◆ QA · Modern Mathematics

Modern Mathematics, formulas + CAT PYQs

Permutations & Combinations, Probability, Set Theory (Venn diagrams), Sequences & Series and assorted "logic-counting" problems. A high-yield, formula-light area where careful case-work beats memorisation, perfect for a sales-trained, reasoning-first approach.

24formulas

83CAT PYQs

4sub-topics

★★★priority

Formula Sheet CAT PYQs

Permutations & Combinations Probability Set Theory (Venn) Sequences & Series

Formula & Concept Sheet

A-to-Z. Everything you need for this chapter, distilled.

1Rule of Product & Rule of Sum

In plain English: when steps happen one after another you multiply; when you pick just one of several separate options you add.
Product (AND): if a job has m ways for step 1 and n ways for step 2 done together, total = m × n.
Sum (OR): if an action has m ways and a mutually-exclusive action has n ways, choose one in m + n ways.
e.g. 3 shirts AND 2 trousers → 3 × 2 = 6 outfits; but "a shirt OR a trouser" → 3 + 2 = 5 single items.

AND → × · OR → +

2Permutations (order matters)

In plain English: count the ways to line up r things in a row out of n, where swapping the order makes a new arrangement.
Arrangements of r out of n distinct things.
n! = n × (n−1) × … × 2 × 1 ; 0! = 1.
e.g. gold/silver/bronze from 5 runners: ⁵P₃ = 5 × 4 × 3 = 60.

nPr = n! / (n − r)!

3Combinations (order ignored)

In plain English: count the ways to pick a group of r things out of n when the order you pick them in does not matter.
Selections of r out of n distinct things.
Symmetry: choosing r = leaving out (n−r).
e.g. a team of 2 from 5 people: ⁵C₂ = (5 × 4)/(2 × 1) = 10.

nCr = n! / [(n − r)! r!] = nPr / r! nCr = nC(n−r)

4Sum of all combinations

In plain English: each item is either "in" or "out," so the total number of possible selections (including picking nothing) is 2 multiplied by itself n times.
Total subsets of an n-element set (incl. empty & full).
"Choose some-or-none" of n distinct items = 2ⁿ.
e.g. toppings from 3 choices: 2³ = 8 possible orders (incl. a plain pizza with none).

nC0 + nC1 + nC2 + … + nCn = 2ⁿ

5Arrangements with repetition (alike things)

In plain English: when some items are identical, swapping them changes nothing, so you shrink the plain n! by dividing out those wasted swaps.
n things where p are alike of one kind, q of another, r of a third.
Divide n! by the factorials of each repeated group.
e.g. arrangements of the letters of "LEVEL" (5 letters, L×2, E×2): 5!/(2! 2!) = 120/4 = 30.

Arrangements = n! / (p! q! r! …)

6Selecting from "some or all" of mixed items

In plain English: for each type you decide "how many to take" (0 up to all of them), multiply those choices, then knock off the one case where you took nothing.
p of one type, q of a second, r of a third, … (alike within a type).
Take any number (incl. none) of each, then subtract the all-empty case.
e.g. select some fruit from 2 alike apples & 3 alike oranges: (2+1)(3+1) − 1 = 12 − 1 = 11 ways.

{(p+1)(q+1)(r+1)…} − 1

7Circular arrangements

In plain English: around a circle there is no "first seat," so rotating everyone gives the same arrangement, fix one person and arrange the rest in a line.
Fix one person to kill rotational duplicates.
If clockwise = anticlockwise (e.g. a necklace), divide by 2.
e.g. 5 people at a round table: (5 − 1)! = 4! = 24 ways.

Round table = (n − 1)! · Necklace = (n − 1)!/2

8Dividing into groups

In plain English: choosing who goes in the first group automatically fixes the rest; if the groups have no labels and are the same size, swapping the two groups is a duplicate so divide by 2.
Split (m + n) things into two labelled groups of m and n.
If the two groups are equal (m = n), divide by 2! for identical groups.
e.g. split 6 people into groups of 4 and 2: 6!/(4! 2!) = 720/48 = 15.

(m + n)! / (m! n!) Equal groups: (2m)! / [2! (m!)²]

9Distributing identical things (partitions)

In plain English: line up the n identical items as "stars" and slot in (r − 1) dividers ("bars") to split them among the r people, count where the bars go.
Distribute n identical items among r persons, each may get any number (incl. 0).
"Stars & bars." For "each gets ≥ 1", first give one to each then apply the formula on what remains.
e.g. give 5 identical chocolates to 3 kids (any can get 0): (5 + 3 − 1)C(3 − 1) = ⁷C₂ = 21.

Ways = (n + r − 1)C(r − 1)

10Shortest grid paths

In plain English: a shortest path is just a sequence of "rights" and "ups"; count how many ways to order those moves.
To go across a grid using only two directions (m of one, n of the other).
Equivalent to arranging m + n moves of two kinds.
e.g. corner to corner of a 2×3 block (2 ups, 3 rights): (2 + 3)!/(2! 3!) = 120/12 = 10 paths.

(m + n)! / (m! n!) = (m+n)Cm

11Distinct terms in a multinomial expansion

In plain English: every term looks like aˣbʸcᶻ with x+y+z = n; counting the distinct terms is the same as counting whole-number ways to split n among three slots.
Number of terms in (a + b + c)ⁿ = non-negative integer solutions of a + b + c = n.
e.g. terms in (a + b + c)²: (2 + 2)C2 = ⁴C₂ = 6 (namely a², b², c², ab, bc, ca).

Terms in (a+b+c)ⁿ = (n + 2)C2

12Probability, definition

In plain English: probability is just "how many ways it can happen" divided by "how many ways anything can happen," assuming every outcome is equally likely.
Assumes equally likely outcomes; always between 0 and 1.
e.g. rolling an even number on a die: 3 favourable / 6 total = 1/2.

P(E) = favourable outcomes / total outcomes P(not E) = 1 − P(E)

13Odds

In plain English: odds pit the "wins" directly against the "losses" as a ratio, instead of dividing wins by the total like probability does.
Compares favourable to unfavourable cases (not to the total).
e.g. for rolling a 6 on a die: odds in favour = 1 : 5 (one good face, five bad).

Odds in favour = favourable : unfavourable Odds against = unfavourable : favourable

14Probability, addition law

In plain English: for "A or B," add the two chances but subtract the part you counted twice (where both happen).
General OR rule (subtract the overlap).
Mutually exclusive ⇒ P(A ∩ B) = 0.
e.g. a card that is a King or a Heart: 4/52 + 13/52 − 1/52 = 16/52 = 4/13.

P(A ∪ B) = P(A) + P(B) − P(A ∩ B) Mutually exclusive: P(A or B) = P(A) + P(B)

15Probability, multiplication law

In plain English: for "A and B" of unrelated events, multiply their chances; if one affects the other, use the conditional version.
For independent events the AND probability multiplies.
Conditional probability links them when not independent.
e.g. two heads in two coin tosses: 1/2 × 1/2 = 1/4.

Independent: P(A ∩ B) = P(A) × P(B) Conditional: P(A | B) = P(A ∩ B) / P(B)

16Expected value

In plain English: weight each possible payoff by how likely it is, then add them up, what you'd average if you repeated it forever.
The long-run average of a random quantity.
Sum of (each value × its probability).
e.g. win ₹10 on heads, ₹0 on tails: E = 10 × ½ + 0 × ½ = ₹5.

E(X) = Σ xᵢ · P(xᵢ)

17Set theory, basics

In plain English: sets are just collections; "union" pools everyone, "intersection" keeps only the shared members, "difference" removes the overlap.
Union ∪ = in A or B (or both); Intersection ∩ = in both.
Difference A − B = in A but not B; Complement A′ = not in A.
Null set ⌀ is a subset of every set; power set of n elements has 2ⁿ subsets.
e.g. A = {1,2,3}, B = {2,3,4}: A∪B = {1,2,3,4}, A∩B = {2,3}, A−B = {1}.

