Number System Aptitude Questions and Answers:


Overview:


Questions and Answers Type:MCQ (Multiple Choice Questions).
Main Topic:Quantitative Aptitude.
Quantitative Aptitude Sub-topic:Number System Aptitude Questions and Answers.
Number of Questions:10 Questions with Solutions.

  1. Find the remainder of the expression \(\frac{125 \times 245 \times 255}{10}\)?

    1. 2
    2. 3
    3. 5
    4. 9


Answer: (c) 5

Solution: According to basic remainder theorem,$$\frac{125 \times 245 \times 255}{10}$$ $$= \frac{5 \times 5 \times 5}{10} $$ $$ = \frac{125}{10} = 5 \ (Remainder)$$

  1. Find the remainder of the expression \(\frac{9^{99}}{10}\)?

    1. 2
    2. 3
    3. 6
    4. 9


Answer: (d) 9

Solution: According to polynomial remainder theorem, $$\frac{9^{99}}{10} = \frac{9^{98} \times 9}{10}$$ $$= \frac{(9^2)^{49} \times 9}{10} $$ $$= \frac{(81)^{49} \times 9}{10} = \frac{(1)^{49} \times 9}{10} $$ $$ = \frac{9}{10} = 9 \ (Remainder)$$

  1. Find the factors of composite number \(850\)?

    1. 10
    2. 12
    3. 15
    4. 18


Answer: (b) 12

Solution: \(850 = 2 \times 5 \times 5 \times 17 = 2^1 \times 5^2 \times 17^1\)

\(factors \ of \ composite \ number\)

\(= (1 + 1) \ (2 + 1) \ (1 + 1) = 2 \times 3 \times 2 = 12\)

  1. Find the smallest positive number which is exactly divisible by 12, 16, and 28?

    1. 286
    2. 336
    3. 368
    4. 372


Answer: (b) 336

Solution:

Step(1): Factorize the numbers into their prime factors.$$ 12 = 2 \times 2 \times 3 = 2^{2} \times 3^{1} $$ $$ 16 = 2 \times 2 \times 2 \times 2 = 2^{4} $$ $$ 28 = 2 \times 2 \times 7 = 2^{2} \times 7^{1} $$ Step(2): Collect all the distinct factors with their maximum available power.$$ = 2^{4}, \ 3^{1}, \ 7^{1} $$ Step(3): Multiply the collected factors.$$ = 2^{4} \times 3^{1} \times 7^{1} $$ $$ = 16 \times 3 \times 7 = 336 $$ Here, 336 is the smallest positive number which is exactly divisible by 12, 16, and 28.


  1. Find the LCM of \(\frac{9}{2}\) and \(\frac{7}{3}\)?

    1. 1
    2. 72
    3. 63
    4. 54


Answer: (c) 63

Solution: \(LCM \ of \ fractions = \frac{LCM \ of \ Numerators}{HCF \ of \ Denominators}\) $$ LCM = \frac{LCM \ of \ (9,7)}{HCF \ of \ (2,3)} $$ $$ = \frac{63}{1} = 63 $$

  1. Which one of the following is the HCF of \(\frac{8}{7}\), 5 and \(\frac{5}{4}\)?

    1. \(\frac{1}{28}\)
    2. \(\frac{1}{27}\)
    3. \(\frac{1}{20}\)
    4. \(\frac{1}{15}\)


Answer: (a) \(\frac{1}{28}\)

Solution: \(HCF \ of \ fractions = \frac{HCF \ of \ Numerators}{LCM \ of \ Denominators}\)
$$ HCF = \frac{HCF \ of \ (8,5,5)}{LCM \ of \ (7,1,4)} $$ $$ = \frac{1}{28} $$

  1. Find the HCF of 27, 30, 60?

    1. 2
    2. 3
    3. 5
    4. 6


Answer: (b) 3

Solution:

Step(1): Factorize the numbers into their prime factors.$$ 27 = 3 \times 3 \times 3 = 3^{3} $$ $$ 30 = 2 \times 3 \times 5 = 2^{1} \times 3^{1} \times 5^{1} $$ $$ 30 = 2 \times 2 \times 3 \times 5 = 2^{2} \times 3^{1} \times 5^{1} $$ Step(2): Collect all the common factors with their minimum available power.$$ = 3^{1} = 3 $$


  1. Find the LCM of 24, 27, 30?

    1. 1080
    2. 1120
    3. 1224
    4. 1236


Answer: (a) 1080

Solution:

Step(1): Factorize the numbers into their prime factors.$$ 24 = 2 \times 2 \times 2 \times 3 = 2^{3} \times 3^{1} $$ $$ 27 = 3 \times 3 \times 3 = 3^{3} $$ $$ 30 = 2 \times 3 \times 5 = 2^{1} \times 3^{1} \times 5^{1} $$ Step(2): Collect all the distinct factors with their maximum available power.$$ = 2^{3}, \ 3^{3}, \ 5^{1} $$ Step(3): Multiply the collected factors.$$ = 2^{3} \times 3^{3} \times 5^{1} $$ $$ = 8 \times 27 \times 5 = 1080 $$


  1. If 5A94 is divisible by 9, then find the value of the smallest natural number A?

    1. 3
    2. 6
    3. 7
    4. 9


Answer: (d) 9

Solution: According to the condition, 5A94 is divisible by 9, if and only if 5 + A + 9 + 4 = 18 + A, is divisible by 9. Here A = 9, fulfill the condition.

  1. If 1A593 is divisible by 11, then find the value of the smallest natural number A?

    1. 0
    2. 2
    3. 3
    4. 7


Answer: (a) 0

Solution: According to the condition, 1A593 is divisible by 11 if and only if (1 + 5 + 3) - (A + 9) = 9 - (A + 9), is divisible by 11. Here A = 0, fulfill the condition.