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Arun Sharma Test: Number System- 1 - CAT MCQ


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15 Questions MCQ Test - Arun Sharma Test: Number System- 1

Arun Sharma Test: Number System- 1 for CAT 2025 is part of CAT preparation. The Arun Sharma Test: Number System- 1 questions and answers have been prepared according to the CAT exam syllabus.The Arun Sharma Test: Number System- 1 MCQs are made for CAT 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Arun Sharma Test: Number System- 1 below.
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Arun Sharma Test: Number System- 1 - Question 1
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Which of the following is a Real Number but not a Rational Number?

Detailed Solution for Arun Sharma Test: Number System- 1 - Question 1

(a) −√4 = −2 is a Real Number as well as a Rational Number
(b) −√3 is a Real number and an irrational number
(c) √−4 is an imaginary number

Arun Sharma Test: Number System- 1 - Question 2
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Convert  in p/q form

Detailed Solution for Arun Sharma Test: Number System- 1 - Question 2

► All the digits written once = 2116
► All the digits without Bar written once = 21
► Number of digits with bar after decimal = 2
► Number of digits without bar after decimal = 0
► Rational form = 

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Arun Sharma Test: Number System- 1 - Question 3
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√16 will not fall under which of the following categories?

Detailed Solution for Arun Sharma Test: Number System- 1 - Question 3

► √16 = 4, which is a natural number, rational number as well as a real number.

Arun Sharma Test: Number System- 1 - Question 4
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(16a + 6) will fall under which of the following categories, a is an Integer
Detailed Solution for Arun Sharma Test: Number System- 1 - Question 4

(16a + 6) will always be even, whether the a is odd or even.

  • For a = 1 (odd)
    (16 * 1) + 6 = 16 + 6 = 22 (even)
  • For a = 2 (even)
    (16* 2) + 6 = 32 + 6 = 38 (even)
Arun Sharma Test: Number System- 1 - Question 5
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If (7a - 3) is odd then 'a' would be, a is an Integer
Detailed Solution for Arun Sharma Test: Number System- 1 - Question 5

Since -3 is odd, the parity of 7a - 3 depends on the parity of 7a:

  • If 7a is even (i.e., a is even), then even - odd = odd.

  • If 7a is odd (i.e., a is odd), then odd - odd = even.

  • 7a - 3 is odd when 7a is even.

  • 7a is even when a is even.

Thus, for 7a - 3 to be odd, a must be an even integer

Arun Sharma Test: Number System- 1 - Question 6
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1837 × 1712 × 1839 will be
Detailed Solution for Arun Sharma Test: Number System- 1 - Question 6

  1. Notice symmetry around 1838:

    • 1837 = 1838 − 1

    • 1839 = 1838 + 1

  2. Use identity: (a − 1)(a + 1) = a² − 1
    So, 1837 × 1839 = 1838² − 1

  3. Now the full product:
    1837 × 1712 × 1839 = 1712 × (1838² − 1) = 1712×1838² − 1712

  4. Compute 1838² quickly:

    • 1838² = (1800 + 38)² = 1800² + 2·1800·38 + 38²

    • = 3,240,000 + 136,800 + 1,444 = 3,378,244

  5. Multiply 1712 × 3,378,244:

    • Break 1712 = 1600 + 112

    • 3,378,244 × 1600 = 3,378,244 × 16 × 100

      • 3,378,244 × 16 = 3,378,244×(2⁴) = 54,051,904

      • Then ×100 ⇒ 5,405,190,400

    • 3,378,244 × 112 = 3,378,244 × (7 × 16)

      • ×16 = 54,051,904 (from above)

      • ×7 ⇒ 378,363,328

    • Sum: 5,405,190,400 + 378,363,328 = 5,783,553,728

  6. Subtract 1712:
    5,783,553,728 − 1,712 = 5,783,552,016

Therefore, the value is 5,783,552,016 ⇒ Option A.

