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. |

- Find the remainder of the expression \(\frac{125 \times 245 \times 255}{10}\)?
- 2
- 3
- 5
- 9

Answer: (c) 5

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

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

- Find the remainder of the expression \(\frac{9^{99}}{10}\)?
- 2
- 3
- 6
- 9

Answer: (d) 9

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

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

- Find the factors of composite number \(850\)?
- 10
- 12
- 15
- 18

Answer: (b) 12

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

\(factors \ of \ composite \ number\)

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

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

\(factors \ of \ composite \ number\)

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

- Find the smallest positive number which is exactly divisible by 12, 16, and 28?
- 286
- 336
- 368
- 372

Answer: (b) 336

Solution:

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.

- Find the LCM of \(\frac{9}{2}\) and \(\frac{7}{3}\)?
- 1
- 72
- 63
- 54

Answer: (c) 63

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

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

- Which one of the following is the HCF of \(\frac{8}{7}\), 5 and \(\frac{5}{4}\)?
- \(\frac{1}{28}\)
- \(\frac{1}{27}\)
- \(\frac{1}{20}\)
- \(\frac{1}{15}\)

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

Solution: \(HCF \ of \ fractions = \frac{HCF \ of \ Numerators}{LCM \ of \ Denominators}\)

$$ HCF = \frac{HCF \ of \ (8,5,5)}{LCM \ of \ (7,1,4)} $$ $$ = \frac{1}{28} $$

Solution: \(HCF \ of \ fractions = \frac{HCF \ of \ Numerators}{LCM \ of \ Denominators}\)

$$ HCF = \frac{HCF \ of \ (8,5,5)}{LCM \ of \ (7,1,4)} $$ $$ = \frac{1}{28} $$

- Find the HCF of 27, 30, 60?
- 2
- 3
- 5
- 6

Answer: (b) 3

Solution:

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.$$ = 3^{1} = 3 $$

- Find the LCM of 24, 27, 30?
- 1080
- 1120
- 1224
- 1236

Answer: (a) 1080

Solution:

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.$$ = 2^{3} \times 3^{3} \times 5^{1} $$ $$ = 8 \times 27 \times 5 = 1080 $$

- If 5A94 is divisible by 9, then find the value of the smallest natural number A?
- 3
- 6
- 7
- 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.

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.

- If 1A593 is divisible by 11, then find the value of the smallest natural number A?
- 0
- 2
- 3
- 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.

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.

Lec 1: Introduction to Number System
Lec 2: Factors of Composite Number
Questions and Answers-1
Lec 3: Basic Remainder Theorem
Questions and Answers-2
Lec 4: Polynomial Remainder Theorem
Questions and Answers-3
Questions and Answers-4
Questions and Answers-5
Lec 5: LCM of Numbers
Questions and Answers-6
Lec 6: HCF of Numbers
Questions and Answers-7
Questions and Answers-8
Lec 7: Divisibility Rules of Numbers
Questions and Answers-9
Questions and Answers-10
Questions and Answers-11
Questions and Answers-12
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