Speed & Distance - Online Quiz


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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.

Questions and Answers

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

A - 20.65 m

B - 206.5 m

C - 2065 m

D - 20650 m

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:

A - 3 km/hr

B - 4km/hr

C - 4.5 km/hr

D - 9 km/hr

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?

A - 5 hrs.

B - 6 hrs.

C - 7 hrs.

D - 12 hrs.

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?

A - 150 m

B - 250 m

C - 300 m

D - 400 m

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)2m = √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

A - 1/r = (p-q)

B - r = (p-q)

C - 1/r = (p+q)

D - r = (p+q)

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:

A - 25 ares

B - 30 ares

C - 50 ares

D - 60 ares

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 )m2= 5000 m2
= 5000/100 ares = 50 ares {1 are= 100 m2}

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:

A - 9 hours

B - 7 hours

C - 8 hours

D - 13/2 hours

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:

A - 445 kms

B - 444 kms

C - 440 kms

D - 450 kms

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:

A - 10 km/hr

B - 20 km/hr

C - 25 km/hr

D - 30 km/hr

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

A - 60 km/hr

B - 66.67 km/hr

C - 69.23 km/hr

D - 65 km/hr

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

aptitude_speed_distance.htm
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