Test Level 2: Speed, Time and Distance - 2 - CAT MCQ

Let the speed of B be V_B m/s and that of C be V_C m/s.
Now, in the same time as A runs d metres, B runs (d - 20) metres and this time is (d - 20)/V_B ..(i)
In the same time that A runs d metres, C runs (d - 28) metres and this time is (d - 28)/V_C ...(ii)
Equating the 2 times from the expressions (i) and (ii), we get (d - 20)/V_B = (d - 28)/V_C..(iii)
Also, in the time that B runs d m, C runs (d - 10) m, which is (d/V_B) = (d - 10)/V_C ...(iv)
Now, from equation (iv), we get V_B = V_Cd/(d - 10)..(v)
Plugging in this value of V_B from equation (v) into equation (iii), we get ((d - 20)(d - 10))/(V_Cd) = (d - 28)/V_C
Solving for d, we get d = 100 metres
Thus, answer option (3) is correct.

Let us assume, he changed his speed at a distance of d km from Hyderabad.
So, total time taken for the journey (both forward and return) = +  +  + 
Since we don't know d, we can't find the total time and hence, we can't find the average speed.

Let P be the speed of boat P and Q be that of boat Q.
Let s be the speed of stream.
We have P - s = 40/10 = 4 km/hr and P + s = 40/5 = 8 km/hr
Solving, we get P = 6 and s = 2
Again, Q - s = 30/5 = 6
Since s = 2, Q = 8 km/hr
Therefore, downstream speed of Q = 2 + 8 = 10 km/hr
Hence, time taken = 30/(10 + 4) = 30/14 = 15/7 hours = 2 hrs 8 mins (approx.)
Time at which both boats meet each other = 12:08 pm

Let the speed of the train be x kmph and that of the bike be y kmph.
Then, according to the question,
(180/x) + (120/y) = 6 ---------(1)
(200/x) + (100/y) = (6 hrs + 15 min) = 25/4 ---------(2)
On solving (1) and (2), we get
x = 40 and y = 80
Therefore, ratio = 40 : 80 = 1 : 2

Let the speed of Vikram be 4vm/min.
Speed of Mani is 3vm/min.
As Vikram takes 30 minutes to cover the distance, distance to office = d m = 4v × 30 m ...(i)
Let the two meet at t minutes after 7:20.
By that time Mani has traveled for (t + 20) minutes and has covered a distance of d₁m = 3v(t + 20) m ...(ii)
Vikram has traveled a distance of d₂ m = 4vt m ...(iii)
Between the two of them, they together have covered a distance of the total one way distance from home to office or vice versa.
Thus, d m = d₁ m + d₂ m ...(iv)
Plugging in the values of d, d₁ and d₂ from equations (i), (ii) and (iii) into (iv) in terms of v and t, and then solving for t, we get t = 9 (approx.)
Thus, they cross each other at (20 + 9) min after 7:00 a.m. or at 7:29 a.m.
Thus, option 3 is correct.

The distance between the rescue vessel and the ship, which is 12 km, has to be covered in 16 minutes (the ship can stay afloat for only 20 minutes and it takes 4 minutes to evacuate the people aboard the ship).
Therefore, the ship and the rescue vessel should move towards each other at a combined speed of 12 × (60/16) km/hr = 45 km/hr.
The ship is moving at a speed of 8 km/hr. Therefore, the rescue vessel should move at a speed of 45 - 8 = 37 km/hr.

