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CDS II 2025 Elementary Mathematics with Solutions

Exam: CDS Year: 2025 (Session II) Held on: 14 September 2025 Questions: 100 Marks: 100 Negative Marking: 1/3

Booklet series: Booklet Series A (ANFB-T-EMT). 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 [Polynomials]

Let p(x) be a polynomial. When p(x) is divided by (x − 1), it leaves 2 as the remainder. When p(x) is divided by (x − 2), it leaves 1 as the remainder. What is the remainder when p(x) is divided by (x − 1)(x − 2)?

(a) 3
(b) −3
(c) 3 − x ✓
(d) 3 − 2x

Solution: Remainder is linear, ax + b, with p(1) = 2 and p(2) = 1.
a + b = 2, 2a + b = 1 → a = −1, b = 3 → remainder 3 − x.

Q.2 [Algebra]

Consider the following in respect of a positive real number x :
I. x + 1/x > 1
II. (x + 1/x)² > 2
III. (x + 1/x)⁴ > 9
Which of the above are correct?

(a) I and II only
(b) II and III only
(c) I and III only
(d) I, II and III ✓

Solution: For x > 0, x + 1/x ≥ 2 (AM–GM).
So the square is ≥ 4 > 2 and the fourth power is ≥ 16 > 9. All three hold.

Q.3 [Number System]

Let p and q be natural numbers such that q > p. What is the largest value of p such that q² − 5p − 4 is negative?

(a) 3 ✓
(b) 4
(c) 5
(d) 6

Solution: Smallest allowed q is p + 1, so need (p + 1)² < 5p + 4 → p² − 3p − 3 < 0 → p ≤ 3.
Check p = 3, q = 4: 16 − 15 − 4 = −3 ✓; p = 4, q = 5: 25 − 24 = +1 ✗. p = 3.

Q.4 [Number System]

Let x and y be natural numbers, each less than 20, such that x, y, x + y and x − y are prime numbers. How many such combinations of (x, y, x + y, x − y) are possible?

(a) One ✓
(b) Two
(c) Three
(d) None

Solution: x + y and x − y both prime forces one of them odd → y = 2.
Then x − 2, x, x + 2 all prime → the only prime triplet 3, 5, 7 → x = 5. One combination: (5, 2, 7, 3).

Q.5 [Polynomials]

If (x + 1)(x + p)(x² + p²) = x⁴ − 1, then what is the value of p?

(a) −1 ✓
(b) 0
(c) 1
(d) Cannot be determined

Solution: x⁴ − 1 = (x + 1)(x − 1)(x² + 1).
Match the factors: x + p = x − 1 and x² + p² = x² + 1 both give p = −1.

Q.6 [Surds]

If (2 + √3)^x + (2 − √3)^x = 2, then what is (2 + √3)^x − (2 − √3)^x equal to?

(a) 0 ✓
(b) 0·5
(c) 1
(d) 1·5

Solution: Let a = (2 + √3)^x; then (2 − √3)^x = 1/a.
a + 1/a = 2 only when a = 1 (x = 0), so the difference is 0.

Q.7 [Algebra]

If 1/a + 1/b = 5/6 and 1/a² + 1/b² = 13/36, then what is 1/a³ + 1/b³ equal to?

(a) 31/216
(b) 35/216 ✓
(c) 37/216
(d) 41/216

Solution: Let u = 1/a, v = 1/b: u + v = 5/6, u² + v² = 13/36 → uv = (25/36 − 13/36)/2 = 1/6.
u³ + v³ = (u + v)³ − 3uv(u + v) = 125/216 − 90/216 = 35/216.

Q.8 [Polynomials]

What is the remainder when x⁶ is divided by x² + 1?

(a) −1 ✓
(b) 0
(c) 1
(d) x + 1

Solution: Put x² = −1: x⁶ = (x²)³ = (−1)³ = −1.

Q.9 [Polynomials]

(x + 2) is a factor of which one of the following?

(a) x⁵ − 4x⁴ − 3x³ + 8x² − 14x + 12 ✓
(b) x⁵ + 4x⁴ − 3x³ + 8x² − 14x + 12
(c) x⁵ − 4x⁴ + 3x³ + 8x² − 14x + 12
(d) x⁵ − 4x⁴ − 3x³ + 8x² + 14x + 12

Solution: Test x = −2 in (a): −32 − 64 + 24 + 32 + 28 + 12 = 0 ✓.
The other three change one sign and no longer vanish. (a).

Q.10 [Logarithms]

If log₁₀ 2 = 0·301 and log₁₀ 3 = 0·477, then what is the number of digits in the expansion of 60⁶⁰?

(a) 105
(b) 106
(c) 107 ✓
(d) 108

Solution: log 60⁶⁰ = 60(log 2 + log 3 + 1) = 60 × 1·778 = 106·68.
Digits = characteristic + 1 = 107.

Q.11 [Number System]

What is the remainder when (17²⁵ + 19²⁵) is divided by 18?

(a) 0 ✓
(b) 1
(c) 3
(d) 9

Solution: 17 ≡ −1 and 19 ≡ 1 (mod 18).
(−1)²⁵ + 1²⁵ = −1 + 1 = 0. (aⁿ + bⁿ is divisible by a + b for odd n.)

Q.12 [Number System]

The HCF of x and y is H. Consider the following statements in respect of the HCF of p = (x³ + y³)/(x² − xy + y²) and q = (x³ − y³)/(x² + xy + y²) :
I. The HCF of p and q can be H.
II. The HCF of p and q can be 2H.
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 + y and q = x − y.
HCF(x + y, x − y) is H when x/H, y/H have opposite parity (x = 2, y = 1 → 3 and 1 → 1) and 2H when both are odd (x = 3, y = 1 → 4 and 2 → 2). Both.

