# 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

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

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

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

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

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

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

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

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