CDS (Defence) · Previous Year Questions
Mathematics
100 CDS Mathematics questions from past papers (2025–2025), each with the answer and a full solution. Arithmetic, algebra, geometry, trigonometry, mensuration and statistics.
CDS 2025100 questions
- Q1
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)?
- A3
- B−3
- C3 − x
- D3 − 2x
Show answer & solution
Correct answer: (C) 3 − x
By the Remainder Theorem, p(1) = 2 and p(2) = 1. Dividing by a quadratic leaves a remainder of degree at most 1, so write it as ax + b. Then a + b = 2 and 2a + b = 1, which gives a = −1 and b = 3. The remainder is 3 − x, option (c).
- Q2
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?
- A3
- B4
- C5
- D6
Show answer & solution
Correct answer: (A) 3
q² is smallest when q = p + 1, so we need (p + 1)² < 5p + 4 for at least one valid q. This gives p² − 3p − 3 < 0, so p < (3 + √21)/2 ≈ 3.79. Check p = 3, q = 4: 16 − 15 − 4 = −3 < 0, which works. Check p = 4, q = 5: 25 − 20 − 4 = 1 > 0, which fails, and a larger q only makes it bigger. The largest p is 3, option (a).
- Q3
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?
- AI and II only
- BII and III only
- CI and III only
- DI, II and III
Show answer & solution
Correct answer: (D) I, II and III
For x > 0, AM ≥ GM gives x + 1/x ≥ 2. I: x + 1/x ≥ 2 > 1, so it is true. II: (x + 1/x)² ≥ 4 > 2, so it is true. III: (x + 1/x)⁴ ≥ 16 > 9, so it is true. All three hold: I, II and III, option (d).
- Q4
If log₁₀ 2 = 0.301 and log₁₀ 3 = 0.477, then what is the number of digits in the expansion of 60⁶⁰?
- A105
- B106
- C107
- D108
Show answer & solution
Correct answer: (C) 107
log 60 = log 2 + log 3 + log 10 = 0.301 + 0.477 + 1 = 1.778. log 60⁶⁰ = 60 × 1.778 = 106.68. The number of digits is the integer part plus 1, which is 106 + 1 = 107, option (c).
- Q5
(x + 2) is a factor of which one of the following?
- Ax⁵ − 4x⁴ − 3x³ + 8x² − 14x + 12
- Bx⁵ + 4x⁴ − 3x³ + 8x² − 14x + 12
- Cx⁵ − 4x⁴ + 3x³ + 8x² − 14x + 12
- Dx⁵ − 4x⁴ − 3x³ + 8x² + 14x + 12
Show answer & solution
Correct answer: (A) x⁵ − 4x⁴ − 3x³ + 8x² − 14x + 12
By the Factor Theorem, (x + 2) is a factor exactly when the polynomial equals 0 at x = −2. For option (a): (−32) − 4(16) − 3(−8) + 8(4) − 14(−2) + 12 = −32 − 64 + 24 + 32 + 28 + 12 = 0. So (x + 2) divides x⁵ − 4x⁴ − 3x³ + 8x² − 14x + 12, option (a).
- Q6
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?
- AOne
- BTwo
- CThree
- DNone
Show answer & solution
Correct answer: (A) One
If x and y were both odd primes, x + y would be even and greater than 2, so it could not be prime. One of them must therefore be 2, and since x − y > 0, y = 2. Now x − 2, x and x + 2 must all be prime. Among any three numbers spaced 2 apart, one is divisible by 3, so that one must equal 3 itself. This forces x − 2 = 3, so x = 5. The only set is (5, 2, 7, 3). That is one combination, option (a).
- Q7
If (x + 1)(x + p)(x² + p²) = x⁴ − 1, then what is the value of p?
- A−1
- B0
- C1
- DCannot be determined
Show answer & solution
Correct answer: (A) −1
Factorise x⁴ − 1 = (x² − 1)(x² + 1) = (x + 1)(x − 1)(x² + 1). Compare with (x + 1)(x + p)(x² + p²). Taking p = −1 gives (x + 1)(x − 1)(x² + 1), which matches exactly, since p² = 1. So p = −1, option (a).
- Q8
If (2 + √3)^x + (2 − √3)^x = 2, then what is (2 + √3)^x − (2 − √3)^x equal to?
- A0
- B0.5
- C1
- D1.5
Show answer & solution
Correct answer: (A) 0
Since (2 + √3)(2 − √3) = 1, let t = (2 + √3)^x. Then (2 − √3)^x = 1/t. The condition becomes t + 1/t = 2, so (t − 1)² = 0 and t = 1, which means x = 0. The required value is t − 1/t = 1 − 1 = 0, option (a).
- Q9
What is the remainder when x⁶ is divided by x² + 1?
- A−1
- B0
- C1
- Dx + 1
Show answer & solution
Correct answer: (A) −1
When dividing by x² + 1, replace x² with −1. Then x⁶ = (x²)³ = (−1)³ = −1. The remainder is −1, option (a).
- Q10
If 1/a + 1/b = 5/6 and 1/a² + 1/b² = 13/36, then what is 1/a³ + 1/b³ equal to?
- A31/216
- B35/216
- C37/216
- D41/216
Show answer & solution
Correct answer: (B) 35/216
Let u = 1/a and v = 1/b, so u + v = 5/6 and u² + v² = 13/36. Then 2uv = (u + v)² − (u² + v²) = 25/36 − 13/36 = 12/36, so uv = 1/6. u³ + v³ = (u + v)³ − 3uv(u + v) = 125/216 − 3·(1/6)·(5/6) = 125/216 − 90/216 = 35/216, option (b).
- Q11
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?
- A0
- B1
- C7
- D9
Show answer & solution
Correct answer: (A) 0
XYZ − ZYX = (100X + 10Y + Z) − (100Z + 10Y + X) = 99(X − Z). D is a multiple of 99, so the remainder is 0, option (a).
- Q12
What is the HCF of x³ + y³ + 3xy − 1 and (x + y)⁴ − 1?
- Ax + y
- Bx + y + 1
- Cx + y − 1
- D1
Show answer & solution
Correct answer: (C) x + y − 1
Write x³ + y³ + 3xy − 1 as x³ + y³ + (−1)³ − 3·x·y·(−1). By the identity a³ + b³ + c³ − 3abc = (a + b + c)(…), it has the factor x + y − 1. Also (x + y)⁴ − 1 = (x + y − 1)(x + y + 1)[(x + y)² + 1]. The other factor of the first expression, x² + y² + 1 − xy + x + y, has no factor in common with the second expression. So the HCF is x + y − 1, option (c).
