◆ QA · Modern Math

Sequences & Series , formulas + CAT PYQs

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

4CAT PYQs

Modern Mathchapter

Formulas PYQs (4)↩ 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

Sequences & Series, 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.