Answer: Option A
Explanation:
Mayank’s speed = 20 km /hr
= (20*\(\frac{5}{18}\)) m/sec =\(\frac{50}{9}\)m/sec
So, time taken to cover 400 m
(400*\(\frac{9}{50}\))sec= 72 sec= 1\(\frac{1}{5}\)min
2. A cyclist covers a distance of 750 m in 2 min 30 sec. what is the speed in km /hr of the cyclist ?
Answer: Option C
Explanation:
By applying a formula,
Speed =\(\frac{Distance}{Time}\)
Speed = \(\frac{750}{150}\)m/sec =5m/sec
=5*\(\frac{18}{5}\)km / hr =18 km / hr
3. A Jackal takes 4 leaps for every 5 leaps of goat but 3 leaps of a Jackal are equal to 4 leaps of the goat. compare their speeds
Answer: Option D
Explanation:
Let the distance covered in 1 leap of the jackal be x and that covered in 1 leap of the goat be y.
Then, 3x = 4y ⇒ x =\(\frac{4}{3}\)y
4x= \(\frac{16}{3}\)y
So, Ratio speed of jackal and goat = Ratio of distances covered by them in the same time,
⇒ 4x : 5y =\(\frac{16}{3}\)y: 5y
16 : 15
4. While covering a distance of 24 km, a man noticed that after walking for 1 hour and 40 minutes, the distance covered by him was \(\frac{5}{7}\) of the remaining distance. What was his speed in meters per second?
Answer: Option C
Explanation:
Let the speed be x km/hr.
Then, distance covered in 1 hr. 40 min.
i.e , 1\(\frac{2}{3}\)hrs= \(\frac{5x}{3}\)km
Remaining distance
(24-\(\frac{5x}{3}\))Km.
7x = 72 – 5x
12x = 72
x = 6
Hence, speed = 6 km/hr
(6*\(\frac{5}{18}\))m/sec
=1\(\frac{2}{3}\)m/sec
5. A boy travelled from the home to the college at the rate of 25 kmph and walked back at the rate of 4 kmph. If the whole journey took 5 hours 48 minutes, Find the distance of the college from the home.
Answer: Option D
Explanation:
Average speed =\(\frac{2xy}{x+y}\)km/hr
= \(\frac{ 2 * 25 * 4}{25 + 4}\)km/hr
= \(\frac{ 200}{29}\)km/hr
Distance travelled in 5 hours 48 minutes i.e,5\(\frac{4}{5}\)hrs
(\(\frac{200}{29}\) * \(\frac{29}{5}\))km = 40 km.
So, Distance of the college from the home
\(\frac{40}{2}\)= 20km
Answer: Option D
Explanation:
Realtive speed of policeman = (10 – 8) km/hr = 2 km/hr.
Time =\(\frac{Distance}{Speed}\)
Time taken by policeman to cover 100 m
= (\(\frac{100}{1000}\)* \(\frac{1}{2}\))hr= \(\frac{1}{20}\)hr
In \(\frac{1}{20}\)hr the thief covers a distance of
= (8*\(\frac{1}{20}\))= 400m.
2. I walk a certain distance and ride back taking a total time of 37 minutes. I could walk both ways in 55 minutes. How long would it take me to ride both ways?
Answer: Option D
Explanation:
Let the distance be x, then,
(Time taken to walk x km) + (Time taken to ride x km) = 37min.
⇒ (Time taken to walk 2x km) + (Time taken to ride 2x km) = 74 min.
But, time taken to walk 2x km = 55 min.
So, Time to ride 2x km = (74 – 55) min ⇒ 19 min.
Hence, option D is correct.
3. A motor-cycle covers 40 km with a speed of 20 km/hr. find the speed of the motor-cycle for the next 40 km journey so that the average speed of the whole journey will be 30 km/hr.
Answer: Option C
Explanation:
Average speed of whole journey = 30 km/hr
x = speed of first 40 km = 20 km/hr
30= \(\frac{2 * 20 * y}{20 + y}\)
600 + 30y = 40y ⇒ 10y = 600
y= \(\frac{600}{10}\)= 60 km/hr
4. A man rides at the rate of 18 km/hr, but stops for 6 minutes to change horses at the end of every 7th km. The time that he will take to cover a distance of 90 km is
Answer: Option B
Explanation:
Number of stoppages
= \(\frac{90}{7}\)= 12.8,
it means there is 12 stoppages.
Distance to be covered = 90 km, Speed = 18,
Time for each stoppage = 6 mins.
By the short trick approach, we get
= \(\frac{90}{18}\) + 12 * 6
5 hrs + 72 mins
Convert mins to hours then,
= 6 hrs 12 mins.
5. Walking at 3 km/hr . Pintu reaches his school 5 minutes late. If he walks at 4 km per hour he will be 5 minutes early. The distance of Pintu’s from his house is
Answer: Option B
Explanation:
Speed1 = 3 km/hr, Speed2 = 4 km/hr
Time1 = 5 mins late, Time2 = 5 min early
Reqd. Distance
\(\frac{3 * 4}{4 – 3}\) * \(\frac{5 + 5}{60}\)= 2 km
Answer: Option A
Explanation:
Distance between Agra and Delhi = 300 km
Relative speed = 38 + 37 = 75 km/hr
Time taken to cross each other = \(\frac{Distance}{Speed}\)
= \(\frac{300}{75}\)= 4 hours.
2. Raj and Prem walk in opposite direction at the rate of 3 km and 2 km per hour respectively. How far will they be from each other after 2 hrs?
Answer: Option C
Explanation:
Distance covered per hour = Relative speed × Time
= (3 + 2) * 1 = 5 km
Distance covered in 2 hours = 5 × 2 = 10 km.
3. Walking \(\frac{2}{3}\) times of his usual speed a man gets late by 20 minutes. The usual time taken by him to cover that distance:
Answer: Option C
Explanation:
New Speed =\(\frac{2}{3}\) of the usual speed
New time taken =\(\frac{3}{2}\)of the usual time
So, (\(\frac{3}{2}\)of the usual time)– (usual time) = 20 min
=\(\frac{1}{2}\)of the usual time = 20 min
usual time = 40 mins.
4. A car travels the first one-third of a certain distance with a speed of 10 km/hr, the next one-third distance with a speed of 20 km/hr and the last one-third distance with a speed of 60 km/hr. The average speed of the car for the whole journey is
Answer: Option A
Explanation:
Let the total distance covered = LCM of (10, 20, 60) = 60.
As per the question,
Distance covered by the car with each speed
\(\frac{1}{3}\)*60= 20 km
\(\frac{20}{10}\)+\(\frac{20}{20}\)+\(\frac{20}{60}\)= \(\frac{60}{avg. speed}\)
average speed = 18 km/hr
5. A student goes to school at the rate of 2\(\frac{1}{2}\)km/hr and reaches 6 minutes late. If he travels at the speed of 3 km/hr he is 10 minutes early. What is the distance to the school?
Answer: Option B
Explanation:
Where,
‘a’ is the extra of time taken by first speed = 6 mins
‘b’ is the less of time taken by second speed = 10 mins
a + b = 10 + 6 = 16 mins = \(\frac{4}{15}\)hrs
By the short trick approach, we get
Reqd. distance = 4km
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