- Q13
If x⁴ = x² + 1, where x > 0, then what is 2x⁴ equal to?
- A2 + √3
- B3 + √5
- C5 − 2√3
- D3 − √5
Show answer & solution
Correct answer: (B) 3 + √5
Write the equation as (x²)² − x² − 1 = 0, a quadratic in x². Since x² > 0, x² = (1 + √5)/2. Then x⁴ = x² + 1 = (3 + √5)/2. So 2x⁴ = 3 + √5, option (b).
- Q14
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.
- AI only
- BII only
- CBoth I and II
- DNeither I nor II
Show answer & solution
Correct answer: (B) II only
Write n = 2k. Then x = 2k(2k + 1)(2k + 2) = 4k(k + 1)(2k + 1). k(k + 1) is even, so 8 divides x. Three consecutive integers include a multiple of 3, so 3 divides x as well. Hence 24 always divides x. I is false: for n = 2, x = 24, which is not divisible by 48. II is true: x² is divisible by 24² = 576, and 576 is a multiple of 144. Only II is correct, option (b).
- Q15
What is the LCM of x⁴ + x²y² + y⁴, x³y + y⁴ and x⁴y² − x³y³?
- Ax³y³(x⁶ − y⁶)
- Bx³y²(x⁶ − y⁶)
- Cx³y(x⁶ − y⁶)
- Dxy(x⁶ − y⁶)
Show answer & solution
Correct answer: (B) x³y²(x⁶ − y⁶)
Factorise each expression: x⁴ + x²y² + y⁴ = (x² + xy + y²)(x² − xy + y²) x³y + y⁴ = y(x + y)(x² − xy + y²) x⁴y² − x³y³ = x³y²(x − y) The LCM takes each factor to its highest power: x³y²(x + y)(x − y)(x² + xy + y²)(x² − xy + y²). This equals x³y²(x³ + y³)(x³ − y³) = x³y²(x⁶ − y⁶), option (b).
- Q16
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?
- AI only
- BII only
- CBoth I and II
- DNeither I nor II
Show answer & solution
Correct answer: (C) Both I and II
Using the sum and difference of cubes, p = x + y and q = x − y. Write x = Ha and y = Hb with HCF(a, b) = 1. Any common divisor of a + b and a − b divides 2a and 2b, so it divides 2. This means HCF(p, q) is either H or 2H. Both cases occur. For x = 2, y = 1: HCF(3, 1) = 1 = H. For x = 3, y = 1: HCF(4, 2) = 2 = 2H. Both statements are correct, option (c).
- Q17
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?
- A6
- B5
- C4
- D3
Show answer & solution
Correct answer: (A) 6
Any common divisor of 12n + 2 and 8n + 1 also divides 2(12n + 2) − 3(8n + 1). That combination equals 24n + 4 − 24n − 3 = 1. So the two numbers are coprime for every n, and all of n = 1, 2, 3, 4, 5, 6 work. That is 6 values, option (a).
- Q18
What is the remainder when (17²⁵ + 19²⁵) is divided by 18?
- A0
- B1
- C3
- D9
Show answer & solution
Correct answer: (A) 0
Modulo 18, 17 ≡ −1 and 19 ≡ 1. So 17²⁵ + 19²⁵ ≡ (−1)²⁵ + 1²⁵ = −1 + 1 = 0. Another way to see it: for odd n, aⁿ + bⁿ is divisible by a + b = 36, which is a multiple of 18. The remainder is 0, option (a).
- Q19
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)?
- A4
- B6
- C8
- D12
Show answer & solution
Correct answer: (C) 8
512 = 2⁹, so (p + q)^(p + q) must contain at least nine factors of 2. This means p + q must be even. p + q = 2: 2² = 4, not enough. p + q = 4: 4⁴ = 2⁸, not enough. p + q = 6: 6⁶ = 2⁶·3⁶, not enough. p + q = 8: 8⁸ = 2²⁴, which works. The least value is 8, option (c).
- Q20
If (p + q)/(q + r) = (r + s)/(s + p); (q + r) ≠ 0, (s + p) ≠ 0, then which one of the following is correct?
- Ap + q + r + s = 0
- Bp = r
- CEither p + q + r + s = 0 or p = r
- DNone of the above
Show answer & solution
Correct answer: (C) Either p + q + r + s = 0 or p = r
Cross-multiply: (p + q)(s + p) = (r + s)(q + r). Expanding and cancelling qs gives p² − r² + pq + ps − qr − rs = 0. Group the terms: (p − r)(p + r) + q(p − r) + s(p − r) = (p − r)(p + q + r + s) = 0. So either p + q + r + s = 0 or p = r, option (c).
- Q21
What is the digit at hundreds place of the number (25)¹⁰ ?
- A1
- B2
- C5
- D6
Show answer & solution
Correct answer: (D) 6
25² = 625 and 625 × 25 = 15625, which again ends in 625. So every power 25ⁿ with n ≥ 2 ends in 625. 25¹⁰ ends in 625, and its hundreds digit is 6, option (d).
- Q22
If x − 1/x = 2, x > 0; then what is x² − 1/x² equal to?
- A6
- B4√2
- C4
- D2√2
Show answer & solution
Correct answer: (B) 4√2
(x + 1/x)² = (x − 1/x)² + 4 = 4 + 4 = 8. Since x > 0, x + 1/x = 2√2. Then x² − 1/x² = (x − 1/x)(x + 1/x) = 2·2√2 = 4√2, option (b).
- Q23
If ((a − b)/2)x² − ((a + b)/2)x + b = 0, then what are the roots of this equation?
- A1, b/(a − b)
- B1, 2b/(a − b)
- C1/2, b/(a + b)
- D1/2, 2b/(a + b)
Show answer & solution
Correct answer: (B) 1, 2b/(a − b)
Multiply through by 2: (a − b)x² − (a + b)x + 2b = 0. x = 1 is a root, because (a − b) − (a + b) + 2b = 0. The product of the roots is 2b/(a − b), so the other root is 2b/(a − b). The roots are 1 and 2b/(a − b), option (b).
