Quantitative Aptitude

Quantitative Aptitude in GATE is not a mathematics paper. It is a short set of arithmetic and data-handling questions carrying part of the 15-mark General Aptitude section, and it is examined at roughly Class 10 level with a strict time budget.

That combination — easy content, tight clock — produces a specific failure mode. Candidates who know every formula still lose marks because they set the problem up slowly, or set it up in a form that forces heavy arithmetic.

Almost every question in this area is one idea in disguise: find the quantity that stays constant, and express everything else as a multiple of it.

In a ratio problem the constant is the common multiplier. In a work problem it is the total job. In a speed problem it is the distance. In a mixture problem it is the amount of pure substance. Naming that constant first is what turns a three-step problem into a one-line one.

The second principle is that percentages should be converted to fractions and ratios to whole numbers before any calculation begins. Working with 37.5 per cent is slow; working with 3/8 is not. The on-screen calculator makes brute force possible but never fast.

1. Percentages and Percentage Change

A percentage is a fraction with denominator 100, and the fastest route through most questions is to stop treating it as one.

Successive percentage changes do not add. Applying a change of $a$ per cent and then $b$ per cent gives a net change of

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

A 20 per cent rise followed by a 20 per cent fall gives $20-20-400/100=-4$ per cent, a net loss. This asymmetry is examined constantly, usually disguised as a discount followed by a tax.

A percentage increase and the percentage decrease that undoes it are different numbers. If a quantity rises by $x$ per cent, restoring it requires a fall of $\frac{100x}{100+x}$ per cent. Raising by 25 per cent needs a fall of 20 per cent to return.

The reason is the base changes. This single observation resolves most percentage confusions: always ask what the percentage is a percentage of.

2. Ratio, Proportion and Partnership

Write a ratio $a:b:c$ as $ak,bk,ck$ for an unknown multiplier $k$. That multiplier is the constant the whole problem turns on, and finding it is usually the entire solution.

If two quantities are in the ratio 3 : 5 and their difference is 24, then $5k-3k=24$, so $k=12$ and the quantities are 36 and 60. No equation-solving beyond one line is required.

Combining two ratios that share a term requires scaling to make the shared term equal. Given $A:B=2:3$ and $B:C=4:5$, scale the first by 4 and the second by 3 to get $A:B:C=8:12:15$.

In partnership problems, profit divides in the ratio of capital multiplied by time invested. A partner contributing twice the capital for half the time earns the same share as one contributing half the capital for twice the time.

Proportion questions come in two forms and mixing them is a standard trap. In direct proportion the ratio stays fixed; in inverse proportion the product stays fixed. More workers finishing sooner is inverse; more workers producing more is direct.

3. Averages, Mixtures and Alligation

An average is a total divided by a count, so the reliable move is to work with totals rather than averages.

If the average of 10 numbers is 42 and one number is corrected from 25 to 55, the total rises by 30, so the average rises by 3. Reasoning through the total takes one step; reasoning through the average directly takes several.

Alligation solves any two-component mixture in one line. If two components with values $x$ and $y$ are mixed to give a mean value $m$, the ratio in which they are mixed is

$\frac{q_1}{q_2}=\frac{y-m}{m-x}$

The quantities are in the inverse ratio of their distances from the mean. Mixing rice at 40 and 60 rupees per kilogram to average 45 gives distances 15 and 5, so the ratio is 3 : 1 in favour of the cheaper.

Alligation applies to anything that averages, not just physical mixtures: average speeds over equal times, average marks across two groups, interest rates on split investments.

For repeated replacement, if a vessel holds $V$ of liquid and $r$ is removed and replaced with water $n$ times, the remaining original liquid is

$L=V{\left(1-\frac{r}{V}\right)}^n$

4. Time, Speed and Distance

The relation is $d=st$, and the constant to fix is whichever of the three does not change.

Average speed is total distance over total time, never the average of the speeds. For equal distances at speeds $u$ and $v$, the average speed is the harmonic mean

$s_{av}=\frac{2uv}{u+v}$

For equal times at those speeds, the average is the ordinary mean $(u+v)/2$. Confusing the two cases is the single commonest error in this topic.

Relative speed reduces every two-body problem to a one-body problem. Moving in the same direction, subtract the speeds; in opposite directions, add them.

A train of length $L$ crossing a stationary pole covers $L$; crossing a platform of length $P$ covers $L+P$. Two trains crossing each other cover the sum of their lengths at their relative speed.

For boats, speed downstream is $b+c$ and upstream is $b-c$, where $b$ is the boat's speed in still water and $c$ the current. Adding and subtracting the two given speeds recovers $b$ and $c$ immediately.

5. Time and Work

Set the total work to the LCM of the given times. This converts every fraction into an integer rate and removes almost all the arithmetic.