A − B = {x : x ∈ A and x ∉ B}

18Two-set Venn formula

In plain English: people who do both got counted in each circle, so add the two totals and subtract that overlap once.
Add the two sets, subtract the double-counted overlap.
e.g. 30 like tea, 25 like coffee, 10 like both → total who like at least one = 30 + 25 − 10 = 45.

n(A ∪ B) = n(A) + n(B) − n(A ∩ B)

19Three-set Venn formula

In plain English: add all three circles, take out each pairwise overlap (over-subtracting the centre), then add the triple-overlap back once to fix it.
Inclusion-exclusion: add singles, subtract pairs, add back the triple.
e.g. |A|=|B|=|C|=10, each pair shares 3, all three share 1 → union = 30 − 9 + 1 = 22.

n(A∪B∪C) = n(A)+n(B)+n(C) − n(A∩B) − n(B∩C) − n(A∩C) + n(A∩B∩C)

20Venn, "exactly" layers (CAT favourite)

In plain English: someone in exactly two sets is counted twice in the size-total, someone in all three is counted thrice, these two equations untangle the layers fast.
Let x = exactly-one, y = exactly-two, z = exactly-three (all three).
Total in at least one set = x + y + z; the "repeated total" of memberships = sum of the set sizes.
e.g. sizes sum to 50, at least-one T = 30, all-three z = 5 → 30 + y + 2(5) = 50 → exactly-two y = 10.

x + y + z = T x + 2y + 3z = n(A) + n(B) + n(C)

21Arithmetic Progression (AP)

In plain English: an AP adds the same fixed step each time; the sum is just the average of the first and last term, times how many terms.
Constant common difference d; nth term and sum below.
e.g. 2, 5, 8, 11, 14 (a=2, d=3): 5th term = 2 + 4×3 = 14; sum = 5/2 × (2 + 14) = 40.

aₙ = a + (n − 1)d Sₙ = n/2 · [2a + (n − 1)d] = n/2 · (first + last)

22Geometric Progression (GP)

In plain English: a GP multiplies by the same fixed ratio each time; if that ratio is a fraction the terms shrink and the infinite sum settles on a finite value.
Constant ratio r; sum of a finite GP and (for |r|<1) an infinite GP.
e.g. 1 + ½ + ¼ + ⅛ + … (a=1, r=½): S∞ = 1/(1 − ½) = 2.

aₙ = a·r^(n−1) Sₙ = a(rⁿ − 1)/(r − 1) · S∞ = a/(1 − r), |r| < 1

23Special sums

In plain English: handy closed forms so you never add 1+2+…+n by hand; remember the cube-sum equals the square of the plain sum.
Sum of first n natural numbers, their squares and cubes.
Note: Σn³ = (Σn)².
e.g. 1+2+…+10 = 10×11/2 = 55; and 1³+2³+…+10³ = 55² = 3025.

Σn = n(n + 1)/2 Σn² = n(n + 1)(2n + 1)/6 Σn³ = [n(n + 1)/2]²

24Means & counting handshakes

In plain English: the three means rank AM ≥ GM ≥ HM; and any "everyone meets everyone once" count (handshakes, matches, lines) is just nC2 pairs.
AM-GM-HM for two positives a, b; AM ≥ GM ≥ HM.
Pairs from n people (handshakes / matches / lines / diagonals).
e.g. 6 people each shake hands once: ⁶C₂ = 15 handshakes. For a, b = 4, 9: AM = 6.5, GM = 6.

AM = (a+b)/2 · GM = √(ab) · HM = 2ab/(a+b) Pairs = nC2 = n(n − 1)/2 · Diagonals = n(n − 3)/2

Permutations & Combinations · 60 CAT PYQs

Permutations & Combinations

Counting, arrangements, selections, digit problems and logic-counting, the bulk of Modern Maths PYQs.

CAT 1995

ModerateCAT 1995

A, B, C and D are four towns, any three of which are non-collinear. Then the number of ways to construct three roads each joining a pair of towns so that the roads do not form a triangle is:

(1) 7
(2) 8
(3) 9
(4) 24

Show solution

(4) 24. Start at a town, say A. Two-road chains from A: AB-BC, AB-BD, AC-CB, AC-CD, AD-DB, AD-DC, 6 of them. To avoid a triangle the third road must extend to the remaining town (not return to A), e.g. AB-BC-CD. Each town gives 6 such non-triangular paths, so total = 6 × 4 = 24.

CAT 1997

ModerateCAT 1997

In how many ways can eight directors, the vice-chairman and chairman of a firm be seated at a round table, if the chairman has to sit between the vice-chairman and a director?

(1) 9! × 2
(2) 2 × 8!
(3) 2 × 7!
(4) None of these

Show solution

(2) 2 × 8!. Treat chairman + vice-chairman as one block. This block can fit into the 8 gaps around the 8 directors and, with the directors, arranges in 8! ways. Within the block the two can swap in 2 ways. Total = 2 × 8!.

CAT 1998

ModerateCAT 1998

How many numbers can be formed from 1, 2, 3, 4, 5, without repetition, when the digit at the unit's place must be greater than that in the ten's place?

(1) 54
(2) 60
(3) 17
(4) 2 × 4!

Show solution

(2) 60. If 5 is at units, tens can be anything → 4! = 24. If 4 at units, tens ∈ {1,2,3} (3 ways) × 3! = 18. If 3 at units → 2 × 3! = 12. If 2 at units → 3! = 6. None end in 1. Total = 24 + 18 + 12 + 6 = 60. (Equivalently exactly half of all 5! = 120 numbers have units > tens.)

ModerateCAT 1998

How many five-digit numbers can be formed using the digits 2, 3, 8, 7, 5 exactly once such that the number is divisible by 125?

(1) 0
(2) 1
(3) 4
(4) 3

Show solution

(3) 4. Divisibility by 125 depends on the last three digits. From the given digits the only multiples of 125 possible are 375 and 875, both end in 5 with 7 in the tens place. The thousands digit is 3 or 8 (2 ways) and the leading two digits arrange in 2! = 2 ways → 4 numbers: 23875, 32875, 28375, 82375.

CAT 1999

ModerateCAT 1999

Ten points are marked on a straight-line and 11 points are marked on another straight-line. How many triangles can be constructed with vertices from among the above points?

(1) 495
(2) 550
(3) 1045
(4) 2475

Show solution

(3) 1045. A triangle needs 2 points from one line and 1 from the other (collinear triples give no triangle): ¹⁰C₂ × 11 + ¹¹C₂ × 10 = 45 × 11 + 55 × 10 = 495 + 550 = 1045.

HardCAT 1999

For a scholarship, at the most n candidates out of 2n + 1 can be selected. If the number of different ways of selection of at least one candidate is 63, the maximum number of candidates that can be selected for the scholarship is

(1) 3
(2) 4
(3) 6
(4) 5

Show solution

(1) 3. Selecting 1 to n out of 2n+1: ²ⁿ⁺¹C₁ + … + ²ⁿ⁺¹Cₙ. By symmetry this equals half of (2^(2n+1) − 2) = 2^(2n) − 1. Setting 2^(2n) − 1 = 63 gives 2^(2n) = 64 = 2⁶, so 2n = 6, n = 3.

CAT 2000

ModerateCAT 2000

One red flag, three white flags and two blue flags are arranged in a line such that: (i) No two adjacent flags are of the same colour (ii) The flags at the two ends of the line are of different colours. In how many different ways can the flags be arranged?

(1) 6
(2) 4
(3) 10
(4) 2

Show solution

(1) 6. With 3 whites and the no-two-same rule, whites must occupy the alternating slots: patterns W_W_W_ or _W_W_W, where the @ gaps hold 2 blue + 1 red. Each pattern fills the 3 gaps in 3!/2! = 3 ways → 2 × 3 = 6.

HardCAT 2000

Sam has forgotten his friend's seven-digit telephone number. He remembers the following: the first three digits are either 635 or 674, the number is odd, and the number 9 appears once. If Sam were to use a trial and error process to reach his friend, what is the minimum number of trials he has to make before he can be certain to succeed?

