1. The distance between two stations A and B is 220 km. A train leaves A towards B at an average speed of 80 km/hr. After half a hour, another train leaves B towards A at an average speed of 100 km/hr. The distance of the point where the two trains meet, from A is :
a. 120 km
b. 130 km
c. 140 km
d. 150 km
2. A cart has to cover a distance of 80 km in 10 hours. If it covers half of the journey in (3/5)th time, what should be its speed to cover the remaining distance in the time left ?
a. 8 km/hr
b. 20 km/hr
c. 6.4 km/hr
d. 10 km/hr.
3. Ravi started cycling along the boundaries of a square field from corner point A. After half an hour, he reached the corner point C, diagonally opposite to A. If his speed was 8 km/hr. what is the area of the field in square km ?
a. 64
b. 4
c. 8
d. Cannot be determined
4. A man goes uphill with an average speed of 35 km/hr and comes down with an average speed of 45 km/hr. The distance travelled in both the cases being the same, the average speed for the entire journey is :
a. 38.38 km/hr
b. 39.38 km/hr
c. 40 km/hr
d. None of these
5. A man walking at 3 km/hr crosses a square field diagonally in 2 min. The area of the field is :
a. 2500 m2
b. 3000 m2
c. 5000 m2
d. 6000 m2
6. A certain distance is covered at a certain speed. If half of this distance is covered in double the time, the ratio of the two speeds is :
a. 4 : 1
b. 1 : 4
c. 2 : 1
d. 1 : 2
7. If a boy takes as much time in running 10 m as a car takes in covering 25 m, the distance covered by the boy during the time the car covers 1 km. is :
a. 400 m
b. 40 m
c. 250 m
d. 650 m
8. The ratio between the rates of walking of P and Q is 2 : 3. If the time taken by Q to cover a certain distance is 36 minutes, the time taken by P to cover that much distance is :
a. 24 min
b. 54 min
c. 48 min
d. 21.6 min
9. A man, on tour, travels first 160 km at 64 km/hr and the next 160 km at 80 km/hr. the average speed for the first 320 km of the tour, is :
a. 35.55 km/hr
b. 71.11 km/hr
c. 36 km/hr
d. 72 km/hr
10. A man travels 35 km partly at 4 km/hr and at 5 km/hr. If he covers former distance at 5 km/hr and later distance at 4 km/hr, he could cover 2 km more in the same time. The time taken to cover the whole distance at original rate is
a. 9 hours
b. 7 hours
c. 412 hours
d. 8 hours
Answers
Answers
1. A
2. D
Distance left = (12×80) km=40 km
Time left = [(1-35)×10]=25×10=4hrs
Speed required = (40÷4) km/hr= 10 km/hr.
3. B
As his speed is 8kmph, in half an hour he covers 4 kmph. But 4 km is equal to two sides of the square. So side of the square = 2 km
Area of the field = 2 km x 2 km = 4 km
4. B
Average speed = (2×35×45/35+45) km/hr = 39/38 km/hr.
5. C
Speed = (3×58) m/sec = (56) m/sec.
Distance covered in (2×60)sec. = (56×2×60)m=100m
Length of diagonal = 100 m
So, area = 12×(diagonal)2 = (12×100×100)m2 = 5000m2
6. A
Let x km be covered in y hrs. Then,
Ist speed = (xy) km/hr.
2nd speed = (x2÷2y)km/hr = (x4y) km/hr
Ratio of speed = xy:x4y=1:14=4:1
7. A
25 : 10 :: 1000 : x or x = 10×100025 = 400 m
8. B
Ratio of times taken = 12:13
12:13 = x : 36 or 12×36:13×x?18=13×x or x = 54 min.
9.B
Average speed = (2×64×8064+80) km/hr = (2×64×80144)km/hr.
=71.11 km/hr.