If A finishes a job in 12 days and B in 18 days, take the work as 36 units. Then A does 3 units per day and B does 2. Together they do 5 units per day, so they finish in 36/5 = 7.2 days.

The same trick handles pipes and cisterns, with an outlet pipe contributing a negative rate.

Efficiency and time are inversely proportional. If A is twice as efficient as B, A takes half the time. A statement about efficiency is therefore a statement about rate, and the LCM method absorbs it directly.

For work done in alternate days or in shifts, compute the work completed per full cycle, find how many complete cycles fit, then finish the remainder explicitly. Assuming the last cycle completes is a standard error, since the job usually finishes partway through it.

6. Interest, Profit and Loss

Simple interest is linear and compound interest is exponential.

$SI=\frac{Prn}{100}A=P{\left(1+\frac{r}{100}\right)}^n$

The difference between compound and simple interest over two years has a closed form that saves considerable time:

$D=P{\left(\frac{r}{100}\right)}^2$

Over three years the difference is $P{\left(\frac{r}{100}\right)}^2\left(3+\frac{r}{100}\right)$.

For profit and loss, profit percentage is always computed on cost price unless the question says otherwise, and this is the assumption most often violated by careless reading.

If an article is sold at a profit of $p$ per cent, then $SP=CP\left(1+\frac{p}{100}\right)$. When a discount is offered on a marked price, the chain is marked price, then discount, then selling price, then profit against cost price.

Two articles sold at the same price, one at $x$ per cent profit and one at $x$ per cent loss, always produce a net loss of $x^2/100$ per cent. The result is independent of the price, and the intuition that the two cancel is wrong for the same reason successive percentage changes do not cancel.

7. Permutations, Combinations and Probability

Permutation counts arrangements where order matters; combination counts selections where it does not.

$P(n,r)=\frac{n!}{(n-r)!}C(n,r)=\frac{n!}{r!(n-r)!}$

The decision test is simple: if swapping two chosen items produces a different outcome, it is a permutation.

Three standard techniques cover most GA questions.

Treat items that must stay together as one block. Arranging 5 people with 2 particular people adjacent gives $4!\times 2!=48$ arrangements, since the block can be internally ordered two ways.

Count the complement when a condition says "at least". The number of ways with at least one defective item is the total minus the number with none.

For circular arrangements, fix one position. Arranging $n$ people around a table gives $(n-1)!$ because rotations of the same arrangement are identical.

Probability of an event is favourable outcomes over total outcomes, both counted in the same way. For "at least one" probability questions, the complement is almost always faster: the probability of at least one success is one minus the probability of no successes.

8. Mensuration and Basic Geometry

GATE keeps this elementary, and a small table covers nearly everything asked.

Volumes: cube $a^3$, cuboid $lbh$, cylinder $\pi r^{2}h$, cone $\frac{1}{3}\pi r^{2}h$, sphere $\frac{4}{3}\pi r^3$.

Scaling is examined more often than the formulas themselves. If every linear dimension of a solid is multiplied by $k$, area scales by $k^2$ and volume by $k^3$. Doubling a sphere's radius multiplies its volume by eight, not by two.

For right triangles, the Pythagorean triples 3-4-5, 5-12-13, 8-15-17 and 7-24-25 and their multiples cover most computed answers and are worth recognising on sight.

9. Data Interpretation

Data interpretation supplies a chart or table and several questions. It is the highest-yield part of the section because the reading effort is shared across the questions.

Read the axes, the units and any footnote before reading the question. A chart in thousands with one series as a percentage rather than an absolute is the standard trap, and it is invisible if the axes are skipped.

Most DI questions ask for a ratio, a percentage change or a maximum, and none of these requires exact arithmetic. Comparing 4820/5310 with 3960/4520 does not need division to four places; the second fraction is farther from 1, and that is enough.

Percentage change is always computed against the earlier value:

$\Delta \%=\frac{V_2-V_1}{V_1}\times 100$

Take the base from the question, not from the nearest column. "Growth from 2019 to 2022" uses the 2019 value as the base, even when the table's leftmost column is 2018.

Approximate deliberately and check whether the options permit it. If the options are 12, 18, 24 and 31 per cent, one significant figure decides the answer; if they are 18.2, 18.6 and 19.1, it does not.

10. Number Properties and Series

Divisibility rules are worth holding exactly, since they turn a factorisation question into an inspection.

If $N=p^a{q}^b{r}^c$ with $p,q,r$ prime, the number of divisors is $(a+1)(b+1)(c+1)$. This is asked directly and is also the fastest route to questions about perfect squares, which require every exponent even.

For two numbers, the product of the HCF and LCM equals the product of the numbers. Given any three of the four, the fourth follows immediately.

