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. |

- If \(x = 0.125\) then find the value of \(\sqrt{9x^2 - 12x + 4} + 5x\)?
- 3.25
- 2.75
- 2.50
- 2.25

Answer: (d) 2.25

Solution: $$ \sqrt{9x^2 - 12x + 4} + 5x $$ $$ = \sqrt{(2 - 3x)^2} + 5x $$ $$ = (2 - 3x) + 5x $$ $$ = 2 + 2x $$ $$ = 2 \ (1 + x) $$ $$ = 2 \ (1 + 0.125) $$ $$ = 2 \times 1.125 $$ $$ = 2.25 $$

Solution: $$ \sqrt{9x^2 - 12x + 4} + 5x $$ $$ = \sqrt{(2 - 3x)^2} + 5x $$ $$ = (2 - 3x) + 5x $$ $$ = 2 + 2x $$ $$ = 2 \ (1 + x) $$ $$ = 2 \ (1 + 0.125) $$ $$ = 2 \times 1.125 $$ $$ = 2.25 $$

- If \(5 \sqrt{2} + \sqrt{72} = 66\) then find the value of \(\sqrt{50} + 3 \sqrt{2}\)?
- 45
- 48
- 52
- 54

Answer: (b) 48

Solution: $$ 5 \sqrt{2} + \sqrt{72} = 66 $$ $$ 5 \sqrt{2} + \sqrt{36 \times 2} = 66 $$ $$ 5 \sqrt{2} + 6 \sqrt{2} = 66 $$ $$ 11 \sqrt{2} = 66 $$ $$ \sqrt{2} = 6 $$ Now the value of \(\sqrt{50} + 3 \sqrt{2}\). $$ = \sqrt{25 \times 2} + 3 \sqrt{2} $$ $$ = 5 \sqrt{2} + 3 \sqrt{2} $$ $$ = 8 \sqrt{2} $$ $$ = 8 \times 6 = 48 $$

Solution: $$ 5 \sqrt{2} + \sqrt{72} = 66 $$ $$ 5 \sqrt{2} + \sqrt{36 \times 2} = 66 $$ $$ 5 \sqrt{2} + 6 \sqrt{2} = 66 $$ $$ 11 \sqrt{2} = 66 $$ $$ \sqrt{2} = 6 $$ Now the value of \(\sqrt{50} + 3 \sqrt{2}\). $$ = \sqrt{25 \times 2} + 3 \sqrt{2} $$ $$ = 5 \sqrt{2} + 3 \sqrt{2} $$ $$ = 8 \sqrt{2} $$ $$ = 8 \times 6 = 48 $$

- There are three consecutive road crossings at which traffic lights change after every 30 seconds, 40 seconds, and 50 seconds respectively. If the lights change simultaneously at 6:30:00 hours, then at what time will they again change simultaneously?
- 6:50:00 hours
- 6:45:00 hours
- 6:40:00 hours
- 6:35:00 hours

Answer: (c) 6:40:00 hours

Solution: LCM of 30, 40, and 50 is 600.

Hence after every 600 seconds, lights will change simultaneously. $$ 600 \ seconds = \frac{600}{60} \ minutes $$ $$ = 10 \ minutes $$ So after 10 minutes at 6:40:00 hours, the lights will change again simultaneously.

Solution: LCM of 30, 40, and 50 is 600.

Hence after every 600 seconds, lights will change simultaneously. $$ 600 \ seconds = \frac{600}{60} \ minutes $$ $$ = 10 \ minutes $$ So after 10 minutes at 6:40:00 hours, the lights will change again simultaneously.

- Three men A, B, and C start to run at the same time in the same direction to run around a circular stadium. A completes a round in 160 seconds, B in 180 seconds, and C in 140 seconds, all starting from the same point. After what time will they meet again at the starting point?
- 168 minutes
- 160 minutes
- 158 minutes
- 156 minutes

Answer: (a) 168 minutes

Solution: LCM of 160, 180, and 140 is 10080.

Hence they will meet after 10080 seconds again at the starting point. $$ 10080 \ seconds = \frac{10080}{60} \ minutes $$ $$ = 168 \ minutes $$

Solution: LCM of 160, 180, and 140 is 10080.

Hence they will meet after 10080 seconds again at the starting point. $$ 10080 \ seconds = \frac{10080}{60} \ minutes $$ $$ = 168 \ minutes $$

- What fraction must be subtracted from the sum of \(\frac{1}{5}\) and \(\frac{1}{3}\) to have an average of \(\frac{1}{9}\) of all three fractions?
- \(\frac{1}{3}\)
- \(\frac{1}{5}\)
- \(\frac{1}{9}\)
- \(\frac{1}{6}\)

Answer: (b) \(\frac{1}{5}\)

Solution: Let the fraction subtracted is \(x\) then. $$ \frac{1}{5} + \frac{1}{3} - x = 3 \times \frac{1}{9} $$ $$ x = \frac{1}{5} + \frac{1}{3} - \frac{1}{3} $$ $$ x = \frac{1}{5} $$

Solution: Let the fraction subtracted is \(x\) then. $$ \frac{1}{5} + \frac{1}{3} - x = 3 \times \frac{1}{9} $$ $$ x = \frac{1}{5} + \frac{1}{3} - \frac{1}{3} $$ $$ x = \frac{1}{5} $$

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
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