To find the time taken, we use the fundamental formula relating distance, speed, and time: Time = \frac{Distance}{Speed} Given:
Distance = 450 km
Speed = 90 km/h
Substitute the given values into the formula:
Time = \frac{450}{90} = 5 \text{ hours} Thus, the correct option is B.
First, find the speed in meters per second using the formula Speed = \frac{Distance}{Time} Given:
Distance = 100 m
Time = 10 s
Speed = \frac{100}{10} = 10 \text{ m/s} To convert speed from meters per second (m/s) to kilometers per hour (km/h), multiply by \frac{18}{5}: Speed = 10 \times \frac{18}{5} = 2 \times 18 = 36 \text{ km/h} Thus, the correct option is B.
First, convert the time from minutes to hours because the speed is given in kilometers per hour (km/h):
Time = \frac{15}{60} = \frac{1}{4} \text{ hours} Next, use the formula relating distance, speed, and time:
Distance = Speed \times Time Substitute the given values into the formula:
Distance = 72 \times \frac{1}{4} = 18 \text{ km} Thus, the correct option is A.
When equal distances are covered at two different speeds v_1 and v_2, the average speed is given by the formula:
Average\ Speed = \frac{2 \times v_1 \times v_2}{v_1 + v_2} Given:
v_1 = 40 \text{ km/h} v_2 = 60 \text{ km/h} Substitute the given values into the formula: Average\ Speed = \frac{2 \times 40 \times 60}{40 + 60} = \frac{4800}{100} = 48 \text{ km/h} Thus, the correct option is C.
When two bodies move towards each other, their relative speed is the sum of their individual speeds.
Relative\ Speed = v_1 + v_2 Given:
v_1 = 40 \text{ km/h} v_2 = 60 \text{ km/h} Relative\ Speed = 40 + 60 = 100 \text{ km/h}
The time taken to meet is given by dividing the total distance by the relative speed:
Time = \frac{Distance}{Relative\ Speed} = \frac{300}{100} = 3 \text{ hours} Thus, the correct option is D.
First, convert the speed of the train from kilometers per hour (km/h) to meters per second (m/s) by multiplying by 5/18:
Speed = 72 \times \frac{5}{18} = 4 \times 5 = 20 \text{ m/s} The total distance to be covered when crossing a platform is the sum of the length of the train and the length of the platform:
Total\ Distance = Length\ of\ train + Length\ of\ platform = 120 + 180 = 300 \text{ meters} Now, calculate the time taken using the formula:
Time = \frac{Total\ Distance}{Speed} = \frac{300}{20} = 15 \text{ seconds} Thus, the correct option is C.
First, convert the head start time of 10 minutes into hours:
Time = \frac{10}{60} = \frac{1}{6} \text{ hours} Calculate the distance covered by the thieves in this 10-minute head start: Distance = Speed \times Time = 60 \times \frac{1}{6} = 10 \text{ km}
Since both vehicles are moving in the same direction, their relative speed is the difference between their speeds:
Relative\ Speed = 80 - 60 = 20 \text{ km/h} The time taken by the police to catch up with the thieves is:
Time = \frac{Distance\ gap}{Relative\ Speed} = \frac{10}{20} = 0.5 \text{ hours} The distance from the starting point where the police catch the thieves can be calculated using the police car’s speed and time:
Distance = 80 \times 0.5 = 40 \text{ km} Thus, the correct option is B.
Let the speed of the stream be s \text{ km/h}.
Given:
Speed of the boat in still water = 18 \text{ km/h}
Downstream speed v_d = 18 + s
Upstream speed v_u = 18 - s
The total time taken for both downstream and upstream journeys is given by the sum of their individual times:
Total\ Time = \frac{Distance}{v_d} + \frac{Distance}{v_u} Substitute the given values into the equation:
\frac{35}{18 + s} + \frac{35}{18 - s} = 4
Factor out 35 and combine the fractions using a common denominator:
35 \times \left( \frac{(18 - s) + (18 + s)}{(18 + s)(18 - s)} \right) = 4 \frac{35 \times 36}{18^2 - s^2} = 4 \frac{1260}{324 - s^2} = 4
Solve for s^2:
324 - s^2 = \frac{1260}{4} 324 - s^2 = 315 s^2 = 324 - 315 = 9 s = \sqrt{9} = 3 \text{ km/h} Thus, the correct option is C.
When two bodies move in opposite directions on a circular track, their relative speed is the sum of their individual speeds.
Relative\ Speed = v_1 + v_2 Given:
v_1 = 6 \text{ m/s} v_2 = 4 \text{ m/s} Relative\ Speed = 6 + 4 = 10 \text{ m/s}
The time taken to meet for the first time anywhere on the track is given by dividing the total circumference by the relative speed:
Time = \frac{Circumference}{Relative\ Speed} = \frac{1200}{10} = 120 \text{ seconds} Thus, the correct option is A.