- Q24
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
- B0
- C1/2
- D1
Show answer & solution
Correct answer: (A) −1/2
Factorise c² + bc − a² − ab = (c − a)(c + a) + b(c − a) = (c − a)(a + b + c). The first two terms then combine to (a + b)(a + b + c)/[(c − a)(a + b + c)] = (a + b)/(c − a). Subtract the third term: [2(a + b) − (a + 2b + c)]/[2(c − a)] = (a − c)/[2(c − a)] = −1/2, option (a).
- Q25
If a^b = b^a, then what is [a × (a/b)^(a/b)] / a^(a/b) equal to?
- A1
- Bab
- Cb
- Da^b
Show answer & solution
Correct answer: (A) 1
The expression is a·a^(a/b)/(b^(a/b)·a^(a/b)) = a/b^(a/b). Using a^b = b^a: b^(a/b) = (b^a)^(1/b) = (a^b)^(1/b) = a. So the value is a/a = 1, option (a).
- Q26
If p = (√5 − 2)/(√5 + 2) and q = (√5 + 2)/(√5 − 2), then what is (p/q + q/p) equal to?
- A18
- B8√5
- C322
- D72√5
Show answer & solution
Correct answer: (C) 322
Rationalise: p = (√5 − 2)²/(5 − 4) = 9 − 4√5, and q = 9 + 4√5. Note that pq = 1. So p/q + q/p = (p² + q²)/(pq) = p² + q². p² + q² = (p + q)² − 2pq = 18² − 2 = 322, option (c).
- Q27
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?
- A1003
- B1005
- C1007
- D1011
Show answer & solution
Correct answer: (D) 1011
LCM(4, 6, 7, 9) = 252, so N = 252k + 3. 252 × 3 = 756 is too small, while 252 × 4 = 1008. The smallest 4-digit N is 1008 + 3 = 1011, option (d).
- Q28
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
Show answer & solution
Correct answer: (A) ±9
The left side equals 2(a² + b² + c²) − 2(ab + bc + ca) = 6. So 58 − 2(ab + bc + ca) = 6, giving ab + bc + ca = 26. Then (a + b + c)² = 29 + 2(26) = 81, so a + b + c = ±9, option (a).
- Q29
If [√(p + x) + √(p − x)] / [√(p + x) − √(p − x)] = p, then what is x equal to?
- Ap/(p² + 1)
- B2p/(p² + 1)
- Cp²/(p² + 1)
- D2p²/(p² + 1)
Show answer & solution
Correct answer: (D) 2p²/(p² + 1)
Apply componendo–dividendo: √(p + x)/√(p − x) = (p + 1)/(p − 1). Square both sides: (p + x)/(p − x) = (p + 1)²/(p − 1)². Apply componendo–dividendo again: p/x = [(p + 1)² + (p − 1)²]/[(p + 1)² − (p − 1)²] = (2p² + 2)/(4p). So x = 2p²/(p² + 1), option (d).
- Q30
If x = 2 + 2^(1/2) + 2^(3/2), then what is x² − 4x − 10 equal to?
- A0
- B1
- C4
- D6
Show answer & solution
Correct answer: (C) 4
2^(1/2) = √2 and 2^(3/2) = 2√2, so x = 2 + 3√2. Then x − 2 = 3√2, and squaring gives x² − 4x + 4 = 18. So x² − 4x = 14, and x² − 4x − 10 = 4, option (c).
- Q31
If √(2 + √(2 + √(2 + √(2 + ...)))) = cosec θ, then what is sin θ equal to?
- A1
- B√3/2
- C1/√2
- D1/2
Show answer & solution
Correct answer: (D) 1/2
Let y be the value of the nested radical. Then y = √(2 + y), so y² − y − 2 = 0, which factors as (y − 2)(y + 1) = 0. Since y > 0, y = 2. cosec θ = 2, so sin θ = 1/2, option (d).
- Q32
If cot θ = √7, then what is (cosec²θ − sec²θ)/(cosec²θ + sec²θ) equal to?
- A1/2
- B1/3
- C2/3
- D3/4
Show answer & solution
Correct answer: (D) 3/4
cosec²θ = 1 + cot²θ = 8, and since tan²θ = 1/7, sec²θ = 1 + 1/7 = 8/7. The ratio is (8 − 8/7)/(8 + 8/7) = (48/7)/(64/7) = 3/4, option (d).
- Q33
If 2 tan θ = sec²θ − 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
Show answer & solution
Correct answer: (A) √2 − 1
Use sec²θ = 1 + tan²θ: 2 tan θ = tan²θ − 1, so tan²θ − 2 tan θ − 1 = 0. tan θ = 1 ± √2. Since θ is acute, tan θ = 1 + √2. cot θ = 1/(√2 + 1) = √2 − 1, option (a).
- Q34
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?
- AI only
- BII only
- CBoth I and II
- DNeither I nor II
Show answer & solution
Correct answer: (C) Both I and II
Assume x, y > 0 with x > y. (x² + y²)² = x⁴ + y⁴ + 2x²y² = 16x²y², so x² + y² = 4xy. Taking logs gives log x + log y + 2 log 2, so I is true. (x² − y²)² = x⁴ + y⁴ − 2x²y² = 12x²y², so x² − y² = 2√3·xy. Taking logs gives log x + log y + log 2 + 0.5 log 3, so II is true. Both are correct, option (c).
- Q35
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.
- AI only
- BII only
- CBoth I and II
- DNeither I nor II
Show answer & solution
Correct answer: (C) Both I and II
Let a = 3x + y and b = x + 5y. The expression is a² + ab − 20b² = (a + 5b)(a − 4b). a + 5b = 8x + 26y = 2(4x + 13y). a − 4b = −x − 19y = −(x + 19y). So both (4x + 13y) and (x + 19y) are factors, option (c).
- Q36
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?
- A1
- B1/2
- C1/3
- D1/4
Show answer & solution
Correct answer: (B) 1/2
Call the numerator N. Each denominator term can be rewritten, for example (x + y)/(x − y) = 2x/(x − y) − 1, and the other two in the same way. So the three fractions add up to 2N − 3, and adding 3 gives a denominator of 2N. The ratio is N/(2N) = 1/2, option (b).
- Q37
If α and β are the roots of the equation log₁₀[998 + √(x² − 18x + 76)] = 3 then what is (α − β)² equal to?
