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CDS II 2026 Elementary Mathematics with Solutions

Exam: CDS Year: 2026 (Session II) Held on: 13 September 2026 Questions: 100 Marks: 100 Negative Marking: 1/3

Booklet series: Booklet Series B (BFVS-T-TME). UPSC shuffles the option order between booklet series, so the option letters below apply to this series only. If you sat a different series, match the answer by its text, not by its letter.

Q.1 [Trigonometry]

If cot θ + cos θ = m, cot θ − cos θ = n, where 0 < θ < π/2, then what is (m² − n²)/√(mn) equal to?

(a) 1
(b) 2
(c) 2√2
(d) 4 ✓

Solution: m²−n² = (m+n)(m−n) = 4 cot θ cos θ; mn = cot²θ − cos²θ = cos⁴θ/sin²θ, so √(mn) = cos²θ/sin θ.
Ratio = 4 cot θ cos θ ÷ (cos²θ/sin θ) = 4.

Q.2 [Trigonometry]

If 1/(cosec θ + cot θ) − 1/sin θ = p, then what is 1/sin θ − 1/(cosec θ − cot θ) equal to?

(a) − p
(b) − 1/p
(c) p ✓
(d) 1/p

Solution: (cosec θ + cot θ)(cosec θ − cot θ) = 1, so each reciprocal just flips the sign of cot θ.
p = (cosec θ − cot θ) − cosec θ = −cot θ; second expression = cosec θ − (cosec θ + cot θ) = −cot θ = p.

Q.3 [Trigonometry]

If 8 sin θ − cos θ = 4, where π/6 < θ < π/3, then what is sin θ equal to?

(a) 3/5 ✓
(b) 5/13
(c) 4/5
(d) 12/13

Solution: cos θ = 8 sin θ − 4 into sin²θ + cos²θ = 1: 65s² − 64s + 15 = 0 → s = 3/5 or 5/13.
π/6 < θ < π/3 needs 0.5 < sin θ < 0.866, so sin θ = 3/5.
Trap: 5/13 is the other root — the range kills it.

Q.4 [Heights & Distances]

A person walking along a straight road observes that at two consecutive kilometre-stones the angles of elevation of a hill in front of him are 30° and 60° respectively. What is the height of the hill?

(a) 3√3 km
(b) 2√3 km
(c) √3 km
(d) √3/2 km ✓

Solution: From the two stones: x = h cot 60° = h/√3 and x + 1 = h cot 30° = h√3.
Subtract: h(√3 − 1/√3) = 1 → 2h/√3 = 1 → h = √3/2 km.

Q.5 [Heights & Distances]

A tower subtends an angle α at a point P on the same level as the foot of the tower. Q is a point vertically above P and PQ = h. If the angle of depression of the foot of the tower measured from Q is β, then what is the height of the tower?

(a) h tan α tan β
(b) h cot α tan β
(c) h tan α cot β ✓
(d) h cot α cot β

Solution: From Q: d = h cot β. From P: H = d tan α.
H = h tan α cot β.
Trap: β pairs with h (seen from Q), α with H (seen from P).

Q.6 [Trigonometry]

If 5 sin θ + 12 cos θ = 13, where 0 < θ < π/2, then what is tan θ + cot θ equal to?

(a) 194/65
(b) 169/60 ✓
(c) 313/156
(d) 371/120

Solution: 5² + 12² = 13², so 5 sin θ + 12 cos θ = 13 is the maximum case: sin θ = 5/13, cos θ = 12/13.
tan θ + cot θ = 1/(sin θ cos θ) = 169/60.

Q.7 [Trigonometry]

What is the ratio of the greatest value of sin²x + 2 (where 0 ≤ x ≤ π/2) to its least value?

(a) 2
(b) 3/2 ✓
(c) 2
(d) 1/2

Solution: On [0, π/2], sin²x runs from 0 to 1, so sin²x + 2 runs from 2 to 3.
Ratio = 3/2.

Q.8 [Trigonometry]

If A + B + C = π (A, B, C > 0) and the angle C is obtuse, then which of the following is/are correct?
I. sin A · sin B < 1
II. tan A · tan B > 1

(a) I only ✓
(b) II only
(c) Both I and II
(d) Neither I nor II

Solution: C obtuse ⇒ A + B < π/2, both acute.
I: sin A, sin B < 1 ⇒ product < 1 ✓. II: A < π/2 − B ⇒ tan A < cot B ⇒ tan A tan B < 1 ✗.
I only.

Q.9 [Trigonometry]

If (cos 35° + cos 55°) = p, then what is sin 35° · cos 35° equal to?

(a) (1 − p²)/2
(b) (p² − 1)/2 ✓
(c) (1 + p²)/2
(d) (p − 1)/2

Solution: cos 55° = sin 35°, so p = sin 35° + cos 35°.
Square: p² = 1 + 2 sin 35° cos 35° ⇒ sin 35° cos 35° = (p² − 1)/2.

Q.10 [Trigonometry]

If p = sec θ + tan θ and q = cosec θ − cot θ, then what is (p − q − pq) equal to?

(a) − 1
(b) 0
(c) 1 ✓
(d) 2

Solution: Test θ = 45°: p = √2 + 1, q = √2 − 1, pq = 1.
p − q − pq = 2 − 1 = 1 (the identity holds for every θ).

Q.11 [Sets]

Consider the following statements:
I. The set of all birds living on the Earth is an infinite set.
II. The set of all real numbers between 0 and 10 is a finite set.
Which of the statements given above is/are correct?

(a) I only
(b) II only
(c) Both I and II
(d) Neither I nor II ✓

Solution: I: the birds on Earth are many but finite ✗. II: (0, 10) contains infinitely many reals ✗.
Neither I nor II.

Q.12 [Sets]

Consider the following sets:
I. The set of even prime numbers
II. {x ∈ ℝ : x³ + 1 = 0}
III. {n ∈ ℤ : n² < 1}
How many of the above are null sets?

(a) None ✓
(b) One
(c) Two
(d) All the three

Solution: I = {2}, II = {−1} (x³ = −1), III = {0} (0² < 1). None is empty → None.
Trap: forgetting 0 in III.

Q.13 [Sets]

Consider the following statements:
I. If A is a subset of U and U is the Universal set, then the complement of the set A is also a subset of U.
II. The complement of a Universal set is a singleton set.
Which of the statements given above is/are correct?

