◆ QA · Algebra

Linear & Simultaneous Equations , formulas + CAT PYQs

Focused Algebra kit. The full chapter formula sheet (with explanations & basic examples) is tucked below; every CAT PYQ for Linear & Simultaneous Equations is here.

29CAT PYQs

Algebrachapter

Formulas PYQs (29)↩ Full Algebra chapter

Algebra, formula sheet

Show the full Algebra formula sheet (explanations + basic examples)

1Polynomials & zeroes

Plain English: a polynomial is just a sum of x-powers; its "zeroes" are the x-values that make it equal 0.
A polynomial of degree n: aₙxⁿ + aₙ₋₁xⁿ⁻¹ + … + a₁x + a₀
k is a zero of p(x) if p(k) = 0. Zeroes are the x-coordinates where y = p(x) cuts the x-axis.
Max zeroes = degree: linear → 1, quadratic → 2, cubic → 3, degree n → n.
e.g. p(x) = x² − 9 has degree 2, so at most 2 zeroes: x = 3 and x = −3.

2Sum & product of roots

Plain English: you can read the sum and product of the roots straight off the coefficients, no need to solve.
Quadratic ax²+bx+c: α + β = −b/a, αβ = c/a
Cubic ax³+bx²+cx+d: α+β+γ = −b/a, αβ+βγ+γα = c/a, αβγ = −d/a
Build the equation: x² − (sum)x + (product) = 0.
e.g. x² − 5x + 6 = 0: sum = 5, product = 6 ⇒ roots 2 and 3 (2+3=5, 2×3=6).

3Discriminant & nature of roots

Plain English: the discriminant D is a single number that tells you how many real roots a quadratic has, before solving.
For ax²+bx+c (a≠0): D = b² − 4ac
D > 0 → two distinct real roots; D = 0 → equal real roots; D < 0 → no real roots (complex).
D a perfect square (a,b,c rational) → roots are rational.
Roots: x = (−b ± √D)/2a
e.g. x² + x + 1: D = 1 − 4 = −3 < 0 ⇒ no real roots.

4Sum of squares of roots (trick)

Plain English: you can get α²+β² from the sum and product alone, never solve for the roots first.
α² + β² = (α+β)² − 2αβ
To minimise a sum-of-squares-of-roots expression in a parameter, complete the square, minimum is at the vertex.
e.g. if α+β = 3 and αβ = 2, then α²+β² = 9 − 4 = 5.

5Algebraic identities

Plain English: memorised expand/factor templates that turn ugly expressions into products (or vice-versa) instantly.
(a±b)² = a² ± 2ab + b²
a² − b² = (a+b)(a−b)
(a+b+c)² = a²+b²+c² + 2(ab+bc+ca)
a³ ± b³ = (a±b)(a² ∓ ab + b²)
a³+b³+c³ − 3abc = (a+b+c)(a²+b²+c² − ab−bc−ca)
e.g. 97×103 = (100−3)(100+3) = 100² − 3² = 9991.

6Linear equations in two variables

Plain English: comparing the coefficient ratios tells you whether two lines cross once, never, or lie on top of each other.
a₁x+b₁y+c₁=0 and a₂x+b₂y+c₂=0.
Unique solution (intersecting): a₁/a₂ ≠ b₁/b₂
No solution (parallel): a₁/a₂ = b₁/b₂ ≠ c₁/c₂
Infinite solutions (coincident): a₁/a₂ = b₁/b₂ = c₁/c₂
e.g. x+y=2 and 2x+2y=5: ratios 1/2 = 1/2 ≠ 2/5 ⇒ parallel, no solution.

7Inequalities, basic rules

Plain English: inequalities behave like equations, except multiplying or dividing by a negative reverses the arrow.
Adding/subtracting keeps direction; multiplying by a negative flips the sign.
If X > Y > 0 then 1/X < 1/Y.
For x > 0: x + 1/x ≥ 2 (equality at x = 1).
e.g. −2x > 6 ⇒ divide by −2 and flip ⇒ x < −3.

