3. Shortcuts & Speed Tricks

Trick 1 — The criss-cross diagram is always faster than forming equations. For any two-ingredient mean-price question, skip the algebra entirely: write C_1 and C_2 on top, M in the middle, subtract diagonally, and read off the ratio. Example: Milk at Rs 40/L and Rs 60/L mixed to give Rs 45/L → ratio = (60-45):(45-40)=15:5=3:1 in under 5 seconds.

Trick 2 — Equal quantities always give the simple average as the mean price. If a question tells you two ingredients are mixed in equal amounts, don’t even set up alligation — just average the two prices directly. Example: Equal quantities of rice at Rs 30/kg and Rs 50/kg mixed → mean price =(30+50)/(2)= Rs 40/kg instantly.

Trick 3 — Multiply away decimals before subtracting. When prices involve paise or decimals (e.g., Rs 8.20), multiply every value in the alligation by 10 or 100 first so you work with whole numbers, then simplify the final ratio. This avoids decimal-arithmetic slips under time pressure.

Trick 4 — Memorise the replacement formula as a single expression and just plug in. Don’t re-derive P(1-x/P)^n every time — treat it as a formula. The moment you see “withdrawn and replaced with water, repeated n times,” identify P, x, and n and compute directly. Example: P=50, x=10, n=2 → left =50(0.8)^2=50(0.64)=32.

Trick 5 — When x/P is a clean fraction, use fractions instead of decimals. If a third of a mixture is removed each time, work with ((2)/(3))^n rather than converting to 0.666…, since fraction powers stay exact and are easier to multiply with the total volume.

Trick 6 — Alligation works on percentages exactly like it works on prices. Whenever you see “X% concentration mixed with Y% concentration to give Z% concentration,” treat X, Y, Z exactly as you would treat three prices in the criss-cross diagram — no separate method needed.

Trick 7 — Alligation is a general “weighted average splitter,” so it applies to average speed problems when averaging is time-weighted** (not distance-weighted).** If a vehicle travels at speed s_1 for some time and speed s_2 for another time, and the time-weighted average speed for the whole trip is s, then the ratio of times is exactly the alligation ratio: t_1:t_2 = (s_2- s):( s - s_1). Example: speeds 40 and 60 km/hr, average speed (by time) 52 km/hr → ratio of times =(60-52):(52-40)=8:12=2:3. Caution: this shortcut does NOT apply when the two distances (not times) are equal — see Common Mistake in Section 4 and Question B14.

Trick 8 — Alligation applies to average marks/scores problems too. If class A has average marks m_1 and class B has average marks m_2, and the combined average is m, then the ratio of the number of students is n_A:n_B = (m_2- m):( m-m_1). Example: Class A average 60, Class B average 80, combined average 68 → n_A:n_B=(80-68):(68-60)=12:8=3:2.

Trick 9 — For “mixture of mixtures,” never average the ratios directly — always convert to actual quantities first, then add. A very common shortcut mistake is to average 3:1 and 2:3 directly (e.g., by adding numerators and denominators) — this only works by coincidence when the total volumes of both vessels happen to be equal in the exact right way; in general you must convert each ratio into actual quantity of each component based on the given total volume, then sum. When the two vessel volumes ARE stated to be equal, adding the raw quantities from each and reducing is safe and fast — but always verify equal volumes before using the shortcut.

Trick 10 — For dishonest-trader / free-ingredient questions, treat the free item’s cost as zero and let the given selling price double as the “cost price of the pure item” for computing mean CP via Profit%. This turns a seemingly wordy Profit-Loss + Mixture question into a two-line alligation the moment you spot the “gains X% by selling at cost price of the pure item” phrasing.

Trick 11 — For the “fraction of solution replaced” concentration problems (Section 2.3’s percentage variant), use the direct formula f=(a-c)/(a-b) instead of setting up an equation from scratch. Here a is the original concentration, b is the concentration of the liquid used to replace part of it, and c is the resulting concentration. Example: 40% whisky partly replaced with 19% whisky to give 26% → f = (40-26)/(40-19)=(14)/(21)=(2)/(3), i.e., two-thirds of the original whisky was replaced — no equation-writing needed at all.

Trick 12 — When a “replace mixture with pure X, ratio changes” problem gives you the removed volume and both the before-and-after ratios, always express everything as a fraction of one variable (the original total) before equating — never try to guess numeric quantities. Set total =kx using the LCM of the ratio sums as the multiplier, write the “removed” and “added back” quantities in terms of x, then solve the single linear equation that results. This is what makes Questions like B11, B12 and B15 solvable in one clean pass instead of getting tangled in unknowns.