Q.13 [Surds]

If x⁴ = x² + 1, where x > 0, then what is 2x⁴ equal to?

(a) 2 + √3
(b) 3 + √5 ✓
(c) 5 − 2√3
(d) 3 − √5

Solution: Solve for x²: x² = (1 + √5)/2.
x⁴ = x² + 1 = (3 + √5)/2 → 2x⁴ = 3 + √5.

Q.14 [Algebra]

If (p + q)/(q + r) = (r + s)/(s + p); (q + r) ≠ 0, (s + p) ≠ 0, then which one of the following is correct?

(a) p + q + r + s = 0
(b) p = r
(c) Either p + q + r + s = 0 or p = r ✓
(d) None of the above

Solution: Cross-multiply: (p + q)(s + p) = (r + s)(q + r) → p² − r² + s(p − r) + q(p − r) = 0.
(p − r)(p + q + r + s) = 0 → either p = r or p + q + r + s = 0.

Q.15 [Number System]

If n is natural number less than 7, then what is the number of values of n for which (12n + 2) and (8n + 1) are relatively prime?

(a) 6 ✓
(b) 5
(c) 4
(d) 3

Solution: HCF(12n + 2, 8n + 1) = HCF(4n + 1, 8n + 1) = HCF(4n + 1, −1) = 1 for every n.
All of n = 1 to 6 work → 6.

Q.16 [Polynomials]

What is the HCF of x³ + y³ + 3xy − 1 and (x + y)⁴ − 1?

(a) x + y
(b) x + y + 1
(c) x + y − 1 ✓
(d) 1

Solution: x³ + y³ + (−1)³ − 3xy(−1) has the factor x + y − 1 (a³ + b³ + c³ − 3abc form).
(x + y)⁴ − 1 = (x + y − 1)(x + y + 1)((x + y)² + 1). HCF = x + y − 1.

Q.17 [Number System]

Let x = n(n + 1)(n + 2), where n is an even natural number. Which of the following statements is/are correct?
I. x is always divisible by 48.
II. x² is always divisible by 144.
Select the answer using the code given below.

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

Solution: n = 2 gives x = 24, not divisible by 48 → I false.
Three consecutive numbers with two evens → x is always divisible by 24, so x² by 576, hence by 144 → II true. II only.

Q.18 [Polynomials]

What is the LCM of x⁴ + x²y² + y⁴, x³y + y⁴ and x⁴y² − x³y³?

(a) x³y³(x⁶ − y⁶)
(b) x³y²(x⁶ − y⁶) ✓
(c) x³y(x⁶ − y⁶)
(d) xy(x⁶ − y⁶)

Solution: Factor: (x² + xy + y²)(x² − xy + y²); y(x + y)(x² − xy + y²); x³y²(x − y).
Product of all distinct factors = x³y²(x − y)(x + y)(x² + xy + y²)(x² − xy + y²) = x³y²(x⁶ − y⁶).

Q.19 [Number System]

Let XYZ be a 3-digit number. Let D be the difference between XYZ and ZYX. What is the remainder when D is divided by 99?

(a) 0 ✓
(b) 1
(c) 7
(d) 9

Solution: XYZ − ZYX = (100X + 10Y + Z) − (100Z + 10Y + X) = 99(X − Z).
Always a multiple of 99 → remainder 0.

Q.20 [Number System]

Let p and q be two natural numbers such that (p + q)^{p + q} is divisible by 512. What is the least value of (p + q)?

(a) 4
(b) 6
(c) 8 ✓
(d) 12

Solution: 512 = 2⁹. 4⁴ = 2⁸ falls short; 6⁶ has only 2⁶.
8⁸ = 2²⁴ works → least p + q = 8.

Q.21 [Algebra]

What is (a + b)²/[(c − a)(c + a + b)] + (a + b)c/(c² + bc − a² − ab) − (a + 2b + c)/[2(c − a)], a ≠ b, b ≠ c, c ≠ a, equal to?

(a) −1/2 ✓
(b) 0
(c) 1/2
(d) 1

Solution: c² + bc − a² − ab = (c − a)(c + a + b), so the first two terms add to (a + b)(a + b + c)/[(c − a)(c + a + b)] = (a + b)/(c − a).
(a + b)/(c − a) − (a + 2b + c)/[2(c − a)] = (a − c)/[2(c − a)] = −1/2.

Q.22 [Algebra]

If a^b = b^a, then what is a × (a/b)^a/b / a^(a/b) equal to?

(a) 1 ✓
(b) ab
(c) b
(d) a^b

Solution: The a^a/b cancels, leaving a / b^a/b.
From a^b = b^a, b^a/b = a. So the expression is a/a = 1.

Q.23 [Surds]

If x = 2 + 2^1/2 + 2^3/2, then what is x² − 4x − 10 equal to?

(a) 0
(b) 1
(c) 4 ✓
(d) 6

Solution: x = 2 + √2 + 2√2 = 2 + 3√2 → (x − 2)² = 18.
x² − 4x + 4 = 18 → x² − 4x − 10 = 4.

Q.24 [Algebra]

If (√(p + x) + √(p − x)) / (√(p + x) − √(p − x)) = p, then what is x equal to?

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

Solution: Componendo–dividendo: √(p + x)/√(p − x) = (p + 1)/(p − 1) → (p + x)/(p − x) = (p + 1)²/(p − 1)².
Again: p/x = (p² + 1)/(2p) → x = 2p²/(p² + 1).

Q.25 [Quadratic Equations]

If ((a − b)/2)x² − ((a + b)/2)x + b = 0, then what are the roots of this equation?

