Arithmetic — CAT Quantitative Ability

Arithmetic is the single largest topic in CAT's Quantitative Ability section most years, ahead of Algebra, Geometry & Mensuration, Number System and Modern Maths taken individually. It is also the topic where the on-screen calculator matters least: CAT gives you a basic four-function calculator, but the calculator cannot decide which quantities to multiply, and setup error — not arithmetic error — is what actually costs marks here.

Nearly every arithmetic sub-topic reduces to one habit: convert the words into a ratio or a rate before you touch a number. Percentage is a ratio to 100. Profit and loss is a ratio between cost and selling price. Time and work is a rate of output.

Time, speed and distance is a rate of motion. Once the relationship is written down correctly, the computation is usually short — which is exactly why CAT rewards recognising the relationship fast, not computing it fast.

1. Percentages — the multiplier method

Treat a percentage as a multiplier, not a separate calculation step. A 20% increase is multiplication by $1.20$; a 15% decrease is multiplication by $0.85$. This one habit removes most of the arithmetic that slows candidates down.

$\text{New value}=\text{Original}\times \left(1+\frac{\%\text{ change}}{100}\right)$

Successive percentage changes do not add. For two changes of $a\%$ and $b\%$ applied one after another, the net change is:

$\text{Net}\%=a+b+\frac{ab}{100}$

A price rising 20% and then falling 20% does not return to the original value — the net change is $20-20+\frac{20\times (-20)}{100}=-4\%$, a 4% net fall. This is CAT's most recycled arithmetic trap, precisely because the "up 20, down 20, back to start" intuition feels obviously true and is obviously wrong.

Trap. The percentage that exactly undoes a $x\%$ increase is not $x\%$ — it is $\frac{x}{100+x}\times 100\%$. A 25% increase is undone by a $\frac{25}{125}\times 100=20\%$ decrease, not a 25% decrease. Check: $1.25\times 0.80=1.00$ exactly.

A quick reference for common combinations, since the net figure is rarely the sum a candidate expects:

The gap between the naive sum and the actual net grows with the size of the percentages — at small percentages (a few per cent), the $\frac{ab}{100}$ correction term is nearly negligible and the naive sum is a close approximation; at CAT-typical percentages of 15% and above, ignoring the correction term is a reliable way to lose an easy mark.

Percentage points are not percentage change. If a discount rate moves from 20% to 25%, that is a rise of 5 percentage points, but a 25% increase in the discount rate itself ($5/20=25\%$). Reports that quote "up 5%" when they mean "up 5 percentage points" are a real source of CAT reading-comprehension-style traps inside QA stems — read the units of the stated change before applying a formula.

2. Profit, loss and discount

$\text{Profit\%}=\frac{SP-CP}{CP}\times 100,\text{Loss\%}=\frac{CP-SP}{CP}\times 100$

Marked price (MP) is what is printed before any discount; selling price (SP) is what the customer actually pays:

$SP=MP\times \left(1-\frac{\text{discount}\%}{100}\right)$

Two successive discounts combine exactly like two successive percentage changes — a 10% discount followed by a 20% discount is a net discount of $10+20-\frac{10\times 20}{100}=28\%$, not 30%. Shops advertise "10% + 20% off" precisely because it sounds like 30% while actually being 28%.

A distinct and frequently tested trick is profit through under-weighing: a trader claims to sell at cost price but uses a false weight (say, 900 g passed off as 1 kg). The trader's real profit is not zero — it is the saved material, expressed as a fraction of what was actually given:

$\text{Profit\%}=\frac{\text{true weight of 1 kg}-\text{false weight used}}{\text{false weight used}}\times 100$

Using 900 g for a claimed kilogram at "cost price" yields a genuine profit of $\frac{1000-900}{900}\times 100\approx 11.11\%$ — a trader who never marks up a rupee is still making money.