If the person had not moved towards the source of the gunfire, he would have heard the second shot 12 minutes after the first shot.
Since the person is actually moving towards the source, the shot is now heard after 10 minutes.
It means the sound would have taken 2 more minutes to reach the initial position of the person, but this very distance was travelled by the train in 10 minutes.
It means that speed of the train is 1/5 of the speed of sound.
Or 330/5 m/s = 66 m/s

Let the speed of boat P in still water be M.
Let the speed of water be x.
Therefore;
While going upstream:
M - x = 40/10 = 4 …................................. (1)
While going downstream:
M + x = 40/5 = 8 …................................. (2)
From (1) and (2), we get
M = 6, x = 2
For boat Q:
Let the speed of boat Q be S.
While going upstream:
S - 2 = 30/5 = 6
Therefore, S = 8
Distance between points B and C is 10 km.
Boats P and Q started from A and C, respectively.
Let the distance covered by boat P from point A be y and that by boat Q be (30 - y).
The time taken by both the boats is the same as they have started at the same time, so
y/4 = (30 - y)/10 (as boat P is going upstream and boat Q is going downstream)
Hence, y = 8.6
Therefore, the distance of their meeting point from B is 10 + (30 - 8.6) = 10 + 21.4 = 31.4 km. 

AG₁ = 5 min at 30 km/hr = 2.5 km
G₃B = 2.5 km
G₁G₃ = 20 - 2.5 - 2.5 = 15 km
Time for AG₁ = 5 min
Time for G₁G₃ = 15 km at 60 km/hr = 15 min
Time for G₃ A = 17.5 km at 60 km/hr = 17.5 min
Total time = 15 + 17.5 + 5 = 37.5 min
1 min is taken in transferring the patient into and out of the ambulance.
Hence, (40 – 37.5 – 1) = 1.5 min remain.

By travelling at 10 kmph higher than the original speed, the train is able to make up 16 minutes while traveling 80 km. This condition is only satisfied at an initial speed of 50 (and a new speed of 60 kmph).

In 1 hours 15 minutes an individual will be able to cover 25% more than his speed per hour.
The relationship between the original speed and the new speed is best represented as below:
Original speed  speed per 75 minutes  New speed.
Thus, to go from the new speed to the original speed the process would be:
New speed Speed per 75 minutes  We need to use this process to check the option.
Only the first option satisfies this condition. (at 16 kmph it would take 6 hours while at 12 kmph it would take 8 hours).

The relative speed is 20 kmph. Also, the pedestrian should take 7:30 hours more than the cyclist.
Using option (a) the speeds of the two people are 4km/hr and 16 km/hr respectively.
At this speed, the respective times would be 10 hrs and 2:30 hours, giving the required answer.

The distance between the motorists will be shown on the hypotenuse.
Using the 3,4,5 Pythagoras triplet and the condition that the two speeds are 6 kmph different from each other, you will get the triplet as: 18, 24, 30.
Hence, the slower motorist travelled at 18 kmph.

Since the second ant covers 7/120 of the distance in 2 hours 30 minutes, we can infer that is covers 8.4/120 = 7% of the distance in 3 hours.
Thus, in 3 hours both ants together cover 15% of the distance → 5% per hour → they will meet in 20 hours.
Also, ratio of speeds = 8 : 7.
So, the second ant would cover 700 ft to the meeting point in 20 hours and its speed would be 35 feet/hr.

The requisite conditions are met on a Pythagoras triplet 6,8,10.
Since the racetrack only consists of the legs of the right triangle the length must be 6 + 8 = 14 km.

When Rahim starts, Karim would have covered 40 km.
Also, their relative speed is 12 kmph and the distance between the two would get to 0 in 40/12 = 3.33 hours.
Distance covered = 28 × 3.33 = 93.33 km. 

The ant would cover 7 × 8 = 56 meters in 16 hours.
Further, it would require 7/12 of the 17th hour to reach the top.
Thus time required = 16 hours 35 minutes

The movement of the ant in the two cases would be 3, 7, 11, 15, 19, 23 and 1, 9, 17, 25, 33, 41.
It can be seen that after 3 seconds the difference is 6mm, after 4 seconds, the difference is 16mm and after 5 seconds the difference is 30 mm.
Thus, it is clearly seen that the ant moved for 4 seconds.

When the signal happened distance left was 150 km.
150/(s) – 150/(s + 15) = 1/2 hours → s = 60.

The total distance the bird would travel would be dependent on the time that the cars crash with each other.
Also, the speed of the bird is the same as the relative speed of the cars.
Hence, the answer to question 41 will be 12 km.