Quick Verification / Shortcut

  • Use pairing trick: (a−1)(a+1) = a²−1 with a=1838 to reduce three-number product to a simple square minus a small correction.

Arun Sharma Test: Number System- 1 - Question 7
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How many two-digit numbers are divisible by 3 or 5?
Detailed Solution for Arun Sharma Test: Number System- 1 - Question 7

  1. Two-digit numbers: 10 to 99 (inclusive) → total 90 numbers.

  2. Count multiples of 3 between 10 and 99:

    • First: 12, Last: 99

    • Count = (99 − 12)/3 + 1 = 87/3 + 1 = 29 + 1 = 30

  3. Count multiples of 5 between 10 and 99:

    • First: 10, Last: 95

    • Count = (95 − 10)/5 + 1 = 85/5 + 1 = 17 + 1 = 18

  4. Subtract overlap (multiples of 15):

    • Between 10 and 99: first 15, last 90

    • Count = (90 − 15)/15 + 1 = 75/15 + 1 = 5 + 1 = 6

  5. By inclusion–exclusion:

    • Total = 30 + 18 − 6 = 42

Answer: 42 ⇒ Option C.

Quick Verification 

  • Approx check: About 1/3 are multiples of 3 (~30) and about 1/5 are multiples of 5 (~18), subtract ~1/15 overlap (~6) ⇒ 42.

Arun Sharma Test: Number System- 1 - Question 8
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Largest 3 digit composite number is
Detailed Solution for Arun Sharma Test: Number System- 1 - Question 8

  1. Composite Number Definition: A composite number has more than two factors (divisors other than 1 and itself).

  2. Largest Three-Digit Number: 999.

  3. Divisibility Check:

    • Sum of digits: 9 + 9 + 9 = 27 (divisible by 3).

    • 999 ÷ 3 = 333 (so 3 and 333 are factors).

    • 999 ÷ 9 = 111 (multiple factors confirm it is composite).

  4. Verification: No three-digit number greater than 999 exists.

Final Answer:
999

Arun Sharma Test: Number System- 1 - Question 9
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The sum of 7 consecutive natural numbers is 1617. Find how many of these are prime numbers?
Detailed Solution for Arun Sharma Test: Number System- 1 - Question 9

Given:
The sum of seven consecutive natural numbers = 1617

Calculation:
Let the numbers be n, n+1, n+2, n+3, n+4, n+5, n+6 respectively
=> 7n + 21 = 1617
=> 7n = 1596
=> n = 228

The numbers are 228, 229, 230, 231, 232, 233, 234
Out of these, 229 and 233 are prime numbers

Therefore, the required prime numbers count is 2.

Arun Sharma Test: Number System- 1 - Question 10
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What is the difference between 517th Whole number and 516th Natural Number?
Detailed Solution for Arun Sharma Test: Number System- 1 - Question 10

► 517th Whole number = 516
► 516th Natural Number = 516

  • 516 - 516 = 0
Arun Sharma Test: Number System- 1 - Question 11
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Four bells ring simultaneously at starting and an interval of 6 sec, 12 sec, 15 sec and 20 sec respectively. How many times they ring together in 2 hours?
Detailed Solution for Arun Sharma Test: Number System- 1 - Question 11

GIVEN:

Bells ring together every LCM of their intervals (6, 12, 15, 20 s). LCM is 60 s (common multiple).

Step 1: Find LCM

  • Prime factors: 6=2×3, 12=2²×3, 15=3×5, 20=2²×5.

  • LCM = 2² × 3 × 5 = 60 s.

Step 2: Total time

  • 2 hours = 7200 s.

Step 3: Count rings

  • They ring together at t=0 and every 60 s up to 7200 s.

  • Intervals: 7200 / 60 = 120.

  • Total (including start): 120 + 1 = 121.

The answer is C: 121.