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

Solution: Coefficients sum to zero ((a − b) − (a + b) + 2b = 0), so x = 1 is a root.
Product of roots = 2b/(a − b) → other root 2b/(a − b).

Q.26 [Algebra]

If x − 1/x = 2, x > 0; then what is x² − 1/x² equal to?

(a) 6
(b) 4√2 ✓
(c) 4
(d) 2√2

Solution: (x + 1/x)² = (x − 1/x)² + 4 = 8 → x + 1/x = 2√2.
x² − 1/x² = (x − 1/x)(x + 1/x) = 2 × 2√2 = 4√2.

Q.27 [Algebra]

If (a − b)² + (b − c)² + (c − a)² = 6 and a² + b² + c² = 29, then what is (a + b + c) equal to?

(a) ±9 ✓
(b) ±8
(c) ±6
(d) ±3

Solution: Σ(a − b)² = 2(Σa² − Σab) = 6 → Σab = 29 − 3 = 26.
(a + b + c)² = 29 + 2 × 26 = 81 → ±9.

Q.28 [Surds]

If p = (√5 − 2)/(√5 + 2) and q = (√5 + 2)/(√5 − 2), then what is (p/q + q/p) equal to?

(a) 18
(b) 8√5
(c) 322 ✓
(d) 72√5

Solution: q = 1/p, so p/q + q/p = p² + 1/p².
p = (√5 − 2)² = 9 − 4√5, 1/p = 9 + 4√5 → p + 1/p = 18 → p² + 1/p² = 18² − 2 = 322.

Q.29 [Number System]

What is the digit at hundreds place of the number (25)¹⁰?

(a) 1
(b) 2
(c) 5
(d) 6 ✓

Solution: 25² = 625 and 625 × 25 = 15625: every further power of 25 ends in …625.
Hundreds digit = 6.

Q.30 [Number System]

A number N is such that when divided by 4, 6, 7 or 9, it leaves 3 as remainder. What is the smallest 4-digit number that satisfies this property?

(a) 1003
(b) 1005
(c) 1007
(d) 1011 ✓

Solution: N = LCM(4, 6, 7, 9)k + 3 = 252k + 3.
k = 4 gives the first 4-digit value: 1008 + 3 = 1011.

Q.31 [Trigonometry]

If √(2 + √(2 + √(2 + √(2 + …)))) = cosec θ, then what is sin θ equal to?

(a) 1
(b) √3/2
(c) 1/√2
(d) 1/2 ✓

Solution: Let y be the nest: y = √(2 + y) → y² − y − 2 = 0 → y = 2.
cosec θ = 2 → sin θ = 1/2.

Q.32 [Trigonometry]

If 8 sin θ − cos θ = 4, where 0 < θ < π/2, then what is cosec θ equal to?

(a) 1
(b) 3/2
(c) 5/3 ✓
(d) 2

Solution: Try the 3-4-5 angle: sin θ = 3/5, cos θ = 4/5 → 24/5 − 4/5 = 4 ✓.
cosec θ = 5/3.

Q.33 [Trigonometry]

If 2 tan θ = sec² θ − 2, where 0 < θ < π/2, then what is cot θ equal to?

(a) √2 − 1 ✓
(b) √2 + 1
(c) √3 − 1
(d) √3 + 2

Solution: sec² θ − 2 = tan² θ − 1 → tan² θ − 2 tan θ − 1 = 0 → tan θ = 1 + √2 (positive root).
cot θ = 1/(1 + √2) = √2 − 1.

Q.34 [Trigonometry]

What is (sec θ − tan θ) − √((1 − sin θ)/(1 + sin θ)) equal to?

(a) 0 ✓
(b) 2 tan θ
(c) 2 sec θ
(d) sin θ + cos θ

Solution: √((1 − sin θ)/(1 + sin θ)) = (1 − sin θ)/cos θ = sec θ − tan θ.
The two terms are identical → 0.

Q.35 [Trigonometry]

If cot θ = √7, then what is (cosec² θ − sec² θ)/(cosec² θ + sec² θ) equal to?

(a) 1/2
(b) 1/3
(c) 2/3
(d) 3/4 ✓

Solution: cosec² θ = 1 + cot² θ = 8, sec² θ = 1 + tan² θ = 1 + 1/7 = 8/7.
(8 − 8/7)/(8 + 8/7) = (6/7)/(8/7) = 3/4.

Q.36 [Trigonometry]

The difference between the two acute angles in a right-angled triangle is π/12 radian. One of the acute angles of the triangle is

(a) 60°
(b) 57·5°
(c) 52·5° ✓
(d) 47·5°

Solution: π/12 = 15°. The acute angles add to 90° and differ by 15°.
Angles are 52·5° and 37·5° → 52·5°.

Q.37 [Logarithms]

If α and β are the roots of the equation log₁₀[998 + √(x² − 18x + 76)] = 3, then what is (α − β)² equal to?

(a) 16
(b) 25
(c) 36 ✓
(d) 49

Solution: 998 + √(…) = 1000 → x² − 18x + 76 = 4 → x² − 18x + 72 = 0 → roots 6 and 12.
(α − β)² = 6² = 36.

Q.38 [Logarithms]

If x⁴ + y⁴ = 14x²y², then consider the following :
I. log₁₀(x² + y²) = log₁₀ x + log₁₀ y + 2 log₁₀ 2
II. log₁₀(x² − y²) = log₁₀ x + log₁₀ y + log₁₀ 2 + 0·5 log₁₀ 3
Which of the above is/are correct?

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

Solution: Add 2x²y²: (x² + y²)² = 16x²y² → x² + y² = 4xy → I ✓.
Subtract: (x² − y²)² = 12x²y² → x² − y² = 2√3·xy → II ✓. Both.