(a) I only ✓
(b) II only
(c) Both I and II
(d) Neither I nor II

Solution: I: A′ = U − A ⊆ U ✓. II: U′ = ∅, not a singleton ✗.
I only.

Q.14 [Algebra]

What should be added to (x² − 1)/(x² + 1) to get (2x³ + 3x² − 1)/((x + 2)(x² + 1))?

(a) (x² − 1)/((x+2)(x²+1))
(b) (x² + 2)/((x+2)(x²+1))
(c) (x + 1)/(x + 2) ✓
(d) (x − 1)/(x + 2)

Solution: Required = [2x³ + 3x² − 1 − (x² − 1)(x + 2)] / ((x+2)(x²+1)).
Numerator = x³ + x² + x + 1 = (x+1)(x²+1) → (x + 1)/(x + 2).

Q.15 [Algebra]

If x + y + z = 0, then what is x(y + z)² + y(z + x)² + z(x + y)² equal to?

(a) xyz
(b) 2xyz
(c) 3xyz ✓
(d) 6xyz

Solution: y + z = −x etc., so the expression = x³ + y³ + z³.
With x + y + z = 0, x³ + y³ + z³ = 3xyz.

Q.16 [Trigonometry]

If tan 6θ · tan 3θ = 1, where 0 < θ < 30°, then what is θ equal to?

(a) 5°
(b) 10° ✓
(c) 15°
(d) No such value exists

Solution: tan A tan B = 1 ⇒ A + B = 90°: 9θ = 90° → θ = 10° (in range; tan 60° tan 30° = 1 ✓).

Q.17 [Trigonometry]

If (1 − tan x)/(1 + tan x) = 1 − 2 tan x/(1 + tan²x), where 0 ≤ x < π/2, x ≠ π/4, then what is (sin x + cos x) equal to?

(a) 1/2
(b) 3/4
(c) 1 ✓
(d) √2

Solution: RHS = 1 − sin 2x = (cos x − sin x)²; LHS = (cos x − sin x)/(cos x + sin x).
With u = cos x − sin x ≠ 0 (x ≠ π/4): u(cos x + sin x) = 1 → cos 2x = 1 → x = 0.
sin x + cos x = 1.

Q.18 [Trigonometry]

If cos²x + cos⁴x = 1, where 0 < x < π/2, then what is (sin²x + sin³x) + (sin³x + sin⁴x) equal to?

(a) (21 + 12√3)/16
(b) (3 + 2√2)/4
(c) 9/8
(d) 1 ✓

Solution: Let c = cos²x: c + c² = 1 → sin²x = c², so sin x = c.
Expression = sin²x(1 + sin x)² = [c(1 + c)]² = (c + c²)² = 1.

Q.19 [Trigonometry]

If sin⁴x + cos⁴x = 1, where 0 ≤ x ≤ π/2, then what is sin x · cos x equal to?

(a) 0 ✓
(b) 1/2
(c) 3/4
(d) 1

Solution: sin⁴x + cos⁴x = 1 − 2 sin²x cos²x = 1 → sin x cos x = 0 (x = 0 or π/2).

Q.20 [Trigonometry]

If 3 cos θ + 4 sin θ = 4, where 0 ≤ θ < π/2, then what is 4 cos θ + 3 sin θ equal to?

(a) 117/25 ✓
(b) 3
(c) 107/25
(d) 4

Solution: cos θ = (4 − 4 sin θ)/3 into sin² + cos² = 1: 25s² − 32s + 7 = 0 → s = 7/25 (s = 1 is θ = π/2, excluded), c = 24/25.
4 cos θ + 3 sin θ = 96/25 + 21/25 = 117/25.

Q.21 [Algebra]

If 1/(x + k) + 1/(x + 2k) + 1/(x + 5k) = 1/k, then what is the solution of the equation?

(a) k/3
(b) k/2
(c) 2k/3
(d) k ✓

Solution: Test x = k: 1/2k + 1/3k + 1/6k = 6/6k = 1/k ✓ → x = k.

Q.22 [Algebra]

Consider the polynomial p(x) = x⁵ + 2x⁴ + x³ − x² − 2x − 1:
I. x² + 2x + 1 is a factor of p(x).
II. x² + x + 1 divides p(x).
Which of the statements given above is/are correct?

(a) I only
(b) II only
(c) Both I and II ✓
(d) Neither I nor II

Solution: p(x) = x³(x² + 2x + 1) − (x² + 2x + 1) = (x + 1)²(x − 1)(x² + x + 1).
Both given factors divide p(x) → Both I and II.

Q.23 [Quadratic Equations]

If 5 is a root of the equation x² + px − 55 = 0 and the quadratic equation x² + px + k = 0 has equal roots, then what is the value of k?

(a) 3
(b) 6
(c) 9 ✓
(d) 12

Solution: x = 5: 25 + 5p − 55 = 0 → p = 6.
Equal roots: 36 − 4k = 0 → k = 9.

Q.24 [Time, Speed & Distance]

Had I walked 0·5 km/hr faster, I would have taken 8 minutes less to walk 2 km. What is my original speed?

(a) 2·5 km/hr ✓
(b) 2·75 km/hr
(c) 3 km/hr
(d) 3·25 km/hr

Solution: 8 min = 2/15 h: 2/v − 2/(v + 0.5) = 2/15 → 2v² + v − 15 = 0 → (2v − 5)(v + 3) = 0.
v = 2.5 km/hr.

Q.25 [Quadratic Equations]

If the roots of the equation x² + mx + n = 0 are increased by the same quantity k, then they become the roots of the equation x² + nx + m = 0. What is the value of (m + n)?

(a) − 4 ✓
(b) − 2
(c) 2
(d) 4

Solution: Roots α, β: α + β = −m, αβ = n. Shifted by k: −m + 2k = −n and n − mk + k² = m.
k = (m − n)/2 gives (n − m)(m + n + 4) = 0; n = m means no shift, so m + n = −4.

Q.26 [Quadratic Equations]

If α and β are the roots of the equation 2x² − 2(n + 1)x + (n² + n + 1) = 0, then what is (2α³ + 2β³) equal to?

(a) n³ − 1
(b) 1 − n³
(c) − (n³ + 1) ✓
(d) − (2n³ + 1)

Solution: α + β = n + 1, αβ = (n² + n + 1)/2.
2(α³ + β³) = 2(α+β)³ − 6αβ(α+β) = 2(n+1)³ − 3(n²+n+1)(n+1) = −(n³ + 1).