3. Ratio, proportion and partnership

A ratio $a:b$ scales freely, so treat $a$ and $b$ as $ak$ and $bk$ for some common multiplier $k$ whenever a second condition pins down $k$.

Two quantities are in direct proportion when their ratio stays constant (double one, double the other) and in inverse proportion when their product stays constant (double one, halve the other). Speed-and-time at a fixed distance, and workers-and-days at a fixed job, are the two inverse-proportion relationships that recur constantly through this chapter — recognising a relationship as inverse rather than direct, before setting up an equation, prevents a large class of setup errors.

The mean proportional between $a$ and $b$ is $\sqrt{ab}$ (the value $x$ such that $a:x=x:b$), and the third proportional to $a$ and $b$ is the value $c$ such that $a:b=b:c$, giving $c=b^2/a$. Both appear occasionally as standalone questions and more often buried inside a longer ratio problem.

Partnership profits split in the ratio of capital multiplied by the time it was invested, not capital alone. A partner who invests more but for a shorter period can earn an equal or smaller share:

$\text{Profit share ratio}=\sum ({\text{capital}}_i\times {\text{time}}_i)$

If A invests ₹4,000 for the full 12 months while B invests ₹6,000 but only for 8 months (joining 4 months late), the capital-months are $4000\times 12=48,000$ for A and $6000\times 8=48,000$ for B — an exact 1:1 split despite B investing 50% more capital, because A's money worked for 50% longer.

4. Averages, mixtures and alligation

The average of a set is the total divided by the count — but CAT rarely asks for a plain average; it asks for a weighted one, and alligation is the fast route to it.

Alligation rule: to mix a cheaper item (price $c$) and a dearer item (price $d$) into a blend priced at $m$, the required ratio of cheaper to dearer quantities is:

$\frac{\text{cheaper qty}}{\text{dearer qty}}=\frac{d-m}{m-c}$

To blend tea at ₹60/kg and ₹90/kg into a mixture worth ₹70/kg: ratio $=\frac{90-70}{70-60}=\frac{20}{10}=2:1$ — two parts of the cheaper tea to one part of the dearer. This single cross-rule replaces slower simultaneous-equation setups for every "mix two things to hit a target average" question, including mixtures of milk-and-water, alloys, and even average-speed and average-marks problems recast as a mixture.

Replacement problems (repeatedly removing a mixture and replacing it with pure solvent) use a distinct exponential formula, not alligation:

$\text{Final quantity of original substance}=P{\left(1-\frac{x}{P}\right)}^n$

for $n$ removals of $x$ units each from an initial quantity $P$.

5. Time and work

Set the total work equal to the LCM of the individual completion times, so every rate becomes a whole number — this single habit removes the fractions that make time-work slow under a clock.

If A alone takes 10 days and B alone takes 15 days, set total work $=\text{LCM}(10,15)=30$ units. A's rate is $30/10=3$ units/day, B's is $30/15=2$ units/day, and together they do 5 units/day, finishing in $30/5=6$ days.

Pipes and cisterns use the same method with signed rates: a filling pipe contributes a positive rate, an emptying pipe (a leak) contributes a negative one, and the net rate acts on the same LCM-sized tank.

A tank filled by pipe A in 10 hours, by pipe B in 15 hours, with an outlet leak C that alone empties it in 30 hours, opened together: total $=\text{LCM}(10,15,30)=30$; rates $+3,+2,-1$ give a net of $4$ units/hour, so the tank fills in $30/4=7.5$ hours — slower than A and B alone would manage, because the leak is actively working against them.

6. Time, speed and distance

$\text{Speed}=\frac{\text{Distance}}{\text{Time}},1\text{km/h}=\frac{5}{18}\text{m/s}$

For a fixed distance, speed and time are inversely proportional — this converts many TSD questions into a percentage-change question in disguise: raising speed by a factor of $5/4$ (a 25% rise) cuts time to $4/5$ of its original value (a 20% fall), which is the same $1.25\times 0.80=1.00$ relationship from Section 1.

