Quantitative Aptitude
SPEED TIME AND DISTANCE MCQs
Time And Distance, Time & Distance, Speed Time & Distance
Total Questions : 1223
| Page 118 of 123 pages
Answer: Option A. -> 17:13
NO EXPLANATION IS AVAILABLE FOR THIS QUESTION!
NO EXPLANATION IS AVAILABLE FOR THIS QUESTION!
Answer: Option D. -> 8 kmph
HE CROSSES 0.4 KM IN 3/60TH OF AN HOUR. SPEED = 0.4/(1/20) = 8 KMPH
HE CROSSES 0.4 KM IN 3/60TH OF AN HOUR. SPEED = 0.4/(1/20) = 8 KMPH
Answer: Option C. -> 7.5s
60 KM/HR = 60 * 5/18 = 16.67 M/S
SPEED = DISTANCE/TIME; V = D/T
16.67 = 125/T
T = 7.5S
60 KM/HR = 60 * 5/18 = 16.67 M/S
SPEED = DISTANCE/TIME; V = D/T
16.67 = 125/T
T = 7.5S
Answer: Option D. -> 1 hr 12 min
NEW SPEED = 6/7 OF USUAL SPEED
SPEED AND TIME ARE INVERSELY PROPORTIONAL.
HENCE NEW TIME = 7/6 OF USUAL TIME
HENCE, 7/6 OF USUAL TIME – USUAL TIME = 12 MINUTES
=> 1/6 OF USUAL TIME = 12 MINUTES
=> USUAL TIME = 12 X 6 = 72 MINUTES
= 1 HOUR 12 MINUTES
NEW SPEED = 6/7 OF USUAL SPEED
SPEED AND TIME ARE INVERSELY PROPORTIONAL.
HENCE NEW TIME = 7/6 OF USUAL TIME
HENCE, 7/6 OF USUAL TIME – USUAL TIME = 12 MINUTES
=> 1/6 OF USUAL TIME = 12 MINUTES
=> USUAL TIME = 12 X 6 = 72 MINUTES
= 1 HOUR 12 MINUTES
Answer: Option C. -> 10
SPEED OF THE BUS EXCLUDING STOPPAGES = 54 KMPH
SPEED OF THE BUS INCLUDING STOPPAGES = 45 KMPH
LOSS IN SPEED WHEN INCLUDING STOPPAGES = 54 – 45 = 9KMPH
=> IN 1 HOUR, BUS COVERS 9 KM LESS DUE TO STOPPAGES
HENCE, TIME THAT THE BUS STOP PER HOUR = TIME TAKEN TO COVER 9 KM
=DISTANCE/SPEED=9/54 HOUR=1/6 HOUR = 60/6 MIN=10 MIN
SPEED OF THE BUS EXCLUDING STOPPAGES = 54 KMPH
SPEED OF THE BUS INCLUDING STOPPAGES = 45 KMPH
LOSS IN SPEED WHEN INCLUDING STOPPAGES = 54 – 45 = 9KMPH
=> IN 1 HOUR, BUS COVERS 9 KM LESS DUE TO STOPPAGES
HENCE, TIME THAT THE BUS STOP PER HOUR = TIME TAKEN TO COVER 9 KM
=DISTANCE/SPEED=9/54 HOUR=1/6 HOUR = 60/6 MIN=10 MIN
Answer: Option B. -> 40 seconds
THE FIRST POLE IS CROSSED AT THE ZEROTH SECOND. THE TIMER STARTS AFTER THE CAR CROSSES THE FIRST POLE. LET THE DISTANCE BETWEEN EACH POLE BE X M. TO REACH THE 16TH POLE, THE DISTANCE TRAVELLED BY THE CAR IS 15X. THE SPEED OF THE CAR IS 15X/30 = X/2 M/S TO REACH THE 21ST POLE, IT HAS TO TRAVEL A DISTANCE OF 20X. TIME TAKEN
= 20X÷(X/2)
= 40 SECONDS
THE FIRST POLE IS CROSSED AT THE ZEROTH SECOND. THE TIMER STARTS AFTER THE CAR CROSSES THE FIRST POLE. LET THE DISTANCE BETWEEN EACH POLE BE X M. TO REACH THE 16TH POLE, THE DISTANCE TRAVELLED BY THE CAR IS 15X. THE SPEED OF THE CAR IS 15X/30 = X/2 M/S TO REACH THE 21ST POLE, IT HAS TO TRAVEL A DISTANCE OF 20X. TIME TAKEN
= 20X÷(X/2)
= 40 SECONDS
Answer: Option A. -> 3 hours 20 minutes
LET HIS USUAL SPEED BE X KMPH, THE DISTANCE TO HIS OFFICE BE Y KM AND HIS USUAL TRAVEL TIME BE T HRS. Y = XT = (5/6)X * (T+(40/60)) SOLVING THE EQUATION,
T = 3.33 HRS = 3 HRS 20 MINUTES
LET HIS USUAL SPEED BE X KMPH, THE DISTANCE TO HIS OFFICE BE Y KM AND HIS USUAL TRAVEL TIME BE T HRS. Y = XT = (5/6)X * (T+(40/60)) SOLVING THE EQUATION,
T = 3.33 HRS = 3 HRS 20 MINUTES
Answer: Option C. -> 16 km
NO EXPLANATION IS AVAILABLE FOR THIS QUESTION!
NO EXPLANATION IS AVAILABLE FOR THIS QUESTION!
Answer: Option D. -> 7.2
NO EXPLANATION IS AVAILABLE FOR THIS QUESTION!
NO EXPLANATION IS AVAILABLE FOR THIS QUESTION!
Answer: Option C. -> 66 km/hr
THE RELATIVE SPEED OF A AND B IS 6 KM/HR = 1.67 M/S
AS THE CAR PASSES A AFTER 10S, THE DISTANCE BETWEEN A AND B AFTER 10S (I.E. AT 11TH SECOND) IS THE DISTANCE COVERED BY CAR IN 1 SECOND.
THEREFORE, AT T = 11, D = 1.67 * 11
D = 18.33 M
V = D/T = 18.33/1 = 18.33M/S
V = 66 KM/HR
THE RELATIVE SPEED OF A AND B IS 6 KM/HR = 1.67 M/S
AS THE CAR PASSES A AFTER 10S, THE DISTANCE BETWEEN A AND B AFTER 10S (I.E. AT 11TH SECOND) IS THE DISTANCE COVERED BY CAR IN 1 SECOND.
THEREFORE, AT T = 11, D = 1.67 * 11
D = 18.33 M
V = D/T = 18.33/1 = 18.33M/S
V = 66 KM/HR
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