- A16
- B25
- C36
- D49
Show answer & solution
Correct answer: (C) 36
log₁₀(…) = 3 means 998 + √(x² − 18x + 76) = 1000, so √(x² − 18x + 76) = 2. Squaring: x² − 18x + 76 = 4, so x² − 18x + 72 = 0, with roots 6 and 12. (α − β)² = (12 − 6)² = 36, option (c).
- Q38
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
- A60°
- B57.5°
- C52.5°
- D47.5°
Show answer & solution
Correct answer: (C) 52.5°
The two acute angles add up to 90°, and their difference is π/12 = 15°. So the angles are (90 + 15)/2 = 52.5° and 37.5°. One of them is 52.5°, option (c).
- Q39
What is (sec θ − tan θ) − √((1 − sin θ)/(1 + sin θ)) equal to?
- A0
- B2 tan θ
- C2 sec θ
- Dsin θ + cos θ
Show answer & solution
Correct answer: (A) 0
Multiply inside the root by (1 − sin θ)/(1 − sin θ): √((1 − sin θ)²/(1 − sin²θ)) = (1 − sin θ)/cos θ, taking cos θ > 0. (1 − sin θ)/cos θ = sec θ − tan θ. So the expression is (sec θ − tan θ) − (sec θ − tan θ) = 0, option (a).
- Q40
If 8 sin θ − cos θ = 4, where 0 < θ < π/2, then what is cosec θ equal to?
- A1
- B3/2
- C5/3
- D2
Show answer & solution
Correct answer: (C) 5/3
From the equation, cos θ = 8 sin θ − 4. Substitute into sin²θ + cos²θ = 1: s² + (8s − 4)² = 1, which gives 65s² − 64s + 15 = 0. So s = (64 ± 14)/130, which is 3/5 or 5/13. For s = 5/13, cos θ = −12/13 < 0, which is not allowed in (0, π/2). So sin θ = 3/5, with cos θ = 4/5. cosec θ = 5/3, option (c).
- Q41
Let cosec θ − sin θ = p and sec θ − cos θ = q. What is (p sin θ + q cos θ) equal to?
- A−1
- B0
- C1
- D2
Show answer & solution
Correct answer: (C) 1
p = 1/sin θ − sin θ = cos²θ/sin θ, and q = 1/cos θ − cos θ = sin²θ/cos θ. Then p sin θ + q cos θ = cos²θ + sin²θ = 1, option (c).
- Q42
Let p sin²α + q cos²α = m, q sin²β + p cos²β = n; p ≠ m, n and q ≠ m, n. If α and β are complementary angles, then which one of the following is correct?
- Amn − 1 = 0
- Bmn + 1 = 0
- Cm + n = 0
- Dm − n = 0
Show answer & solution
Correct answer: (D) m − n = 0
If β = 90° − α, then sin β = cos α and cos β = sin α. So n = q cos²α + p sin²α, which is exactly the expression for m. Hence m = n, that is m − n = 0, option (d).
- Q43
Let p sin²α + q cos²α = m, q sin²β + p cos²β = 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))
Show answer & solution
Correct answer: (C) (m − q)(n − q)/((m − p)(n − p))
Divide the first equation by cos²α: p tan²α + q = m(1 + tan²α), so tan²α = (m − q)/(p − m). Divide the second by cos²β: q tan²β + p = n(1 + tan²β), so tan²β = (n − p)/(q − n). The ratio is (m − q)(q − n)/[(p − m)(n − p)]. The two sign changes cancel, leaving (m − q)(n − q)/[(m − p)(n − p)], option (c).
- Q44
Let sin α / sin β = 4√2/3 and cos α / cos β = 2√3/9. What is tan²α equal to?
- A8
- B6
- C4
- D3
Show answer & solution
Correct answer: (A) 8
Substitute sin α = (4√2/3) sin β and cos α = (2√3/9) cos β into sin²α + cos²α = 1 = sin²β + cos²β: (32/9) sin²β + (12/81) cos²β = sin²β + cos²β. This gives (23/9) sin²β = (69/81) cos²β, so tan²β = 1/3. tan α/tan β = (4√2/3) ÷ (2√3/9) = 6√2/√3, so (tan α/tan β)² = 24. tan²α = 24 × 1/3 = 8, option (a).
- Q45
For the following two (02) items: Let sinθ + cosθ = p and secθ + cosecθ = q, where p ≠ 1. What is tanθ + cotθ equal to?
- Ap/q
- Bq/p
- C2p/q
- D2q/p
Show answer & solution
Correct answer: (B) q/p
tan θ + cot θ = (sin²θ + cos²θ)/(sin θ cos θ) = 1/(sin θ cos θ). Since q = (sin θ + cos θ)/(sin θ cos θ) = p/(sin θ cos θ), we have 1/(sin θ cos θ) = q/p. So tan θ + cot θ = q/p, option (b).
- Q46
Let (1 + sin θ)/cos θ = p + √(p² + 1). What is tan θ equal to?
- Ap
- B√(p² + 1)
- C1/√(p² + 1)
- Dp/√(p² + 1)
Show answer & solution
Correct answer: (A) p
Using the common preamble (1 + sin θ)/cos θ = p + √(p² + 1): this means sec θ + tan θ = p + √(p² + 1). Since sec²θ − tan²θ = 1, sec θ − tan θ = √(p² + 1) − p. Subtract the second equation from the first: 2 tan θ = 2p, so tan θ = p, option (a).
- Q47
Let sin α / sin β = 4√2/3 and cos α / cos β = 2√3/9. What is tan²β equal to?
- A1/√2
- B3/√2
- C1/3
- D2/3
Show answer & solution
Correct answer: (C) 1/3
Substitute sin α = (4√2/3) sin β and cos α = (2√3/9) cos β into sin²α + cos²α = 1 = sin²β + cos²β: (32/9) sin²β + (12/81) cos²β = sin²β + cos²β. This gives (32/9 − 1) sin²β = (1 − 12/81) cos²β, which is (23/9) sin²β = (69/81) cos²β. tan²β = (69/81)·(9/23) = 1/3, option (c).
- Q48
Let (1 + sin θ)/cos θ = p + √(p² + 1). What is sec θ equal to?
- Ap
- B√(p² + 1)
- C1/√(p² + 1)
- Dp/√(p² + 1)
Show answer & solution
Correct answer: (B) √(p² + 1)
(1 + sin θ)/cos θ = sec θ + tan θ = p + √(p² + 1). Since sec²θ − tan²θ = 1, sec θ − tan θ = 1/(p + √(p² + 1)) = √(p² + 1) − p. Add the two equations: 2 sec θ = 2√(p² + 1), so sec θ = √(p² + 1), option (b).
