Ranking and Order tests ask you to figure out where a person or object sits in a sequence — by weight, speed, height, marks, or literally their position in a row. The question gives you a set of comparative statements, and you have to build a single ordered chain from them.
Think of it like a queue outside a government office. Someone tells you: "Raju is ahead of Priya, but behind Mohan." Another person says: "Priya is not last." From these fragments, you reconstruct the exact queue. That is ranking and ordering in a nutshell.
Two broad question types show up in SSC MTS:
Type 1 — Simple row/queue problems. A person's rank from the top and total count are given, and you derive rank from the bottom (or vice versa). These use one formula and take under 30 seconds if you know it cold.
Type 2 — Constraint-based ordering. A group of people or objects are compared across several clues. No single clue gives you the full picture. You chain the clues together to build the complete order. The six-person magazine problem and the seven-boys weight problem (both from real MTS papers) are classic examples.
The core skill here is not imagination — it is systematic elimination. You don't need to guess or visualize creatively. You write partial chains, merge them, and the order falls out. Every question of this type has exactly one valid arrangement that satisfies all clues simultaneously. Your job is to find it without leaving any clue unused.
One analogy that helps: treat each clue as a piece of a jigsaw. A single piece tells you almost nothing. But once you identify which pieces share an edge, the picture assembles quickly. The mistake most test-takers make is staring at the pieces one by one instead of looking for the connectors.
If you know a person's rank from the top and the total number of people in the row:
And the reverse:
Example: In a class of 40 students, Asha ranks 14th from the top. Her rank from the bottom is 40 - 14 + 1 = 27.
Example: A person ranks 8th from the front and 12th from the back in a row. Total people = 8 + 12 - 1 = 19.
These are mechanical — get them right every single time by writing the formula once on your rough sheet before applying it.
Here is a systematic four-step method. Do not skip steps, even when the problem looks simple.
Step 1: Extract all direct comparisons.
Go through each clue and write a simple inequality or sequence fragment. Use > for "heavier/faster/taller/ahead of". Example: "B5 is less than only one boy" → B5 is 2nd from the top → write ? > B5.
Step 2: Find anchor statements. An anchor is a clue that fixes an absolute position — "first", "last", "second", "not first or last". These are gold. Lock them in immediately because they constrain everything else.
Step 3: Build partial chains.
Connect your fragments wherever they share a common element. If you have A > B and B > C, merge to A > B > C. Keep extending until you have one chain (or discover a branch that eliminates an option).
Step 4: Verify every clue against your final order. This takes 20 seconds and saves you from silly errors. Check each original statement against your arrangement. If even one clue is violated, your order is wrong.
These are traps for the careless reader. Here is how to decode them:
Take 5 extra seconds to parse these phrases. A wrong read here cascades through the entire solution.
When a clue says "there were as many readers between P and T as between R and P," it means the gap (number of people between them) is equal. If P is at position 3 and the gap on each side is 2, then T is at position 1 and R is at position 5 (or the reverse). You need to test both directions and see which one satisfies all other constraints.
If you build an order and two people seem interchangeable, re-read the clues. Nine times out of ten, you missed a clue or read one loosely. Go back to Step 4 and check every clue. SSC MTS problems at this level always have a unique answer.
Before reading the question, write on your rough sheet: Total = Top + Bottom - 1. The moment you see a row-ranking question, apply it directly. No re-derivation needed. Standard method (re-deriving each time): ~40 seconds. Using the written formula: ~10 seconds. Four such questions in a paper saves you over 2 minutes.
In any constraint-ordering problem, scan all clues and circle the one that says "first", "last", "not first", "only one above/below". Place that person on the chain before touching any other clue. This eliminates 2-3 possible arrangements instantly. Without this, you may test 6+ permutations. With it, you typically test 1-2. Step reduction: from ~8 steps to ~4 steps on a 6-person problem.
The phrase "less than only N people" directly maps to rank from top: rank = N + 1. Write this as a mental rule: "less than only 1" = rank 2, "less than only 2" = rank 3, "less than only 0" = rank 1 (top). Similarly "more than only 1" = rank 2 from the bottom. Without this pattern, students re-read the phrase 3-4 times and still misplace the person. With it: instant placement, zero re-reads.