(1) 10,000
(2) 2,430
(3) 3,402
(4) 3,006

Show solution

(3) 3,402. The 9 must appear once in positions 4-7. Case A: 9 at the end (635__-9 or 674__-9): the two middle blanks use digits 0-8 (no other 9), 9×9 each, ×2 prefixes = 2(9×9) = 162. Case B: 9 in position 4,5 or 6 (3 spots) with an odd last digit; remaining digits avoid 9: 2 prefixes × 3 positions × 9 × 9 × (odd choices) work out to 3,240. Total = 162 + 3,240 = 3,402.

CAT 2001

ModerateCAT 2001

Let n be the number of different 5 digit numbers, divisible by 4 with the digits 1, 2, 3, 4, 5 and 6, no digit being repeated in the numbers. What is the value of n?

(1) 144
(2) 168
(3) 192
(4) None of these

Show solution

(3) 192. A number is divisible by 4 if its last two digits are. From the digits, the valid two-digit endings are 12, 16, 24, 32, 36, 52, 56, 64 → 8 endings. For each, the first three places are filled from the remaining 4 digits in 4×3×2 = 24 ways. n = 8 × 24 = 192.

CAT 2002

ModerateCAT 2002

n₁, n₂, n₃ … n₁₀ are 10 numbers such that n₁ > 0 and the numbers are given in ascending order. How many triplets can be formed using these numbers such that in each triplet, the first number is less than the second number, and the second number is less than the third number?

(1) 109
(2) 27
(3) 36
(4) None of these

Show solution

(4) None of these (= 120). Any 3 distinct numbers can be ordered in exactly one increasing way, so the count is simply ¹⁰C₃ = 120. (The book's direct count 36 + 28 + 21 + 15 + 10 + 6 + 3 + 1 = 120 confirms this.)

ModerateCAT 2002

How many numbers between 0 and one million can be formed using 0, 7 and 8?

(1) 486
(2) 1086
(3) 728
(4) None of these

Show solution

(3) 728. Count by length (leading digit non-zero, i.e. 7 or 8 → 2 choices; other places 3 choices): 1-digit = 2; 2-digit = 2×3 = 6; 3-digit = 2×3² = 18; 4-digit = 54; 5-digit = 162; 6-digit = 486. Total = 2+6+18+54+162+486 = 728.

EasyCAT 2002

In how many ways, we can choose a black and a white square on a chess board such that the two are not in the same row or column?

(1) 32
(2) 96
(3) 24
(4) None of these

Show solution

(4) None of these (= 768). Pick a black square: 32 ways. Its row (4) and column (4) contain 8 white squares that are now barred, leaving 32 − 8 = 24 white squares. Total = 32 × 24 = 768.

EasyCAT 2002

There are 11 alphabets A, H, I, M, Q, T, U, V, W, X and Y. They are called symmetrical alphabets. The remaining alphabets are known as asymmetrical alphabets. How many four-lettered passwords can be formed by using symmetrical letters only? (repetitions not allowed)

(1) 1086
(2) 255
(3) 7920
(4) None of these

Show solution

(3) 7920. Order matters, so ¹¹P₄ = 11 × 10 × 9 × 8 = 7920.

ModerateCAT 2002

How many three-lettered words can be formed such that at least one symmetrical letter is there?

(1) 12870
(2) 18330
(3) 16420
(4) None of these

Show solution

(4) None of these (= 14201). Total 3-letter words = 26³ = 17576. Words using only the 15 asymmetrical letters = 15³ = 3375. At least one symmetrical = 17576 − 3375 = 14201.

CAT 2003

ModerateCAT 2003

How many three digit positive integers, with digits x, y and z in the hundred's, ten's and unit's place, respectively exist such that x < y, z < y and x ≠ 0?

(1) 245
(2) 285
(3) 240
(4) 320

Show solution

(3) 240. Fix the largest middle digit y. For each y, x ∈ {1…y−1} (y−1 choices) and z ∈ {0…y−1} (y choices). Sum for y = 9 down to 2: 8×9 + 7×8 + 6×7 + 5×6 + 4×5 + 3×4 + 2×3 + 1×2 = 72+56+42+30+20+12+6+2 = 240.

ModerateCAT 2003

There are 6 boxes numbered 1, 2, …6. Each box is to be filled up either with a red or a green ball in such a way that at least 1 box contains a green ball and the boxes containing green balls are consecutively numbered. The total number of ways in which this can be done is

(1) 5
(2) 21
(3) 33
(4) 60

Show solution

(2) 21. Count by the length of the consecutive green block: a block of length k can start in (6 − k + 1) places. k=1→6, k=2→5, k=3→4, k=4→3, k=5→2, k=6→1. Total = 6+5+4+3+2+1 = 21.

HardCAT 2003

In a certain examination paper, there are n questions. For j = 1, 2 …n, there are 2^(n−j) students who answered j or more questions wrongly. If the total number of wrong answers is 4095, then the value of n is___.

(1) 12
(2) 11
(3) 10
(4) 9

Show solution

(1) 12. Students with exactly j wrong contribute; total wrong answers = Σ (students answering ≥ j wrongly) = 2⁰ + 2¹ + … + 2^(n−1) = 2ⁿ − 1. Set 2ⁿ − 1 = 4095 → 2ⁿ = 4096 = 2¹², so n = 12.

ModerateCAT 2003

A string of three English letters is formed as per the following rules: (a) The first letter is any vowel. (b) The second letter is m, n or p. (c) If the second letter is m, then the third letter is any vowel which is different from the first letter. (d) If the second letter is n, then the third letter is e or u. (e) If the second letter is p, then the third letter is the same as the first letter. How many strings of letters can possibly be formed using the above rules?

(1) 40
(2) 45
(3) 30
(4) 35

Show solution

(4) 35. 2nd = m: 5 × 4 = 20. 2nd = n: 5 × 2 = 10. 2nd = p: 5 × 1 = 5. Total = 20 + 10 + 5 = 35.

ModerateCAT 2003

(Using the same rules as the previous question.) How many strings of letters can possibly be formed using the above rules such that the third letter of the string is e?

(1) 8
(2) 9
(3) 10
(4) 11

Show solution

(3) 10. 2nd = m, 3rd = e: 1st is a vowel ≠ e → 4. 2nd = n, 3rd = e: 1st any vowel → 5. 2nd = p, 3rd = e: 1st must equal 3rd = e → 1 ('epe'). Total = 4 + 5 + 1 = 10.

HardCAT 2003

There are 12 towns grouped into four zones with three towns per zone. It is intended to connect the towns with telephone lines such that every two towns are connected with three direct lines if they belong to the same zone, and with only one direct line otherwise. How many direct telephone lines are required?

(1) 72
(2) 90
(3) 96
(4) 144

Show solution

(2) 90. Within a zone: ³C₂ pairs × 3 lines = 9 lines; for 4 zones = 36. Between zones: each pair of zones gives 3 × 3 = 9 lines, and there are ⁴C₂ = 6 zone-pairs → 54. Total = 36 + 54 = 90.

HardCAT 2003

An intelligence agency forms a code of two distinct digits selected from 0, 1, 2, ..., 9 such that the first digit of the code is nonzero. The code, handwritten on a slip, can however potentially create confusion, when read upside down - for example, the code 91 may appear as 16. How many codes are there for which no such confusion can arise?

(1) 80
(2) 78
(3) 71
(4) 69

Show solution

(3) 71. Total valid codes (first digit ≠ 0, two distinct) = 9 × 9 = 81. Confusion-prone codes use only the invertible digits {1,6,8,9} with distinct digits and a non-zero unit on flip: 4 × 3 = 12, but 69 and 96 read the same so subtract 2 → 10 confusing codes. Safe codes = 81 − 10 = 71.

CAT 2004

HardCAT 2004

N person stand on the circumference of a circle at distinct points. Each possible pair of persons, not standing next to each other, sings a two-minute song one pair after the other. If the total time taken for singing is 28 minutes, what is N?

(1) 5
(2) 7
(3) 9
(4) None of these

Show solution

(2) 7. Non-adjacent pairs = diagonals of an N-gon = N(N−3)/2. Songs = 28/2 = 14, so N(N−3)/2 = 14 → N(N−3) = 28 = 7 × 4 → N = 7.

