Answer: Option C
Explanation: Let Distance = D km, Speed = S km/h
Time = \(\frac{Distance}{Speed}\) – (1)
40 = \(\frac{Distance}{Speed}\)
After reducing speed,
Distance = Distance – 5 km, Speed = S × \(\frac{14}{15}\)km/h
Time = \(\frac{(Distance – 5)}{\frac{14}{15} ×{Speed}}\) (2)
After solving equation 1 and 2
\(\frac{Distance}{Speed}\) = \(\frac{(Distance – 5)}{\frac{14}{15} ×{Speed}}\)
14 Distance = 15 (Distance – 5)
14 Distance = 15 Distance – 75
Distance = 75 km.
2. A person travelled 132 km by auto, 852 km by train and 248 km by bike. It took 21 hours in all. If the speed of train is 6 times the speed of auto and 1.5 times speed of bike, what is the speed of train?
Answer: Option C
Explanation: Let the speed of auto be km\({h}^{1}\) So, the speed of train will be 6x and that of bike will be
= \(\frac{6×}{1.5}\) = 4x
As per the given information, ⇒ \(\frac{132}{×}\) + \(\frac{852}{6×}\) + \(\frac{248}{4×}\) = 21
∴ × = \(\frac{336}{21}\) = 16
∴ Speed of the train = 6x = 6 × 16 = 96 km\({h}^{1}\)
3. A car drove from Agra to Delhi without stopping. It covered the first 50 miles of its journey at an average speed of 25 mph. What was the car’s average speed (in mph), for the remaining 130 miles if its overall average speed was 45 mph?
Answer: Option D
Explanation:
Average speed =\(\frac{total distance }{total time}\)
Total distance = 50 + 130 = 180 miles
∴ Total time = \(\frac{180}{45}\)= 4 hours.
Time spent for the first 50 miles = \(\frac{50}{25}\)= 2 hours
∴ Time spent on the remaining journey = 4 − 2 = 2 hours
∴ Average speed for the remaining 130 miles = \(\frac{130}{2}\) = 65 mph
4. A motor car does a journey in 16 hours, covering the first half at 30 Km/hr and the second half at 50 Km/hr. What is the distance covered?
C. 500 km
D. 600 km
Answer: Option D
Explanation: Since different speeds are traveled for equal distance, the average speed can be found out
Average speed = \(\frac{2 × 30 × 50}{30 + 50}\) = \(\frac{300}{8}\) km/h
Distance covered = \(\frac{300}{8}\)× 16 = 600 km
5. A and B are two stations 1200 km apart. Train X starts from station A towards station B at 60 kmph 2 hours before train Y starts from station B towards station A. If they meet at a point 360 km from station B, What is the speed of Train Y?
Answer: Option D
Explanation: Q’s 8 days work =[ \(\frac{1}{20}\) x 8] work ⇒ \(\frac{2}{5}\)The distance covered by X in 2 hours is 2 × 60 = 120 km
The distance remaining is 1080 km
This distance travelled before meeting is directly proportional to their speed
They meet 360 km from B. Hence, train Y must’ve travelled 360 km
X has travelled 720 km and Y has travelled only 360 km in the same time
This means that speed of X is twice of Y
Since speed of X is 60 kmph therefore speed of Y is 30 kmph
D. 620 km
Answer: Option B
Explanation: Distance covered by Shaan in 3 hours = 3 × 50 = 150 km
∴ Remaining distance = 710 – 150 = 560 km Relative speed = 50 + 30 = 80 kmph
∴ They meet after = \(\frac{560}{80}\) = 7 hours
Now, Shaan covers the total distance in (7 + 3 )= 10 hours
∴ Distaance covered by Shaan in 10 hours = 50 × 10 = 500 km
∴ Distance between R and N = 500 km.
2. The distance between two places A and B is 370 km. the \({1}^{st}\) car departs from place A to B, at a speed of 80 kmph at 10 am and \({2}^{nd}\) car departs from place B to A at a speed of 50 kmph at 1 pm. At what time both cars meet each other?
Answer: Option B
Explanation: Total distance = 370 km
Now, the distance covered by first car in (10 am to 1 pm =) 3 hours = 80 × 3 = 240 km
Remaining distance = 370 – 240 = 130 km
Relative speed = 80 + 50 = 130 kmph
Now, they cover 130 km in \(\frac{130}{130 = }\)\(\frac{11}{1 }\) hours = 60 mins
So, they meet 60 mins after 1 pm. So, required answer = 2 : 00 pm
3. A man takes 5 hours 45 minutes to walk to a certain place and ride back. He would have saved 2 hours had he ridden both ways. The time he would take to walk both ways is
B. 7 hours 30 minutes
C. 7 hours 45 minutes
D. 11 hours 45 minutes
Answer: Option C
Explanation: 1 Bike + 1 Walk = 5 hrs 45 mins
2 Bike = 3 hrs 45 mins
Hence 1 way Bike journey takes 1 hrs 52.5 mins
So, 1 way walk should take (5 hr 45 mins) – (1 hrs 52.5 mins) = 3 hr 52.5 mins
2way walk would take: 3 hr 52.5 mins 2 = 7 hr 45 mins
4. A and B start at the same time with speeds of 40 km/hr and 50 km/hr respectively. If in covering the journey A takes 15 minutes longer than B, the total distance of the journey is
Answer: Option C
Explanation: Let x be the total distance of the journey.
