CAT (Common Admission Test) · Data Interpretation & Logical Puzzles

Caselets and Data Sets

Text-heavy data puzzles requiring conditions to be organized before solving.

A caselet buries every number and every bound inside prose. The skill is not fancy arithmetic — it is refusing to compute until the sentences have become a table, a short list of equations, and a clear note of what is fixed versus what is still free.

CAT (Common Admission Test)
Hard level
6 concepts
17 practice questions

1Convert the prose into a table before you compute

In a caselet the data is written as sentences, not as a chart. The first job is to extract every number and every named group into a compact table or list — who, how many, and which condition attaches to which quantity. Structuring the data is most of the work; the arithmetic that follows is usually short.

Skipping the table is the common trap. One buried constraint left in the paragraph produces an answer that looks finished and is simply inconsistent with a sentence you never transcribed.

Figure. The caselet numbers become a partitioned headcount before any formula: tea-only 40, both 20, coffee-only 30, neither 10. The Both segment is the double-count the prose hides; Neither is the residual after the union of 90.

How to organise a caselet

List every named quantityUnderline each group or variable the prose mentions — tea drinkers, both, total headcount — and give each a row.
Park every number next to its nameCopy the figure exactly; do not simplify or combine yet. A number without a name is still unusable.
Only then choose a relationInclusion-exclusion, a total-minus-parts subtraction, or a pair of equations — pick the tool after the table exists, not before.

Extracted counts from the tea/coffee caselet
Quantity	Value
Like tea	60
Like coffee	50
Like both	20
Total people	100
Like neither (asked)	10

Tea, coffee, and neither

In a group of 100 people, 60 like tea, 50 like coffee, and 20 like both. How many like neither?

Extracted counts: tea = 60, coffee = 50, both = 20, total = 100table ready
At least one = 60 + 50 − 2090
Neither = 100 − 9010
People who like neither10

Pro tip. Write the four counts in a table first, then apply inclusion-exclusion mechanically. The trap is starting from "60 and 50" and forgetting to subtract the double-counted 20.

In a group of 80 people, 45 like tea, 35 like coffee, and 15 like both. How many like neither?

At least one = 45 + 35 − 15 = 65, so neither = 80 − 65 = 15. Adding without subtracting the overlap gives 20 left over from 100-style arithmetic on the wrong total; subtracting both twice gives 5.

2Assign variables and turn each sentence into an equation

Once the table exists, unknowns that are not given as numbers become symbols. Each sentence translates into one equation or inequality: "total items are 50" is p + n = 50; "total revenue is Rs 900" is a second equation in the same symbols.

Caselets of this shape reduce to a small linear system. Substitute or eliminate; do not invent a third relation the prose never stated.

Figure. The solved split of the 50 items: 30 pens and 20 notebooks. Length is count, not revenue — the revenue equation is what forced this split rather than any other pair adding to 50.

How to build the system

Name the unknownsOne symbol per missing count or amount — pens p, notebooks n.
One sentence, one equationWalk the paragraph in order; each complete claim becomes one line. Skip nothing that contains a number or a comparison.
Substitute and solveExpress one variable from the simplest equation and substitute into the other. Check both originals with the answer.

Pens and notebooks

A shop sold pens and notebooks. Total items = 50, total revenue = Rs 900. Each pen costs Rs 10, each notebook Rs 30. How many pens were sold?

p + n = 50, so n = 50 − pn = 50 − p
10p + 30(50 − p) = 90010p + 1500 − 30p = 900
−20p = 900 − 1500−20p = −600
p = (−600)/(−20)30 pens

Pro tip. After p = 30, recover n = 20 and check revenue: 10\times 30 + 30\times 20 = 900. If the check fails, the mistranslated sentence is almost always the revenue equation.

Total items 40; pens Rs 15 each, notebooks Rs 25 each; revenue Rs 700. Pens sold are

p + n = 40 and 15p + 25n = 700. Substitute n = 40 − p: 15p + 25(40 − p) = 700 → 15p + 1000 − 25p = 700 → −10p = −300 → p = 30. Guessing equal split gives 20; swapping the prices in the algebra gives 10.

3Find a missing part from the total

When every category is named and the total is known, the missing count is total minus the sum of the known parts. No system is required — only a complete list of parts that really partition the total.

The trap is subtracting from a total that is not the right whole, or omitting a category that still belongs in the sum. If the parts can overlap, subtraction alone is wrong and you need inclusion-exclusion instead.