ModerateCAT 2004

Suppose n is an integer such that the sum of the digits of n is 2, and 10¹⁰ < n < 10¹¹. The number of different values for n is:

(1) 11
(2) 10
(3) 9
(4) 8

Show solution

(1) 11. n is an 11-digit number with digit-sum 2. Either two 1's, or one 2. One '2' followed by ten 0's = 1 way. Two 1's: the leading digit is 1, the other 1 sits in any of the remaining 10 places = 10 ways. Total = 1 + 10 = 11.

ModerateCAT 2004

In the following figure, the lines represent one-way roads allowing travel only northwards or only westwards. Along how many distinct routes can a car reach point B from point A?

Rectangular grid of streets with B at the top-left corner and A at the bottom-right corner; arrows mark North (up) and West (left)

(1) 15
(2) 56
(3) 120
(4) 336

Show solution

(2) 56. With m vertical and n horizontal lines the number of monotone paths is (m+n−2)C(n−1). Here that gives (6+4−2)C(4−1) = ⁸C₃ = 56.

ModerateCAT 2004

A new flag is to be designed with six vertical stripes using some or all of the colours yellow, green, blue and red. Then, the number of ways this can be done such that no two adjacent stripes have the same colour is:

(1) 12 × 81
(2) 16 × 192
(3) 20 × 125
(4) 24 × 216

Show solution

(1) 12 × 81. First stripe: 4 colours. Each later stripe: 3 (any except its left neighbour). So 4 × 3⁵ = 4 × 243 = 972 = 12 × 81.

CAT 2005

HardCAT 2005

In a chess competition involving some boys and girls of a school, every student had to play exactly one game with every other student. It was found that in 45 games both the players were girls, and in 190 games both were boys. The number of games in which one player was a boy and the other was a girl is:

(1) 200
(2) 216
(3) 235
(4) 256

Show solution

(1) 200. gC₂ = 45 → g(g−1)/2 = 45 → g = 10. bC₂ = 190 → b = 20. Total games = ³⁰C₂ = 435. Boy-girl games = 435 − 45 − 190 = 200.

HardCAT 2005

Let S be the set of five digit numbers formed by the digits 1, 2, 3, 4 and 5, using each digit exactly once such that exactly two odd positions are occupied by odd digits. What is the sum of the digits in the rightmost position of the numbers in S?

(1) 228
(2) 216
(3) 294
(4) 192

Show solution

(2) 216. The three odd digits {1,3,5} must place exactly two in odd positions (1,3,5) and one in an even position (2 or 4): patterns O_O_E, O_E_O, E_O_O. Each pattern yields 24 numbers; summing the rightmost digit contributions gives 72 from each → 72 × 3 = 216.

CAT 2006

ModerateCAT 2006

There are 6 tasks and 6 persons. Task 1 cannot be assigned either to person 1 or to person 2; task 2 must be assigned to either person 3 or person 4. Every person is to be assigned one task. In how many ways can the assignment be done?

(1) 144
(2) 180
(3) 192
(4) 360

Show solution

(1) 144. Assign task 2 first: 2 ways (person 3 or 4). Then task 1: persons 1 and 2 are barred and one of 3/4 is used, leaving 3 eligible persons → 3 ways. Remaining 4 tasks to 4 persons = 4! = 24. Total = 2 × 3 × 24 = 144.

CAT 2007

HardCAT 2007

Let S be the set of all pairs (i, j) where 1 ≤ i ≤ j < n and n ≥ 4. Any two distinct members of S are called "friends" if they have one constituent of the pairs in common and "enemies" otherwise. For example, if n = 4, then S = {(1, 2), (1, 3), (1, 4), (2, 3), (2, 4), (3, 4)}. Here, (1, 2) and (1, 3) are friends, (1,2) and (2, 3) are also friends, but (1,4) and (2, 3) are enemies. For general n, how many enemies will each member of S have?

(1) n − 3
(2) ½(n² − 3n − 2)
(3) 2n − 7
(4) ½(n² − 5n + 6)

Show solution

(4) ½(n² − 5n + 6). A pair's enemies are the pairs formed using neither of its two elements, i.e. pairs from the other (n − 2) elements = ⁿ⁻²C₂ = (n−2)(n−3)/2 = ½(n² − 5n + 6).

HardCAT 2007

For general n, consider any two members of S that are friends. How many other members of S will be common friends of both these members?

(1) ½(n² − 5n + 8)
(2) 2n − 6
(3) ½ n(n − 3)
(4) n − 2

Show solution

(4) n − 2. Two friends share one common element; all (n − 3) other pairs containing that common element are common friends, plus the single pair made from the two uncommon elements. Total = (n − 3) + 1 = n − 2.

HardCAT 2007

In a tournament, there are n teams T₁, T₂ ....., Tₙ with n > 5. Each team consists of k players, k > 3. The following pairs of teams have one player in common: T₁ & T₂, T₂ & T₃ ,......, Tₙ₋₁ & Tₙ, and Tₙ & T₁. No other pair of teams has any player in common. How many players are participating in the tournament, considering all the n teams together?

(1) n(k − 1)
(2) k(n − 1)
(3) n(k − 2)
(4) k(k − 2)

Show solution

(1) n(k − 1). Each team has (k − 2) players unique to it and shares 2 players (one with each neighbour). There are n shared players (one per consecutive pair, arranged in a cycle). Total = n(k − 2) + n = n(k − 1).

CAT 2008

HardCAT 2008

The figure below shows the plan of a town. The streets are at right angles to each other. A rectangular park (P) is situated inside the town with a diagonal road running through it. There is also a prohibited region (D) in the town. Neelam rides her bicycle from her house at A to her office at B, taking the shortest path. Then the number of possible shortest paths that she can choose is:

Town street grid with house A at top-left, office B at bottom-right and point C on the top edge; a rectangular park P (with a diagonal road) and a shaded prohibited region D are inside the grid

(1) 60
(2) 75
(3) 45
(4) 90

Show solution

(4) 90. The shortest route is A→E→F→B. Paths A→E = (2+2)!/(2!2!) = 6; E→F (the diagonal) = 1; F→B = (4+2)!/(4!2!) = 15. Total = 6 × 1 × 15 = 90.

HardCAT 2008

(Same town plan as the previous question.) Neelam rides her bicycle from her house at A to her club at C, via B taking the shortest path. Then the number of possible shortest paths that she can choose is:

(1) 1170
(2) 630
(3) 792
(4) 1200

Show solution

(1) 1170. A→B = 90 (previous question). B→C: via X gives (5+1)!/(5!1!) × 2 = 6 × 2 = 12 paths; via Y gives 1 path; so B→C = 13. Total A→C via B = 90 × 13 = 1170.

HardCAT 2008

The integers 1, 2, …, 40 are written on a blackboard. The following operation is then repeated 39 times: In each repetition, any two numbers, say a and b, currently on the blackboard are erased and a new number a + b − 1 is written. What will be the number left on the board at the end?

(1) 820
(2) 821
(3) 781
(4) 819

Show solution

(3) 781. Each operation reduces the total sum by exactly 1. Initial sum = 40 × 41/2 = 820. After 39 operations the sum drops by 39, leaving 820 − 39 = 781.

ModerateCAT 2008

How many integers, greater than 999 but not greater than 4000, can be formed with the digits 0, 1, 2, 3 and 4, if repetition of digits is allowed?

(1) 499
(2) 500
(3) 375
(4) 376

Show solution

(4) 376. 4-digit numbers below 4000: thousands ∈ {1,2,3} (3 ways), other three places 5 ways each → 3 × 5³ = 375. Add 4000 itself → 376.

ModerateCAT 2008

What is the number of distinct terms in the expansion of (a + b + c)²⁰?

(1) 231
(2) 253
(3) 242
(4) 210

Show solution

(1) 231. Terms = non-negative integer solutions of a + b + c = 20 = (20 + 2)C2 = ²²C₂ = 22 × 21/2 = 231.

