◆ QA · Algebra

Inequalities & Modulus , formulas + CAT PYQs

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

31CAT PYQs

Algebrachapter

Formulas PYQs (31)↩ 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.

Inequalities & Modulus, CAT PYQs

Inequalities & Modulus

ModerateCAT 1995

Let x < 0, 0 < y < 1, z > 1. Which of the following may be false?

(1) (x² − z²) has to be positive.
(2) yz can be less than one.
(3) xy can never be zero.
(4) (y² − z²) is always negative.

Show solution

(1). x² can be small (x near 0) while z² > 1, so x² − z² need not be positive, (1) may be false. The others are always true given the ranges.

ModerateCAT 1996

Which of the following values of x do not satisfy the inequality (x² − 3x + 2 > 0) at all?

(1) 1 ≤ x ≤ 2
(2) −1 ≥ x ≥ −2
(3) 0 ≤ x ≤ 2
(4) 0 ≥ x ≥ −2

Show solution

(1) 1 ≤ x ≤ 2. Factor: (x−1)(x−2) > 0 holds for x < 1 or x > 2. So values with 1 ≤ x ≤ 2 never satisfy the inequality.

HardCAT 1999

If |r − 6| = 11 and |2q − 12| = 8, what is the minimum possible value of q/r?

(1) −2/5
(2) 2/17
(3) 10/17
(4) None of these

Show solution

(4) None of these. |r − 6| = 11 ⇒ r = 17 or r = −5. |2q − 12| = 8 ⇒ q = 10 or q = 2. To minimise q/r take q = 10, r = −5 ⇒ q/r = −10/5 = −2, which is not among the listed options, so the answer is (4).

ModerateCAT 2000

If x > 2 and y > −1, then which of the following statements is necessarily true?

(1) xy > −2
(2) −x < 2y
(3) xy < −2
(4) −x > 2y

Show solution

(2) −x < 2y. x > 2 ⇒ −x < −2; y > −1 ⇒ 2y > −2. So −x < −2 < 2y ⇒ −x < 2y.

ModerateCAT 2000

If x² + y² = 0.1 and |x − y| = 0.2, then |x| + |y| is equal to

(1) 0.3
(2) 0.4
(3) 0.2
(4) 0.6

Show solution

(2) 0.4. |x−y|² = x²+y² − 2xy ⇒ 0.04 = 0.1 − 2xy ⇒ xy = 0.03. Then (|x|+|y|)² = x²+y² + 2|xy| = 0.1 + 0.06 = 0.16 ⇒ |x|+|y| = 0.4.

ModerateCAT 2001

m is the smallest positive integer such that for any integer n > m, the quantity n³ − 7n² + 11n − 5 is positive. What is the value of m?

(1) 4
(2) 5
(3) 8
(4) None of these

Show solution

(4) None of these (m = 6). Factor: n³−7n²+11n−5 = (n−1)²(n−5). (n−1)² ≥ 0, so the sign follows (n−5); it is positive only for n > 5. The smallest such m is 6.

ModerateCAT 2001

If a, b, c and d are four positive real numbers such that abcd = 1, what is the minimum value of (1 + a)(1 + b)(1 + c)(1 + d)?

(1) 4
(2) 1
(3) 16
(4) 18

Show solution

(3) 16. Each 1 + x ≥ 2√x by AM-GM; product ≥ 16·√(abcd) = 16. Minimum (a=b=c=d=1) gives exactly 16.

ModerateCAT 2001

If x > 5 and y < −1, then which of the following statements is true?

(1) (x+4y) > 1
(2) x > −4y
(3) −4x < 5y
(4) None of these

Show solution

(4) None of these. Counter-example x = 6, y = −26 violates (1), (2) and (3). So none is necessarily true.

ModerateCAT 2001

x and y are real numbers satisfying the conditions 2 < x < 3 and −8 < y < −7. Which of the following expressions will have the least value?

(1) x²y
(2) xy²
(3) 5xy
(4) None of these

Show solution

(3) 5xy. y < 0, so xy² > 0 (not minimum). Between x²y and 5xy (both negative): for 2 < x < 3, x² < 5x, and since y < 0, x²y > 5xy. So 5xy is least.

ModerateCAT 2003

The function f(x) = |x − 2| + |2.5 − x| + |3.6 − x|, where x is a real number, attains a minimum at

(1) x = 2.3
(2) x = 2.5
(3) x = 2.7
(4) None of these

Show solution

(2) x = 2.5. A sum of absolute values |x − aᵢ| is minimised at the median of the points. Median of {2, 2.5, 3.6} is 2.5.

ModerateCAT 2003

If the product of n positive real numbers is unity, then their sum is necessarily

(1) a multiple of n
(2) equal to n + 1/n
(3) never less than n
(4) a positive integer

Show solution

(3) never less than n. By AM ≥ GM, sum/n ≥ (product)^(1/n) = 1 ⇒ sum ≥ n.

