CDS-II 2025 — Elementary Mathematics

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CDS-II 2025 — Elementary Mathematics — solved practice questions

8 CDS Previous Year Papers questions with step-by-step solutions. Attempt each, then reveal the answer.

1. Q1easy

   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).

2. Q2easy

   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).

3. Q3medium

   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).

4. Q4easy

   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).

5. Q5easy

   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).

6. Q6medium

   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).

7. Q7easy

   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).

8. Q8easy

   (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).

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