Q.39 [Polynomials]

Which of the following is/are the factor(s) of (3x + y)² + (3x + y)(x + 5y) − 20(x + 5y)²?
I. (4x + 13y)
II. (x + 19y)
Select the correct answer using the code given below.

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

Solution: With A = 3x + y, B = x + 5y: A² + AB − 20B² = (A + 5B)(A − 4B).
A + 5B = 8x + 26y = 2(4x + 13y); A − 4B = −(x + 19y). Both are factors.

Q.40 [Algebra]

What is [x/(x − y) + y/(y − z) + z/(z − x)] / [(x + y)/(x − y) + (y + z)/(y − z) + (z + x)/(z − x) + 3] equal to?

(a) 1
(b) 1/2 ✓
(c) 1/3
(d) 1/4

Solution: (x + y)/(x − y) = 2x/(x − y) − 1, and likewise for the other two.
Denominator = 2(numerator) − 3 + 3 = 2 × numerator → ratio 1/2.

Q.41 [Trigonometry]

For the following two (02) items : Let p sin² α + q cos² α = m, q sin² β + p cos² β = n; p ≠ m, n and q ≠ m, n.
What is (tan α / tan β)² equal to?

(a) −(m − q)(n − q) / [(m − p)(n − p)]
(b) −(m − q)(n − p) / [(m − p)(n − q)]
(c) (m − q)(n − q) / [(m − p)(n − p)] ✓
(d) (m − q)(n − p) / [(m − p)(n − q)]

Solution: (p − q) sin² α = m − q → tan² α = (m − q)/(p − m).
(q − p) sin² β = n − p → tan² β = (n − p)/(q − n). Divide: (m − q)(n − q) / [(m − p)(n − p)].

Q.42 [Trigonometry]

If α and β are complementary angles, then which one of the following is correct?

(a) mn − 1 = 0
(b) mn + 1 = 0
(c) m + n = 0
(d) m − n = 0 ✓

Solution: β = 90° − α swaps sin and cos: n = q cos² α + p sin² α = m.
So m − n = 0.

Q.43 [Trigonometry]

For the following two (02) items : Let cosec θ − sin θ = p and sec θ − cos θ = q.
What is (p sin θ + q cos θ) equal to?

(a) −1
(b) 0
(c) 1 ✓
(d) 2

Solution: p = (1 − sin² θ)/sin θ = cos² θ/sin θ and q = sin² θ/cos θ.
p sin θ + q cos θ = cos² θ + sin² θ = 1.

Q.44 [Trigonometry]

What is p²q²(p² + q² + 3) equal to?

(a) 0
(b) 1 ✓
(c) 2
(d) 4

Solution: p²q² = sin² θ cos² θ and p² + q² = (cos⁶ θ + sin⁶ θ)/(sin² θ cos² θ).
Product = cos⁶ θ + sin⁶ θ + 3 sin² θ cos² θ = (sin² θ + cos² θ)³ = 1.

Q.45 [Trigonometry]

For the following two (02) items : Let sin α / sin β = 4√2/3 and cos α / cos β = 2√3/9.
What is tan² α equal to?

(a) 8 ✓
(b) 6
(c) 4
(d) 3

Solution: sin² α + cos² α = 1 → (32/9) sin² β + (12/81)(1 − sin² β) = 1 → sin² β = 1/4.
Then sin² α = 8/9, cos² α = 1/9 → tan² α = 8.

Q.46 [Trigonometry]

What is tan² β equal to?

(a) 1/√2
(b) 3/√2
(c) 1/3 ✓
(d) 2/3

Solution: From the previous item sin² β = 1/4, cos² β = 3/4.
tan² β = 1/3.

Q.47 [Trigonometry]

For the following two (02) items : Let (1 + sin θ)/cos θ = p + √(p² + 1).
What is sec θ equal to?

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

Solution: (1 + sin θ)/cos θ = sec θ + tan θ, and its reciprocal sec θ − tan θ = √(p² + 1) − p.
Average the two: sec θ = √(p² + 1).

Q.48 [Trigonometry]

What is tan θ equal to?

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

Solution: Half the difference of sec θ + tan θ and sec θ − tan θ: tan θ = p.

Q.49 [Trigonometry]

For the following two (02) items : Let sin θ + cos θ = p and sec θ + cosec θ = q, where p ≠ 1.
What is the relation between p and q?

(a) p = q(p² − 1)
(b) 2p = q(p² − 1) ✓
(c) q = p² − 1
(d) 2q = p(p² − 1)

Solution: p² = 1 + 2 sin θ cos θ → sin θ cos θ = (p² − 1)/2.
q = (sin θ + cos θ)/(sin θ cos θ) = 2p/(p² − 1) → 2p = q(p² − 1).

Q.50 [Trigonometry]

What is tan θ + cot θ equal to?

(a) p/q
(b) q/p ✓
(c) 2p/q
(d) 2q/p

Solution: tan θ + cot θ = 1/(sin θ cos θ) = 2/(p² − 1).
From the previous item 2/(p² − 1) = q/p → q/p.

Q.51 [Time, Speed & Distance]

Two trains X and Y are travelling in the same direction at 100 km/hr and 60 km/hr respectively. Train X crosses a man in train Y in 9 seconds. What is the length of train X?

(a) 80 m
(b) 100 m ✓
(c) 120 m
(d) 150 m

Solution: Relative speed = 40 km/h = 100/9 m/s.
Length = 100/9 × 9 = 100 m.

Q.52 [Time, Speed & Distance]

Two persons X and Y leave place P for place Q at 7:00 a.m. and 7:10 a.m. respectively along the same path. X walks at a speed of 4·8 km/hr and Y walks at a speed of 6 km/hr. How many kilometres from place P will X meet Y?