Average speed for equal distances is the harmonic mean of the two speeds, never the simple average. Travelling a route at 60 km/h and returning the same route at 40 km/h gives an average speed of

$\bar{v}=\frac{2uv}{u+v}=\frac{2\times 60\times 40}{100}=48\text{km/h}$

not $\frac{60+40}{2}=50$. The harmonic mean is always pulled toward the slower speed, because the return leg — being slower — takes proportionally more time and so counts for more of the average.

Relative speed governs every "two moving objects" question, and the rule is simply which direction they move: speeds add when moving toward each other or in opposite directions, and speeds subtract when moving in the same direction. Trains crossing a stationary object use the train's own length; two trains crossing each other use the sum of both lengths, travelled at their relative speed.

Boats and streams are relative speed with the current itself as a mover: downstream speed is boat-speed-plus-current, upstream is boat-speed-minus-current, and the two individual speeds are recovered as half the sum and half the difference of the downstream and upstream speeds.

Races apply the same relative-speed idea to a fixed track rather than an open road. "A gives B a start of 20 m in a 200 m race" means B only has to run 180 m while A runs the full 200 m; both finish in the same time if they are truly evenly matched at that handicap.

A "start of 20 seconds" is different from a "start of 20 m" — a time head start converts to a distance head start only once the slower runner's own speed is known, and confusing the two is the standard error. A dead heat is simply the case where the handicap given exactly equals the speed difference, so both finish simultaneously with neither "winning."

7. Simple and compound interest

$SI=\frac{P\times R\times T}{100},CI=P\left[{\left(1+\frac{R}{100}\right)}^T-1\right]$

For 2 years, the difference between compound and simple interest has a direct shortcut that avoids computing either in full:

$CI-SI(\text{for 2 years})=P{\left(\frac{R}{100}\right)}^2$

If the CI−SI difference over 2 years is ₹25 at 5% per annum, the principal follows directly: $P\times (0.05{)}^2=25\Rightarrow P=10,000$ — without ever computing the simple or compound interest amounts individually. This shortcut exists because the 2-year CI−SI gap is exactly the interest earned on the first year's interest, which is $P\times \frac{R}{100}\times \frac{R}{100}$.

Worked Examples

Example 1 (percentage — easy). A number is increased by 40% and then decreased by 25%. What is the net percentage change?

Net $=40-25+\frac{40\times (-25)}{100}=15-10=5\%$ increase. The two changes do not cancel because they are unequal-and-opposite on different bases — the 25% fall is computed on the already-inflated number, so it removes more absolute value than the 40% rise added on the smaller original base only when the percentages are equal in magnitude; here they are not, and a net rise survives.

Example 2 (profit-loss with false weight — hard). A trader marks goods 25% above cost price, then offers a 10% discount, and also uses a weight of 900 g for every claimed kilogram. Find the trader's actual profit percentage.

Let the true cost of 1 kg be ₹100. Marked price $=100\times 1.25=125$. After the 10% discount, the selling price for a "kilogram" is $125\times 0.90=112.50$. But the customer only receives 900 g, whose true cost is $100\times 0.9=90$. Actual profit $=112.50-90=22.50$ on a true cost of ₹90, giving a profit percentage of $\frac{22.5}{90}\times 100=25\%$ — a full 25% profit despite an advertised 10% discount, because the false weight silently reduces the trader's real cost far more than the discount reduces the price.

Example 3 (partnership — medium). A starts a business with ₹5,000. After 6 months, B joins with ₹8,000, and after a further 3 months, C joins with ₹9,000. At the end of the year, the profit is ₹9,900. Find B's share.