Figure. The five departments fill the bar of 400. IT is the residual segment after the four known lengths are laid down — exactly what total-minus-parts computes.

How to recover a missing part

Confirm a partitionThe known parts plus the unknown must cover the total with no overlap and no omission.
Add the known partsSum every stated category that sits inside the total.
Subtract from the totalMissing = total − sum of known parts.

Five departments, one blank

A company has 400 employees in five departments: Sales 120, HR 40, Ops 90, Finance 70, and IT unknown. How many are in IT?

Known parts = 120 + 40 + 90 + 70320
IT = 400 − 32080
Check: 320 + 80400
IT headcount80

Pro tip. Departments partition the company, so subtraction is valid. The same pattern fails on "liked tea or coffee" counts, where people in both would be removed twice if you treated the groups as a partition.

A school of 250 students has four houses: 60, 55, 70, and the rest in House D. House D has

60 + 55 + 70 = 185, so D = 250 − 185 = 65. Adding only two houses before subtracting gives 75; copying the 55 figure gives a distractor that ignores the total.

4Percentage wording becomes a multiplier

Caselet sentences hide percentages inside comparison words. "A is 20% more than B" means A = 1.2B, not A = B + 20 and not "A is 20% of B". "C is 10% less than A" means C = 0.9A.

Translate each comparison into a factor before chaining. The base of each percentage is the noun after "than" or "of" — miss the base and every later absolute value is wrong.

Figure. B at Rs 25,000, A a fifth longer at Rs 30,000, then C cut back to Rs 27,000 — still longer than B, because the 10% cut used A's base rather than B's.

How to translate the wording

Find the baseThe quantity after "than" or "of" is the base the percentage sits on.
Write the factorMore than → 1 + x/100; less than → 1 - x/100.
Chain in orderApply each factor to the value the previous sentence produced; do not jump back to the original base unless the prose says so.

A above B, then C below A

B earns Rs 25,000. A earns 20% more than B. C earns 10% less than A. What does C earn?

A = 1.2 × 25,000Rs 30,000
C = 0.9 × 30,000Rs 27,000
Check factors: 1.2 × 0.91.08
C as a factor of B: 1.08 × 25,000Rs 27,000

Pro tip. C is 8% more than B, not 10% more — the 10% cut acts on A's larger base. Chaining the factors (1.2 × 0.9 = 1.08) catches that without re-deriving A.

B = Rs 400. A is 25% more than B. A equals

Rs 500
Rs 425
Rs 100

A = 1.25 × 400 = 500. Adding 25 to 400 gives 425; taking 25% of 400 as the answer itself gives 100. "More than B" multiplies B, it does not replace B.

5At least and at most pin the extreme

Words like "at least", "at most", "maximum", "minimum" and "only" are not flavour — they are inequalities. "A scores at least 40" is A \ge 40; "at most 55" is A \le 55. Together with a sum, each bound on one variable becomes a bound on the other.

CAT questions that look under-determined are often decided entirely by these extremes: the maximum of one quantity is what remains when the other is pushed to its minimum allowed value.

Figure. Same total 90 in both bars. B is longest when A sits at its floor of 40, and shortest when A sits at its ceiling of 55 — the question's "maximum B" is the first bar's B segment.

How to use a bound

Circle every bounding wordAt least / at most / maximum / minimum / only — each becomes \ge, \le, or equality.
Write the tight pairIf A + B = S and A \ge a_{\min}, then B \le S - a_{\min}. The max of B uses the min of A.
Answer the extreme askedMaximum → push the other variable to its lowest legal value; minimum → push it to its highest.

Maximum B from bounds on A

A and B together score 90 marks. A scores at least 40 and at most 55. What is the maximum B can score?

A + B = 90, so B = 90 − AB falls as A rises
Max B when A is minimised: A = 40A = 40
B = 90 − 4050
Maximum B50

Pro tip. The upper bound A ≤ 55 decides the *minimum* B (35), not the maximum. Max and min swap which end of A's range you use — circle the word in the question before picking an end.

x + y = 100 with x ≥ 60 and y ≥ 25. The maximum possible value of x is

y ≥ 25 forces x ≤ 75, so the maximum x is 75. Reading only x ≥ 60 gives the minimum of x, not the maximum; ignoring y's bound suggests 100.

6Answer what is uniquely fixed; leave the rest

A caselet need not determine every variable. Some questions ask for a combination that is fixed even while the individual pieces still move — B - C from A + B and A + C, or one variable from a total once the other group is known as a sum.

