represented by a negative value, the overall multiplier is:

$$\left(1+\frac{a}{100}\right)\left(1+\frac{b}{100}\right)$$

For fixed expenditure, this logic is especially useful. If price rises by $25\%$, consumption must be multiplied by:

$$\frac{1}{1.25}=0.8$$

So consumption must fall by $20\%$, not $25\%$.

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Ratios: values are scaled, not equal to the displayed parts

A ratio $4:3$ does not mean the two quantities are literally $4$ and $3$. It means they are proportional to $4$ and $3$. Write:

$$A=4k,B=3k$$

The multiplier $k$ carries the actual scale.

A mixed ratio-and-percentage problem

Suppose the ratio of boys to girls taking an exam is $4:3$. The overall pass percentage is $80\%$, and the pass percentage among girls is $90\%$. Find the pass percentage among boys.

Let the number of boys and girls be:

$$4k\text{ and }3k$$

Total candidates:

$$7k$$

Total students who passed:

$$80\%\text{ of }7k=5.6k$$

Girls who passed:

$$90\%\text{ of }3k=2.7k$$

Therefore boys who passed:

$$5.6k-2.7k=2.9k$$

The pass percentage among boys is:

$$\frac{2.9k}{4k}\times 100=72.5\%$$

The structure matters more than the arithmetic:

1. Turn the ratio into counts.
2. Convert overall percentages into actual totals.
3. Subtract the known subgroup contribution.
4. Divide by the correct subgroup total.

A frequent error is to calculate $90\%$ of the entire $7k$. The $90\%$ applies only to the girls, whose count is $3k$.

Combining ratios safely

If you are given:

$$A:B=3:4$$

and

$$B:C=5:2$$

make the shared quantity $B$ equal in both forms. The least common multiple of $4$ and $5$ is $20$:

$$A:B=15:20$$ $$B:C=20:8$$

Therefore:

$$A:B:C=15:20:8$$

Do not combine ratios merely by writing $3:4:2$. The two appearances of $B$ would represent different numbers of parts.

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Averages: convert immediately to totals

The average of $n$ values is:

$$\text{Average}=\frac{\text{Sum}}{n}$$

For GATE problems, the more useful rearrangement is:

$$\text{Sum}=n\times \text{Average}$$

Averages themselves usually do not add meaningfully; totals do.

Adding or replacing one observation

Suppose the average score of $8$ students is $72$. One score of $72$ was entered incorrectly; the actual score was $90$. Find the corrected average.

Original total:

$$8\times 72=576$$

Corrected total:

$$576-72+90=594$$

Corrected average:

$$\frac{594}{8}=74.25$$

The count remains $8$, so only the total needs repair.

A clean way to prevent sign errors is to say the correction aloud:

Remove the incorrect value; add the correct value.

If a value was recorded as $13$ instead of $33$, the total was short by $20$. Add $20$; do not subtract it.

Weighted averages

If two groups have different sizes, their combined average is not the simple average of their two averages. It is a weighted average:

$$\bar{x}=\frac{n_1{\bar{x}}_1+n_2{\bar{x}}_2}{n_1+n_2}$$

Suppose $30$ students have average score $60$, and $20$ students have average score $75$.

$$\bar{x}=\frac{30(60)+20(75)}{30+20}$$ $$=\frac{1800+1500}{50}=66$$

A useful range check follows immediately:

$$60<66<75$$

The combined average must lie between the two group averages. It should be closer to $60$ because the group with average $60$ is larger.

This is the same mathematical idea used in price mixtures and concentration problems: values are weighted by their quantities.

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Mixtures: track components, not labels

A mixture question is a weighted-average question in which the average may be price per kilogram, percentage concentration, purity, or some other per-unit measure.

The most reliable method is component balance:

$$\text{Amount of component after mixing}=\text{Amount contributed by each source}$$

For a sugar solution, the conserved component is sugar. For a milk-and-water mixture, track milk and water separately. For two rice types, track cost.