Q.27 [Number System]

The product of two numbers is 1050. The quotient when the larger number is divided by the smaller number is 4 and the remainder is 10. What is the sum of the two numbers?

(a) 70
(b) 75
(c) 80
(d) 85 ✓

Solution: L = 4s + 10 and s(4s + 10) = 1050 → 2s² + 5s − 525 = 0 → s = 15, L = 70.
Sum = 85.

Q.28 [Polynomials]

What is the HCF of x³ + x²y − 10xy² + 8y³ and x³ + x²y − 4xy² − 4y³?

(a) (x − y)
(b) (x − 2y) ✓
(c) (x + 2y)
(d) (x + 4y)

Solution: x = y kills the first: (x − y)(x + 4y)(x − 2y). x = −y kills the second: (x + y)(x − 2y)(x + 2y).
Common factor: (x − 2y).

Q.29 [Surds]

What is (√2 + √3 + √5)(√2 + √3 − √5)(√2 − √3 + √5)(− √2 + √3 + √5) equal to?

(a) 60
(b) 40
(c) 24 ✓
(d) 12

Solution: Pair the brackets: [(√2+√3)² − 5][5 − (√2−√3)²] = (2√6)(2√6) = 24.

Q.30 [Polynomials]

The LCM and HCF of two polynomials p(x) and q(x) are (x + k)²(x² − 7x + 6) and (x + k) respectively. If [x³ + (2k − 1)x² + k(k − 2)x − k²] is one of the polynomials, then which of the following statements is/are correct?
I. The other polynomial is (x + k)(x − 6) where k ≠ − 6.
II. (x + k) cannot be the HCF if k = − 6.

(a) I only
(b) II only
(c) Both I and II ✓
(d) Neither I nor II

Solution: Given polynomial = (x + k)²(x − 1); LCM × HCF = p × q gives q(x) = (x + k)(x − 6) ✓ (I).
If k = −6, p = (x − 6)²(x − 1) and q = (x − 6)², so the HCF would be (x − 6)², not (x + k) ✓ (II).
Both I and II.

Q.31 [Number System]

Two positive integers x and y are in the ratio 17 : 19. If the LCM of the two numbers is 1615, then what is (x + y) equal to?

(a) 36
(b) 180 ✓
(c) 360
(d) Cannot be determined due to insufficient data

Solution: x = 17a, y = 19a, LCM = 323a = 1615 → a = 5.
x + y = 85 + 95 = 180.

Q.32 [Ratio & Proportion]

If p/q = q/r = r/s = k, then what is (p³ + q³ + r³)/(q³ + r³ + s³) equal to?

(a) p/s ✓
(b) r/q
(c) s/p
(d) q/r

Solution: p = kq, q = kr, r = ks ⇒ numerator = k³(q³ + r³ + s³), so the ratio = k³ = (p/q)(q/r)(r/s) = p/s.

Q.33 [Polynomials]

If x² − 4x + 3 is a factor of x⁴ + px² + q, then what is (p − q) equal to?

(a) − 19 ✓
(b) − 17
(c) 17
(d) 19

Solution: Roots 1 and 3 of x⁴ + px² + q: 1 + p + q = 0 and 81 + 9p + q = 0 → p = −10, q = 9.
p − q = −19.

Q.34 [Quadratic Equations]

Suppose p and q are positive integers where q is not a perfect square. If 1/(p + √q) is a root of a quadratic equation having integer coefficients, then which of the following statements is/are correct?
I. The sum of the roots is 2p times the product of the roots.
II. The sum of the reciprocals of the roots is equal to 2p.

(a) I only
(b) II only
(c) Both I and II ✓
(d) Neither I nor II

Solution: Integer coefficients ⇒ conjugate roots 1/(p ± √q): sum = 2p/(p² − q), product = 1/(p² − q).
I: sum = 2p × product ✓. II: 1/α + 1/β = 2p ✓. Both I and II.

Q.35 [Algebra]

If 1/(2b + 2c) + 1/(2c + 2a) = 1/(a + b), then what is a² + b² + c² equal to?

(a) 9c²
(b) 4c²
(c) 3c² ✓
(d) 2c²

Solution: Cross-multiply: (a + b + 2c)(a + b) = 2(b + c)(c + a) → a² + b² = 2c².
a² + b² + c² = 3c².

Q.36 [Time, Speed & Distance]

A train completely crossed two persons X and Y moving in the same direction as the train in 8 seconds and 8·1 seconds respectively. X is walking at a uniform speed of 3 km/hr and Y is walking at a uniform speed of 4 km/hr. What is the speed of the train?

(a) 81 km/hr
(b) 82 km/hr
(c) 83 km/hr
(d) 84 km/hr ✓

Solution: Same train length both times: 8(v − 3) = 8.1(v − 4) → 0.1v = 8.4 → v = 84 km/hr.

Q.37 [Number System]

A number is formed by three digits, each less by unity than the digit that follows it. If 15 is added to the number, then the sum is 30 times the sum of the digits of the number. What is the product of the digits of the number?

(a) 6
(b) 24
(c) 60 ✓
(d) 120

Solution: Digits a, a+1, a+2: number = 111a + 12, digit sum = 3a + 3.
111a + 27 = 30(3a + 3) → a = 3; digits 3, 4, 5 → product 60.

Q.38 [Logarithms]

Consider the following statements in respect of common logarithms:
I. The logarithm of a number greater than 100 but less than 1000 lies between 2 and 3.
II. The logarithm of a positive number less than unity is negative.

(a) I only
(b) II only
(c) Both I and II ✓
(d) Neither I nor II

Solution: I: log 100 = 2, log 1000 = 3 and log is increasing ✓. II: 0 < N < 1 ⇒ log N < 0 ✓.
Both I and II.

Q.39 [Logarithms]

What is the number of zeros immediately after the decimal point in (0·5)¹⁰⁰⁰? (Given that log₁₀2 = 0·30103)

(a) 300
(b) 301 ✓
(c) 302
(d) 303

Solution: log(0.5)¹⁰⁰⁰ = −1000 × 0.30103 = −301.03 = −302 + 0.97, i.e. 9.3 × 10⁻³⁰².
Characteristic −302 ⇒ 301 zeros after the decimal point.

Q.40 [Surds]

If x = 11 + 2√30, then what is the value of x^(1/2) + x^(−1/2)?