First, find the time taken by each runner to complete one full lap around the circular track using the formula Time = \frac{Distance}{Speed}
For runner X:
Time_X = \frac{600}{5} = 120 \text{ seconds} For runner Y:
Time_Y = \frac{600}{3} = 200 \text{ seconds}
The time at which they will meet again at the starting point for the first time is the least common multiple (LCM) of their individual lap completion times:
Time = \text{LCM}(120, 200)
Prime factorization of 120 and 200:
120 = 2^3 \times 3 \times 5 200 = 2^3 \times 5^2 Taking the highest power of each prime factor:
\text{LCM} = 2^3 \times 3 \times 5^2 = 8 \times 3 \times 25 = 600 \text{ seconds} Thus, the correct option is B.
When two trains move in opposite directions, their relative speed is the sum of their individual speeds:
Relative\ Speed = v_1 + v_2 Given:
v_1 = 54 \text{ km/h} v_2 = 36 \text{ km/h} Relative\ Speed = 54 + 36 = 90 \text{ km/h}
Convert the relative speed from kilometers per hour (km/h) to meters per second (m/s) by multiplying by \frac{5}{18}:
Relative\ Speed\ in\ m/s = 90 \times \frac{5}{18} = 5 \times 5 = 25 \text{ m/s}
The total distance to be covered when two trains pass each other is the sum of their lengths:
Total\ Distance = Length\ of\ train\ 1 + Length\ of\ train\ 2 = 150 + 200 = 350 \text{ meters} Now, calculate the time taken using the formula:
Time = \frac{Total\ Distance}{Relative\ Speed} = \frac{350}{25} = 14 \text{ seconds} Thus, the correct option is B.
When two trains move in the same direction, their relative speed is the difference between their individual speeds:
Relative\ Speed = v_1 - v_2 Given:
v_1 = 72 \text{ km/h} v_2 = 54 \text{ km/h} Relative\ Speed = 72 - 54 = 18 \text{ km/h}
Convert the relative speed from kilometers per hour (km/h) to meters per second (m/s) by multiplying by \frac{5}{18}:
Relative\ Speed\ in\ m/s = 18 \times \frac{5}{18} = 5 \text{ m/s} The total distance to be covered when the faster train passes the slower train is the sum of their lengths:
Total\ Distance = Length\ of\ train\ 1 + Length\ of\ train\ 2 = 180 + 220 = 400 \text{ meters} Now, calculate the time taken using the formula:
Time = \frac{Total\ Distance}{Relative\ Speed} = \frac{400}{5} = 80 \text{ seconds} Thus, the correct option is C.
Let the distance traveled on foot be x km. Then the remaining distance traveled by bicycle is (60 – x) km.
Using the formula:
Time = \frac{Distance}{Speed} we can write the time taken for each part:
Time taken on foot = \frac{x}{10} hours
Time taken by bicycle = \frac{60 - x}{20} hours
The total time taken for the entire journey is 4 hours:
\frac{x}{10} + \frac{60 - x}{20} = 4
Multiply the entire equation by 20 to clear the denominators:
2x + (60 - x) = 80 x + 60 = 80 x = 80 - 60 = 20 \text{ km} Thus, the correct option is B.
Let the total number of steps or length of the escalator be L. The speed of the man walking on the stationary escalator is:
v_m = \frac{L}{30} The speed of the moving escalator when the man is stationary is:
v_e = \frac{L}{20} When the man walks up the moving escalator, his effective speed is the sum of his walking speed and the escalator’s speed:
v_{total} = v_m + v_e = \frac{L}{30} + \frac{L}{20}
Finding a common denominator (60):
v_{total} = \frac{2L + 3L}{60} = \frac{5L}{60} = \frac{L}{12} The time taken to cover the length L at this combined speed is:
Time = \frac{L}{v_{total}} = \frac{L}{\frac{L}{12}} = 12 \text{ seconds} Thus, the correct option is C.
When A runs 100 meters, B runs 100 - 10 = 90 meters. Therefore, the ratio of the distances covered by A and B is:
\frac{A}{B} = \frac{100}{90} = \frac{10}{9} When B runs 100 meters, C runs 100 - 20 = 80 meters. Therefore, the ratio of the distances covered by B and C is:
\frac{B}{C} = \frac{100}{80} = \frac{5}{4} To find the ratio of the distances covered by A and C, multiply the two ratios:
\frac{A}{C} = \frac{A}{B} \times \frac{B}{C} = \frac{10}{9} \times \frac{5}{4} = \frac{50}{36} = \frac{25}{18} This means that when A covers 25 meters, C covers 18 meters.
