Quantitative Aptitude
SPEED TIME AND DISTANCE MCQs
Time And Distance, Time & Distance, Speed Time & Distance
Total Questions : 1223
| Page 120 of 123 pages
Answer: Option D. -> 4:3
IF THE RATIO OF THE SPEEDS OF TWO OBJECTS IS A:B, THEN THE TIME TAKEN BY THEM TO COVER THE SAME DISTANCE IS
B:A HENCE, THE ANSWER IS 4:3
IF THE RATIO OF THE SPEEDS OF TWO OBJECTS IS A:B, THEN THE TIME TAKEN BY THEM TO COVER THE SAME DISTANCE IS
B:A HENCE, THE ANSWER IS 4:3
Answer: Option B. -> 10
DUE TO STOPPAGES, THE BUS TRAVELS ONLY 45 KMS IN AN HOUR (9 KMS LESS). TO COVER A DISTANCE OF 9 KM AT A SPEED OF 54 KMPH, TIME TAKEN
= 9/54 = 1/6 HRS = 10 MINS.
DUE TO STOPPAGES, THE BUS TRAVELS ONLY 45 KMS IN AN HOUR (9 KMS LESS). TO COVER A DISTANCE OF 9 KM AT A SPEED OF 54 KMPH, TIME TAKEN
= 9/54 = 1/6 HRS = 10 MINS.
Answer: Option A. -> 30 kmph
DISTANCE = 24*50/60 = 20 KM NEW SPEED = 20/(40/60)) = 30 KMPH
DISTANCE = 24*50/60 = 20 KM NEW SPEED = 20/(40/60)) = 30 KMPH
Answer: Option B. -> 320
WITH A FULL TANK, IT CAN TRAVEL
360*4/3 = 480 KM WITH 2/3 FULL TANK, IT CAN TRAVEL
480*2/3 = 320 KM
WITH A FULL TANK, IT CAN TRAVEL
360*4/3 = 480 KM WITH 2/3 FULL TANK, IT CAN TRAVEL
480*2/3 = 320 KM
Answer: Option A. -> 17 hr
RELATIVE SPEED = 5.5 – 5 = .5 KMPH (BECAUSE THEY WALK IN THE SAME DIRECTION)
DISTANCE = 8.5 KM
TIME = DISTANCE/SPEED=8.5/.5=17 HR
RELATIVE SPEED = 5.5 – 5 = .5 KMPH (BECAUSE THEY WALK IN THE SAME DIRECTION)
DISTANCE = 8.5 KM
TIME = DISTANCE/SPEED=8.5/.5=17 HR
Answer: Option B. -> 24
LET T BE HIS USUAL TIME TO REACH HIS OFFICE AND V BE HIS USUAL SPEED.
V = D/T ……….(D IS THE DISTANCE AFNAN TRAVELS WHILE GOING TO HIS OFFICE)
VT = D
AT V1 = 4V/5 ; T1 = T + 6
4V/5 = D/(T + 6)
4V/5* (T + 6) = D
4V/5* (T + 6) = VT
ON SOLVING WE GET,
T = 24 MINUTES
LET T BE HIS USUAL TIME TO REACH HIS OFFICE AND V BE HIS USUAL SPEED.
V = D/T ……….(D IS THE DISTANCE AFNAN TRAVELS WHILE GOING TO HIS OFFICE)
VT = D
AT V1 = 4V/5 ; T1 = T + 6
4V/5 = D/(T + 6)
4V/5* (T + 6) = D
4V/5* (T + 6) = VT
ON SOLVING WE GET,
T = 24 MINUTES
Answer: Option B. -> 40 km
X = 7 KM/HR ; Y = 3 KM/HR
DS = 10 KM/HR ; US = 4 KM/HR
DISTANCE (D) IS SAME. THEREFORE, IF TIME TAKEN FOR DOWNSTREAM IS T HOURS, THE TIME FOR UPSTREAM IS (T + 6) HOURS.