8Quadratic inequalities

Plain English: factor it, then "< 0" means between the roots and "> 0" means outside the roots.
(x−m)(x−n) < 0, n > m ⇒ m < x < n (between the roots).
(x−m)(x−n) > 0 ⇒ x < m or x > n (outside the roots).
Sign-of-product / wavy-curve method handles higher degree.
e.g. x² − 5x + 6 < 0 ⇒ (x−2)(x−3) < 0 ⇒ 2 < x < 3.

9Modulus (absolute value)

Plain English: |x| is the distance of x from 0, so it strips the sign and is never negative.
|x| = max(x, −x); −|x| ≤ x ≤ |x|.
|a+b| ≤ |a|+|b| and |a|−|b| ≤ |a−b|; |ab| = |a||b|.
|x| ≤ k ⇒ −k ≤ x ≤ k. |x| ≥ k ⇒ x ≥ k or x ≤ −k.
|f| + |g| = |f+g| only when f, g have the same sign.
e.g. |x| ≤ 3 ⇒ −3 ≤ x ≤ 3; |x − 4| = 2 ⇒ x = 6 or x = 2.

10AM-GM-HM inequality

Plain English: for positive numbers the plain average is always ≥ the geometric average, the go-to tool for "find the minimum".
For positive reals: AM ≥ GM ≥ HM, equality when all equal.
Two numbers: AM = (a+b)/2, GM = √(ab), HM = 2ab/(a+b).
AM × HM = GM²
If a₁a₂…aₙ = 1 then a₁+a₂+…+aₙ ≥ n.
e.g. for a = 2, b = 8: AM = 5 ≥ GM = √16 = 4. ✓

11Maxima & minima of a quadratic

Plain English: a parabola's turning point is at x = −b/2a; that's where the min (opens up) or max (opens down) lives.
ax²+bx+c: vertex at x = −b/2a; extreme value = (4ac − b²)/4a = −D/4a.
a > 0 → opens up → minimum; a < 0 → opens down → maximum.
min/max of max-of-two / min-of-two lines occurs where the two graphs intersect.
e.g. x² − 6x + 5: vertex at x = 3, minimum value = 9 − 18 + 5 = −4.

12Functions, domain, range, even/odd

Plain English: domain is what you may feed in, range is what comes out; even/odd describe the graph's symmetry.
Domain = allowed inputs; range = resulting outputs.
Even: f(−x) = f(x) (graph symmetric about y-axis), e.g. x², |x|.
Odd: f(−x) = −f(x) (symmetric about origin), e.g. x³, 1/x.
Inverse exists only if f is one-to-one.
e.g. f(x) = x³ is odd: f(−2) = −8 = −f(2). ✓

13Functional equations

Plain English: the form of a functional rule reveals the function, "turns + into ×" means exponential, etc.
f(x+y) = f(x)·f(y) ⇒ exponential type, f(x) = aˣ.
f(xy) = f(x)·f(y) ⇒ power/multiplicative; f(1) = 1.
If f(a+x) = f(a−x), the graph is symmetric about x = a; roots pair around a (sum of 4 roots = 4a).
e.g. f(x+y) = f(x)f(y) with f(1) = 3 ⇒ f(2) = f(1)² = 9.

14Graph shifting

Plain English: changes outside f() move the graph vertically; changes inside f() move it horizontally (and oppositely).
f(x)+c → shift up c; f(x)−c → shift down c.
f(x+c) → shift left c; f(x−c) → shift right c.
−f(x) → reflect in x-axis; f(−x) → reflect in y-axis.
e.g. y = (x−2)² is y = x² shifted 2 units right.

15Logarithm, definition

Plain English: log_b x just asks "what power of b gives x?", it's the inverse of raising to a power.
y = log_b x ⇔ x = bʸ (b > 0, b ≠ 1, x > 0).
log_a a = 1; log_a 1 = 0; a^(log_a m) = m.
e.g. log₂8 = 3 because 2³ = 8.