- Q49
Let cosec θ − sin θ = p and sec θ − cos θ = q. What is p²q²(p² + q² + 3) equal to?
- A0
- B1
- C2
- D4
Show answer & solution
Correct answer: (B) 1
With p = cos²θ/sin θ and q = sin²θ/cos θ: p²q² = sin²θ cos²θ, and p² + q² = (cos⁶θ + sin⁶θ)/(sin²θ cos²θ). So p²q²(p² + q² + 3) = cos⁶θ + sin⁶θ + 3 sin²θ cos²θ. This equals (sin²θ + cos²θ)³ = 1, option (b).
- Q50
For the following two (02) items: Let sinθ + cosθ = p and secθ + cosecθ = q, where p ≠ 1. What is the relation between p and q?
- Ap = q(p² − 1)
- B2p = q(p² − 1)
- Cq = p² − 1
- D2q = p(p² − 1)
Show answer & solution
Correct answer: (B) 2p = q(p² − 1)
Write q = 1/sin θ + 1/cos θ = (sin θ + cos θ)/(sin θ cos θ) = p/(sin θ cos θ). Squaring p gives p² = 1 + 2 sin θ cos θ, so sin θ cos θ = (p² − 1)/2. Then q = 2p/(p² − 1), so 2p = q(p² − 1), option (b).
- Q51
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?
- A20%
- B40%
- C50%
- DNone of the above
Show answer & solution
Correct answer: (D) None of the above
Let the marked price be M. After all three discounts, SP = M × 0.9 × 0.8 × 0.75 = 0.54M. This equals 1.08 × CP, so CP = 0.5M. After the first discount only, SP = 0.9M, so profit = (0.9M − 0.5M)/0.5M = 80%. 80% is not among (a)–(c), so the answer is option (d).
- Q52
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?
- A50 : 53
- B51 : 53
- C121 : 126
- D121 : 125
Show answer & solution
Correct answer: (C) 121 : 126
Present salary of X = X × 1.12 × 0.90 = 1.008X. Present salary of Y = Y × 1.10 × 0.88 = 0.968Y. Setting them equal: 1.008X = 0.968Y, so X/Y = 0.968/1.008 = 968/1008 = 121/126. Option (c).
- Q53
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?
- A3 km
- B3.5 km
- C4 km
- D4.5 km
Show answer & solution
Correct answer: (C) 4 km
In the 10 minutes before Y starts, X walks 4.8 × (1/6) = 0.8 km. Y gains on X at 6 − 4.8 = 1.2 km/hr, so Y catches up after 0.8/1.2 = 2/3 hr. In that time Y covers 6 × 2/3 = 4 km from P. They meet 4 km from P. Option (c).
- Q54
What is the solution of the inequalities 5x + 3 < 8x − 9 and 2x + 20 > 5x + 2?
- A4 < x < 6
- B3 < x < 5
- Cx < 3 or x > 5
- Dx < 4 or x > 6
Show answer & solution
Correct answer: (A) 4 < x < 6
From 5x + 3 < 8x − 9 we get 12 < 3x, so x > 4. From 2x + 20 > 5x + 2 we get 18 > 3x, so x < 6. Both hold together when 4 < x < 6. Option (a).
- Q55
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?
- Ap(x − z)/(y − z)
- Bp(y − z)/(x − z)
- Cpx/y
- Dpy/x
Show answer & solution
Correct answer: (A) p(x − z)/(y − z)
Let there be m males, so there are p − m females. Counting the literate people two ways: (y/100)m + (z/100)(p − m) = (x/100)p. This gives m(y − z) = p(x − z). So m = p(x − z)/(y − z). Option (a).
- Q56
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?
- A5 : 8
- B5 : 7
- C6 : 7
- D7 : 8
Show answer & solution
Correct answer: (B) 5 : 7
Take 18 units of each alloy. Alloy X (5 : 4 out of 9) gives Cu 10 and Zn 8. Alloy Y (5 : 13 out of 18) gives Cu 5 and Zn 13. Alloy Z then has Cu 15 and Zn 21, so Cu : Zn = 15 : 21 = 5 : 7. Option (b).
- Q57
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
Show answer & solution
Correct answer: (C) ₹4,021
The loan equals the present value of the three equal installments I: 10000 = I/1.1 + I/1.1² + I/1.1³. The bracket is 0.9091 + 0.8264 + 0.7513 = 2.4869. So I = 10000/2.4869 ≈ ₹4,021. Option (c).
- Q58
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?
- A80 m
- B100 m
- C120 m
- D150 m
Show answer & solution
Correct answer: (B) 100 m
Both trains move the same way, so the relative speed is 100 − 60 = 40 km/hr = 40 × 5/18 = 100/9 m/s. To pass the man in Y, train X has to cover only its own length. Length = (100/9) × 9 = 100 m. Option (b).
- Q59
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?
- A5 : 1
- B4 : 1
- C3 : 1
- D3 : 2
Show answer & solution
Correct answer: (C) 3 : 1
X = 0.75(X + Y) gives 0.25X = 0.75Y, so X = 3Y. With X = Y + 20 this means 3Y = Y + 20, so Y = 10 and X = 30. The ratio X : Y = 3 : 1. Option (c).
- Q60
If one root of the equation 2x² − 5px + 2p² = 0 exceeds the other by 4, then what is the value of p?
- A8/3
- B4/3
- C2/3
- D1/3
Show answer & solution
Correct answer: (A) 8/3
Factorise: 2x² − 5px + 2p² = (2x − p)(x − 2p), so the roots are p/2 and 2p. Their difference is 2p − p/2 = 3p/2 = 4. So p = 8/3. Option (a).
- Q61
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?
- A3.5 cm
- B7 cm
- C10.5 cm
- D14 cm
Show answer & solution
Correct answer: (D) 14 cm
OP is a radius and is perpendicular to AB at C, so OC = 14 − 7 = 7. Then AC² = 14² − 7² = 147. In right triangle ACP: AP² = AC² + PC² = 147 + 49 = 196. So AP = 14 cm. Option (d).
- Q62
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?