Series questions ask for the next term or a missing one. The reliable method is to compute successive differences: a constant first difference means an arithmetic progression, a constant second difference means a quadratic pattern, and a constant ratio means a geometric progression.

If differences reveal nothing, check for alternating patterns, for squares and cubes offset by a constant, and for two interleaved sequences.

11. Worked Examples

Example 1. A shopkeeper marks an item 40 per cent above cost and then offers a 25 per cent discount. What is the profit percentage?

Take the cost price as 100, which is always the right base to assume when no absolute figures are given.

The marked price is 140. A 25 per cent discount removes 35, giving a selling price of 105.

Profit is 5 on a cost of 100, so the profit is 5 per cent.

The same result follows from the successive-change formula: $40-25-(40\times 25)/100=40-25-10=5$ per cent. Note that the two changes are applied to different bases, which is exactly why they do not simply subtract to 15.

Example 2. A car travels from P to Q at 40 km/h and returns at 60 km/h. What is the average speed for the whole journey?

The distance each way is the same, so this is the equal-distance case and the average is the harmonic mean.

$s_{av}=\frac{2\times 40\times 60}{40+60}=\frac{4800}{100}=48\text{ km/h}$

Verify it by taking a convenient distance. Over 120 km each way, the outward trip takes 3 hours and the return takes 2, so 240 km in 5 hours gives 48 km/h.

The answer 50 km/h is the trap, and it would be correct only if the car spent equal times at each speed rather than covering equal distances.

Example 3. A and B together finish a job in 12 days, B and C in 15 days, and A and C in 20 days. How long do all three take together?

Take the total work as 60 units, the LCM of 12, 15 and 20.

Then $A+B=5$ units per day, $B+C=4$, and $A+C=3$.

Adding all three gives $2(A+B+C)=12$, so $A+B+C=6$ units per day.

All three together finish 60 units in 10 days.

The step worth noting is adding the three equations rather than solving for the individual rates. The question asks only for the sum, and the sum is available directly.

Example 4. A vessel contains 40 litres of milk. Ten litres are removed and replaced with water, and this is done twice more. How much milk remains?

Each operation removes a fixed fraction of whatever is present, not a fixed quantity of milk, which is why the answer is not 10 litres.

After each step the milk remaining is multiplied by $(1-10/40)=3/4$.

$L=40\times {\left(\frac{3}{4}\right)}^3=40\times \frac{27}{64}=16.875\text{ litres}$

The mixture removed on the second and third operations already contains water, so less milk leaves each time. Treating the removals as three lots of 10 litres of milk is the standard error.

Example 5. In how many ways can the letters of ENGINE be arranged so that the two Ns are never together?

ENGINE has 6 letters with E twice and N twice.

Total arrangements are $\frac{6!}{2!\times 2!}=\frac{720}{4}=180$.

Now count the arrangements where the two Ns are together by treating NN as a single block. That leaves 5 objects — NN, E, G, I, E — with E repeated twice, giving $\frac{5!}{2!}=60$ arrangements. The block has no internal orderings to count, since both letters are N.

Arrangements with the Ns never together are $180-60=120$.

Counting the complement is faster here than counting directly, which is the general rule whenever a condition is phrased as a prohibition.

Example 6. The population of a town rose by 10 per cent in the first year and fell by 10 per cent in the second. If it is now 29,700, what was it originally?

The net change is $10-10-100/100=-1$ per cent, so the current value is 99 per cent of the original.

$P=\frac{29700}{0.99}=30000$

Check it forward: 30,000 rises to 33,000, then falls by 3,300 to 29,700.

The reason the two changes do not cancel is that the 10 per cent fall is taken on the larger, already-increased figure. Every successive-percentage question in this section reduces to noticing which base each change is applied to.

Summary

Find the quantity that stays constant and express everything as a multiple of it: the ratio multiplier, the total work, the distance, the pure substance.

Convert percentages to fractions before calculating. Successive changes combine as $a+b+ab/100$ and never simply add.

Write a ratio with an unknown multiplier and solve for it in one line. To combine ratios, scale the shared term.

Work with totals rather than averages. Alligation gives any two-component mixture ratio as the inverse ratio of distances from the mean.

Average speed over equal distances is the harmonic mean, over equal times the arithmetic mean. Relative speed reduces two bodies to one.

Set total work to the LCM of the given times so every rate becomes an integer.

Profit percentage is on cost price. Two articles at the same price with equal per cent profit and loss give a net loss of $x^2/100$ per cent.

Use combinations when order does not matter, block adjacent items together, count the complement for "at least" conditions, and fix a seat for circular arrangements.

Linear scaling by $k$ scales area by $k^2$ and volume by $k^3$.

In data interpretation, read the axes and units first, take the base from the question, and approximate as far as the options allow.