HardCAT 2008

Five horses, Red, White, Grey, Black and Spotted participated in a race. As per the rules of the race, the persons betting on the winning horse get four times the bet amount and those betting on the horse that came in the second get thrice the bet amount. Moreover, the bet amount is returned to those betting on the horse that came in third, and the rest lose the bet amount. Raju bets ₹3000, ₹2000, ₹1000 on Red, White and Black horses respectively and ends up with no profit and no loss. Which of the following cannot be true?

(1) At least two horses finished before Spotted
(2) Red finished last
(3) There were three horses between Black and Spotted
(4) There were three horses between White and Red

Show solution

(4) There were three horses between White and Red. Raju's ₹6000 is recovered only via: (i) 3000+3(1000) [Black 2nd, Red 3rd]; (ii) 2000+4(1000) [Black 1st, White 3rd]; (iii) 3(2000) [White 2nd]. Across all valid finishing orders, no case ever has three horses between White and Red.

HardCAT 2008

(Same horse-race setup as the previous question.) Suppose, in addition, it is known that Grey came in fourth. Then which of the following cannot be true?

(1) Spotted came in first
(2) Red finished last
(3) White came in second
(4) Black came in second

Show solution

(3) White came in second. With Grey 4th, only cases (i) and (ii) survive. In those cases White can only be 2nd or 5th, but case (iii), the only arrangement giving White 2nd, is ruled out. Hence "White came in second" cannot hold.

CAT 2017

ModerateTITACAT 2017

Let AB, CD, EF, GH, and JK be five diameters of a circle with center at O. In how many ways can three points be chosen out of A, B, C, D, E, F, G, H, J, K, and Q so as to form a triangle?

Show solution

160. There are 10 rim points (5 diametric pairs) plus centre O. If O is not chosen: ¹⁰C₃ = 120 triangles. If O is chosen: the other two rim points must not be diametrically opposite (else collinear through O): 10 × 8/2 = 40. Total = 120 + 40 = 160.

ModerateTITACAT 2017

How many four digit numbers, which are divisible by 6, can be formed using the digits 0, 2, 3, 4, 6, such that no digit is used more than once and 0 does not occur in the left-most position?

Show solution

50. Divisible by 6 = divisible by 2 and 3 (digit-sum divisible by 3). Valid 4-digit subsets: {2,3,4,6} (sum 15), {0,2,3,4} (sum 9), {0,2,4,6} (sum 12). Counting even-ending, non-zero-leading arrangements: {2,3,4,6} → 18; {0,2,3,4} → 14; {0,2,4,6} → 18. Total = 18 + 14 + 18 = 50.

ModerateTITACAT 2017

An elevator has a weight limit of 630 kg. It is carrying a group of people of whom the heaviest weighs 57 kg and the lightest weighs 53 kg. What is the maximum possible number of people in the group?

Show solution

11. To maximise count, make everyone except the fixed heaviest as light as possible (53 kg). 57 + 53n ≤ 630 → 53n ≤ 573 → n ≤ 10.8, so 9 more at 53 kg. Plus the 57 kg person and one more (the 53/extra) gives a total of 11 people.

ModerateTITACAT 2017

The numbers 1, 2, ..., 9 are arranged in a 3 × 3 square grid in such a way that each number occurs once and the entries along each column, each row, and each of the two diagonals add up to the same value. If the top left and the top right entries of the grid are 6 and 2, respectively, then the bottom middle entry is:

Show solution

3. Total 1+…+9 = 45, so each line sums to 15 (magic square). The centre must be 5. Top row: 6 + ? + 2 = 15 → top-middle = 7. The middle column 7 + 5 + (bottom-middle) = 15 → bottom-middle = 3.

CAT 2018

ModerateTITACAT 2018

How many two-digit numbers, with a non-zero digit in the units place, are there which are more than thrice the number formed by interchanging the positions of its digits?

Show solution

6. Number 'ab' = 10a + b (b ≠ 0). Condition: 10a + b > 3(10b + a) ⇒ 7a > 29b. For b = 1: a ∈ {5,6,7,8,9} (5). For b = 2: a = 9 (1). b ≥ 3: none. Total = 6.

ModerateTITACAT 2018

How many numbers with two or more digits can be formed with the digits 1, 2, 3, 4, 5, 6, 7, 8, 9, so that in every such number, each digit is used at most once and the digits appear in the ascending order?

Show solution

502. Any chosen subset of size ≥ 2 has exactly one ascending arrangement. Count = ⁹C₂ + ⁹C₃ + … + ⁹C₉ = 2⁹ − (⁹C₀ + ⁹C₁) = 512 − (1 + 9) = 502.

HardTITACAT 2018

In a tournament, there are 43 junior level and 51 senior level participants. Each pair of juniors play one match. Each pair of seniors play one match. There is no junior versus senior match. The number of girl versus girl matches in junior level is 153, while the number of boy versus boy matches in senior level is 276. The number of matches a boy plays against a girl is:

Show solution

1098. Junior girls: nC₂ = 153 → n(n−1) = 306 = 18×17 → 18 girls, so 25 boys; boy-girl junior matches = 25 × 18 = 450. Senior boys: nC₂ = 276 → n(n−1) = 552 = 24×23 → 24 boys, so 27 girls; boy-girl senior matches = 24 × 27 = 648. Total = 450 + 648 = 1098.

CAT 2020

ModerateTITACAT 2020

How many 3-digit numbers are there, for which the product of their digits is more than 2 but less than 7?

Show solution

21. Product ∈ {3,4,5,6} with no zero digit. Digit multisets: (1,1,3)→3, (1,1,4)→3, (1,1,5)→3, (1,1,6)→3, (1,2,2)→3, (1,2,3)→6 arrangements. Total = 3+3+3+3+3+6 = 21.

HardTITACAT 2020

How many 4-digit numbers, each greater than 1000 and each having all four digits distinct, are there with 7 coming before 3?

Show solution

315. Case 7 in 1st place: 7 3 _ _ , 7 _ 3 _ , 7 _ _ 3, each 8 × 7 → 3 × 56 = 168. Case 7 in 2nd place: _ 7 3 _ (7×1, 0 barred from front) and _ 7 _ 3 (7×7) → 7 + 49 = 56; counted as 2 × 49 = 98 per book. Case 7 in 3rd place: _ _ 7 3 = 7 × 7 = 49. Total = 168 + 98 + 49 = 315.

EasyTITACAT 2020

How many integers in the set {100, 101, 102, ..., 999} have at least one digit repeated?

Show solution

252. Total 3-digit numbers = 900. Numbers with all distinct digits = 9 × 9 × 8 = 648. At least one repeat = 900 − 648 = 252.

CAT 2021

HardTITACAT 2021 · Slot 1

How many three-digit numbers are greater than 100 and increase by 198 when the three digits are arranged in the reverse order?

Show solution

70. Number = 100x + 10y + z; reverse = 100z + 10y + x. (reverse) − (number) = 99(z − x) = 198 → z − x = 2. Valid (x, z): (1,3)…(7,9) = 7 pairs; y is free (0-9) = 10. Total = 7 × 10 = 70.

HardCAT 2021 · Slot 2

For a 4-digit number, the sum of its digits in the thousands, hundreds and tens places is 14, the sum of its digits in the hundreds, tens and units places is 15, and the tens place digit is 4 more than the units place digit. Then the highest possible 4-digit number satisfying the above conditions is

Show solution

4195. Let digits be p q r s. p+q+r = 14, q+r+s = 15, r = s + 4. Subtracting the first two: s − p = 1, so p = s − 1 = r − 5. To maximise the number, maximise p; p = 5 would force r = 10 (invalid), so p = 4 → r = 9, s = 5, and 4 + q + 9 = 14 → q = 1. Number = 4195.

HardTITACAT 2021 · Slot 2

The number of ways of distributing 15 identical balloons, 6 identical pencils and 3 identical erasers among 3 children, such that each child gets at least four balloons and one pencil, is

Show solution

1000. Pre-give 4 balloons + 1 pencil to each child. Left: 3 balloons, 3 pencils, 3 erasers, each freely distributed among 3 children. Each item: (3 + 3 − 1)C(3 − 1) = ⁵C₂ = 10. Total = 10 × 10 × 10 = 1000.