(a) 2√2
(b) 2√3
(c) 2√5
(d) 2√6 ✓

Solution: 11 + 2√30 = (√6 + √5)², so √x = √6 + √5 and 1/√x = √6 − √5.
Sum = 2√6.

Q.41 [Profit & Loss]

A shopkeeper sold his goods at the cost price. By using false weights, he gained 11 1/9 %. What weight did he use for 1 kg?

(a) 960 gm
(b) 950 gm
(c) 925 gm
(d) 900 gm ✓

Solution: Selling at cost, the gain comes only from the short weight w: (1000 − w)/w = 1/9 → w = 900 gm.
Trap: dividing by 1000 instead of by w.

Q.42 [Quadratic Equations]

If the roots of the equation px² − 40x + 96 = 0 are even integers, then what is the value of p?

(a) 1
(b) 2
(c) 3
(d) 4 ✓

Solution: Roots 2m, 2n: m + n = 20/p, mn = 24/p. p = 4 gives m + n = 5, mn = 6 → roots 4 and 6 ✓ (x² − 10x + 24 = 0).
p = 4.

Q.43 [Number System]

A number N consists of two digits. The digit in the tens place is 3 times the digit in the units place. The digits are reversed and the resulting number is denoted by R. If N × R = 3627, then what is the product of the digits of the number N?

(a) 27 ✓
(b) 18
(c) 12
(d) 3

Solution: Units u, tens 3u ⇒ N ∈ {31, 62, 93}. Only 93 × 39 = 3627 ✓.
Product = 9 × 3 = 27.

Q.44 [Boats & Streams]

A boatman can row to a place (Y) at a distance of 24 km from the starting point (X) and back in 14 hours. If he can row 4 km with the stream in the same time as 3 km against it, what is the speed of the stream?

(a) 0·5 km/hr ✓
(b) 1 km/hr
(c) 2 km/hr
(d) 2·5 km/hr

Solution: 4/d = 3/u → u = 3d/4. 24/d + 24/u = 14 → 56/d = 14 → d = 4, u = 3.
Stream = (d − u)/2 = 0.5 km/hr.

Q.45 [Alligation]

An alloy (X) of gold and silver is taken in the ratio 1 : 2 and another alloy (Y) of gold and silver is taken in the ratio 2 : 3. How many parts of X and Y must be taken to obtain a new alloy Z consisting of gold and silver in the ratio 4 : 7?

(a) 5 : 6
(b) 2 : 1
(c) 3 : 5
(d) 6 : 5 ✓

Solution: Gold fractions: X = 1/3, Y = 2/5, Z = 4/11.
X : Y = (2/5 − 4/11) : (4/11 − 1/3) = 2/55 : 1/33 = 6 : 5.

Q.46 [Number System]

N is a 3-digit number. The middle digit of N is equal to the sum of the other two digits and the sum of the digits is 10. How many such numbers (N) can be possible?

(a) 3
(b) 4
(c) 5 ✓
(d) 6

Solution: b = a + c and a + b + c = 10 → b = 5, a + c = 5 with a ≥ 1: (1,4), (2,3), (3,2), (4,1), (5,0).
5 numbers (154, 253, 352, 451, 550).

Q.47 [Profit & Loss]

The ratio of the cost price of two articles X and Y is 1 : 2. A businessman earns a profit of p% by selling X and a loss of 3p% by selling Y. If he loses 20% in this transaction, then what is the value of p?

(a) 6%
(b) 8%
(c) 10%
(d) 12% ✓

Solution: Costs 1 and 2: net loss = 6p/100 − p/100 = 5p/100 on cost 3 → loss % = 5p/3 = 20 → p = 12%.
Trap: averaging p and 3p directly.

Q.48 [Compound Interest]

A sum of money at the rate of 5% per annum compounded annually becomes n times in 100 years. What is the value of n? (Given log₁₀2 = 0·301, log₁₀3 = 0·477 and log₁₀7 = 0·845)

(a) 98
(b) 99
(c) 100
(d) More than 100 ✓

Solution: log n = 100 log 1.05 = 100(log 105 − 2); log 105 = log 3 + log 5 + log 7 = 0.477 + 0.699 + 0.845 = 2.021.
log n = 2.1 → n ≈ 126 → more than 100.

Q.49 [Variation]

A variable y is directly proportional to a variable quantity xⁿ. Given that when x = 2, y = 20·8 and when x = 3, y = 105·3. Which one of the following is the constant of proportionality if it is greater than 1?

(a) 1·1
(b) 1·3 ✓
(c) 2
(d) 4

Solution: 105.3/20.8 = (3/2)ⁿ = 5.0625 = 1.5⁴ → n = 4.
k = 20.8/2⁴ = 1.3.

Q.50 [Surds]

If x = 3 − 3^(1/3) − 3^(2/3), then what is x(x − 3)(x − 6) equal to?

(a) − 12 ✓
(b) − 9
(c) 0
(d) 12

Solution: a = 3^(1/3), t = a + a², x = 3 − t: x(x − 3)(x − 6) = t(9 − t²).
t² = a² + 3a + 6 → 9 − t² = 3 − 3a − a², and (a + a²)(3 − 3a − a²) = −12 (using a³ = 3).

Q.51 [Statistics]

If the mean of a set of 7 observations is 10 and the mean of a set of 3 observations is 5, then what is the combined mean?

(a) 9
(b) 8·5 ✓
(c) 6·5
(d) 5·5

Solution: Combined mean = (70 + 15)/10 = 8.5.
Trap: (10 + 5)/2 — means must be weighted by counts.

Q.52 [Alligation]

The average weekly wages of male employees in a company is ₹4,200 and that of females is ₹3,200. If the average weekly wage of all employees is ₹4,000, then what is the ratio of male to female employees?

(a) 3 : 2
(b) 5 : 3
(c) 4 : 1 ✓
(d) 3 : 5

Solution: Alligation about 4000: M : F = (4000 − 3200) : (4200 − 4000) = 800 : 200 = 4 : 1.

Q.53 [Statistics]

Consider the observations x, x + 4, 30, 33, 74, 78, 49, 52, 98, 85. If the median of the data is 67, then what is the value of x?