- AI only
- BII only
- CBoth I and II
- DMore information is needed
Show answer & solution
Correct answer: (C) Both I and II
Let AB = c and BC = a. Then AQ² = c² + a²/4 = 73 and CP² = a² + c²/4 = 52. Adding: (5/4)(a² + c²) = 125, so a² + c² = 100. Subtracting: (3/4)(c² − a²) = 21, so c² − a² = 28. That gives c = 8, a = 6 and area = ½ × 8 × 6 = 24. One equation alone has two unknowns, so both statements are needed. Option (c).
- Q63
Two poles are situated 24 m apart and their heights differ by 10 m. What is the distance between their tips?
- A25 m
- B26 m
- C30 m
- DCannot be determined due to insufficient data
Show answer & solution
Correct answer: (B) 26 m
The horizontal gap (24 m) and the height difference (10 m) are the two legs of a right triangle. The distance between the tips is its hypotenuse: √(24² + 10²) = √(576 + 100) = √676 = 26 m. Option (b).
- Q64
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?
- A4.8 m
- B5 m
- C6 m
- D6.4 m
Show answer & solution
Correct answer: (C) 6 m
For poles of heights a and b, the crossing point of the two lines is at height ab/(a + b), whatever the distance between them. Height = (10 × 15)/(10 + 15) = 150/25 = 6 m. Option (c).
- Q65
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?
- A75 square units
- B100 square units
- C125 square units
- D150 square units
Show answer & solution
Correct answer: (D) 150 square units
AB + BC = AC + 10 and the perimeter is 60, so 2AC + 10 = 60. That gives AC = 25 and AB + BC = 35. Squaring: 35² = AB² + BC² + 2·AB·BC = 625 + 2·AB·BC, so AB·BC = 300. Area = ½ × AB × BC = 150 square units. Option (d).
- Q66
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)
- A1078 cm³
- B1617 cm³
- C1927 cm³
- D2156 cm³
Show answer & solution
Correct answer: (D) 2156 cm³
CSA: 2πrh = 616, so rh = 616 × 7/44 = 98. We also have r + h = 21, so r and h are the roots of t² − 21t + 98 = 0, i.e. 7 and 14. Since r < h, r = 7 and h = 14. Volume = (22/7) × 7² × 14 = 2156 cm³. Option (d).
- Q67
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?
- Aqr/p
- Bpr/q
- Cpq/r
- Dp + q − r
Show answer & solution
Correct answer: (C) pq/r
From right triangles ABD and ACD: AD² = AB² − BD² = AC² − CD². So AB² − AC² = BD² − CD². Factorising: (AB + AC)(AB − AC) = (BD − CD)(BD + CD), which reads pq = r × BC because D lies on BC. Therefore BC = pq/r. Option (c).
- Q68
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?
- AOne
- BTwo
- CThree
- DAll four
Show answer & solution
Correct answer: (D) All four
An exterior angle E of a regular polygon must satisfy n = 360°/E, with n a whole number ≥ 3. 360/4 = 90, 360/5 = 72, 360/6 = 60 and 360/8 = 45 are all whole numbers. So all four angles are possible. Option (d).
- Q69
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?
- AI only
- BII only
- CBoth I and II
- DNeither I nor II
Show answer & solution
Correct answer: (C) Both I and II
By the midpoint theorem, ZX ∥ AC (so ZX ∥ AY) and XY ∥ AB (so XY ∥ AZ). So AZXY is a parallelogram, and I is true. The three midpoints split ABC into 4 congruent triangles, each 1/4 of its area. AZXY is made of 2 of them, so its area is 1/2 of ABC, and II is true. Option (c).
- Q70
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?
- A288 square units
- B292 square units
- C296 square units
- DCannot be determined due to insufficient data
Show answer & solution
Correct answer: (B) 292 square units
336 = 6 × 7 × 8, so the sides are 6, 7 and 8. TSA = 2(6·7 + 7·8 + 6·8) = 2(42 + 56 + 48) = 2 × 146 = 292 square units. Option (b).
- Q71
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?
- A9
- B10
- C11
- D12
Show answer & solution
Correct answer: (D) 12
Interior − exterior = 120° and interior + exterior = 180°. Subtracting gives 2 × exterior = 60°, so exterior = 30°. Number of sides = 360/30 = 12. Option (d).
- Q72
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)
- A90 cm
- B100 cm
- C105 cm
- D110 cm
Show answer & solution
Correct answer: (C) 105 cm
Arc length = rθ, with θ = 30° = π/6 radians. So 55 = r × π/6, giving r = 330/π = 330 × 7/22 = 105 cm. Option (c).
- Q73
The sides of a triangle are 11 cm, 60 cm and 61 cm. What is the area of the triangle formed by joining the mid-points of the sides of the triangle?
- A165 cm²
- B110 cm²
- C82.5 cm²
- D72.5 cm²
Show answer & solution
Correct answer: (C) 82.5 cm²
11² + 60² = 121 + 3600 = 3721 = 61², so the triangle is right-angled. Its area is ½ × 11 × 60 = 330 cm². The triangle joining the midpoints has 1/4 of that area: 330/4 = 82.5 cm². Option (c).
- Q74
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?
- Ap/3
- Bp/2
- C√3p/2
- D√3p/4
Show answer & solution
Correct answer: (C) √3p/2
AC = √(p² + 3p²) = 2p. In a right triangle, the altitude to the hypotenuse is (product of legs)/hypotenuse. BD = (p × √3p)/(2p) = √3p/2. Option (c).
- Q75
An angle θ is exactly one-fourth of its complementary angle. What is the value of angle θ?
- A12°
- B15°
- C18°
- D20°
Show answer & solution
Correct answer: (C) 18°
θ = (90° − θ)/4, so 4θ = 90° − θ and 5θ = 90°. Therefore θ = 18°. Option (c).
- Q76
In a triangle ABC, ∠A = 30°, AB = 7 cm and AC = 12 cm. What is the area of the triangle ABC?
- A21 cm²
- B21√3 cm²
- C42 cm²
- D42√3 cm²
Show answer & solution
Correct answer: (A) 21 cm²
Area = ½ × AB × AC × sin A = ½ × 7 × 12 × sin 30° = ½ × 84 × ½ = 21 cm². Option (a).
- Q77
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?
- Aπx = 2y
- B2πx = y
- Cπx = 4y
- Dπx = 8y
Show answer & solution
Correct answer: (D) πx = 8y
The square in a unit circle has diagonal 2, so its side is √2 and x = 2. The circle in a unit square has radius ½, so y = π/4. Then πx = 2π and 8y = 2π. So πx = 8y. Option (d).
