- Aptitude Test Preparation
- Aptitude - Home
- Aptitude - Overview
- Quantitative Aptitude

- Aptitude Useful Resources
- Aptitude - Questions & Answers

- Selected Reading
- UPSC IAS Exams Notes
- Developer's Best Practices
- Questions and Answers
- Effective Resume Writing
- HR Interview Questions
- Computer Glossary
- Who is Who

# Speed & Distance - Online Quiz

Following quiz provides Multiple Choice Questions (MCQs) related to **Speed & Distance**. You will have to read all the given answers and click over the correct answer. If you are not sure about the answer then you can check the answer using **Show Answer** button. You can use **Next Quiz** button to check new set of questions in the quiz.

Q 1 - A train goes at 82.6km/hr. What number of meters will it go in 15 minutes?

### Answer : D

### Explanation

82.6 km/hr = (82.6*5/18)m/sec = 413/18 m/sec Distance covered in 15 min = (413/18*15 *60) m =20650 m

Q 2 - A man covers half of his adventure at 6km/hr and the staying half at 3 km/hr. His normal rate is:

### Answer : B

### Explanation

Average speed = 2xy/(x+y) km/hr = 2*6*3/(6+3)km/hr = 4 km/hr

Q 3 - Two train approach one another at 30 km/hr and 27 km/hr from two spot 342 km separated. After how long will they meet?

### Answer : B

### Explanation

Suppose the two trains meet after x hours. Then, 30x+27 x= 342 ⇒ 57 x = 342 ⇒ x = 342/57 = 6. So the two trains will meet after 6 hours.

Q 4 - Two auto begins in the meantime from one point and moves along two streets at right edges to one another. Their rates are 36 km/hr and 48 km/hr individually. After 15 sec, the contrast between them will be?

### Answer : B

### Explanation

36 km/hr = (36*5/18)m/sec= 10 m/sec. Distance covered in 15 sec. A= (10*15) m = 150 m 48 km/hr = (48*5/18) m/sec = 40/3 m/sec. Distance covered in 15 sec. = B = (40/3 *15) m = 200 m Distance between A and B = AB= √ (150)^{2}+ (200)^{2}m = √62500 m = 250 m

Q 5 - A kid is running at a velocity of p km/hr to cover a separation of 1 km. in any case, , because of dangerous ground , his velocity is lessened by q km/hr (p>q). On the off chance that he takes r hours to cover the separation, them

### Answer : A

### Explanation

Actual speed = (p-q) km/hr, time taken = r hrs. Distance = (speed*time) ∴ 1 = (p-q) r ⇒1/r = (p-q)

Q 6 - A man strolling at 3 km/hr crosses a square field corner to corner in 2 minutes. The zone of the field is:

### Answer : C

### Explanation

Speed =(3*5/18)m/sec. = 5/6 m/sec Distance covered in 2 min. = (5/6* 2* 60) m = 100 m Length of the diagonal of the square field = 100 m Area = 1/2 * (diagonal)^{2}= (1/2 *100 *100 )m^{2}= 5000 m^{2}= 5000/100 ares = 50 ares {1 are= 100 m^{2}}

Q 7 - A man ventures 35 km halfway at 4 km/hr and at 5 km/hr. in the event that he covers previous separation at 5 km/hr and later separation at 4 km/hr, he could cover 2 km more in the same time. The time taken to cover the entire separation at unique rate is:

### Answer : C

### Explanation

Suppose the man covers first distance in x hrs and the second distance in y hrs. then, 4x+5 y = 35 ... (a) And 5x+4 y = 37 ...(b) On solving (a) and (b), we get x= 5, y = 3 Total time taken = (5+3) = 8 hrs.

Q 8 - Two trains begin in the meantime from Aligarh and Delhi and continue towards One another at 16 km/hr and 21 km/hr individually. When they meet, it is found that one train has voyage 60 km more than the other. The separation between the two stations is:

### Answer : B

### Explanation

Suppose they meet after x hours. Then , 21x-16 x= 60 ⇒5x =60 ⇒ x= 12 Required distance = (16*12+ 21*12) km = 444 km

Q 9 - A auto covers four progressive 3 km extends at 10 km/hr, 20 km/hr, 30 km/hr and 60 km/hr separately. Its normal rate over this separation is:

### Answer : B

### Explanation

Total distance = (3*4) km = 12 kms Total time taken = (3/10+3/20+3/30+ 3/60) = (36+18+12+6)/120 hrs. = 72/120 hrs = 3/5 hrs. Average speed = 12/ (3/5) km/hr = (12*5)/3 = 20 km/hr

Q 10 - A train voyaged separations of 10 km, 20 km and 30 km at rates of 50 km/hr, 60 km/hr and 90 km/hr separately. The normal rate of the train was

### Answer : C

### Explanation

Total distance covered = (10+20+30) km = 60 km Total time taken = (10/50 + 20/60+ 30/90) hr = (1/5+1/3+1/3) hr = 13/15 hr. Average speed = (60* 15/13) km/hr = (900/13) km/hr = 69.23 km/hr