(a) 66
(b) 65 ✓
(c) 63
(d) Cannot be determined due to insufficient data

Solution: 10 values ⇒ 5th + 6th = 134. No two known values sum to 134, so x and x + 4 are the middle pair: 2x + 4 = 134 → x = 65.
Check: 30, 33, 49, 52, 65, 69, 74, … → median 67 ✓

Q.54 [Averages]

A cyclist pedals from her house to her office at a speed of 3 kmph and back from the office to her house at 6 kmph. What is the average speed?

(a) 5 kmph
(b) 4·5 kmph
(c) 4 kmph ✓
(d) 3·5 kmph

Solution: Equal distances ⇒ harmonic mean: 2 × 3 × 6/(3 + 6) = 4 kmph.
Trap: 4.5 (the arithmetic mean).

Q.55 [Statistics]

If the heights (in cm) of 9 students of a class are 150, 165, 145, 149, 150, 147, 152, 144 and 148, then what is the algebraic sum of the heights measured from their arithmetic mean?

(a) − 5
(b) 0 ✓
(c) 5
(d) 150

Solution: Σ(xᵢ − x̄) = Σxᵢ − n x̄ = 0 for any data set → 0. No arithmetic needed.

Q.56 [Statistics]

Consider the grouped data — Class 0–10: 4, 10–20: 8, 20–30: 15, 30–40: 10, 40–50: 3. What is the mode of the above distribution?

(a) 25·83 ✓
(b) 25·43
(c) 25·00
(d) 24·73

Solution: Modal class 20–30: l = 20, f₁ = 15, f₀ = 8, f₂ = 10, h = 10.
Mode = 20 + (7/12) × 10 = 25.83.

Q.57 [Statistics]

If the mean of 15 observations, namely x₁, x₂, x₃, …, x₁₅ is 2·5, then what is the value of Σ(i=1 to 15) [5(4xᵢ + 1)]?

(a) 400
(b) 525
(c) 625
(d) 825 ✓

Solution: Σxᵢ = 37.5. Σ 5(4xᵢ + 1) = 5(4 × 37.5 + 15) = 5 × 165 = 825.
Trap: Σ1 over 15 terms is 15.

Q.58 [Statistics]

Consider the discrete grouped data — x: 12, 14, 10, 16, 18 with f: 6, 8, 10, 9, 3. What is the arithmetic mean of the above distribution?

(a) 13·69
(b) 13·39 ✓
(c) 13·19
(d) 13·09

Solution: Σf = 36, Σfx = 72 + 112 + 100 + 144 + 54 = 482.
Mean = 482/36 = 13.39.

Q.59 [Statistics]

Consider the following incomplete frequency distribution with a total frequency of 100 — Class 0–10: 10, 10–20: x, 20–30: 20, 30–40: y. If the mean of the frequency distribution is 25, then what are the missing frequencies?

(a) x = 15, y = 55
(b) x = 50, y = 20
(c) x = 20, y = 45
(d) x = 25, y = 45 ✓

Solution: x + y = 70; mean: 550 + 15x + 35y = 2500 → 3x + 7y = 390.
Substituting x = 70 − y: 4y = 180 → y = 45, x = 25 → x = 25, y = 45.

Q.60 [Data Interpretation]

The data gives investment (₹ in crores) by a governmental organization — Science, Technology and Environment: 7,200; Transport and Communication: 3,000; Rural Development: 1,500; Industry and Minerals: 3,800; Agricultural Services: 2,500.
I. The percentage investment in the 'Science, Technology and Environment' category is 40%.
II. The angle (in degrees) for 'Agricultural Services' in a pie diagram is 50°.

(a) I only
(b) II only
(c) Both I and II ✓
(d) Neither I nor II

Solution: Total = 18,000. I: 7200/18000 = 40% ✓. II: 2500/18000 × 360° = 50° ✓.
Both I and II.

Q.61 [Geometry — Circles]

[Items 61–63] The centre O of a circle inside a triangle ABC is at a distance of 13 cm from each of the vertices of the triangle. The diameter of the circle is 10 cm and the circle touches only two sides of the triangle, AB and AC.

If x is the perimeter, in cm, of the triangle, then which one of the following is correct?

(a) 63 cm < x < 65 cm
(b) 65 cm < x < 67 cm ✓
(c) 67 cm < x < 69 cm
(d) 69 cm < x < 71 cm

Solution: O is the circumcentre (R = 13); the radius-5 circle touches AB and AC ⇒ chords at distance 5: AB = AC = 2√(169 − 25) = 24.
With A(0, 13): B, C lie on y = −119/13, BC = 240/13 ≈ 18.46. Perimeter ≈ 66.5 → 65 < x < 67.

Q.62 [Geometry — Triangles]

[Item 62 of the same set] If y is the area in cm² of the triangle, then which one of the following is correct?

(a) 190 cm² < y < 200 cm²
(b) 200 cm² < y < 210 cm² ✓
(c) 210 cm² < y < 220 cm²
(d) 220 cm² < y < 230 cm²

Solution: Height from A to BC = 13 + 119/13 = 288/13, BC = 240/13.
Area = ½ × 240/13 × 288/13 ≈ 204.5 → 200 < y < 210.

Q.63 [Geometry — Triangles]

[Item 63 of the same set] Consider the following statements:
I. ∠ABC lies between 60° and 90°.
II. If z is the distance in cm from the centre O of the circle to the midpoint of BC, then 7 cm < z < 8 cm.

(a) I only ✓
(b) II only
(c) Both I and II
(d) Neither I nor II

Solution: cos B = (120/13)/24 = 5/13 → B ≈ 67.4° ✓ (I).
O to the midpoint of BC = 119/13 ≈ 9.15, not in (7, 8) ✗ (II). I only.

Q.64 [Geometry — Circles]

[Items 64–66] A circle with centre O passes through the vertex A of an equilateral triangle ABC and touches BC at its midpoint M. The circle cuts AB at D and AC at E.

What is AD : DB equal to?

(a) 2 : 1
(b) 3 : 2
(c) 3 : 1 ✓
(d) 4 : 3

Solution: O lies on the altitude AM and OA = OM ⇒ AM is a diameter. Tangent–secant at B: BM² = BD × BA → (a/2)² = BD × a → BD = a/4.
AD : DB = 3a/4 : a/4 = 3 : 1.

Q.65 [Geometry — Circles]

[Item 65 of the same set] If x is the area of the circle and y is the area of the triangle such that z = (x/y)², then which one of the following is correct?

