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Question 1 of 5
1. Question
A train running at the speed of 72 km/h goes past a pole in 15 s. What is the length of the train?
Correct
Sol.(c)
Speed of train = Length of train / Time taken to cross the stationary object
Hence length of train = Speed of train × Time taken to cross the stationary object
= 72 × 1000 × 15/3600 = 300 m.
Incorrect
Sol.(c)
Speed of train = Length of train / Time taken to cross the stationary object
Hence length of train = Speed of train × Time taken to cross the stationary object
= 72 × 1000 × 15/3600 = 300 m.

Question 2 of 5
2. Question
Two cars A and B start simultaneously from a certain place at the speed of 30 km/h and 45 km/h, respectively. The car B reaches the destination 2 h earlier than A. What is the distance between the starting point and destination?
Correct
Sol.(b)
Let the time taken by car B to reach destination be x h.
So, the time taken by car A to reach destination is (x + 2) h.
Now, S_{1} (when distance is constant) T_{1} = S_{2}, (when distance is constant) T_{2}
30 × (x + 2)= 45 × x
30x + 60 = 45x
15x = 60
15 = 4h
Now, distance between starting point and destination S_{2}T_{2}
= 45 × 4 = 180 km.
Incorrect
Sol.(b)
Let the time taken by car B to reach destination be x h.
So, the time taken by car A to reach destination is (x + 2) h.
Now, S_{1} (when distance is constant) T_{1} = S_{2}, (when distance is constant) T_{2}
30 × (x + 2)= 45 × x
30x + 60 = 45x
15x = 60
15 = 4h
Now, distance between starting point and destination S_{2}T_{2}
= 45 × 4 = 180 km.

Question 3 of 5
3. Question
A man cycles with a speed of 10 km/h and reaches his office at 1 pm. However, when he cycles with a speed of 15 km/h, he reaches his office at 11 am. At what speed should he cycle, so that he reaches his office at 12 noon?
Correct
Sol.(b)
Let t h be the time taken by the man to reach his office at the speed of 15 km/h.
Then, time taken to reach office at the speed of 10 km/h= (t + 2)h.
Now, s_{1} × t_{1}= s_{2} × t_{2}
10 × (t + 2) = 15 × t
10t + 20 = 15t
5t = 20
Hence, t = 4h
Now, distance covered to reach office = s_{1} x t_{1}
= 10 × (4 + 2) = 10 × 6 = 60 km
Now, speed required to reach office at 12 noon
i.e. in 5 h = Distance/Time
Hence, Speed = 60/5 = 12 km/h.
Incorrect
Sol.(b)
Let t h be the time taken by the man to reach his office at the speed of 15 km/h.
Then, time taken to reach office at the speed of 10 km/h= (t + 2)h.
Now, s_{1} × t_{1}= s_{2} × t_{2}
10 × (t + 2) = 15 × t
10t + 20 = 15t
5t = 20
Hence, t = 4h
Now, distance covered to reach office = s_{1} x t_{1}
= 10 × (4 + 2) = 10 × 6 = 60 km
Now, speed required to reach office at 12 noon
i.e. in 5 h = Distance/Time
Hence, Speed = 60/5 = 12 km/h.

Question 4 of 5
4. Question
If a body covers a distance at the rate of x km/h and another equal distance at the rate of y km/h, then the average speed (in km/h) is
Correct
Sol: (c )
If a body covers a distance at the rate of x km/h and another equal distance at the rate of y km/h, then
Average speed = (Otal distance travelled)/(Total time taken) = (z+z)/[(z/x)+(z/y)]
[let a body covers z distance in both cases]
= [2z/[z(x+y/xy)] = (2xy)/(x+y)
Incorrect
Sol: (c )
If a body covers a distance at the rate of x km/h and another equal distance at the rate of y km/h, then
Average speed = (Otal distance travelled)/(Total time taken) = (z+z)/[(z/x)+(z/y)]
[let a body covers z distance in both cases]
= [2z/[z(x+y/xy)] = (2xy)/(x+y)

Question 5 of 5
5. Question
A sailor sails a distance of 48 km along the flow of a river in 8 h. If it takes 12 h to return the same distance, then the speed of the flow of the river is
Correct
Sol : (b)
Let speed of the flow of water be v km/h and rate of sailing of sailer be u km/h.
Then,
u+v= 48/8 Þ u+v = 6 ………………………………(1)
And uv = 48/12 Þ uv = 4 ………………….(2)
On solving Eqs. (1) and (2), we get
v=1 km/h
Incorrect
Sol : (b)
Let speed of the flow of water be v km/h and rate of sailing of sailer be u km/h.
Then,
u+v= 48/8 Þ u+v = 6 ………………………………(1)
And uv = 48/12 Þ uv = 4 ………………….(2)
On solving Eqs. (1) and (2), we get
v=1 km/h
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