HardCAT 2003

If x, y, z are distinct positive reals, then [x²(y+z) + y²(x+z) + z²(x+y)]/xyz would be

(1) greater than 4
(2) greater than 5
(3) greater than 6
(4) None of these

Show solution

(3) greater than 6. The expression simplifies to (x/y + y/x) + (y/z + z/y) + (x/z + z/x). Each pair ≥ 2 by AM-GM, total ≥ 6; distinct values make it strictly > 6.

HardCAT 2003

Given that −1 ≤ v ≤ 1, −2 ≤ u ≤ −0.5 and −2 ≤ z ≤ −0.5 and w = vz/u, then which of the following is necessarily true?

(1) −0.5 ≤ w ≤ 2
(2) −4 ≤ w ≤ 4
(3) −4 ≤ w ≤ 2
(4) −2 ≤ w ≤ −0.5

Show solution

(2) −4 ≤ w ≤ 4. Max w = (vz)/u with v=1, z=−2, u=−0.5 ⇒ (−2)/(−0.5) = 4. Min w = with v=−1, z=−2, u=−0.5 ⇒ (2)/(−0.5) = −4. So −4 ≤ w ≤ 4.

ModerateCAT 2003

A real number x satisfying 1 − 1/n < x ≤ 3 + 1/n for every positive integer n, is best described by

(1) 1 < x < 4
(2) 0 < x ≤ 4
(3) 0 < x ≤ 4
(4) 1 ≤ x ≤ 3

Show solution

(3) 0 < x ≤ 4. For n = 1 the bounds are widest: 0 < x ≤ 4. As n increases they tighten toward 1 < x ≤ 3, which lies inside (0, 4]. So x is best described by 0 < x ≤ 4.

ModerateCAT 2005

If R = (30⁶⁵ − 29⁶⁵)/(30⁶⁴ + 29⁶⁴), then

(1) 0 < R ≤ 0.1
(2) 0.1 < R ≤ 0.5
(3) 0.5 < R ≤ 1.0
(4) R > 1.0

Show solution

(4) R > 1.0. Write R = (30⁶⁵ − 29⁶⁵)/(30⁶⁴ + 29⁶⁴). Since 30⁶⁵ = 30·30⁶⁴ dominates, R is close to 30; in any case R > 1.0.

ModerateCAT 2006

What values of x satisfy x²ᐟ³ + x¹ᐟ³ − 2 ≤ 0?

(1) −8 ≤ x ≤ 1
(2) −1 ≤ x ≤ 8
(3) 1 < x < 8
(4) 1 ≤ x ≤ 8

Show solution

(1) −8 ≤ x ≤ 1. Let y = x^(1/3): y² + y − 2 ≤ 0 ⇒ (y+2)(y−1) ≤ 0 ⇒ −2 ≤ y ≤ 1 ⇒ −8 ≤ x ≤ 1.

ModerateCAT 2017TITA

For how many integers n, will the inequality (n − 5)(n − 10) − 3(n − 2) ≤ 0 be satisfied?

Show solution

11. Expand: n² − 18n + 56 ≤ 0 ⇒ (n−4)(n−14) ≤ 0 ⇒ 4 ≤ n ≤ 14. That is 11 integers.

ModerateCAT 2017

Let f(x) = 2x − 5 and g(x) = 7 − 2x. Then |f(x) + g(x)| = |f(x)| + |g(x)| if and only if

(1) 5/2 < x < 7/2
(2) x ≤ 5/2 or x ≥ 7/2
(3) x < 5/2 or x ≥ 7/2
(4) 5/2 ≤ x ≤ 7/2

Show solution

(4) 5/2 ≤ x ≤ 7/2. |a|+|b| = |a+b| requires a, b same sign (or zero). 2x−5 ≥ 0 and 7−2x ≥ 0 ⇒ 5/2 ≤ x ≤ 7/2.

ModerateCAT 2017

The area of the closed region bounded by the equation |x| + |y| = 2 in the two-dimensional plane is

(1) 4π sq. units
(2) 4 sq. units
(3) 8 sq. units
(4) 2π sq. units

Show solution

(3) 8 sq units. |x|+|y| = 2 is a square with vertices (±2,0),(0,±2). Side = √(2²+2²) = 2√2, area = (2√2)² = 8.

ModerateCAT 2017

Let m and n be natural numbers such that n is even and 0.2 < m/20, n/m, n/11 < 0.5. Then m − 2n equals

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

Show solution

(3) 1. 0.2 < m/20 < 0.5 ⇒ 4 < m < 10. 0.2 < n/11 < 0.5 ⇒ 2.2 < n < 5.5, and n even ⇒ n = 4. 0.2 < n/m < 0.5 ⇒ 8 < m < 20; combined with 4 < m < 10 gives m = 9. So m − 2n = 9 − 8 = 1.