10*T = 4*(T + 6)
6T = 24 ; T = 4 HOURS
D = 10*4 = 40 KM
X = 7 KM/HR ; Y = 3 KM/HR
DS = 10 KM/HR ; US = 4 KM/HR
DISTANCE (D) IS SAME. THEREFORE, IF TIME TAKEN FOR DOWNSTREAM IS T HOURS, THE TIME FOR UPSTREAM IS (T + 6) HOURS.
10*T = 4*(T + 6)
6T = 24 ; T = 4 HOURS
D = 10*4 = 40 KM
Answer: Option A. -> (7 , 3) km/hr
IF X: SPEED OF BOATS MAN IN STILL WATER
Y: SPEED OF THE RIVER
DOWNSTREAM SPEED (DS) = X + Y
UPSTREAM SPEED (US) = X – Y
X = (DS + US) / 2
Y = (DS – US) / 2
IN THE ABOVE PROBLEM DS = 10 ; US = 4
X = (10 + 4) / 2 = 14/2 = 7 KM/HR
Y = (10 – 4)/2 = 6/2 = 3 KM/HR
IF X: SPEED OF BOATS MAN IN STILL WATER
Y: SPEED OF THE RIVER
DOWNSTREAM SPEED (DS) = X + Y
UPSTREAM SPEED (US) = X – Y
X = (DS + US) / 2
Y = (DS – US) / 2
IN THE ABOVE PROBLEM DS = 10 ; US = 4
X = (10 + 4) / 2 = 14/2 = 7 KM/HR
Y = (10 – 4)/2 = 6/2 = 3 KM/HR
Question 1199. Taimoor left for his school at 6 am by foot and walked at the rate of 2 kmph. After reaching his school he found it was closed and immediately turned back and started walking back to his home at 3 kmph. If he reached his home at 9 am, find the distance between his school and his home.
Answer: Option D. -> 3.6 km
AVERAGE SPEED OF TAIMOOR = (2*2*3)/(2 + 3) = 2.4 KM/HR
IF D IS HE DISTANCE BETWEEN THE SCHOOL AND TAIMOOR’S HOME, THEN TOTAL DISTANCE WALKED BY TAIMOOR = 2D, AND THE TOTAL TIME TAKEN IS 3 HOURS
2.4 = 2D/3
7.2 = 2D
D = 3.6 KM
AVERAGE SPEED OF TAIMOOR = (2*2*3)/(2 + 3) = 2.4 KM/HR
IF D IS HE DISTANCE BETWEEN THE SCHOOL AND TAIMOOR’S HOME, THEN TOTAL DISTANCE WALKED BY TAIMOOR = 2D, AND THE TOTAL TIME TAKEN IS 3 HOURS
2.4 = 2D/3
7.2 = 2D
D = 3.6 KM
Answer: Option C. -> 7
RELATIVE SPEED = SPEED OF A + SPEED OF B (∴ THEY WALK IN OPPOSITE DIRECTIONS)
= 2 + 3 = 5 ROUNDS PER HOUR
=> THEY CROSS EACH OTHER 5 TIMES IN 1 HOUR AND 2 TIMES IN 1/2 HOUR
TIME DURATION FROM 8 AM TO 9.30 AM = 1.5 HOUR
HENCE THEY CROSS EACH OTHER 7 TIMES BEFORE 9.30 AM
RELATIVE SPEED = SPEED OF A + SPEED OF B (∴ THEY WALK IN OPPOSITE DIRECTIONS)
= 2 + 3 = 5 ROUNDS PER HOUR
=> THEY CROSS EACH OTHER 5 TIMES IN 1 HOUR AND 2 TIMES IN 1/2 HOUR
TIME DURATION FROM 8 AM TO 9.30 AM = 1.5 HOUR
HENCE THEY CROSS EACH OTHER 7 TIMES BEFORE 9.30 AM
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