16Logarithm laws

Plain English: logs turn multiplication into addition, division into subtraction, and powers into multipliers.
log_a(xy) = log_a x + log_a y
log_a(x/y) = log_a x − log_a y
log_a(xᵐ) = m·log_a x
log_(aⁿ)(xᵐ) = (m/n)·log_a x
Change of base: log_a x = (log x)/(log a); log_a x = 1/log_x a
e.g. log₂40 = log₂(8×5) = log₂8 + log₂5 = 3 + log₂5.

17Indices (laws of exponents)

Plain English: same base, add exponents when multiplying, subtract when dividing, multiply when raising a power to a power.
pᵐ·pⁿ = pᵐ⁺ⁿ; pᵐ/pⁿ = pᵐ⁻ⁿ; (pᵐ)ⁿ = pᵐⁿ
pⁿ·qⁿ = (pq)ⁿ; (p/q)ⁿ = pⁿ/qⁿ
p⁻ⁿ = 1/pⁿ; p⁰ = 1; p^(1/n) = ⁿ√p
e.g. 2³·2⁴ = 2⁷ = 128; 8^(2/3) = (∛8)² = 2² = 4.

18Surds & rationalisation

Plain English: a surd is an unresolved root like √2; "rationalising" clears it from a denominator using the conjugate.
√(ab) = √a·√b; √(a/b) = √a/√b.
Rationalise a/(b+√c) by multiplying top & bottom by the conjugate (b−√c).
If a+√b is a root of a rational quadratic, so is its conjugate a−√b.
e.g. 1/(√3 − 1) × (√3 + 1)/(√3 + 1) = (√3 + 1)/2.

19Arithmetic Progression (AP)

Plain English: an AP adds the same step d each time; its sum is just "how many terms × the average of first and last".
Constant difference d. nth term: Tₙ = a + (n−1)d
Sum: Sₙ = n/2 · [2a + (n−1)d] = n/2 · (first + last)
Arithmetic mean of a, b: A = (a+b)/2. Middle term = average of an odd count of AP terms.
e.g. 2, 5, 8, …: T₄ = 2 + 3×3 = 11; sum of first 4 = 4/2·(2+11) = 26.

20Geometric Progression (GP)

Plain English: a GP multiplies by the same ratio r each time; if |r| < 1 the infinite sum settles to a finite value.
Constant ratio r. nth term: Tₙ = a·rⁿ⁻¹
Sum: Sₙ = a(rⁿ − 1)/(r − 1), r ≠ 1.
Infinite sum (|r| < 1): S∞ = a/(1 − r)
Geometric mean: G = √(ab).
e.g. 1 + ½ + ¼ + … = 1/(1 − ½) = 2.

21Harmonic Progression (HP)

Plain English: an HP is just an AP flipped, take reciprocals and you're back to a normal AP.
a, b, c… in HP ⇔ 1/a, 1/b, 1/c… in AP.
Harmonic mean of a, b: H = 2ab/(a+b)
nth term of HP = 1/(nth term of the corresponding AP).
e.g. 1, ½, ⅓, ¼ is an HP (reciprocals 1, 2, 3, 4 form an AP).

22Standard summation formulas

Plain English: ready-made closed forms for adding up the first n numbers, their squares, and their cubes.
Σn = n(n+1)/2
Σn² = n(n+1)(2n+1)/6
Σn³ = [n(n+1)/2]²
Telescoping: 1/(k·(k+1)) = 1/k − 1/(k+1).
e.g. 1 + 2 + … + 10 = 10×11/2 = 55.

23Common terms of two APs

Plain English: numbers shared by two APs themselves form an AP whose step is the LCM of the two steps.
Common terms of two APs form a new AP with common difference = LCM of the two differences.
Find the first common term, then count multiples of the LCM up to the smaller upper limit.
e.g. 2,5,8,… and 3,7,11,…: first common term 11, new step = LCM(3,4) = 12 ⇒ 11, 23, 35, …

24Recurrence & tₙ from Sₙ

Plain English: if you know the running total Sₙ, each term is just this total minus the previous total.
If Sₙ given: aₙ = Sₙ − Sₙ₋₁ (and a₁ = S₁).
Alternating-sum sequences: subtract consecutive defining equations to isolate a term.
e.g. Sₙ = n² ⇒ a₅ = S₅ − S₄ = 25 − 16 = 9.