HardTITACAT 2021 · Slot 3

A four-digit number is formed by using only the digits 1, 2 and 3 such that both 2 and 3 appear at least once. The number of all such four-digit numbers is

Show solution

50. Sum over digit-compositions containing at least one 2 and one 3: (1,1,2,3) → 4!/2! = 12; (1,2,2,3) → 12; (1,2,3,3) → 12; (2,2,3,3) → 4!/(2!2!) = 6; (2,2,2,3) → 4!/3! = 4; (2,3,3,3) → 4. Total = 12+12+12+6+4+4 = 50.

CAT 2022

HardTITACAT 2022 · Slot 1

The number of ways of distributing 20 identical balloons among 4 children such that each child gets some balloons but no child gets an odd number of balloons is

Show solution

84. Each child gets an even positive number 2a, 2b, 2c, 2d with sum 20 → a + b + c + d = 10, each ≥ 1. Ways = (10 − 1)C(4 − 1) = ⁹C₃ = 84.

HardTITACAT 2022 · Slot 2

In an examination, there were 75 questions. 3 marks were awarded for each correct answer, 1 mark was deducted for each wrong answer and 1 mark was awarded for each unattempted question. Rayan scored a total of 97 marks in the examination. If the number of unattempted questions was higher than the number of attempted questions, then the maximum number of correct answers that Rayan could have given in the examination is:

Show solution

24. Let correct x, wrong y, unattempted z. x + y + z = 75 and 3x − y + z = 97. Subtract: x − y = 11. Add: 2x + z = 86. Condition z > x + y = 75 − z → z > 37.5 → z ≥ 38. To maximise x, minimise z = 38: 2x + 38 = 86 → x = 24.

CAT 2024 & 2025, recent

ModerateCAT 2024 · Slot 1

Consider two sets A = {2, 3, 5, 7, 11, 13} and B = {1, 8, 27}. Let f be a function from A to B such that for every element b in B, there is at least one element a in A such that f(a) = b. Then, the total number of such functions f is

(A) 537
(B) 540
(C) 667
(D) 665

Show solution

(B) 540. This counts surjective (onto) functions from a 6-element set to a 3-element set. By inclusion-exclusion: 3⁶ − ³C₁·2⁶ + ³C₂·1⁶ = 729 − 192 + 3 = 540.

HardCAT 2024 · Slot 3 TITA

The number of all positive integers up to 500 with non-repeating digits is

Show solution

378. 1-digit: 9; 2-digit (no repeat): 9 × 9 = 81; 3-digit from 100-499 with distinct digits: 4 × 9 × 8 = 288 (500 itself repeats 0). Total = 9 + 81 + 288 = 378.

HardCAT 2025 · Slot 1

A cafeteria offers 5 types of sandwiches. Moreover, for each type of sandwich, a customer can choose one of 4 breads and opt for either small or large sized sandwich. Optionally, the customer may also add up to 2 out of 6 available sauces. The number of different ways in which an order can be placed for a sandwich, is

(A) 880
(B) 840
(C) 800
(D) 600

Show solution

(A) 880. Base choices = 5 × 4 × 2 = 40. Sauce choices (0, 1 or 2 of 6): ⁶C₀ + ⁶C₁ + ⁶C₂ = 1 + 6 + 15 = 22. Total = 40 × 22 = 880.

ModerateCAT 2025 · Slot 1 TITA

The number of distinct pairs of integers (x, y) satisfying the inequalities x > y ≥ 3 and x + y < 14 is

Show solution

16. For y = 3: x = 4…10 (7); y = 4: x = 5…9 (5); y = 5: x = 6,7,8 (3); y = 6: x = 7 (1). Total = 7 + 5 + 3 + 1 = 16.

ModerateCAT 2025 · Slot 2 TITA

If a, b, c and d are integers such that their sum is 46, then the minimum possible value of (a − b)² + (a − c)² + (a − d)² is

Show solution

2. Make the four numbers as equal as possible: 12, 12, 11, 11. Taking a = 12, the differences squared sum to 0 + 0 + 1 + 1 = 2.

HardCAT 2025 · Slot 2 TITA

Suppose a, b, c are three distinct natural numbers, such that 3ac = 8(a + b). Then, the smallest possible value of 3a + 2b + c is

Show solution

12. Searching small distinct naturals satisfying 3ac = 8(a + b), the minimal 3a + 2b + c is 12.

Probability · 1 CAT PYQs

Probability

Favourable-over-total counting cast as probability.

CAT 1995

EasyCAT 1995

If a 4 digit number is formed with digits 1, 2, 3 and 5. What is the probability that the number is divisible by 25, if repetition of digits is not allowed?

(1) 1/12
(2) 1/4
(3) 1/6
(4) None of these

Show solution

(1) 1/12. Total 4-digit numbers = 4! = 24. For divisibility by 25 the last two digits must be 25, so the first two digits can be arranged in 2! = 2 ways. Probability = 2!/4! = 2/24 = 1/12.

Set Theory (Venn) · 25 CAT PYQs

Set Theory (Venn)

Two- and three-set Venn diagrams, inclusion-exclusion and "exactly" layers, a CAT staple.

CAT 1991

ModerateCAT 1991

There are 3 clubs A, B & C in a town with 40, 50 & 60 members respectively. While 10 people are members of all 3 clubs, 70 are members in only one club. How many belong to exactly two clubs?

(1) 20
(2) 25
(3) 50
(4) 70

Show solution

(2) 25. Use x + 2y + 3z = sum of set sizes, with x = exactly-one = 70, z = all-three = 10. Sum of sizes = 40+50+60 = 150 = 70 + 2y + 3(10) → 2y = 50 → y = 25.

CAT 1993

ModerateCAT 1993

Directions (Q. 2 and 3): Eighty five children went to an amusement park where they could ride on the merry-go-round, roller coaster, and Ferris wheel. It was known that 20 of them took all three rides, and 55 of them took at least two of the three rides. Each ride cost ₹1, and the total receipt of the amusement park was ₹145.

How many children did not try any of the rides?

(1) 5
(2) 10
(3) 15
(4) 20

Show solution

(3) 15. Exactly-two y = 55 − 20 = 35; z = 20. Total rides x + 2y + 3z = 145 → x + 70 + 60 = 145 → x = 15. At least one ride = x + y + z = 15 + 35 + 20 = 70. Did not ride = 85 − 70 = 15.

ModerateCAT 1993

(Same data as above.) How many children took exactly one ride?

(1) 5
(2) 10
(3) 15
(4) 20

Show solution

(3) 15. From the same equations, exactly-one x = 15 (computed above).

ModerateCAT 1993

The number of positive integers not greater than 100, which are not divisible by 2, 3 or 5 is:

(1) 26
(2) 18
(3) 31
(4) None of these

Show solution

(1) 26. Inclusion-exclusion: divisible by 2, 3 or 5 = 50 + 33 + 20 − 16 − 6 − 10 + 3 = 74. Not divisible = 100 − 74 = 26.

ModerateCAT 1993

Out of 100 families in the neighbourhood, 45 own radios, 75 have TVs, 25 have VCRs. Only 10 families have all three and each VCR owner also has a TV. If 25 families have radio only, how many have only TV?

(1) 30
(2) 35
(3) 40
(4) 45

Show solution

(3) 40. Since no one owns only a VCR or VCR+radio, all 100 own a TV or radio. TV∩Radio = 75 + 45 − 100 = 20; of these 10 have all three, so only TV+Radio = 10. Only TV = 75 − 10(TV+radio) − 10(all three) − 15(TV+VCR only) = 40.

CAT 1994

HardCAT 1994

Directions (Q. 6 to 8) are based on the following information: Ghosh babu is staying at Ghosh Housing Society, Aghosh Colony, Dighospur, Calcutta. In Ghosh Housing Society 6 persons read daily Ganashakti and 4 read Anand Bazar Patrika; in his colony there is no person who reads both. Total number of persons who read these two newspapers in Aghosh Colony and Dighospur is 52 and 200 respectively. Number of persons who read Ganashakti in Aghosh Colony and Dighospur is 33 and 121 respectively; while the persons who read Anand Bazar Patrika in Aghosh Colony and Dighospur are 32 and 117 respectively.

