# 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. If $$x = 0.125$$ then find the value of $$\sqrt{9x^2 - 12x + 4} + 5x$$?

1. 3.25
2. 2.75
3. 2.50
4. 2.25

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

1. If $$5 \sqrt{2} + \sqrt{72} = 66$$ then find the value of $$\sqrt{50} + 3 \sqrt{2}$$?

1. 45
2. 48
3. 52
4. 54

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 $$\sqrt{50} + 3 \sqrt{2}$$. $$= \sqrt{25 \times 2} + 3 \sqrt{2}$$ $$= 5 \sqrt{2} + 3 \sqrt{2}$$ $$= 8 \sqrt{2}$$ $$= 8 \times 6 = 48$$

1. 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?

1. 6:50:00 hours
2. 6:45:00 hours
3. 6:40:00 hours
4. 6:35:00 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.

1. 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?

1. 168 minutes
2. 160 minutes
3. 158 minutes
4. 156 minutes

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

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

1. 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?

1. $$\frac{1}{3}$$
2. $$\frac{1}{5}$$
3. $$\frac{1}{9}$$
4. $$\frac{1}{6}$$

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

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