Capital-months: A $=5000\times 12=60,000$; B $=8000\times 6=48,000$ (invested for the remaining 6 months); C $=9000\times 3=27,000$ (invested for the remaining 3 months). Ratio $60,000:48,000:27,000=20:16:9$, total $45$ parts. B's share $=\frac{16}{45}\times 9900=\text{₹}3,520$.

Example 4 (alligation with an embedded profit — medium). In what ratio must tea at ₹40/kg be mixed with tea at ₹56/kg so that the mixture, sold at ₹55.20/kg, gives a profit of 20% on the cost of the mixture?

The selling price is not the alligation target — the mixture's cost price is, and the 20% profit must be stripped out first: $\text{CP of mixture}=\frac{55.20}{1.20}=46$/kg. Now apply alligation on the two component costs against this ₹46 target: ratio of cheaper to dearer $=\frac{56-46}{46-40}=\frac{10}{6}=5:3$. The most common error is applying alligation directly to the ₹55.20 selling price, which silently bakes the profit margin into the mixing ratio and gives a wrong answer.

Example 5 (time and work with pipes — medium). Pipe A fills a tank in 12 hours, pipe B fills it in 18 hours, and an outlet pipe C empties a full tank in 9 hours. If all three are opened together, how long does it take to fill the tank, and is it possible at all?

LCM$(12,18,9)=36$. Rates: A $=+3$, B $=+2$, C $=-4$ units/hour. Net rate $=3+2-4=+1$ unit/hour, so the tank fills in $36/1=36$ hours — very slowly, but it does fill, since the combined inflow (5 units/hour) exceeds the single outlet's drain (4 units/hour).

Example 6 (TSD, average speed — easy). A cyclist covers a distance at 15 km/h and returns the same distance at 10 km/h. Find the average speed for the entire journey.

Average speed $=\frac{2\times 15\times 10}{15+10}=\frac{300}{25}=12$ km/h — below the simple average of 12.5 km/h, since the slower return leg occupies more of the total time.

Example 7 (TSD, two trains — hard). A train 130 m long moving at 45 km/h crosses another train 170 m long, moving in the opposite direction at 63 km/h, completely in how many seconds?

Relative speed (opposite directions, so speeds add) $=45+63=108$ km/h $=108\times \frac{5}{18}=30$ m/s. Total distance to be covered is the sum of both lengths: $130+170=300$ m. Time $=\frac{300}{30}=10$ seconds.

Example 8 (compound interest — medium). The difference between the compound interest and the simple interest on a sum for 2 years at 8% per annum is ₹192. Find the sum.

Using the 2-year shortcut directly: $P{\left(\frac{8}{100}\right)}^2=192\Rightarrow P\times 0.0064=192\Rightarrow P=30,000$. This principal can be recovered in one line without computing the simple or compound interest amounts separately.

Summary

Convert every arithmetic word problem into a ratio or rate before computing anything — CAT tests setup, not calculation, since a basic calculator is provided.

A percentage change is a multiplier; successive changes combine as $a+b+\frac{ab}{100}$, not by simple addition, and the percentage that undoes an $x\%$ rise is $\frac{x}{100+x}\times 100\%$, not $x\%$.

Profit and loss run on $CP$, $MP$ and $SP$; successive discounts combine like successive percentage changes, and a false weight generates real profit even at an advertised cost price.

Partnership profit splits by capital-time products, not capital alone. Alligation's cross-rule $\frac{d-m}{m-c}$ solves every two-item weighted-mixture question directly.

Time and work is fastest when the total job is set to the LCM of the individual times, turning every rate into a whole number; pipes use the same method with negative rates for leaks.

Average speed for equal distances is the harmonic mean $\frac{2uv}{u+v}$, always below the simple average; relative speed adds for opposite directions and subtracts for the same direction, and boats-and-streams recovers still-water and current speed from half-sum and half-difference of the downstream and upstream speeds.

For 2-year compound interest problems, the CI−SI gap equals $P(R/100{)}^2$ directly — a fast route to the principal without computing either interest amount in full.