25Integer / Diophantine solutions

Plain English: once you spot one whole-number solution, all the rest come by stepping x and y in fixed jumps.
ax + by = c with one integer solution (x₀, y₀): all others are x₀ + (b/g)t, y₀ − (a/g)t, where g = gcd(a,b).
Bound the count using the given ranges on x and y.
e.g. 2x + 3y = 12: (x,y) = (3,2) works; next is (0,4), then (6,0), x jumps by 3, y by 2.

26When does Aᴮ = 1?

Plain English: a power equals 1 in exactly three situations, check all three or you'll miss cases.
Base = 1 (any exponent), or
Exponent = 0 (base ≠ 0), or
Base = −1 with an even exponent.
e.g. (−1)⁴ = 1 (base −1, even power); 7⁰ = 1 (zero power); 1⁹⁹ = 1 (base 1).

27Three terms in AP / GP

Plain English: centering three terms on a middle value makes their sum (AP) or product (GP) collapse to one symbol.
Three in AP: take a−d, a, a+d (their sum = 3a).
Three in GP: take a/r, a, ar (product = a³).
Three consecutive integers as roots: n−1, n, n+1.
e.g. three numbers in AP summing to 18 ⇒ middle = 6, so 6−d, 6, 6+d.

28|x − a| as distance (modulus sums)

Plain English: read |x−a| as "distance from a", and sums of such distances are smallest when x sits among the points.
|x−a| = distance of x from a on the number line.
|x−p|+|x−q| is minimised for any x between p and q; minimum value = |p−q|.
|x−p| = |x−q| at the midpoint x = (p+q)/2.
e.g. |x−2| + |x−7| ≥ 5, achieved for any x in [2, 7].

29Sum of squares identity trick

Plain English: squares can't be negative, so if a bunch of squares add to 0 every single one must be 0.
If a sum of squares equals 0, each square = 0: e.g. (x−2y)² + (y−z)² = 0 ⇒ x = 2y and y = z.
Group given expressions into perfect squares to pin exact values.
e.g. (a−3)² + (b+1)² = 0 forces a = 3 and b = −1.

30Cauchy / vector identity

Plain English: this identity links two "sum-of-squares" products to two cross-terms, handy when three of the four pieces are given.
(a²+b²)(x²+y²) = (ax+by)² + (ay−bx)².
Useful when given a²+b², x²+y² and ax+by to find ay−bx.
e.g. (1²+2²)(3²+4²) = 5·25 = 125 = 11² + 2² = (1·3+2·4)² + (1·4−2·3)².

Practice questions generated · up to 100

Original easy-hard warm-up drills (not CAT PYQs). Pick the levels, generate a set, reveal answers.

Linear & Simultaneous Equations, CAT PYQs

Linear & Simultaneous Equations

ModerateCAT 1999

The number of positive integer valued pairs (x, y) satisfying 4x − 17y = 1 and x ≤ 1000 is

(1) 59
(2) 57
(3) 55
(4) 58

Show solution

(1) 59. A base solution is (x, y) = (13, 3); general x increases by 17 each step. Counting x ≤ 1000 from x = 13 onward in steps of 17 gives 59 pairs.

ModerateCAT 2000

If x³ − ax² + bx − a = 0 has three real roots, then it must be the case that

(1) b = 1
(2) b ≠ 1
(3) a = 1
(4) a ≠ 1

Show solution

(2) b ≠ 1. Test b = 1: the factorisation becomes (x − a)(x² + 1), and x² + 1 = 0 has no real roots, so only one real root, contradiction. Hence b ≠ 1.

ModerateCAT 2002

If x² + 5y² + z² = 2y(2x + z), then which of the following statements are necessarily true? I. x = 2y II. x = 2z III. 2x = z

(1) Only I
(2) Only II
(3) Only III
(4) Only I and II

Show solution

(4) Only I and II. Rearrange: (x² − 4xy + 4y²) + (y² − 2yz + z²) = 0 ⇒ (x − 2y)² + (y − z)² = 0. Each square is 0 ⇒ x = 2y and y = z ⇒ x = 2z. So I and II hold.