Before launching a full solve, ask which quantities are uniquely determined by the stated conditions. If the stem only needs a fixed combination, compute that combination and stop.

Figure. A + B = 50 and A + C = 40. Subtract: (A+B)−(A+C) = B−C = 50−40 = 10. A is still free — A = 10 gives B = 40, C = 30, same difference 10. The stem asked for the combination, not the individuals.

How to spot a fixed combination

Write what is askedName the exact expression — B - C, z alone, "must be true" — not "find everything".
See what cancelsSubtract or add the given equations so the free variables drop out and the asked expression remains.
Stop when the stem is answeredIf individuals are still free, that is allowed — the question never required them.

B − C without finding A

In a caselet, A + B = 50 and A + C = 40. What is B − C?

(A + B) − (A + C)B − C
50 − 4010
B − C10
A still free (e.g. A = 10 → B = 40, C = 30)same difference 10

Pro tip. Any legal A gives the same B − C. Spending time solving for A is optional bookkeeping — the stem never asked for it.

P + Q = 30 and P + R = 22. The value of Q − R is

8
52
Cannot be determined

(P + Q) − (P + R) = Q − R = 30 − 22 = 8. Adding the equations gives 52 and still leaves P free; "cannot be determined" would be right for Q alone, not for Q − R.

Notes

Caselet - Convert Text to a Table: The data is buried in prose; extract every number and condition into a table or equations before answering. Structuring the data is 70% of the work.
Identify Variables and Relations: Assign symbols to unknowns and translate each sentence into an equation or constraint; caselets often reduce to a small system of linear equations.
Track Constraints Carefully: Note maxima, minima, and 'at least/at most' conditions - these bound the possible values and often decide the answer.
Partial Information Sets: Some questions can be answered even when the full data set is not fully determined; identify what is uniquely fixed versus what remains variable.
Common trap: Starting calculations before organising all conditions - missing one buried constraint leads to an inconsistent or wrong solution.

Formulas

Translate 'sum of A and B is S' into A + B = S; 'A is twice B' into A = 2B.
Total = sum of parts; use it to find a missing category by subtraction.
Percentage relations: 'A is 20% more than B' means A = 1.2B.
Average = total / count; rearrange to find total when average and count are given.
'At least k' sets a lower bound; 'at most k' sets an upper bound on a variable.

Exam traps & shortcuts

Rewrite the caselet as a compact table or list of equations before solving anything.
Assign variables to unknowns and convert each sentence into one equation systematically.
Circle bounding words (at least, maximum, only) - they usually pin the final answer.
Check which quantities are uniquely determined; you can often answer without solving everything.

Reference tables

Translate the wording first; compute second. Each row is one sentence shape that caselets reuse.

Sentence → relation cheatsheet
Wording in the caselet	Relation	Watch for
Sum of A and B is S	A + B = S	S is a total, not a product
A is twice B	A = 2B	Twice is a factor, not +2
A is 20% more than B	A = 1.2B	Base is B
A is 10% less than B	A = 0.9B	Base is still B
Average of n values is M	total = M \times n	Rearrange either way
At least k	x \ge k	Pins a minimum
At most k	x \le k	Pins a maximum
Total of a partition known	missing = total − known parts	Parts must not overlap

Before solving everything, mark what the stem actually needs. Combinations in the left column are often fixed while the individuals on the right still move.

Fixed versus still free
Often uniquely fixed	Still free until more data arrives
B - C from A+B and A+C	A, B, C individually
One part from total − other parts (partition)	How that part splits further
Max/min of B from bounds on A and A+B	The exact interior value of A
Neither from inclusion-exclusion counts	How the "at least one" group splits beyond the given overlap

Recap

Read only this the night before a DILR set.

Table first: Extract every number and name before calculating. Structuring is most of the work; arithmetic is the short end.
One sentence, one equation: Assign symbols to unknowns and translate each claim. Check the system against every sentence, including the buried one.
Partition vs overlap: Missing = total − known parts only when parts cannot overlap. Overlapping groups need inclusion-exclusion.
Percent factors: "20% more than B" is 1.2B. Chain factors in the order the prose chains the people.
Bounds flip extremes: Max of one variable uses the min of the other under a fixed sum. Circle at least / at most before picking an end.
Stop at fixed: If B − C is asked and A cancels, compute the difference and leave A alone.

Practise Caselets and Data Sets

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