- Q78
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?
- A1 : 1
- B1 : 2
- C2 : 1
- D3 : 2
Show answer & solution
Correct answer: (C) 2 : 1
Equal volumes with the same radius r: (1/3)πr²h = (2/3)πr³. So h = 2r, and h : r = 2 : 1. Option (c).
- Q79
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?
- A9/π cm
- B18/(π + 2) cm
- C18/π cm
- DNone of the above
Show answer & solution
Correct answer: (D) None of the above
(√3/4)a² = 36√3 gives a² = 144, so a = 12 and the wire is 36 cm long. A semicircle's boundary is the arc plus the diameter: πr + 2r = 36, so r = 36/(π + 2) cm. (Using the arc only gives 36/π.) Neither value is listed in (a)–(c), so the answer is option (d).
- Q80
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?
- A250√2/√(3π) m³
- B500√3/√π m³
- C1000√3/√(2π) m³
- D1000√3/√π m³
Show answer & solution
Correct answer: (A) 250√2/√(3π) m³
The semi-vertical angle is 30°, so with slant height l: r = l/2 and h = √3l/2. CSA = πrl = πl²/2 = 100, so l² = 200/π. Then r² = 50/π and h = (√3/2)·√(200/π) = 5√6/√π. V = (1/3)π r² h = (1/3)π(50/π)(5√6/√π) = 250√6/(3√π). This equals 250√2/√(3π) m³. Option (a).
- Q81
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?
- AThe Question can be answered by using one of the Statements alone, but cannot be answered using the other Statement alone
- BThe Question can be answered by using either Statement alone
- CThe Question can be answered by using both the Statements together, but cannot be answered using either Statement alone
- DThe Question can be answered even without using any of the Statements
Show answer & solution
Correct answer: (A) The Question can be answered by using one of the Statements alone, but cannot be answered using the other Statement alone
Let half the base be b and the height h. Then b² + h² = 100 and bh = 48. So (b + h)² = 196 and (h − b)² = 4, giving (b, h) = (6, 8) or (8, 6). That makes BC = 12 or 16. Statement I (even) fits both, so it is not enough. Statement II (h > b) picks h = 8, b = 6, so BC = 12. Only II works. Option (a).
- Q82
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?
- AThe Question can be answered by using one of the Statements alone, but cannot be answered using the other Statement alone
- BThe Question can be answered by using either Statement alone
- CThe Question can be answered by using both the Statements together, but cannot be answered using either Statement alone
- DThe Question can be answered even without using any of the Statements
Show answer & solution
Correct answer: (A) The Question can be answered by using one of the Statements alone, but cannot be answered using the other Statement alone
The only factor pair of 252 made of a number and its digit reversal is 12 × 21, so N is 12 or 21. Statement I: 12 + 21 = 33 holds either way, so N is still not fixed. Statement II: N > M forces N = 21. So one statement (II) is enough and the other is not. Option (a).
- Q83
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?
- AThe Question can be answered by using one of the Statements alone, but cannot be answered using the other Statement alone
- BThe Question can be answered by using either Statement alone
- CThe Question can be answered by using both the Statements together, but cannot be answered using either Statement alone
- DThe Question can be answered even without using any of the Statements
Show answer & solution
Correct answer: (D) The Question can be answered even without using any of the Statements
Put A = B − C into A + B + C = 180°: (B − C) + B + C = 180°, so 2B = 180° and B = 90°. Then A + C = 90°, so A < 90° and A is acute. The given condition alone answers the question. Option (d).
- Q84
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 circum-radius 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?
- AThe Question can be answered by using one of the Statements alone, but cannot be answered using the other Statement alone
- BThe Question can be answered by using either Statement alone
- CThe Question can be answered by using both the Statements together, but cannot be answered using either Statement alone
- DThe Question can be answered even without using any of the Statements
Show answer & solution
Correct answer: (D) The Question can be answered even without using any of the Statements
In a right triangle the hypotenuse is a diameter of the circumcircle. Since AC = 20 cm is given, R = 20/2 = 10 cm. Neither statement is needed. Option (d).
- Q85
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?
- AThe Question can be answered by using one of the Statements alone, but cannot be answered using the other Statement alone
- BThe Question can be answered by using either Statement alone
- CThe Question can be answered by using both the Statements together, but cannot be answered using either Statement alone
- DThe Question can be answered even without using any of the Statements
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Correct answer: (B) The Question can be answered by using either Statement alone
Area = AB × BC × sin 60° = (√3/2)·ab = 7√3, so ab = 14. Statement I (both sides prime) forces the sides to be 2 and 7. Statement II (natural numbers > 1) also forces 2 and 7, since 1 × 14 is ruled out. Either way the perimeter is 2(2 + 7) = 18. Option (b).
- Q86
A Question is given followed by two Statements I and II. Consider the Question and the Statements. Question: What is the remainder when x^(2n) − y^(2n) + 1 is divided by x^n + y^n, 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?
- AThe Question can be answered by using one of the Statements alone, but cannot be answered using the other Statement alone
- BThe Question can be answered by using either Statement alone
- CThe Question can be answered by using both the Statements together, but cannot be answered using either Statement alone
- DThe Question can be answered even without using any of the Statements
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Correct answer: (D) The Question can be answered even without using any of the Statements
x^(2n) − y^(2n) = (x^n − y^n)(x^n + y^n), which is always divisible by x^n + y^n for any natural n. So x^(2n) − y^(2n) + 1 always leaves remainder 1. Whether n is odd or even makes no difference, so no statement is needed. Option (d).
- Q87
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?
- AThe Question can be answered by using one of the Statements alone, but cannot be answered using the other Statement alone
- BThe Question can be answered by using either Statement alone
- CThe Question can be answered by using both the Statements together, but cannot be answered using either Statement alone
- DThe Question can be answered even without using any of the Statements
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Correct answer: (C) The Question can be answered by using both the Statements together, but cannot be answered using either Statement alone
The last digit of (54D)^100 is the last digit of D^100. It is 1 only for D = 1, 3, 7, 9, since all their powers cycle with a length dividing 4. Statement I (D > 5) leaves {7, 9}, which is not unique. Statement II (multiple of 3) leaves {3, 9}, also not unique. Together they give D = 9. Option (c).
- Q88
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?