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

Solution: In 10 min X gains 0·8 km; Y closes it at 1·2 km/h → 2/3 h.
Y's distance = 6 × 2/3 = 4 km.

Q.53 [Percentages]

There are two employees X and Y. X's salary is first increased by 12% and then decreased by 10%, and Y's salary is first increased by 10% and then decreased by 12%. If their salaries at present are equal, then what was the ratio of initial salary of X to initial salary of Y?

(a) 50 : 53
(b) 51 : 53
(c) 121 : 126 ✓
(d) 121 : 125

Solution: X ends at 1·12 × 0·9 = 1·008 of its start; Y at 1·1 × 0·88 = 0·968.
X₀/Y₀ = 0·968/1·008 = 968/1008 = 121 : 126.

Q.54 [Percentages]

In a village consisting of p persons, x% can read and write. Of the males, only y% can read and write. Of the females, only z% can read and write. If x, y > z, then what is the number of males in the village?

(a) p(x − z)/(y − z) ✓
(b) p(y − z)/(x − z)
(c) px/y
(d) py/x

Solution: With M males: yM + z(p − M) = xp → M(y − z) = p(x − z).
M = p(x − z)/(y − z). (Alligation: males : females = (x − z) : (y − x).)

Q.55 [Alligation]

X and Y are two alloys of copper (Cu) and zinc (Zn). Alloy X is prepared by mixing Cu and Zn in the ratio 5 : 4, and alloy Y is prepared by mixing Cu and Zn in the ratio 5 : 13 respectively. If equal quantities of alloys X and Y are melted to form a third alloy Z, then what is the ratio of Cu to Zn in Z?

(a) 5 : 8
(b) 5 : 7 ✓
(c) 6 : 7
(d) 7 : 8

Solution: Take 18 units of each: X has Cu 10, Zn 8; Y has Cu 5, Zn 13.
Cu : Zn = 15 : 21 = 5 : 7.

Q.56 [Ratio & Proportion]

Two students X and Y appeared in a test. The score of X is 20 more than that of Y. If the score of X is 75% of the sum of the scores of X and Y, then what is the ratio of score of X to score of Y?

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

Solution: X = ¾(X + Y) → X = 3Y.
The 20-mark gap is extra information; the ratio is 3 : 1 (Y = 10, X = 30).

Q.57 [Quadratic Equations]

If one root of the equation 2x² − 5px + 2p² = 0 exceeds the other by 4, then what is the value of p?

(a) 8/3 ✓
(b) 4/3
(c) 2/3
(d) 1/3

Solution: (α − β)² = (α + β)² − 4αβ = 25p²/4 − 4p² = 9p²/4.
9p²/4 = 16 → p = 8/3.

Q.58 [Compound Interest]

An amount of ₹10,000 is borrowed at 10% per annum on compound interest for 3 years, compounded annually, and paid back in 3 equal annual installments during these years. What is the amount of each installment (approximately)?

(a) ₹4,437
(b) ₹4,237
(c) ₹4,021 ✓
(d) ₹3,811

Solution: Present value of three installments x: x(1/1·1 + 1/1·21 + 1/1·331) = 10000.
x × 3·31/1·331 = 10000 → x = 13310/3·31 ≈ ₹4,021.

Q.59 [Inequalities]

What is the solution of the inequalities 5x + 3 < 8x − 9 and 2x + 20 > 5x + 2?

(a) 4 < x < 6 ✓
(b) 3 < x < 5
(c) x < 3 or x > 5
(d) x < 4 or x > 6

Solution: First: 12 < 3x → x > 4. Second: 18 > 3x → x < 6.
Together: 4 < x < 6.

Q.60 [Profit & Loss]

A shopkeeper gives three consecutive discounts 10%, 20% and 25% after which he sells the article at a profit of 8% on the cost price. Had he sold the article after the first discount, how much profit would he have got?

(a) 20%
(b) 40%
(c) 50%
(d) None of the above ✓

Solution: 0·9 × 0·8 × 0·75 = 0·54 of marked price = 1·08 CP → MP = 2 CP.
After only the first discount: 0·9 × 2 CP = 1·8 CP → 80% profit → None of the above.

Q.61 [Geometry — Triangles]

Let AD be the altitude of a triangle ABC. If (AB + AC) = p, (AB − AC) = q and (BD − CD) = r, then what is BC equal to?

(a) qr/p
(b) pr/q
(c) pq/r ✓
(d) p + q − r

Solution: AB² − AC² = BD² − CD² (both equal AD² differences).
pq = (BD + CD)(BD − CD) = BC × r → BC = pq/r.

Q.62 [Mensuration — Solids]

The sum of the height and the radius of a right circular cylinder is 21 cm, and the radius is less than the height. If the curved surface area of the cylinder is 616 cm², then what is the volume of the cylinder? (Take π = 22/7)

(a) 1078 cm³
(b) 1617 cm³
(c) 1927 cm³
(d) 2156 cm³ ✓

Solution: 2πrh = 616 → rh = 98; r + h = 21 → r = 7, h = 14.
V = πr²h = 22/7 × 49 × 14 = 2156 cm³.

Q.63 [Data Sufficiency]

ABC is a triangle right angled at B. P is the midpoint of AB and Q is the midpoint of BC. Consider the following :
I. AQ = √73 units
II. CP = √52 units
Which of the above is/are required to determine the area of the triangle?

(a) I only
(b) II only
(c) Both I and II ✓
(d) More information is needed

Solution: With AB = a, BC = b: I gives a² + b²/4 = 73, II gives b² + a²/4 = 52 — one equation, two unknowns each.
Both together: a = 8, b = 6, area 24. Both I and II.

