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Practice Test 8
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SSC CGL Divisibility and Remainders Practice Test 8

15 SSC CGL divisibility and remainders questions with chapter-wise explanations and exam-style options.

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SSC CGL Practice Session

SSC CGL Divisibility and Remainders Practice Test 8

15 SSC CGL divisibility and remainders questions with chapter-wise explanations and exam-style options.

Questions
13
Marking
+2 correct, -0.5 wrong
Format
15-question ad-supported session
Preview all 13 questions in SSC CGL Divisibility and Remainders Practice Test 8 (no login required)
  1. Divisibility and Remainders

    1. A number is divisible by both 4 and 6. It must be divisible by:

    • A. 10
    • B. 12 (Correct)
    • C. 18
    • D. 24

    Explanation: LCM(4,6)=12. The number must be divisible by 12.

  2. Divisibility and Remainders

    2. Unit digit of 2^100 is:

    • A. 2
    • B. 4
    • C. 6 (Correct)
    • D. 8

    Explanation: Powers of 2 cycle: 2,4,8,6 (period 4). 100÷4=25 remainder 0. Unit digit = 6.

  3. Divisibility and Remainders

    3. Number of prime factors of 2^3 × 3^2 × 5 is:

    • A. 3 (Correct)
    • B. 4
    • C. 5
    • D. 6

    Explanation: Number of distinct prime factors = 3 (i.e., 2, 3, and 5).

  4. Divisibility and Remainders

    4. Remainder when 17 × 19 × 21 is divided by 4:

    • A. 0
    • B. 1
    • C. 2
    • D. 3 (Correct)

    Explanation: 17≡1, 19≡3, 21≡1 (mod 4). Product≡1×3×1=3 (mod 4).

  5. Divisibility and Remainders

    5. HCF of two numbers is 8 and LCM is 384. One number is 48. Other is:

    • A. 56
    • B. 64 (Correct)
    • C. 72
    • D. 80

    Explanation: Other = HCF×LCM/one number = 8×384/48 = 64.

  6. Divisibility and Remainders

    6. Sum of all prime numbers between 1 and 20 is:

    • A. 55
    • B. 58
    • C. 60
    • D. 77 (Correct)

    Explanation: Primes: 2+3+5+7+11+13+17+19=77.

  7. Divisibility and Remainders

    7. Find least value of k so that 7k5 is divisible by 3:

    • A. 0
    • B. 1 (Correct)
    • C. 2
    • D. 3

    Explanation: Digit sum = 7+k+5=12+k. For divisibility by 3, 12+k must be divisible by 3. k=0 → 12 ✓. But if k must be minimum positive: k=0 qualifies. Answer: 0.

  8. Divisibility and Remainders

    8. Unit digit of 13^73 + 7^55:

    • A. 0
    • B. 2
    • C. 4
    • D. 6 (Correct)

    Explanation: The unit digit of 13^73 is the unit digit of 3^73, which is 3. The unit digit of 7^55 is also 3. So the sum has unit digit 6.

  9. Divisibility and Remainders

    9. How many prime numbers are between 40 and 60?

    • A. 3
    • B. 4 (Correct)
    • C. 5
    • D. 6

    Explanation: Primes: 41, 43, 47, 53 = 4 primes.

  10. Divisibility and Remainders

    10. The greatest number that divides 248, 318, and 458 leaving remainder 8 each time:

    • A. 10 (Correct)
    • B. 15
    • C. 20
    • D. 25

    Explanation: Subtract 8: 240, 310, 450. HCF = HCF(240,310,450)=10.

  11. Divisibility and Remainders

    11. LCM of 2/3, 4/9, and 8/27 is:

    • A. 2/27
    • B. 8/3 (Correct)
    • C. 2/9
    • D. 8/27

    Explanation: LCM of fractions = LCM of numerators/HCF of denominators = LCM(2,4,8)/HCF(3,9,27) = 8/3.

  12. Divisibility and Remainders

    12. The product of two consecutive even numbers is 3968. The numbers are:

    • A. 60,62
    • B. 62,64 (Correct)
    • C. 64,66
    • D. 58,60

    Explanation: 62×64=3968. Verified: 62×64=3968.

  13. Divisibility and Remainders

    13. Total factors of 48:

    • A. 8
    • B. 10 (Correct)
    • C. 12
    • D. 14

    Explanation: 48=2^4×3. Factors=(4+1)(1+1)=10.