ModerateCAT 2018TITA

The smallest integer such that n³ − 11n² + 32n − 28 > 0 is

Show solution

8. Factor: (n−2)²(n−7) > 0. (n−2)² ≥ 0, so sign follows (n−7); positive only for n > 7. Smallest integer is 8.

ModerateCAT 2018TITA

Let S be the set of all points (x, y) in the x-y plane such that |x| + |y| ≤ 2 and |x| ≥ 1. Then, the area, in square units, of the region represented by S equals

Show solution

2. |x|+|y| ≤ 2 is a square of diagonal 4 (area 8). Removing the strip |x| < 1 leaves two triangles, each of base 2 and height 1; total area = 2·(½·1·2) = 2.

ModerateCAT 2020

The number of real-valued solutions of the equation 2ˣ + 2⁻ˣ = 2 − (x − 2)² is:

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

Show solution

(2) 0. LHS = 2ˣ + 2⁻ˣ ≥ 2 (AM-GM), RHS = 2 − (x−2)² ≤ 2. Equality needs both = 2: RHS = 2 ⇒ x = 2, but then LHS = 2² + 2⁻² > 2. No solution.

ModerateCAT 2020TITA

The area of the region satisfying the inequalities |x| − y ≤ 1, y ≥ 0 and y ≤ 1 is

Show solution

3. The region is a trapezium bounded by y = 0, y = 1 and the V-shaped |x| − y = 1; computing its area gives 3 square units.

HardCAT 2020

In how many ways can a pair of integers (x, a) be chosen such that x² − 2|x| + |a − 2| = 0?

(1) 7
(2) 6
(3) 5
(4) 4

Show solution

(1) 7. Let y = |x|: y² − 2y + |a−2| = 0. For integer y, the root pairs (0,2) and (1,1) give |a−2| = 0 or 1 ⇒ a = 2, 1, 3 with corresponding x = 0, ±1, ±1. Counting valid (x, a) pairs gives 7.

HardCAT 2021 · Slot 1

The number of integers n that satisfy the inequalities |n − 60| < |n − 100| < |n − 20| is

(1) 19
(2) 18
(3) 20
(4) 21

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(1) 19. |n−60| < |n−100| ⇒ n < 80 (closer to 60). |n−100| < |n−20| ⇒ n > 60 (closer to 100). So 60 < n < 80 ⇒ integers 61…79, i.e. 19.

ModerateCAT 2021 · Slot 3

If 3x + 2|y| + y = 7 and x + |x| + 3y = 1, then x + 2y is

(1) −4/3
(2) 1
(3) 0
(4) 8/3

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(3) 0. Test the sign cases. The consistent case is x > 0, y < 0: then |y| = −y and |x| = x, so the equations become 3x − y = 7 and 2x + 3y = 1, giving x = 2, y = −1. Hence x + 2y = 2 + 2(−1) = 0.

HardCAT 2021 · Slot 3TITA

The number of distinct pairs of integers (m, n) satisfying |1 + mn| < |m + n| < 5 is

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12. Casework on small |m + n| (which must be 2, 3, or 4) with the constraint |1 + mn| < |m + n| yields 12 valid integer pairs.

HardCAT 2022 · Slot 1

The largest real value of a for which |x + a| + |x − 1| = 2 has an infinite number of solutions of x is

(1) 2
(2) −1
(3) 0
(4) 1

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(4) 1. Infinite solutions occur when x cancels out. Case 1: x + a < 0 and x − 1 ≥ 0 ⇒ −a − x + x − 1 = 2 ⇒ a = −3. Case 2: x + a ≥ 0 and x − 1 < 0 ⇒ x + a − x + 1 = 2 ⇒ a = 1. The largest value of a is 1.

HardCAT 2022 · Slot 3

If c = 16x/y + 49y/x for some non-zero real numbers x and y, then c cannot take the value:

(1) −60
(2) −50
(3) 60
(4) −70

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(2) −50. Let P = x/y: c = 16P + 49/P ⇒ 16P² − cP + 49 = 0. Real P needs c² ≥ 4·16·49 ⇒ |c| ≥ 56. So c = −50 (|c| < 56) is impossible.

CAT 2024 & 2025, recent

ModerateCAT 2024 · Slot 2

If x and y satisfy the equations |x| + x + y = 15 and x + |y| − y = 20, then (x − y) equals

(A) 20
(B) 15
(C) 5
(D) 10

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(B) 15. Testing the fourth quadrant (x > 0, y < 0): |x| = x and |y| = −y, so the equations become 2x + y = 15 and x − 2y = 20. Solving gives x = 10, y = −5, so x − y = 10 − (−5) = 15.