HardCAT 2003

Which one of the following conditions must p, q and r satisfy so that the following system of linear simultaneous equations has at least one solution, such that p + q + r ≠ 0? x + 2y − 3z = p, 2x + 6y − 11z = q, x − 2y + 7z = r

(1) 5p − 2q − r = 0
(2) 5p + 2q + r = 0
(3) 5p + 2q − r = 0
(4) 5p − 2q + r = 0

Show solution

(1) 5p − 2q − r = 0. Eliminating z from pairs of equations gives 5x+4y = 11p−3q and 5x+4y = (7q+11r)/5. For consistency these must be equal: 11p−3q = (7q+11r)/5 ⇒ 5p − 2q − r = 0.

ModerateCAT 2003

The number of roots common between the two equations x³ + 3x² + 4x + 5 = 0 and x³ + 2x² + 7x + 3 = 0 is

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

Show solution

(1) 0. Subtracting the equations: x² − 3x + 2 = 0 ⇒ x = 1 or 2. Neither value satisfies either original cubic, so there are no common roots.

ModerateCAT 2003

A test has 50 questions. A student scores 1 mark for a correct answer, −1/3 for a wrong answer, and −1/6 for not attempting a question. If the net score of a student is 32, the number of questions answered wrongly by that student cannot be less than

(1) 6
(2) 12
(3) 3
(4) 9

Show solution

(3) 3. With R right, W wrong, N not attempted: R + W + N = 50 and R − W/3 − N/6 = 32. Solving gives R = (242 + W)/7, so 242 + W must be divisible by 7; the smallest such W is 3.

HardCAT 2005

For which value of k does the following pair of equations yield a unique solution for x such that the solution is positive? x² − y² = 0, (x − k)² + y² = 1

(1) 2
(2) 0
(3) √2
(4) −√2

Show solution

(3) √2. x² = y²; substitute: 2x² − 2kx + (k²−1) = 0. Unique solution ⇒ discriminant 0: 4k² − 8(k²−1) = 0 ⇒ k² = 2 ⇒ k = ±√2. For a positive x, k = √2.

ModerateCAT 2006

The number of solutions of the equation 2x + y = 40 where both x and y are positive integers and x ≤ y is:

(1) 7
(2) 13
(3) 14
(4) 18

Show solution

(2) 13. y = 40 − 2x > 0 and x ≤ y. Valid x runs from 1 to 13 (at x = 13, y = 14 ≥ x; at x = 14, y = 12 < x). So 13 solutions.

HardCAT 2019

The product of two positive numbers is 616. If the ratio of the difference of their cubes to the cube of their difference is 157 : 3, then the sum of the two numbers is:

(1) 58
(2) 85
(3) 50
(4) 95

Show solution

(3) 50. xy = 616. (x³−y³)/(x−y)³ = 157/3 ⇒ 3(x²+xy+y²) = 157(x²+y²−2xy) ⇒ 154(x+y)² = 625xy = 625·616 ⇒ (x+y)² = 625·4 = 2500 ⇒ x+y = 50.

ModerateCAT 2019TITA

The number of solutions to the equation |x|(6x² + 1) = 5x² is

Show solution

5. x = 0 works. Otherwise divide by |x|: 6x² + 1 = 5|x|. Let |x| = y: 6y² − 5y + 1 = 0 ⇒ y = 1/2 or 1/3, giving x = ±1/2, ±1/3. Total with 0 = 5 solutions.

HardCAT 2019

Let a, b, x, y be real numbers such that a² + b² = 81, x² + y² = 121, and ax + by = 99. If k = ay − bx, then

(1) 0 < k ≤ 5/13
(2) k > 5/13
(3) k = 5/13
(4) k = 0

Show solution

(4) k = 0. By the identity (a²+b²)(x²+y²) = (ax+by)² + (ay−bx)²: 81·121 = 99² + k². But 81·121 = 9801 = 99². So k² = 0 ⇒ k = 0.

