Time, Speed and Distance Solved Examples - 2 - CSAT Preparation - UPSC

Question 8: Car A trails car B by 50 meters. Car B travels at 45km/hr. Car C travels from the opposite direction at 54km/hr. Car C is at a distance of 220 meters from Car B. If car A decides to overtake Car B before cars B and C cross each other, what is the minimum speed at which car A must travel? 
A. 36 km/hr
B. 45 km/hr
C. 67.5 km/hr
D. 18 km/hr
Ans: (C)
Sol:
Cars B and C move towards each other, so their relative speed = 45 + 54 = 99 km/hr.
Convert to m/s: 99 × 5/18 = 27.5 m/s.
Distance between B and C = 220 m, so time available before B and C meet = 220 / 27.5 = 8 seconds.
Car A must close a gap of 50 m relative to car B in at most 8 s, so required relative speed = 50 / 8 = 6.25 m/s.
Convert 6.25 m/s to km/hr: 6.25 × 18/5 = 22.5 km/hr.
Thus car A's speed must exceed car B's speed by 22.5 km/hr: required speed = 45 + 22.5 = 67.5 km/hr.
Hence the minimum speed is 67.5 km/hr.

Question 9: A and B stand at distinct points of a circular race track of length 120m. They run at speeds of a m/s and b m/s respectively. They meet for the first time 16 seconds after they start the race and for the second time 40 seconds from the time they start the race. Now, if B had started in the opposite direction to the one he had originally started, they would have meet for the first time after 40 seconds. If B is quicker than A, find B's speed.
A. 3 m/s
B. 4 m/s
C. 5 m/s
D. 8 m/s
Ans: (B)
Sol:
Distance of one lap = 120 m. Time between the first and second meeting = 40 - 16 = 24 s.
Relative speed (in the original scenario) = Lap length / time between consecutive meetings = 120 / 24 = 5 m/s. (1)
If B had started in the opposite direction, they would meet first after 40 s. In that reversed case their relative speed would be 120 / 40 = 3 m/s. (2)

The change from the original to the reversed case switches whether they are moving towards each other or in the same direction. Since reversing B reduces the relative speed from 5 to 3, the original scenario must have been them moving towards each other (so original relative speed = a + b = 5) and reversing B makes them move in the same direction (so relative speed = |b - a| = 3).
Given B is quicker than A, b - a = 3 and a + b = 5. Add the two equations: 2b = 8 ⇒ b = 4 m/s.
Hence B's speed is 4 m/s.

Question 10: City A to City B is a downstream journey on a stream which flows at a speed of 5km/hr. Boats P and Q run a shuttle service between the two cities that are 300 kms apart. Boat P, which starts from City A has a still-water speed of 25km/hr, while boat Q, which starts from city B at the same time has a still-water speed of 15km/hr. When will the two boats meet for the first time? (this part is easy) When and where will they meet for the second time?
A. 7.5 hours and 15 hours
B. 7.5 hours and 18 hours
C. 8 hours and 18 hours
D. 7.5 hours and 20 hours
Ans: (D)
Sol:

Boat P starts from city A and Q starts from city B.
When boat P travels downstream, it will effectively have a speed to 30kmph. Likewise, Q will have an effective speed of 10kmph. The relative speed = 40kmph. So, the two boats will meet for the first time after 300/40 hours (Distance/relative speed) = 7.5 hours (Actually, for this part we do not need the speed of the stream)

The second part is more interesting, because the speed of the boats change when they change direction. Boat P is quicker, so it will reach the destination sooner. Boat P will reach City B in 10 hours 300/30. When boat P reaches city B, boat Q will be at a point 100kms from city B.
After 10 hours, both P and Q will be travelling upstream,
P's speed = 20 km/hr
Q's speed = 10 km/hr
Relative speed = 10km/hr
Q is ahead of P by 100 kms
P will catch up with Q after 10 more hours Distance/RelativeSpeed − 100/10.
So, P and Q will meet after 20 hours at a point 200 kms from city B
The question is " When will the two boats meet for the first time? When and where will they meet for the second time?"
Hence, the answer is 7.5 hours and 20 hours

Question 11: Cities M and N are 600km apart. Bus A starts from city M towards N at 9AM and bus B starts from city N towards M at the same time. Bus A travels the first one-third of the distance at a speed of 40kmph, the second one-third at 50kmph and the third one-third at 60kmph. Bus B travels the first one-third of the total time taken at a speed of 40kmph, the second one-third at 50kmph and the third one-third at 60kmph. When and where will the two buses cross each other?
A. 300 kms from M
B. 280 kms from M
C. 305 kms from M
D. 295 kms from M
Ans: (D)
Sol:
Bus A
Travels 200 km at  40 kmph
the next 200 km at  50 kmph and
the final 200 km at 60 kmph
So, Bus A will be at a distance of 200km from city M after 5 hours, and at a distance of 400km after 9 hours, and reach N after 12 hours and 20 mins
Bus B
Travels at an overall average speed of 50kmph, so it will take 12 hours for the entire trip.
So, Bus B will travel 160 kms in the first 4 hours, 200 kms in the next 4 hours and 240 in the final 4 hours. So, both buses cross each other when they are in their middle legs.
After 5 hours, bus A will be at a position 200 kms from city M. At the same time, bus B will be at a distance 210 kms from city N (4 x 40 + 50).
The distance between them will be 190 kms (600 - 200 - 210). 
Relative speed = Sum of the two speeds = 50 + 50 = 100 kmph.
Time taken = 190/100= 1.9 hours. = 1 hour and 54 minutes. 
So, the two buses will meet after 6 hours and 54 minutes. 
Distance covered by Bus B by the meeting instant = 210 + (50 × 1.9) = 210 + 95 = 305 km from N, so the meeting point is 305 km from N and 600 - 305 = 295 km from M.
Hence, they cross 295 km from city M.

