◆ QA · Modern Math

Set Theory (Venn) , formulas + CAT PYQs

Focused Modern Math kit. The full chapter formula sheet (with explanations & basic examples) is tucked below; every CAT PYQ for Set Theory (Venn) is here.

25CAT PYQs

Modern Mathchapter

Formulas PYQs (25)↩ Full Modern Math chapter

Modern Math, formula sheet

Show the full Modern Math formula sheet (explanations + basic examples)

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

Set Theory (Venn), 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.