ModerateCAT 2017

The number of solutions (x, y, z) to the equation x − y − z = 25, where x, y, and z are positive integers such that x ≤ 40, y ≤ 12, and z ≤ 12 is

(1) 101
(2) 99
(3) 87
(4) 105

Show solution

(2) 99. x = 25 + y + z with 2 ≤ y+z ≤ 15 (since x ≤ 40) and y, z ≥ 1. Counting valid (y, z) pairs: y = 1,2,3 each give 12 values of z; y = 4 gives 11; …down to y = 12 gives 3. Total = 99.

EasyCAT 2020

Let k be a constant. The equations kx + y = 3 and 4x + ky = 4 have a unique solution if and only if

(1) |k| ≠ 2
(2) k = 2
(3) k ≠ 2
(4) |k| = 2

Show solution

(1) |k| ≠ 2. Unique solution requires k/4 ≠ 1/k ⇒ k² ≠ 4 ⇒ k ≠ ±2 ⇒ |k| ≠ 2.

ModerateCAT 2020TITA

The number of pairs of integers (x, y) satisfying x ≥ y ≥ −20 and 2x + 5y = 99 is:

Show solution

17. 2x even ⇒ 5y odd ⇒ y odd. Minimum y = −19; x ≥ y caps y at 13. Odd y from −19 to 13 gives 17 values.

ModerateCAT 2020TITA

If x and y are non-negative integers such that x + 9 = z, y + 1 = z and x + y < z + 5, then the maximum possible value of 2x + y equals

Show solution

23. Adding the equalities: x + y = 2z − 10 < z + 5 ⇒ z < 15 ⇒ z_max = 14, giving x = 5, y = 13. Then 2x + y = 10 + 13 = 23.

ModerateCAT 2021 · Slot 2

Consider the pair of equations: x² − xy − x = 22 and y² − xy + y = 34. If x > y, then x − y equals

(1) 7
(2) 4
(3) 6
(4) 8

Show solution

(4) 8. Add the two equations: x² − 2xy + y² − x + y = 56 ⇒ (x−y)² − (x−y) = 56. Let t = x−y: t² − t − 56 = 0 ⇒ (t−8)(t+7) = 0 ⇒ t = 8 (since x > y).

ModerateCAT 2022 · Slot 1TITA

For natural numbers x, y, and z, if xy + yz = 19 and yz + xz = 51, then the minimum possible value of xyz is:

Show solution

34. From xy + yz = y(x+z) = 19 (prime) ⇒ y = 1, x + z = 19. Then yz + xz = z(x+1)... using z(x+y) = 51 with y = 1: z(x+1) = 51. With x + z = 19, solving gives x = 2, z = 17 ⇒ xyz = 2·1·17 = 34.

HardCAT 2022 · Slot 2

Let r and c be real numbers. If r and −r are roots of 5x³ + cx² − 10x + 9 = 0, then c equals:

(1) 4
(2) −4
(3) −9/2
(4) 9/2

Show solution

(3) −9/2. Let third root α. Sum of roots = α = −c/5. Product = α(−r²) = −9/5. Sum of products pairwise = −r² = −10/5 = −2 ⇒ r² = 2. From product: α·(−2) = −9/5... solving the relations gives c = −9/2.

HardCAT 2023 · Slot 1TITA

The equation x³ + (2r + 1)x² + (4r − 1)x + 2 = 0 has −2 as one of the roots. If the other two roots are real, then the minimum possible non-negative integer value of r is

Show solution

2. Dividing out (x+2) leaves a quadratic in x; requiring its discriminant ≥ 0 gives a bound on r, whose smallest non-negative integer value is 2.

HardCAT 2023 · Slot 2

If p² + q² − 29 = 2pq − 20 = 52 − 2pq, then the difference between the maximum and minimum possible value of (p³ − q³) is

(1) 243
(2) 378
(3) 189
(4) 486

Show solution

(2) 378. From 2pq − 20 = 52 − 2pq ⇒ pq = 18; and p²+q² − 29 = 2pq − 20 ⇒ (p−q)² = 9 ⇒ p−q = ±3. With pq = 18: (p, q) = (6, 3) or (−6, −3). p³−q³ = 189 or −189. Difference = 378.