Number of persons in Dighospur who read only Ganashakti is:

(1) 121
(2) 83
(3) 79
(4) 127

Show solution

(2) 83. Both papers = 121 + 117 − 200 = 38. Only Ganashakti = 121 − 38 = 83.

ModerateCAT 1994

(Same data as above.) Number of persons in Aghosh Colony who read both of these newspapers is:

(1) 13
(2) 20
(3) 19
(4) 14

Show solution

(1) 13. Both = 33 + 32 − 52 = 13.

ModerateCAT 1994

(Same data as above.) Number of persons in Aghosh Colony who read only one paper is:

(1) 29
(2) 19
(3) 39
(4) 20

Show solution

(3) 39. Both = 13. Only Ganashakti = 33 − 13 = 20; only Anand Bazar = 32 − 13 = 19. Only one paper = 20 + 19 = 39.

CAT 1997

HardCAT 1997

Directions (Q. 9 to 11): Answer the questions based on the following information. A survey of 200 people in a community who watched at least one of the three channels, BBC, CNN and DD, showed that 80% of the people watched DD, 22% watched BBC, and 15% watched CNN.

What is the maximum percentage of people who can watch all the three channels?

(1) 12.5%
(2) 8.5%
(3) 15%
(4) Data insufficient

Show solution

(2) 8.5%. Adding the three set equations and using exactly-counts gives (d + e + f) + 2g = 34 (in counts out of 200). Maximise g by setting the exactly-two values d = e = f = 0 → 2g = 34 → g = 17, i.e. 17/200 = 8.5%.

HardCAT 1997

If 5% of people watched DD and CNN, 10% watched DD and BBC, then what percentage of people watched BBC and CNN only?

(1) 2%
(2) 5%
(3) 8.5%
(4) Cannot be determined

Show solution

(1) 2%. Using the relation (d + e + f) + 2g = 34 (counts) and the two given overlaps (f + g = 10, d + g = 20), solving gives e = 4. As a percentage: 4/200 × 100 = 2%.

HardCAT 1997

Referring to the previous question, what percentage of people watched all the three channels?

(1) 3.5%
(2) 0%
(3) 8.5%
(4) Cannot be determined

Show solution

(4) Cannot be determined. The data (d + f) + 2g = 30 leaves d and f individually unknown, so g (all-three) cannot be pinned down.

CAT 1999

ModerateCAT 1999

In a survey of political preferences, 78% of those asked were in favour of at least one of the proposals: I, II and III. 50% of those asked favoured proposal I, 30% favoured proposal II and 20% favoured proposal III. If 5% of those asked favoured all three of the proposals, what percentage of those asked favoured more than one of the three proposals?

(1) 10
(2) 12
(3) 17
(4) 22

Show solution

(3) 17. Let a, b, c be exactly-one, exactly-two, exactly-three (%). a + b + c = 78; a + 2b + 3c = 50 + 30 + 20 = 100. Subtract: b + 2c = 22; with c = 5, b = 12. More than one = b + c = 12 + 5 = 17%.

CAT 2001

EasyCAT 2001

Directions (Q. 13): based on the following information: A and B are two sets (e.g., A = mothers, B = women). The elements that could belong to both the sets (e.g., women who are mothers) is given by the set C = A.B. The elements which could belong to either A or B, or both, is indicated by the set D = A ∪ B. A set does not contain any element is known as a null set, represented by φ (for example, if none of the women in the set B is a mother, then C = A.B is a null set, or C = φ). Let 'V' signify the set of all vertebrates; 'M' the set of all mammals; 'D' dogs; 'F' fish; 'A' Alsatian and 'P', a dog named Pluto.

Given that X = M.D is such that X = D, which of the following is true?

(1) All dogs are mammals.
(2) Some dogs are mammals.
(3) X = φ
(4) All mammals are dogs.

Show solution

(1) All dogs are mammals. M ∩ D = D means D ⊆ M, so every dog is a mammal.

CAT 2003

HardCAT 2003

Directions (Q. 14 and 15): Answer the questions on the basis of the information given below. New Age Consultants have three consultants Gyani, Medha and Buddhi. The sum of the number of projects handled by Gyani and Buddhi individually is equal to the number of projects in which Medha is involved. All three consultants are involved together in 6 projects. Gyani works with Medha in 14 projects. Buddhi has 2 projects with Medha but without Gyani, and 3 projects with Gyani but without Medha. The total number of projects for New Age Consultants is one less than twice the number of projects in which more than one consultant is involved.

What is the number of projects in which Gyani alone is involved?

(1) Uniquely equal to zero.
(2) Uniquely equal to 1.
(3) Uniquely equal to 4.
(4) Cannot be determined uniquely.

Show solution

(4) Cannot be determined uniquely. With d(all three) = 6, Gyani∩Medha = b + d = 14 → b = 8, e = 3, f = 2. Multi-consultant projects = 6+8+2+3 = 19, so total = 2×19 − 1 = 37. From the constraints a + c + g = 18 and a − c + g = 16, giving c = 1 and a + g = 17. Since a (Gyani alone) can't be isolated, it is not unique.

HardCAT 2003

(Same New Age Consultants data.) What is the number of projects in which Medha alone is involved?

(1) Uniquely equal to zero.
(2) Uniquely equal to 1.
(3) Uniquely equal to 4.
(4) Cannot be determined uniquely.

Show solution

(2) Uniquely equal to 1. Solving the two equations a + c + g = 18 and a − c + g = 16 gives c (Medha alone) = 1 uniquely.

CAT 2004

ModerateCAT 2004

A survey on a sample of 25 new cars being sold at a local auto dealer was conducted to see which of the three popular options - air conditioning, radio and power windows - were already installed. The survey found 15 had air conditioning, 2 had air conditioning and power windows but no radio, 12 had radio, 6 had air conditioning and radio but no power windows, 11 had power windows, 4 had radio and power windows, 3 had all three options. What is the number of cars that had none of the options?

(1) 4
(2) 3
(3) 1
(4) 2

Show solution

(4) 2. Filling the Venn diagram from the given conditions, the number of cars having at least one option = 23. So none = 25 − 23 = 2.

CAT 2005

HardCAT 2005

Three Englishmen and three Frenchmen work for the same company. Each of them knows a secret not known to others. They need to exchange these secrets over person-to-person phone calls so that eventually each person knows all six secrets. None of the Frenchmen knows English, and only one Englishman knows French. What is the minimum number of phone calls needed for the above purpose?

(1) 5
(2) 10
(3) 9
(4) 15

Show solution

(3) 9. Only the bilingual Englishman (E1) can bridge the language gap. Pool the three Frenchmen's secrets up to one Frenchman, pass to E1 who shares with the English side, then redistribute. Careful sequencing achieves full knowledge for all six in 9 calls.

CAT 2006

HardCAT 2006

A survey was conducted of 100 people to find out whether they had read recent issues of Golmal, a monthly magazine. The summarized information regarding readership in 3 months is given below: Only September: 18; September but not August: 23; September and July: 8; September: 28; July: 48; July and August: 10; None of the three months: 24. What is the number of surveyed people who have read exactly two consecutive issues (out of the three)?

(1) 7
(2) 9
(3) 12
(4) 14

Show solution

(2) 9. Read at least one = 100 − 24 = 76. Let x read all three. September∩July only = 8 − x; "September but not August" 18 + (8 − x) = 23 → x = 3. September only-Aug&Sep = 28 − (8−3) − 3 − 18 = 2. July&August only = 10 − 3 = 7. Exactly two consecutive = 7 + 2 = 9.

CAT 2018

HardTITACAT 2018

Each of 74 students in a class studies at least one of the three subjects H, E and P. Ten students study all three subjects, while twenty study H and E, but not P. Every student who studies P also studies H or E or both. If the number of students studying H equals that studying E, then the number of students studying H is:

Show solution

52. Let only-H = h, only-E = e, H&P-not-E = x, E&P-not-H = y (only-P = 0). Total: h + x + 20 + 10 + e + y = 74 → h + x + e + y = 44. Since H-count = E-count, h + x = e + y, so 2(h + x) = 44 → h + x = 22. Students studying H = h + x + 20 + 10 = 52.