- AThe Question can be answered by using one of the Statements alone, but cannot be answered using the other Statement alone
- BThe Question can be answered by using either Statement alone
- CThe Question can be answered by using both the Statements together, but cannot be answered using either Statement alone
- DThe Question can be answered even without using any of the Statements
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Correct answer: (B) The Question can be answered by using either Statement alone
Triangle inequality in ABD: 9 − 6 < BD < 9 + 6, i.e. 3 < BD < 15. In BCD: 18 − 6 < BD < 18 + 6, i.e. 12 < BD < 24. So 12 < BD < 15. Statement I (integer > 13) gives BD = 14. Statement II (even integer) also gives BD = 14. Either statement alone is enough. Option (b).
- Q89
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?
- AThe Question can be answered by using one of the Statements alone, but cannot be answered using the other Statement alone
- BThe Question can be answered by using either Statement alone
- CThe Question can be answered by using both the Statements together, but cannot be answered using either Statement alone
- DThe Question can be answered even without using any of the Statements
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Correct answer: (D) The Question can be answered even without using any of the Statements
By the intersecting chords theorem, AP × PB = CP × PD. So CP × PD = 48 directly from the given data, and neither statement is needed. Option (d).
- Q90
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?
- AThe Question can be answered by using one of the Statements alone, but cannot be answered using the other Statement alone
- BThe Question can be answered by using either Statement alone
- CThe Question can be answered by using both the Statements together, but cannot be answered using either Statement alone
- DThe Question can be answered even without using any of the Statements
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Correct answer: (D) The Question can be answered even without using any of the Statements
Let the diagonals be 5k and 12k. Side = √((2.5k)² + (6k)²) = √(42.25k²) = 6.5k. A diagonal is either 5k or 12k, and neither can equal 6.5k. So the answer is always 'No', and neither statement is needed. Option (d).
- Q91
To find the average ratio like price/unit, work done/hour, kilometre/hour under certain conditions, the suitable measure of central tendency applicable is
- Aarithmetic mean
- Bgeometric mean
- Charmonic mean
- Dmode
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Correct answer: (C) harmonic mean
For rates such as price per unit, work per hour or km per hour, taken over equal amounts of the numerator quantity (for example equal distances), the correct average is the harmonic mean. Option (c).
- Q92
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?
- A2.14
- B2.25
- C2.37
- D2.50
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Correct answer: (A) 2.14
Combined mean = Σ(fᵢ × meanᵢ)/Σfᵢ = (45×2 + 40×2.5 + 55×2)/(45 + 40 + 55) = (90 + 100 + 110)/140 = 300/140 ≈ 2.14. Option (a).
- Q93
The arithmetic mean of 200 observations is 60. If 5 is multiplied to each observation, then what will be the new arithmetic mean?
- A500
- B300
- C60
- D40
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Correct answer: (B) 300
Multiplying every observation by a constant multiplies the mean by the same constant. New mean = 5 × 60 = 300. Option (b).
- Q94
Which one of the following is a positional average?
- AArithmetic mean
- BMedian
- CMode
- DGeometric mean
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Correct answer: (B) Median
The median is found from its position: it is the middle item once the data are arranged in order. It is not calculated from the size of every value, as the arithmetic and geometric means are. So it is called a positional average. Option (b).
- Q95
Which measure of central tendency is least affected by the presence of extreme observations in the data?
- AArithmetic mean
- BHarmonic mean
- CGeometric mean
- DMedian
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Correct answer: (D) Median
The median depends only on the middle value(s) of the ordered data, so very large or very small values hardly change it. Every type of mean uses the actual size of every observation. The median is least affected. Option (d).
- Q96
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?
- A20.5
- B22.5
- C30.5
- D32.5
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Correct answer: (D) 32.5
Convert the 'more than' counts to class frequencies: 10–20: 100 − 75 = 25; 20–30: 75 − 60 = 15; 30–40: 60 − 40 = 20; 40–50: 40. The mid-values are 15, 25, 35, 45. Σfx = 375 + 375 + 700 + 1800 = 3250. Mean = 3250/100 = 32.5. Option (d).
- Q97
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?
- A20
- B22
- C25
- D30
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Correct answer: (C) 25
Class frequencies: 0–10: 5; 10–20: 25; 20–30: k; 30–40: 18; 40–50: 7, so N = 55 + k. The median 24 lies in the class 20–30, where cf = 30. 20 + [((55 + k)/2 − 30)/k] × 10 = 24 gives (55 + k)/2 − 30 = 0.4k. Then 55 + k − 60 = 0.8k, so 0.2k = 5 and k = 25. Option (c).
- Q98
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?
- A1
- B2
- C3
- D4
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Correct answer: (C) 3
First group: N = u + v + 160, and the median 20.8 lies in the class 20–25, where cf = u + v + 63 and f = 75. 20 + [(N/2 − cf)/75] × 5 = 20.8 gives N/2 − cf = 12, which simplifies to u + v = 10. Second group: with mid-values 2.5, 7.5, …, 27.5, the mean condition reads 7.5u + 15v + 2820 = 17.3(3u + 2v + 148), i.e. 44.4u + 19.6v = 259.6. With v = 10 − u this gives u ≈ 2.6, and the nearest whole-number frequency is u = 3. Check: u = 3 gives a mean of about 17.24, closer to 17.3 than u = 2 (about 17.38). Answer: u = 3, option (c).
- Q99
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 mean of the distribution?
- A21.625
- B22.225
- C23.225
- D24.625
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Correct answer: (D) 24.625
With k = 25 the class frequencies are 5, 25, 25, 18, 7, so N = 80. The mid-values are 5, 15, 25, 35, 45. Σfx = 25 + 375 + 625 + 630 + 315 = 1970. Mean = 1970/80 = 24.625. Option (d).
- Q100
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 v?
- A5
- B6
- C7
- D8
Show answer & solution
Correct answer: (C) 7
The first group's median condition fixes u + v = 10: N = u + v + 160, median class 20–25 with cf = u + v + 63 and f = 75, and 20 + [(N/2 − cf)/75] × 5 = 20.8 gives N/2 − cf = 12, so u + v = 10. The second group's mean (17.3) then splits this total, giving u ≈ 3 as the nearest whole number. So v = 10 − 3 = 7. Check: u = 3, v = 7 gives a first-group median of 20 + (85 − 73)/75 × 5 = 20.8 exactly. Option (c).