ModerateCAT 2023 · Slot 3

For some real numbers a and b, the system of equations x + y = 4 and (a + 5)x = (b² − 15)y = 8b has infinitely many solutions for x and y. Then, the maximum possible value of ab is

(1) 33
(2) 55
(3) 15
(4) 25

Show solution

(1) 33. Matching coefficients for infinitely many solutions ties a and b through 8b and 4(a+5) = 4(b²−15) = 8b. Solving and maximising ab gives 33.

CAT 2024 & 2025, recent

HardCAT 2024 · Slot 1

The sum of all real values of k for which (1/8)^k × (1/32768)^(1/3) = (1/8) × (1/32768)^(1/k), is

(A) 2/3
(B) 4/3
(C) −2/3
(D) −4/3

Show solution

ModerateCAT 2024 · Slot 1 TITA

If x is a positive real number such that 4 log₁₀x + 4 log₁₀₀x + 8 log₁₀₀₀x = 13, then the greatest integer not exceeding x, is

Show solution

31. Converting bases: (4 + 2 + 8/3)·log₁₀x = (26/3)·log₁₀x = 13 ⇒ log₁₀x = 1.5 ⇒ x = 10^1.5 ≈ 31.62, so ⌊x⌋ = 31.

ModerateCAT 2024 · Slot 3 TITA

The number of distinct integer solutions (x, y) of the equation |x + y| + |x − y| = 2, is

Show solution

8. The equation means max(|x|, |y|) = 1, i.e. the integer points on the boundary of the square [−1,1]², there are 8 such lattice points.

ModerateCAT 2025 · Slot 1

If a − 6b + 6c = 4 and 6a + 3b − 3c = 50, where a, b and c are real numbers, the value of 2a + 3b − 3c is

(A) 18
(B) 20
(C) 15
(D) 14

Show solution

(A) 18. Let u = 3b − 3c. Eq2: 6a + u = 50; eq1: a − 2u = 4. Solving, a = 8, u = 2. So 2a + 3b − 3c = 2a + u = 16 + 2 = 18.

HardCAT 2025 · Slot 1

The number of distinct integers n for which log₍₁/₄₎(n² − 7n + 11) > 0, is

(A) infinite
(B) 0
(C) 2
(D) 1

Show solution

(B) 0. Base 1/4 < 1, so log > 0 needs 0 < n²−7n+11 < 1. The quadratic equals an integer for integer n, so it cannot lie strictly between 0 and 1. Hence no integer works.

ModerateCAT 2025 · Slot 1

In the set of consecutive odd numbers {1, 3, 5, …, 57}, there is a number k such that the sum of all the elements less than k is equal to the sum of all the elements greater than k. Then, k equals

(A) 41
(B) 39
(C) 37
(D) 43

Show solution

(A) 41. The 29 odd terms sum to 29² = 841. For k = 41, terms below sum to 400 and above sum to 400. ✓

HardCAT 2025 · Slot 1

For any natural number k, let aₖ = 3ᵏ. The smallest natural number m for which (a₁)¹ × (a₂)² × … × (a₂₀)²⁰ < a₂₁ × a₂₂ × … × a₍₂₀₊ₘ₎, is

(A) 59
(B) 56
(C) 58
(D) 57

Show solution

(C) 58. LHS exponent = Σk·k for k=1..20 = Σk² = 2870. RHS exponent = Σ(20+j) for j=1..m = 20m + m(m+1)/2. Need 20m + m(m+1)/2 > 2870 → smallest m = 58.

ModerateCAT 2025 · Slot 3 TITA

In an arithmetic progression, if the sum of the fourth, seventh and tenth terms is 99, and the sum of the first fourteen terms is 497, then the sum of the first five terms is

Show solution

65. 4th+7th+10th = 3(a+6d) = 99 → a + 6d = 33. S₁₄ = 7(2a+13d) = 497 → 2a + 13d = 71. Solving: d = 5, a = 3. S₅ = 5/2(6 + 20) = 65.