10. D
Suppose the man covers first distance in x hrs.and second distance in y hrs. Then,
4x+5y=35 and 5x+4y = 37
Solving the equations,
we get x = 5 and y = 3
Total time taken = (5+3)hrs = 8 hrs
a. 120 km
b. 130 km
c. 140 km
d. 150 km
2. A cart has to cover a distance of 80 km in 10 hours. If it covers half of the journey in (3/5)th time, what should be its speed to cover the remaining distance in the time left ?
a. 8 km/hr
b. 20 km/hr
c. 6.4 km/hr
d. 10 km/hr.
3. Ravi started cycling along the boundaries of a square field from corner point A. After half an hour, he reached the corner point C, diagonally opposite to A. If his speed was 8 km/hr. what is the area of the field in square km ?
a. 64
b. 4
c. 8
d. Cannot be determined
4. A man goes uphill with an average speed of 35 km/hr and comes down with an average speed of 45 km/hr. The distance travelled in both the cases being the same, the average speed for the entire journey is :
a. 38.38 km/hr
b. 39.38 km/hr
c. 40 km/hr
d. None of these
5. A man walking at 3 km/hr crosses a square field diagonally in 2 min. The area of the field is :
a. 2500 m2
b. 3000 m2
c. 5000 m2
d. 6000 m2
6. A certain distance is covered at a certain speed. If half of this distance is covered in double the time, the ratio of the two speeds is :
a. 4 : 1
b. 1 : 4
c. 2 : 1
d. 1 : 2
7. If a boy takes as much time in running 10 m as a car takes in covering 25 m, the distance covered by the boy during the time the car covers 1 km. is :
a. 400 m
b. 40 m
c. 250 m
d. 650 m
8. The ratio between the rates of walking of P and Q is 2 : 3. If the time taken by Q to cover a certain distance is 36 minutes, the time taken by P to cover that much distance is :
a. 24 min
b. 54 min
c. 48 min
d. 21.6 min
9. A man, on tour, travels first 160 km at 64 km/hr and the next 160 km at 80 km/hr. the average speed for the first 320 km of the tour, is :
a. 35.55 km/hr
b. 71.11 km/hr
c. 36 km/hr
d. 72 km/hr
10. A man travels 35 km partly at 4 km/hr and at 5 km/hr. If he covers former distance at 5 km/hr and later distance at 4 km/hr, he could cover 2 km more in the same time. The time taken to cover the whole distance at original rate is
a. 9 hours
b. 7 hours
c. 412 hours
d. 8 hours
Answers
Answers
1. A
2. D
Distance left = (12×80) km=40 km
Time left = [(1-35)×10]=25×10=4hrs
Speed required = (40÷4) km/hr= 10 km/hr.
3. B
As his speed is 8kmph, in half an hour he covers 4 kmph. But 4 km is equal to two sides of the square. So side of the square = 2 km
Area of the field = 2 km x 2 km = 4 km
4. B
Average speed = (2×35×45/35+45) km/hr = 39/38 km/hr.
5. C
Speed = (3×58) m/sec = (56) m/sec.
Distance covered in (2×60)sec. = (56×2×60)m=100m
Length of diagonal = 100 m
So, area = 12×(diagonal)2 = (12×100×100)m2 = 5000m2
6. A
Let x km be covered in y hrs. Then,
Ist speed = (xy) km/hr.
2nd speed = (x2÷2y)km/hr = (x4y) km/hr
Ratio of speed = xy:x4y=1:14=4:1
7. A
25 : 10 :: 1000 : x or x = 10×100025 = 400 m
8. B
Ratio of times taken = 12:13
12:13 = x : 36 or 12×36:13×x?18=13×x or x = 54 min.
9.B
Average speed = (2×64×8064+80) km/hr = (2×64×80144)km/hr.
=71.11 km/hr.
10. D
Suppose the man covers first distance in x hrs.and second distance in y hrs. Then,
4x+5y=35 and 5x+4y = 37
Solving the equations,
we get x = 5 and y = 3
Total time taken = (5+3)hrs = 8 hrs
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