Quick Verification/Shortcut

LCM(6,12,15,20)=60 (quick: max powers 2²,3,5).
Rings = (total s / LCM) + 1 (include t=0).
(7200/60) +1=121.
 

Arun Sharma Test: Number System- 1 - Question 12
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The LCM and HCF of 2 numbers are 168 and 6 respectively. If one of the numbers is 24, find the other?
Detailed Solution for Arun Sharma Test: Number System- 1 - Question 12

The formula is:
LCM × HCF = Number 1 × Number 2.

Let the two numbers be "a" and "b", where "a = 24" (given) and "b" is the number we need to find.

Given:

  • LCM = 168

  • HCF = 6

  • a = 24

Substitute these values into the formula:
168 × 6 = 24 × b
1008 = 24 × b.
Now, solve for b:
b = 1008 ÷ 24 = 42.

Arun Sharma Test: Number System- 1 - Question 13
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The smallest 1-digit number to be added to the 6-digit number 910300 so that it is completely divisible by 11 is
Detailed Solution for Arun Sharma Test: Number System- 1 - Question 13

Step-by-step solution

  1. Compute remainder of 910300 when divided by 11

  • 910300 ÷ 11 leaves a remainder of 6.

  1. Make the new number divisible by 11

  • We need (910300 + d) to have remainder 0 on division by 11.

  • Currently remainder is 6, so choose d so that 6 + d becomes a multiple of 11.

  1. Pick the smallest valid d

  • The smallest multiple of 11 greater than or equal to 6 is 11.

  • Set 6 + d = 11 ⇒ d = 5.

Quick check
910300 + 5 = 910305, and 910305 ÷ 11 = 82,755 exactly.

Arun Sharma Test: Number System- 1 - Question 14
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The smallest natural number which is divisible by 16, 88, 12 and 22 is:
Detailed Solution for Arun Sharma Test: Number System- 1 - Question 14

The smallest natural number which is divisible by all given numbers is the Least Common Multiple (LCM).

Calculation:

Prime factorization:
16 = 24
88 = 23 × 11
12 = 22 × 3
22 = 2 × 11

LCM is found by taking the highest power of each prime:
LCM = 24 × 3 × 11
⇒ LCM = 16 × 3 × 11
⇒ LCM = 48 × 11
⇒ LCM = 528

The smallest natural number which is divisible by 16, 88, 12, and 22 is 528.

Arun Sharma Test: Number System- 1 - Question 15
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 How many pairs of integers (x, y) exist such that the product of x, y and HCF(x, y) = 1080?
Detailed Solution for Arun Sharma Test: Number System- 1 - Question 15

  1. Let g = HCF(x, y). Write x = g·d, y = g·h with gcd(d, h) = 1.
    Then x·y·HCF(x, y) = (g d)(g h)g = g³ d h = 1080.

  2. Prime factorization: 1080 = 2³ · 3³ · 5.
    For g³ | 1080, g can only use primes with exponents ≤ 1 (since exponents triple):
    g ∈ {1, 2, 3, 6}.

  3. For each g, N = 1080 / g³ must be split as d·h with gcd(d, h) = 1.
    If N has k distinct primes, the number of ordered coprime splits (d, h) is 2^k
    (each prime-power goes wholly to d or to h).

  • g = 1 ⇒ N = 1080 (primes {2,3,5}, k=3) ⇒ 2³ = 8 ordered pairs

  • g = 2 ⇒ N = 135 = 3³·5 (primes {3,5}, k=2) ⇒ 4 ordered pairs

  • g = 3 ⇒ N = 40 = 2³·5 (primes {2,5}, k=2) ⇒ 4 ordered pairs

  • g = 6 ⇒ N = 5 (primes {5}, k=1) ⇒ 2 ordered pairs

Total ordered pairs = 8 + 4 + 4 + 2 = 18.
Since d ≠ h in all cases (N ≠ 1), unordered pairs = 18 / 2 = 9.

Hence, 9 pairs.

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