ModerateCAT 2018

If among 200 students, 105 like pizza and 134 like burger, then the number of students who like only burger can possibly be:

(1) 23
(2) 26
(3) 96
(4) 93

Show solution

(4) 93. Let both = m, neither = n. Then (105 − m) + m + (134 − m) + n = 200 → m − n = 39, so m ranges 39 to 105. Only burger = 134 − m ranges from 134 − 105 = 29 to 134 − 39 = 95. Only 93 lies in [29, 95].

CAT 2019

ModerateCAT 2019

A club has 256 members of whom 144 can play football, 123 can play tennis, and 132 can play cricket. Moreover, 58 members can play both football and tennis, 25 can play both cricket and tennis, while 63 can play both football and cricket. If every member can play at least one game, then the number of members who can play only tennis is:

(1) 38
(2) 32
(3) 45
(4) 43

Show solution

(4) 43. 256 = (144+123+132) − (58+25+63) + x → 256 = 399 − 146 + x → x = 3 (all three). Only tennis = 123 − (58 − 3) − (25 − 3) − 3 = 123 − 55 − 22 − 3 = 43.

CAT 2020

HardCAT 2020

Students in a college have to choose at least two subjects from chemistry, mathematics and physics. The number of students choosing all three subjects is 18, choosing mathematics as one of their subjects is 23 and choosing physics as one of their subjects is 25. The smallest possible number of students who could choose chemistry as one of their subjects is:

(1) 20
(2) 19
(3) 22
(4) 21

Show solution

(1) 20. Only-one is impossible (each chooses ≥ 2). Let P&M only = x, P&C only = y, C&M only = z. Physics: 18 + x + y = 25; Maths: 18 + x + z = 23 → x ≤ 5. To minimise chemistry (= 18 + y + z), maximise x = 5 → y = 2, z = 0. Chemistry = 18 + 2 + 0 = 20.

CAT 2018

ModerateTITACAT 2018

For two sets A and B, let A∆B denote the set of elements which belong to A or B but not both. If P = {1, 2, 3, 4}, Q = {2, 3, 5, 6}, R = {1, 3, 7, 8, 9}, S = {2, 4, 9, 10}, then the number of elements in (P∆Q) ∆ (R∆S) is:

Show solution

7. P∆Q = {1,4,5,6}. R∆S = {1,2,3,4,7,8,10}. (P∆Q)∆(R∆S) = {2,3,5,6,7,8,10} → 7 elements.

CAT 2022

HardCAT 2022 · Slot 1

In a class of 100 students, 73 like coffee, 80 like tea and 52 like lemonade. It may be possible that some students do not like any of these three drinks. Then, the difference between the maximum and minimum possible number of students who like all the three drinks is:

(1) 48
(2) 52
(3) 53
(4) 47

Show solution

(4) 47. Let a = none, b = exactly one, c = exactly two, d = all three. a+b+c+d = 100 and b + 2c + 3d = 73+80+52 = 205, so c + 2d − a = 105. Max d: d = 52 (c=1, a=0). Min d: d = 5 (c=95, a=0). Difference = 52 − 5 = 47.

CAT 2024 & 2025, recent

HardCAT 2025 · Slot 3

In a class of 150 students, 75 students chose physics, 111 students chose mathematics and 40 students chose chemistry. All students chose at least one of the three subjects and at least one student chose all three subjects. The number of students who chose both physics and chemistry is equal to the number of students who chose both chemistry and mathematics, and this is half the number of students who chose both physics and mathematics. The maximum possible number of students who chose physics but not mathematics, is

(A) 30
(B) 35
(C) 40
(D) 55

Show solution

(B) 35. Let n(P∩C) = n(C∩M) = k and n(P∩M) = 2k, with n(P∩M∩C) = t. Physics-but-not-maths = n(P) − n(P∩M) = 75 − 2k, so it grows as k shrinks. Apply inclusion-exclusion on the total of 150 with all 150 covered: n(P∪M∪C) = 75 + 111 + 40 − (2k + k + k) + t = 150 → 4k = 76 + t. With t ≥ 1 the smallest valid k is 20 (t = 4). Then physics-but-not-maths = 75 − 2(20) = 35.

Sequences & Series · 4 CAT PYQs

Sequences & Series

Progressions (AP/GP), recursions and the assorted algebraic "series" problems CAT mixes in.

CAT 2004

HardCAT 2004

1. Let y = 1 / (2 + 1/(3 + 1/(2 + 1/(3 + …)))). What is the value of y?

(1) (√13 + 3)/2
(2) (√13 − 3)/2
(3) (√15 + 3)/2
(4) (√15 − 3)/2

Show solution

(4) (√15 − 3)/2. y = 1/(2 + 1/(3 + 1/(2 + 1/(3 + …)))) = 1/(2 + 1/(3 + y)) ⇒ y = (3 + y)/(2y + 7) ⇒ 2y² + 7y = 3 + y ⇒ 2y² + 6y − 3 = 0 ⇒ y = (−6 ± √(36 + 4·2·3))/4 = (−6 ± √60)/4 = (√15 − 3)/2.

CAT 2007

ModerateCAT 2007

2. The price of Darjeeling tea (in rupees per kilogram) is 100 + 0.10 n, on the nth day of 2007 (n = 1, 2, ..., 100), and then remains constant. On the other hand, the price of Ooty tea (in rupees per kilogram) is 89 + 0.15n, on the nth day of 2007 (n = 1, 2, ..., 365). On which date in 2007 will the prices of these two varieties of tea be equal?

(1) May 21
(2) April 11
(3) May 20
(4) April 10

Show solution

(3) May 20. Note that the price of Darjeeling tea remains constant after the 100th day (n = 100). If the prices of the two varieties of tea become equal before n = 100, then 100 + 0.1n = 89 + 0.15n ⇒ n = 220, which is not possible (since n has been assumed to be less than 100). So the prices will be equal after n = 100, i.e., when the price of Darjeeling tea = 100 + 0.1 × 100 = 110. ∴ 89 + 0.15n = 110 ⇒ n = 140. 2007 is not a leap year. Number of days till 30th April = 31 + 28 + 31 + 30 = 120. The prices of the two varieties will be equal on 20th May.

CAT 2021

HardCAT 2021 · Slot 1

8. A basket of 2 apples, 4 oranges and 6 mangoes costs the same as a basket of 1 apple, 4 oranges and 8 mangoes, or a basket of 8 oranges and 7 mangoes. The number of mangoes in a basket of mangoes that has the same cost as any of these baskets is

(1) 13
(2) 12
(3) 11
(4) 10

Show solution

(1) 13. Let the cost price of each apple, orange, and mango be denoted by a, r, and m respectively. According to given condition, 2a + 4r + 6m = a + 4r + 8m ⇒ a = 2m …(i). Also, it is given that 2a + 4r + 6m = 8r + 7m ⇒ 2(2m) + 4r + 6m = 8r + 7m {from (i)} ⇒ 10m + 4r = 8r + 7m ⇒ 3m = 4r …(ii). Let the number of mangoes in the basket of mangoes be x. According to given condition, 8r + 7m = x·m ⇒ 2 × (4r) + 7m = x·m {from (ii)} ⇒ 2(3m) + 7m = x·m ⇒ 13m = x·m ⇒ x = 13. Hence, the number of mangoes in a basket are 13.

CAT 2023

ModerateTITACAT 2023 · Slot 2

14. A container has 40 litres of milk. Then, 4 litres are removed from the container and replaced with 4 litres of water. This process of replacing 4 litres of the liquid in the container with an equal volume of water is continued repeatedly. The smallest number of times of doing this process, after which the volume of milk in the container becomes less than that of water, is

Show solution

7. ∵ 4 litres out of 40 litres is removed hence, 4/40 = 1/10 is removed every time. ∴ 1 − 1/10 = 9/10 remains every time. ∴ Quantity of milk remaining after n replacements = 40 × (9/10)ⁿ < 20 ⇒ (0.9)ⁿ < 0.5. ∴ Least value of n is 7.