Write each clue as a short chain using >. When two chains share a person, merge them at that person. Example: Clue A gives B5 > B6, Clue B gives B4 = B6 > B1, B7, B3, Clue C gives B2 > B5. Merge at B5 and B6: B2 > B5 > B6 = B4 > B1, B7, B3. Full order appears in 3 merges instead of trial-and-error across 7! = 5040 permutations. Standard trial: 3-4 minutes. Chain merger: 60-90 seconds.
After building your chain, verify clues in order of complexity — simple inequalities first, "between" clues last. If your chain fails the simplest clue, you catch the error early and don't waste time checking complex ones. This reduces average verification time from ~45 seconds to ~20 seconds by stopping at the first failure rather than running all checks on a wrong arrangement.
Use this decision tree the moment you see a ranking/order question in the exam hall.
Is the question about a row/queue with rank from top or bottom?
Total = Top + Bottom - 1. Plug in known values. Done in under 15 seconds.Is it a comparison-based ordering puzzle (weights, speeds, heights)?
> chains.Stuck at step 4 (two valid-looking arrangements)?
Target: under 90 seconds for 4-6 person ordering problems. Under 20 seconds for simple row-formula problems.
Why this question matters: This is a classic 6-person ordering problem with a "between" clue and a "took from" clue that implicitly fixes the last position. It tests whether you can combine positional and relational clues.
Solving path: Start with Clue I — S is not first or last, so S is in positions 2-5. Clue IV says the last reader "took it from R", meaning R reads just before the last person — R is not last. Clue III says S reads before Q, and Q reads after U. Clue II says the gap between P and T equals the gap between R and P, meaning P is the midpoint of T and R. Test the arrangement where T is first: positions work out as T(1) → S(2) → P(3) → U(4) → R(5) → Q(6). Check all clues: S is not first or last (position 2, valid). Gap P-to-T = 2, gap R-to-P = 2 (valid). S(2) before Q(6), Q after U(4) (valid). Last reader Q took from R(5) (valid). Answer: T first, Q last.
Why this question matters: This tests your ability to decode "less than only one" and use it as an anchor, then chain remaining inequalities correctly.
Solving path: "B5 is less than only one boy" → B5 is 2nd heaviest. "B5 is more than B6" → B6 is below B5. "B4 is neither more nor less than B6" → B4 = B6 (same rank effectively). "B4 is more than B1, B7, B3" → B1, B7, B3 are below B4. Since B5 is 2nd, there is exactly one person above B5. That person is not B6, B4, B1, B7, or B3 (all shown to be below B5). The only remaining person is B2. So B2 is the heaviest.
Why this question matters: This tests careful reading of contradictory-looking clues ("more than only one car" vs "more than four cars") to extract the real intent, and then building the chain.
Solving path: Read carefully — "Speed of C1 is more than only one car" means C1 is 2nd from the bottom (4th from the top out of 5). "C4 is more than C1 but less than C3 and C5" → C3 > C4 > C1 and C5 > C4 > C1. Chain so far: {C3, C5} > C4 > C1 > {one car}. The one car slower than C1 must be C2 (the only one unplaced). So the full order is C5 > C3 > C4 > C1 > C2. Cars slower than C3: C4, C1, C2 — that is 3 cars. Answer: 3.
Misreading "less than only one" as "less than one." The word "only" is doing critical work. "Less than only one person" means one person beats you — you are 2nd. "Less than one person" would mean you are 1st. These are opposite placements. Always re-read such phrases twice.
Forgetting that "between" is symmetric. If the clue says "as many between P and T as between R and P," students often assign T to one side without testing the other direction. Both directions must be checked against remaining clues before committing.
Ignoring the implicit positional clue. In the magazine problem, "the one who read last had taken it from R" tells you R is second-to-last, not that R is last. Many students read this as "R is last" and build the entire arrangement incorrectly.
Using Total = Top + Bottom (without the -1). This is the single most common arithmetic error in row-ranking problems. The -1 corrects for double-counting the person themselves. Memorize the formula with the -1 baked in.
Placing equal-rank elements arbitrarily. When a clue says B4 is "neither more nor less" than B6, students sometimes place B4 above or below B6 to make the chain look cleaner. Treat them as the same level and only differentiate them if a subsequent clue forces it.
Not verifying every clue. Students build an arrangement, feel satisfied, and answer without checking all clues. A single unchecked clue is exactly what question-setters exploit. The 20 seconds spent verifying is never wasted time in this question type.