Question 12: A car of length 4m wants to overtake a trailer truck of length 20m travelling at 36km/hr within 10 seconds. At what speed should the car travel?A. 12 m/s
B. 14.8 m/s
C. 12.4 m/s
D. 7.6 m/s
Ans: (C)
Sol:
Total distance to be covered by the car relative to the truck = 20 + 4 = 24 m.
Speed of truck = 36 km/hr = 36 × (5/18) = 10 m/s.
Let speed of car = S m/s. Relative speed required = S - 10 (since car must be faster).
Time to overtake = distance / relative speed ⇒ 10 = 24 / (S - 10) ⇒ S - 10 = 24 / 10 = 2.4.
S = 12.4 m/s.
Hence the car should travel at 12.4 m/s.

Question 13: Train A travelling at 63 kmph takes 27 to sec to cross Train B when travelling in opposite direction whereas it takes 162 seconds to overtake it when travelling in the same direction. If the length of train B is 500 meters, find the length of Train A.
A. 400 m
B. 810 m
C. 500 m
D. 310 m
Ans: (D)
Sol:
Let the length of Train A be x meters
Let speed of Train B be y kmph
Relative distance = Relative speed * time taken to cross/overtake

Crossing scenario:
Relative speed of 2 trains = 63 + y
Time taken to cross = 27 sec or 27/3600 hrs
Relative distance between 2 trains = Length of Train A + length of train B = (x + 0.5) km
Therefore, x + 0.5 = (63 + y) * 27 / 3600 ----- (1)

Overtaking scenario:
Relative speed of 2 trains = 63 – y
Time taken to overtake = 162 sec or 162/3600 hrs
Relative distance between 2 trains = x + 0.5
Therefore, x + 0.5 = (63 – y) * 162/3600 --- (2)
From (1) and (2), solve for y.
(63 + y) * 27 = (63 – y) * 162
27y + 162 y = 63*162 – 63 *27
189y = 63 * 135 or y = 45 kmph

Substitute in (2) to get x.
x + 0.5 = (63 – 45) * 162/3600
Or x = 0.31 km or 310 meters
The question is "If the length of train B is 500 meters, find the length of Train A."
Hence, the answer is 310 m

Question 14: P cycles at a speed of 4 m/s for the first 8 seconds, 5 m/s for the next 8 seconds, 6 m/s for the next 8 and so on. Q cycles at a constant speed of 6.5 m/s throughout. If P and Q had to cycle for a 400 m race, how much lead in terms of distance, can P give Q and still finish at the same time as Q?
A. 43.4 m
B. 56.6 m
C. 30.8 m
D. P cannot give a lead as Q is always ahead of P
Ans: (C)
Sol:
Speeds of P in consecutive 8 s blocks are: 4, 5, 6, 7, 8, 9, … m/s.
Distance covered by P:

• In first 8 s = 4 × 8 = 32 m
• In first 16 s = 32 + 5 × 8 = 72 m
• In first 24 s = 72 + 6 × 8 = 120 m
• In first 32 s = 120 + 7 × 8 = 176 m
• In first 40 s = 176 + 8 × 8 = 240 m
• In first 48 s = 240 + 9 × 8 = 312 m

Remaining distance = 400 − 312 = 88 m.
Next speed = 10 m/s, so time to cover 88 m = 88 / 10 = 8.8 s.
Total time taken by P = 48 + 8.8 = 56.8 s.
Distance covered by Q in this time = 6.5 × 56.8 = 369.2 m.
Lead P can give Q = 400 − 369.2 = 30.8 m.
Hence, the correct answer is option C. 

Question 15: A bus starts from a bus stop P and goes to another bus stop Q. In between P and Q, there is a bridge AB of certain length. A man is standing at a point C on the bridge such that AC:CB = 1:3. When the bus starts from P and if the man starts running towards A, he will meet the bus at A. But if he runs towards B, the bus will overtake him at B. Which of the following is true?
A. Bus travels 3x times faster than the man
B. Bus travels 2x times faster than the man
C. The bus and the man travel at the same speed
D. 4x the speed of the man is equal to 3x the speed of the bus
Ans: (B)
Sol:
Let the speed of the bus be ‘b’ and man be ‘m’.
Let the distance between P and A be y.
AC:CB = 1:3 implies AC = x and CB = 3x

When bus goes from P to A and the man from C to A, time taken by both are equal and they meet at A.
y/b = x/m or
b/m = y/x --- (1)

When bus goes from P to B and the man from C to B, again time taken by both are equal and they meet at B.
(y+x+3x)/b = 3x/m
b/m = (4x+y)/3x ---(2)

Equating (1) and (2), we get y/x = (4x+y)/3x
3y = 4x + y or y = 2x

Therefore ratio of speeds b/m = 2x/x or 2 : 1
The question is "When the bus starts at P and if the man starts running towards A, he will meet the bus at A. But if he runs towards B, the bus will overtake him at B. Which of the following is true?"
Hence, the Bus travels 2x times faster than the man.