(a) 0·5 < z < 1
(b) 1 < z < 1·5
(c) 1·5 < z < 2 ✓
(d) z > 2

Solution: r = AM/2 = a√3/4: x = 3πa²/16, y = √3a²/4 → x/y = π√3/4 ≈ 1.36.
z ≈ 1.85 → 1.5 < z < 2.

Q.66 [Geometry — Triangles]

[Item 66 of the same set] What is the ratio of the area of the triangle ADO to the area of the quadrilateral BDOM?

(a) 3 : 5 ✓
(b) 4 : 5
(c) 3 : 3
(d) 2 : 3

Solution: B(0,0), A(a/2, a√3/2), M(a/2, 0), O(a/2, a√3/4), D(a/8, a√3/8).
[ADO] = 3√3a²/64, [ABM] = 8√3a²/64 → [BDOM] = 5√3a²/64. Ratio 3 : 5.

Q.67 [Geometry — Triangles]

[Items 67–68] ABC is an equilateral triangle. BP is perpendicular to AC, PR is perpendicular to AB, and PQ is perpendicular to BC.

What is the ratio of AB² : BP² : PR²?

(a) 4 : 3 : 1
(b) 16 : 12 : 3 ✓
(c) 16 : 12 : 5
(d) 5 : 4 : 3

Solution: BP = a√3/2 (altitude); in ΔAPR, PR = AP sin 60° = a√3/4.
a² : 3a²/4 : 3a²/16 = 16 : 12 : 3.

Q.68 [Geometry — Triangles]

[Item 68 of the same set] What is the ratio of the area of Δ PRB to the area of Δ PQC?

(a) 1 : 1
(b) 2 : 1
(c) 3 : 1 ✓
(d) 3 : 2

Solution: PR = PQ = a√3/4, but RB = 3a/4 and QC = a/4.
[PRB] : [PQC] = RB : QC = 3 : 1.

Q.69 [Geometry — Triangles]

[Items 69–70] PQR is a triangle such that QP = QR = 15 cm and PR = 18 cm. PN, QM and RT are the altitudes of the triangle which intersect at O.

What is QT : QO equal to?

(a) 3 : 4
(b) 4 : 5 ✓
(c) 3 : 5
(d) 1 : 2

Solution: QM = 12, area 108, RT = 216/15 = 14.4, QT = √(225 − 207.36) = 4.2.
With P(−9,0), R(9,0), Q(0,12) the orthocentre is O(0, 6.75) → QO = 5.25. QT : QO = 4.2 : 5.25 = 4 : 5.

Q.70 [Geometry — Triangles]

[Item 70 of the same set] What is the ratio of QO to OM?

(a) 11 : 9
(b) 11 : 8
(c) 25 : 27
(d) 7 : 9 ✓

Solution: QO = 5.25, OM = 6.75 → 7 : 9.

Q.71 [Data Sufficiency]

[Data sufficiency] Question: ABCD is a rhombus of side 10 cm. What is the sum of the squares of the diagonals of the rhombus?
Statement I: AC : BD = 3 : 4.
Statement II: AC = 6 cm.

(a) The Question can be answered by using one of the Statements alone, but cannot be answered using the other Statement alone.
(b) The Question can be answered by using either Statement alone.
(c) The Question can be answered by using both the Statements together, but cannot be answered using either Statement alone.
(d) The Question can be answered even without using both the Statements. ✓

Solution: Rhombus: AC² + BD² = 4 × side² = 400, fixed by the side alone.
(d) — no statement needed.

Q.72 [Data Sufficiency]

[Data sufficiency] Question: The diagonal of a rectangle ABCD (AB > BC) is 5√2 cm. What is its perimeter?
Statement I: Length AB is an integer.
Statement II: Length BC is an integer.

(a) The Question can be answered by using one of the Statements alone, but cannot be answered using the other Statement alone.
(b) The Question can be answered by using either Statement alone.
(c) The Question can be answered by using both the Statements together, but cannot be answered using either Statement alone. ✓
(d) The Question can be answered even without using both the Statements.

Solution: AB² + BC² = 50, AB > BC. I alone: (7, 1) or (6, √14) ✗. II alone: (7, 1) or (√41, 3) ✗.
Both: integer pairs are (7, 1) and (5, 5); only (7, 1) has AB > BC → perimeter 16. (c).

Q.73 [Data Sufficiency]

[Data sufficiency] Question: ABC is a triangle right-angled at A and AD is perpendicular to BC. If AD = 7·2 cm, then what is BD × CD equal to?
Statement I: AB : AC = 3 : 4.
Statement II: BC = 15 cm.

(a) The Question can be answered by using one of the Statements alone, but cannot be answered using the other Statement alone.
(b) The Question can be answered by using either Statement alone.
(c) The Question can be answered by using both the Statements together, but cannot be answered using either Statement alone.
(d) The Question can be answered even without using both the Statements. ✓

Solution: Altitude to the hypotenuse: AD² = BD × CD = 51.84.
(d) — neither statement needed.

Q.74 [Data Sufficiency]

[Data sufficiency] Question: ABCD is a quadrilateral with AB = 6 cm, BC = 8 cm, CD = 4√5 cm, DA = 2√5 cm and inscribed in a circle. What is the diameter of the circle?
Statement I: The area of the quadrilateral is 44 cm².
Statement II: ∠ABC = 90°.

(a) The Question can be answered by using one of the Statements alone, but cannot be answered using the other Statement alone.
(b) The Question can be answered by using either Statement alone.
(c) The Question can be answered by using both the Statements together, but cannot be answered using either Statement alone.
(d) The Question can be answered even without using both the Statements. ✓

Solution: A cyclic quadrilateral with four given sides is rigid, so R is already fixed: Brahmagupta gives K = 44 and R = √((ab+cd)(ac+bd)(ad+bc))/4K = 880/176 = 5 → diameter 10.
(d) — neither statement needed.

Q.75 [Data Sufficiency]

[Data sufficiency] Question: ABC is a triangle inscribed in a semi-circle such that AC coincides with the diameter of the semi-circle. Let P be any point on the arc of the semi-circle. What is AP² + CP² + AC² equal to?
Statement I: The radius of the circle is 5 cm.
Statement II: AB = 6 cm and BC = 8 cm.

(a) The Question can be answered by using one of the Statements alone, but cannot be answered using the other Statement alone.
(b) The Question can be answered by using either Statement alone. ✓
(c) The Question can be answered by using both the Statements together, but cannot be answered using either Statement alone.
(d) The Question can be answered even without using both the Statements.