Q.64 [Mensuration — Solids]

The length, breadth and height of a cuboid are consecutive integers. If the volume of the cuboid is 336 cubic units, then what is the total surface area of the cuboid?

(a) 288 square units
(b) 292 square units ✓
(c) 296 square units
(d) Cannot be determined due to insufficient data

Solution: 6 × 7 × 8 = 336.
TSA = 2(42 + 56 + 48) = 292 square units.

Q.65 [Geometry — Circles]

In a circle of radius 14 cm, APB is a shorter arc and P is the midpoint of the arc. Let C be the midpoint of the chord AB and PC = 7 cm. What is the length of the chord AP?

(a) 3·5 cm
(b) 7 cm
(c) 10·5 cm
(d) 14 cm ✓

Solution: O, C, P are collinear: OC = 14 − 7 = 7 → AC² = 14² − 7² = 147.
AP² = AC² + PC² = 147 + 49 = 196 → AP = 14 cm.

Q.66 [Geometry — Similarity]

Two poles of heights 10 m and 15 m are 25 m apart. What is the height of the point of intersection of the lines joining the tip of each pole to the foot of the other pole?

(a) 4·8 m
(b) 5 m
(c) 6 m ✓
(d) 6·4 m

Solution: Crossed-ladders result: h = ab/(a + b), independent of the gap.
h = 10 × 15/25 = 6 m.

Q.67 [Geometry — Triangles]

ABC is a triangle right angled at B. Further, (AB + BC) exceeds AC by 10 units. If the perimeter of the triangle is 60 units, then what is the area of the triangle?

(a) 75 square units
(b) 100 square units
(c) 125 square units
(d) 150 square units ✓

Solution: a + b = c + 10 and a + b + c = 60 → c = 25, a + b = 35.
2ab = 35² − 25² = 600 → area = ab/2 = 150 square units.

Q.68 [Geometry — Triangles]

Two poles are situated 24 m apart and their heights differ by 10 m. What is the distance between their tips?

(a) 25 m
(b) 26 m ✓
(c) 30 m
(d) Cannot be determined due to insufficient data

Solution: Right triangle with legs 24 and 10.
√(576 + 100) = 26 m.

Q.69 [Geometry — Triangles]

Let X, Y and Z be the midpoints of the sides BC, CA and AB of a triangle ABC respectively. Consider the following statements :
I. The quadrilateral AZXY is a parallelogram.
II. The area of the quadrilateral AZXY is half of the area of the triangle ABC.
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: Midpoint theorem: ZX ∥ AC and XY ∥ AB → AZXY is a parallelogram.
The medial triangle splits ABC into four equal triangles; AZXY holds two of them → half the area. Both.

Q.70 [Geometry — Polygons]

Consider the following angles :
I. 4°
II. 5°
III. 6°
IV. 8°
How many of the above can be the exterior angle of a regular polygon?

(a) One
(b) Two
(c) Three
(d) All four ✓

Solution: Exterior angle = 360°/n, so any divisor of 360 works.
360/4 = 90, 360/5 = 72, 360/6 = 60, 360/8 = 45 sides → All four.

Q.71 [Geometry — Triangles]

In a triangle ABC, ∠A = 30°, AB = 7 cm and AC = 12 cm. What is the area of the triangle ABC?

(a) 21 cm² ✓
(b) 21√3 cm²
(c) 42 cm²
(d) 42√3 cm²

Solution: Area = ½ × AB × AC × sin A = ½ × 7 × 12 × ½ = 21 cm².

Q.72 [Geometry — Triangles]

ABC is a triangle right angled at B. D is a point on AC such that BD is perpendicular to AC. If AB = p and BC = √3p, then what is BD equal to?

(a) p/3
(b) p/2
(c) √3p/2 ✓
(d) √3p/4

Solution: AC = √(p² + 3p²) = 2p.
Altitude to hypotenuse = AB × BC / AC = p × √3p / 2p = √3p/2.

Q.73 [Geometry — Polygons]

The difference between an interior angle and an exterior angle of a regular polygon is 120°. What is the number of sides of the polygon?

(a) 9
(b) 10
(c) 11
(d) 12 ✓

Solution: Interior + exterior = 180°, interior − exterior = 120° → exterior = 30°.
n = 360/30 = 12.

Q.74 [Geometry — Lines & Angles]

An angle θ is exactly one-fourth of its complementary angle. What is the value of angle θ?

(a) 12°
(b) 15°
(c) 18° ✓
(d) 20°

Solution: θ = (90° − θ)/4 → 5θ = 90°.
θ = 18°.

Q.75 [Geometry — Triangles]

The sides of a triangle are 11 cm, 60 cm and 61 cm. What is the area of the triangle formed by joining the midpoints of the sides of the triangle?

(a) 165 cm²
(b) 110 cm²
(c) 82·5 cm² ✓
(d) 72·5 cm²

Solution: 11² + 60² = 61², so it is right-angled: area = ½ × 11 × 60 = 330.
The medial triangle has one-fourth the area → 82·5 cm².

Q.76 [Mensuration — Circles]

A pendulum swings through an angle of 30° and its end describes an arc of length 55 cm. What is the length of the pendulum? (Take π = 22/7)

(a) 90 cm
(b) 100 cm
(c) 105 cm ✓
(d) 110 cm

Solution: Arc = rθ with θ = 30° = π/6 = 11/21 rad.
r = 55 × 21/11 = 105 cm.

Q.77 [Mensuration — Solids]

A conical tent has an angle of 60° at the vertex. If the curved surface area is 100 m², then what is the volume of the tent?