Time taken by A – Time taken by B = 15 minutes = \(\frac{1}{4}\) hours
= \(\frac{ ×}{40}\) – \(\frac{ ×}{50}\)= \(\frac{1}{4}\)
⇒ \(\frac{ × }{200}\)= \(\frac{1}{4}\)
∴ x = 50 km
5.A motor starts with the speed of 70 kmph with its speed increasing every two hours by 10 kmph. In how many hours will it cover 345 kms?
Answer: Option B
Explanation: Distance covered in first two hours = 70 × 2 = 140 km
Distance covered in next two hours = 80 × 2 = 160 km
Distance covered in first four hours = 140 + 160 = 300 km
Remaining distance = 345 – 300 = 45 km
Now, this distance will be covered at the rate of 90 km/hr.
∴ time taken = \(\frac{45}{90}\) = \(\frac{1}{ 2}\)hour
Total time = 4 + \(\frac{1}{ 2}\) = 4 \(\frac{1}{ 2}\)hours
B. 72 km
C. 80 km
D. 90 km
Answer: Option C
Explanation: Let onefourth of the distance between P and Q be x km
then
Time taken for the first onefourth distance = \(\frac{×}{10}\) km/hour
and
Time taken for the remaining distance = \(\frac{2×}{12}\) km/hour
Since Total time taken is 7 hours ⇒ 7 = \(\frac{×}{10}\) + \(\frac{3×}{12}\)
On solving, x = 20
Total distance = 4x = 80 km.
2. In the school, Mohan and Govind took a part in the race and the ratio between their speeds is 5 : 7. Mohan losses the race by 360m then what is the length of the track (in km)?
C. 2 km
D. 0.9 km
Answer: Option A
Explanation: t = \(\frac{D}{S}\)
According to the question,
Let Speed = 5x, 7x Distance = y m
\(\frac{y – 360}{5×}\) = \(\frac{y }{7×}\)
\(\frac{y – 360}{5}\) = \(\frac{y }{7}\)
7 (y – 360) = 5y
7y – 5y = 2520
2y = 2520
y = 1260
= 1.26 km.
3. The respective ratio between the speed of a car, a train, and a bus is 5 : 9 : 4.The average speed of the car , the bus and the train is 72 km/hr together. What is the average speed of the car and the train together?
Answer: Option C
Explanation:Let speed of the car, the train, and the bus be 5a Km/hr , 9a Km/hr and 4a Km/hr respectively.
Given, total speed = (72 × 3) Km/hr = 216 Km/hr
⇒ 5a + 9a + 4a = 216
⇒ 18a = 216
⇒ a = 12Km/hr
⇒ speed of car = 5 × 12 = 60 Km/hr
And
speed of train = 9 × 12 = 108 Km/hr
⇒ their average speed = \(\frac{60 + 108}{2} \) = 84 Km/hr
4. Ranvir goes to his office from his house at a speed of 16 km/hr and returns to his home from his office at a speed of 20 km/hr and he takes 4 hour 30 minutes in all. If the distance of his friend’s house from his office is 20% more than the distance of his house from his office, find the distance of his house to his friend’s house.(assuming the office lies between Ranvir’s house and his friend’s house)
Answer: Option D
Explanation: Let the distance of his house to his office = D km
T = \(\frac{D}{S}\)
4 hour 30 minutes =\(\frac{D}{16}\)+\(\frac{D}{20}\)
\(\frac{9}{2}\) = \(\frac{5D + 4D}{80}\)
\(\frac{9}{2}\) = \(\frac{9D}{80}\)
D = 40
The distance from his house to his office = 40 km
The distance from his office to his friend’s house = 40 km × 120% = 48 km
The distance from his house to his friend’s house = 40 + 48 = 88 km
5. Mohit and Anuj took a part in a race. Mohit runs 300m in 50 second and Anuj takes 1 minute to cover the same distance. By what distance will Mohit beat Anuj in 300 m race?
Answer: Option D
Explanation: Mohit runs 300m in 50 sec.
Anuj runs 300m in 60 sec.
Now, In 1 sec Anuj runs 5m
Therefore, in 50 sec Anuj runs 50×5 = 250m
Mohit beats Anuj by 300 – 250 = 50m
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