Solution: ∠APC = 90° ⇒ AP² + CP² = AC², so the expression = 2AC².
I gives AC = 10; II gives AC = √(36 + 64) = 10. Either alone → (b).

Q.76 [Mensuration — Solids]

A thin metallic sheet, 1·92 m² in area, is cut into two equal pieces. One piece is used to make a hollow cube of volume P (in m³) and the other is used to make a hollow cuboid of volume Q (in m³) with dimensions in the ratio 4 : 2 : 1. If no sheet is wasted, which one of the following is correct?

(a) 343Q² = 216P² ✓
(b) 343Q² = 225P²
(c) 49Q² = 64P²
(d) 49Q² = 25P²

Solution: Each piece A = 0.96: cube P = (A/6)^(3/2); cuboid 4k × 2k × k with 28k² = A → Q = 8(A/28)^(3/2).
(Q/P)² = 64(3/14)³ = 216/343 → 343Q² = 216P².

Q.77 [Geometry — Triangles]

ABC is a triangle such that ∠ABC = 120°. If BD is the bisector of ∠B that meets AC at D, then what is the ratio of the area of Δ ABD to the area of Δ CBD?

(a) AB : BC ✓
(b) BC : AB
(c) AB² : BC²
(d) 1 : 1

Solution: Same apex B, bases on AC: [ABD] : [CBD] = AD : DC = AB : BC (angle bisector theorem) → AB : BC.
The 120° is a distractor.

Q.78 [Geometry — Similarity]

Consider the following statements:
I. The areas of two similar triangles are in the ratio of the squares of the corresponding medians.
II. The areas of two similar triangles are in the ratio of the squares of the corresponding heights.

(a) I only
(b) II only
(c) Both I and II ✓
(d) Neither I nor II

Solution: In similar triangles every linear element (side, median, height) scales by k and the area by k².
Both I and II.

Q.79 [Geometry — Similarity]

If two similar scalene triangles ABC and PQR have the same area, then which of the following statements is/are correct?
I. Both the triangles are congruent.
II. Both the triangles have equal perimeters.

(a) I only
(b) II only
(c) Both I and II ✓
(d) Neither I nor II

Solution: Equal areas ⇒ k² = 1 ⇒ k = 1 ⇒ congruent, hence equal perimeters.
Both I and II.

Q.80 [Geometry — Similarity]

In a triangle ABC, AB = 8 cm and D and E are points on AB and AC respectively such that DE is parallel to BC. If BC = 5DE, then what is AD × BD equal to?

(a) 10 cm²
(b) 10·24 cm² ✓
(c) 10·56 cm²
(d) 12 cm²

Solution: DE ∥ BC ⇒ AD/AB = DE/BC = 1/5 → AD = 1.6, BD = 6.4.
AD × BD = 10.24 cm².
Trap: the ratio is to the whole side AB, not to DB.

Q.81 [Geometry — Circles]

In a cyclic quadrilateral ABCD, ∠A = (4x + 3)°, ∠B = (3y + 9)°, ∠C = (4y − 3)°, ∠D = (5x − 10)°. What is ∠A + ∠B equal to?

(a) 200°
(b) 190°
(c) 180°
(d) 170° ✓

Solution: Opposite angles sum to 180°: x + y = 45 and 5x + 3y = 181 → x = 23, y = 22.
∠A + ∠B = 95° + 75° = 170°.

Q.82 [Mensuration — Areas]

The area of a rhombus is 720 cm² and the sum of its diagonals is 98 cm. What is the perimeter of the rhombus?

(a) 160 cm
(b) 164 cm ✓
(c) 168 cm
(d) 172 cm

Solution: d₁d₂ = 1440, d₁ + d₂ = 98 → (d₁ − d₂)² = 9604 − 5760 = 3844 → d₁ = 80, d₂ = 18.
Side = ½√(80² + 18²) = 41 → perimeter 164 cm.

Q.83 [Mensuration — Circles]

A chord of a circle makes an angle of 90° at the centre of the circle. If the area of the major segment is k times the area of the minor segment, then what is the value of k? (Take π = 22/7)

(a) 8
(b) 9
(c) 10 ✓
(d) 11

Solution: r = 1: minor segment = π/4 − 1/2 = 2/7 (π = 22/7); major = 22/7 − 2/7 = 20/7.
k = 10.

Q.84 [Mensuration — Areas]

ABC is a triangle in which (AB + BC) exceeds CA by 10 cm, (BC + CA) exceeds AB by 8 cm and (CA + AB) exceeds BC by 72 cm. What is the area of the triangle?

(a) 196 cm²
(b) 180 cm² ✓
(c) 172 cm²
(d) 160 cm²

Solution: Add the three conditions: a + b + c = 90, s = 45; each condition is 2(s − side): s − a = 36, s − b = 5, s − c = 4.
Area = √(45 × 36 × 5 × 4) = 180 cm².

Q.85 [Mensuration — Solids]

A reservoir is in the form of a cuboid. Its length is 30 m. If 24 kL of water is removed from the reservoir, the water level goes down by 20 cm. What is the width of the reservoir?

(a) 4 m ✓
(b) 3·5 m
(c) 3 m
(d) 2·5 m

Solution: 24 kL = 24 m³, 20 cm = 0.2 m: 30 × w × 0.2 = 24 → w = 4 m.

Q.86 [Mensuration — Solids]

A well is dug with a diameter of 3·5 m and 16 m depth. The earth so excavated is spread in the form of a right circular cone of radius 7 m. What is the height of the cone?

(a) 1·5 m
(b) 2 m
(c) 2·5 m
(d) 3 m ✓

Solution: Earth dug = π(1.75)²(16) = 49π; cone = ⅓π(7²)H = ⅓ × 49π × H → H = 3 m.

Q.87 [Mensuration — Circles]

The chord of a circle is 20 cm and the height of the minor segment is 5 cm. What is the diameter of the circle?

(a) 24 cm
(b) 25 cm ✓
(c) 26 cm
(d) 30 cm

Solution: Half-chord 10, sagitta 5: r² = (r − 5)² + 10² → r = 12.5.
Diameter = 25 cm.

Q.88 [Mensuration — Circles]

On a circular metal plate of uniform thickness, 16 holes, each of diameter 2 cm, are made. If the plate thereby has lost one-ninth of its original weight, then what is the diameter of the plate?