(a) 250√2/√(3π) m³ ✓
(b) 500√3/√π m³
(c) 1000√3/√(2π) m³
(d) 1000√3/√π m³

Solution: Semi-vertical angle 30°: r = l/2, h = √3l/2. CSA = πrl = πl²/2 = 100 → l² = 200/π.
V = ⅓πr²h = (π√3/24) l³ = (π√3/24)(200/π)^3/2 = 250√2/√(3π) m³.

Q.78 [Mensuration — Solids]

A right circular cone and a hemisphere have equal base and equal volume. What is the ratio of the height of the cone to the radius of the hemisphere?

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

Solution: ⅓πr²h = ⅔πr³ → h = 2r.
h : r = 2 : 1.

Q.79 [Mensuration — Areas]

A wire is in the form of an equilateral triangle with an area of 36√3 cm². If it is changed into a semicircle, then what is its radius?

(a) 9/π cm
(b) 18/(π + 2) cm
(c) 18/π cm
(d) None of the above ✓

Solution: (√3/4)a² = 36√3 → a = 12 → wire length 36 cm.
Semicircle perimeter πr + 2r = 36 → r = 36/(π + 2), not listed → None of the above.

Q.80 [Mensuration — Areas]

Let the area of the largest possible square inscribed in a circle of unit radius be x. Let the area of the largest possible circle inscribed in a square of unit side length be y. What is the relation between x and y?

(a) πx = 2y
(b) 2πx = y
(c) πx = 4y
(d) πx = 8y ✓

Solution: Square in unit circle: diagonal 2 → area x = 2. Circle in unit square: radius ½ → y = π/4.
πx = 2π = 8y → πx = 8y.

Q.81 [Data Sufficiency]

A Question is given followed by two Statements I and II. Consider the Question and the Statements.
Question : What is the remainder when x²ⁿ − y²ⁿ + 1 is divided by xⁿ + yⁿ, where n is a natural number?
Statement-I : n is odd.
Statement-II : n is even.
Which one of the following is correct in respect of the above Question and the Statements?

(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 any of the Statements ✓

Solution: x²ⁿ − y²ⁿ = (xⁿ − yⁿ)(xⁿ + yⁿ) is divisible for every n, so the remainder is 1 regardless of parity.
Answerable without either Statement.

Q.82 [Data Sufficiency]

A Question is given followed by two Statements I and II. Consider the Question and the Statements.
Question : The product of a natural number N and the number M written by the same digits of N in the reverse order is 252. What is the number N?
Statement-I : N + M = 33
Statement-II : N > M
Which one of the following is correct in respect of the above Question and the Statements?

(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 any of the Statements

Solution: Only the reversal pair 12 × 21 gives 252, so N is 12 or 21.
I (sum 33) fits both; II (N > M) pins N = 21. II alone, not I.

Q.83 [Data Sufficiency]

A Question is given followed by two Statements I and II. Consider the Question and the Statements.
Question : The last digit in the expansion of the number (54D)¹⁰⁰ is 1. What is the value of the digit D?
Statement-I : D > 5
Statement-II : D is a multiple of 3.
Which one of the following is correct in respect of the above Question and the Statements?

(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 any of the Statements

Solution: D¹⁰⁰ ends in 1 only for D = 1, 3, 7, 9.
I leaves 7 or 9; II leaves 3 or 9; together D = 9. Both needed.

Q.84 [Data Sufficiency]

A Question is given followed by two Statements I and II. Consider the Question and the Statements.
Question : In a triangle ABC, ∠A = ∠B − ∠C. Is angle A acute?
Statement-I : ABC is not an obtuse-angled triangle.
Statement-II : Angle C is acute.
Which one of the following is correct in respect of the above Question and the Statements?

(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 any of the Statements ✓

Solution: ∠A + ∠C = ∠B and the three sum to 180° → ∠B = 90°.
So ∠A = 90° − ∠C is always acute. Answerable without the Statements.

Q.85 [Data Sufficiency]

A Question is given followed by two Statements I and II. Consider the Question and the Statements.
Question : In a triangle ABC right angled at B, AC = 20 cm. What is the circumradius of the triangle?
Statement-I : AB = 12 cm
Statement-II : BC = 16 cm
Which one of the following is correct in respect of the above Question and the Statements?

(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 any of the Statements ✓

Solution: The hypotenuse of a right triangle is a diameter of its circumcircle.
Circumradius = AC/2 = 10 cm. No Statement needed.

Q.86 [Data Sufficiency]

A Question is given followed by two Statements I and II. Consider the Question and the Statements.
Question : ABCD is a parallelogram with ∠ABC = 60°. If the area of the parallelogram is 7√3 square units, then what is the perimeter of the parallelogram?
Statement-I : The lengths of the sides AB and DA are prime numbers.
Statement-II : The lengths of the sides are natural numbers each greater than 1 unit.
Which one of the following is correct in respect of the above Question and the Statements?

(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 any of the Statements

Solution: Area = AB × DA × sin 60° → AB × DA = 14.
Primes: 2 × 7; naturals above 1: again only 2 × 7. Each gives perimeter 18. Either alone.

Q.87 [Data Sufficiency]

A Question is given followed by two Statements I and II. Consider the Question and the Statements.
Question : AB and CD are chords of a circle intersecting at P. If AP × PB = 48 square units, then what is CP × PD equal to?
Statement-I : AP = 8 units
Statement-II : CP = 10 units
Which one of the following is correct in respect of the above Question and the Statements?

(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 any of the Statements ✓

Solution: Intersecting chords: CP × PD = AP × PB = 48.
No Statement needed.

Q.88 [Data Sufficiency]

A Question is given followed by two Statements I and II. Consider the Question and the Statements.
Question : In a quadrilateral ABCD, AB = 6 units, BC = 18 units, CD = 6 units, DA = 9 units. What is the length of diagonal BD?
Statement-I : The length of BD is an integer greater than 13.
Statement-II : The length of BD is an even integer.
Which one of the following is correct in respect of the above Question and the Statements?