Adding a pure component

Consider $50$ kg of a $2\%$ sugar solution. How much pure sugar should be added to make the concentration $10\%$?

Initially, sugar present is:

$$2\%\text{ of }50=1\text{ kg}$$

Let $x$ kg of pure sugar be added. The sugar amount becomes $1+x$, and the total mixture mass becomes $50+x$. The target concentration gives:

$$\frac{1+x}{50+x}=\frac{10}{100}$$ $$1+x=0.1(50+x)$$ $$1+x=5+0.1x$$ $$0.9x=4$$ $$x=\frac{40}{9}\text{ kg}$$

The key detail is that adding pure sugar changes both numerator and denominator. Treating the final total as $50$ kg would be the central mistake.

Alligation: a shortcut derived from weighted average

Alligation is useful when two values are mixed to obtain a known mean and the question asks for the ratio of quantities. It is not a separate law; it is simply weighted-average algebra arranged efficiently.

Suppose an item costing $10$ per unit and another costing $25$ per unit are mixed to obtain a mean cost of $15$ per unit. Let the quantities be $x$ and $y$, respectively.

The weighted-average equation is:

$$\frac{10x+25y}{x+y}=15$$

Multiplying through:

$$10x+25y=15x+15y$$ $$10y=5x$$ $$x:y=2:1$$

The alligation shortcut reaches the same result:

$$\text{Cheaper quantity}:\text{Dearer quantity}=(25-15):(15-10)$$ $$=10:5$$ $$=2:1$$

The direction is easy to reverse under pressure, so use a conceptual check:

• The mean $15$ is closer to $10$ than to $25$.
• Therefore, the cheaper item must be present in the larger quantity.
• A ratio $2:1$ passes that check.

For lower value $L$, higher value $H$, and mean $M$, where:

$$L<M<H$$

the ratio is:

$$\text{Quantity at }L:\text{Quantity at }H=(H-M):(M-L)$$

Alligation should not be used blindly. It applies only after confirming that the two component values and the mean are measured in compatible units: for example, cost per kg with cost per kg, or concentration percentage with concentration percentage.

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Common traps and a final verification pass

Most errors in this topic are interpretation or calculation errors, not difficult mathematics. Before submitting a numerical answer, run a ten-second check.

Topic	Typical error	Verification
Percentage change	Dividing by new value instead of old value	“Increase relative to what I started with?”
Successive change	Adding percentage changes directly	Use multipliers such as $1.20$ and $0.80$
Ratio	Treating $4:3$ as literal quantities	Write $4k$ and $3k$
Group percentage	Applying subgroup rate to total population	Mark the subgroup beside every percentage
Average	Averaging averages without weights	Reconstruct totals first
Mixture	Forgetting that added material changes total amount	Track component and total separately
Alligation	Reversing quantity ratio	Mean closer to a component means that component is less abundant

For your error log, create one compact entry whenever you miss a General Aptitude question:

• Representation error: wrong total, wrong group, or incorrect ratio scaling.
• Method error: chose unweighted average instead of weighted average, or used alligation when component balance was needed.
• Calculation error: correct equation, incorrect arithmetic.
• Decision error: a direct setup was available, but you deferred it unnecessarily.

This distinction matters. A question missed due to “$90\%$ of the wrong group” needs a different repair from a question missed due to multiplication.

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Key takeaways

Percentages, ratios, averages, and mixtures are unified by one habit: identify the whole and preserve the relevant total.

• Use fraction equivalents and decomposition to compute percentages quickly.
• For percentage increase, compare the change with the original value.
• Represent ratios with a common scale factor such as $4k:3k$.
• Convert averages to totals before combining, adding, or correcting values.
• Use weighted averages whenever group sizes differ.
• In mixtures, track the amount of the component being conserved.
• Use alligation as a shortcut for two-component weighted-average ratio problems, then verify that the larger quantity corresponds to the component closer to the mean.

In the next lesson, you will apply the same representation-first approach to work-rate, speed-distance-time, and basic scheduling problems.