(a) 12 cm
(b) 15 cm
(c) 21 cm
(d) 24 cm ✓

Solution: Weight ∝ area: 16π(1²) = πR²/9 → R² = 144 → diameter 24 cm.
Trap: using 2 cm as the hole radius.

Q.89 [Mensuration — Areas]

A square of side length 4 cm has its corners cut away in such a manner so as to form a regular octagon. What is the length of the side of the octagon? (Take √2 = 1·41)

(a) 1·00 cm
(b) 1·64 cm ✓
(c) 1·68 cm
(d) 1·72 cm

Solution: Corner cut x: 4 − 2x = x√2 → x = 4 − 2√2; octagon side = x√2 = 4(√2 − 1) = 4 × 0.41 = 1.64 cm.

Q.90 [Mensuration — Solids]

A tall cylindrical reservoir kept vertically is 20 m in diameter. Water is poured into it at the rate of 264 m³ per hour. What is the rate at which the water level rises in the reservoir per minute? (Take π = 22/7)

(a) 1·0 cm
(b) 1·2 cm
(c) 1·4 cm ✓
(d) 14 cm

Solution: Base area = (22/7)(100) = 2200/7 m²; rise per hour = 264 ÷ 2200/7 = 0.84 m; per minute = 0.014 m = 1.4 cm.
Trap: option (d) is the per-hour figure.

Q.91 [Mensuration — Solids]

How many square metres of canvas (approximately) will be required to make a conical tent 3 m high so that a man 2 m tall may stand anywhere within a radius of 1 m from its centre without stooping?

(a) 36 m²
(b) 40 m² ✓
(c) 44 m²
(d) 48 m²

Solution: Slope height 3(1 − d/R) = 2 at d = 1 → R = 3; slant l = √(9 + 9) = 3√2.
Canvas = πRl = 9√2π ≈ 40 m² (curved surface only).

Q.92 [Coordinate Geometry]

PQRS is a square. M is a point on PS such that PM : MS = 2 : 1 and N is a point on SR such that SN : NR = 1 : 2. If the area of triangle MQN is 10 square units, then what is the perimeter of the square?

(a) 24 units ✓
(b) 28 units
(c) 32 units
(d) 36 units

Solution: Side a, S(0,0): M(0, a/3), N(a/3, 0), Q(a, a). Shoelace: area = 5a²/18 = 10 → a = 6.
Perimeter = 24 units.

Q.93 [Mensuration — Solids]

A cubical wooden block of length 10 cm is sliced to produce a cross-section of maximum area. What is the area of the cross-section?

(a) 100 cm²
(b) 100√2 cm² ✓
(c) 100√3 cm²
(d) 200 cm²

Solution: The largest plane section of a cube is the rectangle through two opposite edges: 10 × 10√2.
Area = 100√2 cm².

Q.94 [Geometry — Triangles]

In a triangle ABC, AD is the bisector of angle A that meets BC at D. If AB = 21 cm, DC = 20 cm and AB : AC = 3 : 4, then what is AC² − BD² equal to?

(a) 441
(b) 559 ✓
(c) 588
(d) 784

Solution: AC = 28; angle bisector: BD/DC = 3/4 → BD = 15.
28² − 15² = 784 − 225 = 559.

Q.95 [Mensuration — Areas]

A square of maximum area is inscribed in an equilateral triangle. If the side of the triangle is equal to (6 + 4√3) cm, then what is the area of the square?

(a) 36 cm² ✓
(b) 27 cm²
(c) 21 cm²
(d) 18 cm²

Solution: Inscribed square s = ah/(a + h) with h = a√3/2 → s = a(2√3 − 3).
a = 6 + 4√3 → s = 6 → area 36 cm².

Q.96 [Mensuration — Solids]

A room is L m long, B m wide and H m high. Further, L > B > H and L, B and H are integers. The length of the longest pole that can be placed in the room is 17 m and the length of the longest pole that can be placed on the floor is 15 m. What is the volume of the room?

(a) 432 m³
(b) 864 m³ ✓
(c) 1296 m³
(d) Cannot be determined due to insufficient data

Solution: L² + B² + H² = 289 and L² + B² = 225 → H = 8; integer L > B > 8 with L² + B² = 225 → (12, 9).
Volume = 12 × 9 × 8 = 864 m³.

Q.97 [Mensuration — Solids]

A spherical metal ball is molten and made into n smaller identical spheres. In this process, the surface area of the smaller balls increases by 900%. What is the value of n?

(a) 729
(b) 900
(c) 1000 ✓
(d) Cannot be determined due to insufficient data

Solution: Surface area becomes 10× (100% + 900%). Volume fixed ⇒ r = R/n^(1/3) ⇒ total area scales by n^(1/3) = 10.
n = 1000.
Trap: reading it as 9× gives 729.

Q.98 [Mensuration — Areas]

ABCD is a parallelogram with ∠ABC = 150° and the sides have integer values. If the area of the parallelogram is 17·5 cm², then consider the following statements:
I. It is possible to have a perimeter equal to 24 cm.
II. It is possible to have a perimeter equal to 72 cm.

(a) I only
(b) II only
(c) Both I and II ✓
(d) Neither I nor II

Solution: Area = ab sin 150° = ab/2 = 17.5 → ab = 35: (5, 7) → perimeter 24 ✓; (1, 35) → 72 ✓.
Both I and II.

Q.99 [Mensuration — Solids]

N number of cubes each of side length equal to 10 cm are joined end to end in a row. If the total surface area of the resulting cuboid is 3800 cm², then what is the value of N?

(a) 8
(b) 9 ✓
(c) 10
(d) 12

Solution: Cuboid 10N × 10 × 10: TSA = 400N + 200 = 3800 → N = 9.

Q.100 [Geometry — Triangles]

In a quadrilateral ABCD, ∠ABC = 90° and ∠ACD = 90°. Let AB = p units, BC = q units, AC = r units, CD = s units, AD = t units, where p < q < r < s < t < 15. If p, q, r, s and t are integers, then what is the area of the quadrilateral?

(a) 24 square units
(b) 36 square units ✓
(c) 48 square units
(d) Cannot be determined due to insufficient data

Solution: p² + q² = r² and r² + s² = t² with t < 15: (3, 4, 5) then (5, 12, 13).
Area = ½(3)(4) + ½(5)(12) = 36 square units.