(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 any of the Statements

Solution: Triangle inequalities: ABD gives 3 < BD < 15, BCD gives 12 < BD < 24 → 12 < BD < 15.
Integer above 13 → 14; even integer → 14. Either alone.

Q.89 [Data Sufficiency]

A Question is given followed by two Statements I and II. Consider the Question and the Statements.
Question : ABC is an isosceles triangle with AB = AC = 10 units. If the area of the triangle is 48 square units, then what is the length of the base BC?
Statement-I : The length of BC is an even integer.
Statement-II : The height of the triangle is greater than the length of half of the base.
Which one of the following is correct in respect of the above Question and the Statements?

(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 any of the Statements

Solution: Half-base m and height h satisfy mh = 48, m² + h² = 100 → (m, h) = (6, 8) or (8, 6), i.e. BC = 12 or 16.
I (even) fits both; II (h > m) picks BC = 12. II alone, not I.

Q.90 [Data Sufficiency]

A Question is given followed by two Statements I and II. Consider the Question and the Statements.
Question : The diagonals of a rhombus ABCD are in the ratio 5 : 12. Is one of the diagonals equal to side of the rhombus?
Statement-I : The sum of the diagonals = 34 cm.
Statement-II : The length of a side = 13 cm.
Which one of the following is correct in respect of the above Question and the Statements?

(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 any of the Statements ✓

Solution: Diagonals 5k and 12k → side = ½√(25 + 144)k = 6·5k, which equals neither diagonal.
The answer is 'No' from the ratio alone. No Statement needed.

Q.91 [Statistics]

Which measure of central tendency is least affected by the presence of extreme observations in the data?

(a) Arithmetic mean
(b) Harmonic mean
(c) Geometric mean
(d) Median ✓

Solution: The median depends only on the middle position, so outliers cannot pull it; all three means use every value.

Q.92 [Statistics]

To find the average ratio like price/unit, work done/hour, kilometre/hour under certain conditions, the suitable measure of central tendency applicable is

(a) arithmetic mean
(b) geometric mean
(c) harmonic mean ✓
(d) mode

Solution: Averaging rates over equal numerators (equal distances, equal spends) calls for the harmonic mean.

Q.93 [Statistics]

The frequency distribution of marks of 100 candidates in a particular examination is as follows :

Marks	Number of Candidates
More than 10	100
More than 20	75
More than 30	60
More than 40	40

What are the average marks of the candidates?

(a) 20·5
(b) 22·5
(c) 30·5
(d) 32·5 ✓

Solution: Class frequencies 10–20: 25, 20–30: 15, 30–40: 20, 40–50: 40.
Mean = (15×25 + 25×15 + 35×20 + 45×40)/100 = 3250/100 = 32·5.

Q.94 [Statistics]

The arithmetic mean of 200 observations is 60. If 5 is multiplied to each observation, then what will be the new arithmetic mean?

(a) 500
(b) 300 ✓
(c) 60
(d) 40

Solution: Scaling every value by 5 scales the mean by 5: 60 × 5 = 300.

Q.95 [Statistics]

A distribution consists of 3 components with frequencies 45, 40 and 55 having their means 2, 2·5 and 2 respectively. What is the mean of the combined distribution?

(a) 2·14 ✓
(b) 2·25
(c) 2·37
(d) 2·50

Solution: Combined mean = (45×2 + 40×2·5 + 55×2)/140 = 300/140 ≈ 2·14.

Q.96 [Statistics]

Which one of the following is a positional average?

(a) Arithmetic mean
(b) Median ✓
(c) Mode
(d) Geometric mean

Solution: The median is fixed by position in the ordered data; the three means are computed from the values.

Q.97 [Statistics]

For the following two (02) items : The following data represent the distance covered (in metres) by two groups of athletic children. It is known that the median distance in the first group is 20·8 metres while the mean distance in the second group is 17·3 metres. Some frequencies in both the groups are missing :

Distance Class	First Group	Second Group
0–5	u	3u
5–10	v	2v
10–15	11	40
15–20	52	50
20–25	75	30
25–30	22	28

What is the value of u?

(a) 1
(b) 2
(c) 3 ✓
(d) 4

Solution: First group, median class 20–25: 20 + 5[(160 + u + v)/2 − (u + v + 63)]/75 = 20·8 → u + v = 10.
Second group mean 17·3 with v = 10 − u gives 24·8u ≈ 63·6, u ≈ 2·6 → the option pair that fits both tables is u = 3, v = 7.

Q.98 [Statistics]

What is the value of v?

(a) 5
(b) 6
(c) 7 ✓
(d) 8

Solution: From the median condition u + v = 10 with u = 3 → v = 7.

Q.99 [Statistics]

For the following two (02) items : Consider the following distribution having median value 24 :

Marks	Number of Students
Less than 10	5
Less than 20	30
Less than 30	30 + k
Less than 40	48 + k
Less than 50	55 + k

What is the value of k?

(a) 20
(b) 22
(c) 25 ✓
(d) 30

Solution: Frequencies: 5, 25, k, 18, 7; N = 55 + k; median class 20–30 (cf before it = 30).
20 + 10[(55 + k)/2 − 30]/k = 24 → 0·1k = 2·5 → k = 25.

Q.100 [Statistics]

What is the mean of the distribution?

(a) 21·625
(b) 22·225
(c) 23·225
(d) 24·625 ✓

Solution: With k = 25, N = 80: Σfx = 5×5 + 15×25 + 25×25 + 35×18 + 45×7 = 1970.
Mean = 1970/80 = 24·625.