To find how much C covers when A covers a full 100-meter race, scale the distance proportionally:
Distance\ of\ C = \frac{18}{25} \times 100 = 18 \times 4 = 72 \text{ meters} Thus, A beats C by:
100 - 72 = 28 \text{ meters} Thus, the correct option is C.
First, find the distance between two consecutive buses. The buses leave every 10 minutes (which is $\frac{10}{60} = \frac{1}{6}$ hours) at a speed of 60 km/h:
Distance = 60 \times \frac{1}{6} = 10 \text{ km}
Let the speed of the man be v km/h. Since the man is moving in the opposite direction towards the terminal, the relative speed between the buses and the man is the sum of their speeds:
Relative\ Speed = 60 + v The man meets the buses at intervals of 8 minutes (which is 8/60 = 2/15 hours). The distance between consecutive buses is covered by this relative speed in 8 minutes:
10 = (60 + v) \times \frac{2}{15}
Solve for v:
60 + v = 10 \times \frac{15}{2} 60 + v = 75 v = 75 - 60 = 15 \text{ km/h} Thus, the correct option is C.
To find the stoppage time per hour, we first calculate the difference between the speed without stoppages and the speed with stoppages:
Difference\ in\ speed = 50 - 40 = 10 \text{ km/h} The stoppage time per hour is the time taken to cover this difference in distance at the speed without stoppages, which can be found using the formula:
Stoppage\ time\ per\ hour = \frac{Difference\ in\ speed}{Speed\ without\ stoppages} \times 60 \text{ minutes} Substitute the given values into the formula:
Stoppage\ time = \frac{10}{50} \times 60 = \frac{1}{5} \times 60 = 12 \text{ minutes} Thus, the correct option is D.
When two persons A and B start simultaneously from two points towards each other and, after meeting, take T_1 and T_2 hours respectively to reach their destinations, the ratio of their speeds is given by the standard formula:
\frac{S_A}{S_B} = \sqrt{\frac{T_B}{T_A}} Given:
Speed of A (S_A) = 45 km/h
Time taken by A after meeting (T_1) = 4 hours
Time taken by B after meeting (T_2) = 9 hours
Substitute the given values into the formula:
\frac{45}{S_B} = \sqrt{\frac{9}{4}} \frac{45}{S_B} = \frac{3}{2} Cross-multiply and solve for S_B:
3 \times S_B = 45 \times 2 3 \times S_B = 90 S_B = \frac{90}{3} = 30 \text{ km/h} Thus, the correct option is C.
First, calculate the time taken for the two trains to collide. Since they are moving towards each other, their relative speed is the sum of their individual speeds:
Relative\ Speed = 50 + 50 = 100 \text{ km/h} The time taken for them to meet over the initial distance of 300 km is:
Time = \frac{Distance}{Relative\ Speed} = \frac{300}{100} = 3 \text{ hours}
The bird flies continuously back and forth between the two trains during this entire 3-hour duration until the collision occurs. Therefore, the total distance covered by the bird is:
Distance\ of\ bird = Speed\ of\ bird \times Time = 100 \times 3 = 300 \text{ km} Thus, the correct option is D.
To find the ratio of their speeds, we must compare the total distance covered by each animal in the same unit of time. Let the number of leaps taken by the dog in a given time be 5, and the number of leaps taken by the hare in the same time be 6.
Given that 4 leaps of the dog are equal in distance to 5 leaps of the hare, we can determine the relative length of their leaps. Let the distance of 1 leap of the dog be d_d and 1 leap of the hare be d_h:
4 \times d_d = 5 \times d_h \frac{d_d}{d_h} = \frac{5}{4} This means we can assume the distance of 1 leap of the dog is 5 units and 1 leap of the hare is 4 units.
Now, calculate the relative distance covered per unit time for both animals:
Speed\ of\ dog = 5 \text{ leaps} \times 5 \text{ units/leap} = 25 \text{ units} Speed\ of\ hare = 6 \text{ leaps} \times 4 \text{ units/leap} = 24 \text{ units} Thus, the ratio of the speed of the dog to the speed of the hare is 25 : 24. Thus, the correct option is D.
First, find the time taken by each runner to complete one full lap around the circular track using the formula Time = \frac{Distance}{Speed}
For runner A:
Time_A = \frac{1800}{3} = 600 \text{ seconds} For runner B:
Time_B = \frac{1800}{5} = 360 \text{ seconds} For runner C:
Time_C = \frac{1800}{9} = 200 \text{ seconds}
The time at which all three runners will meet again at the starting point for the first time is the least common multiple (LCM) of their individual lap completion times:
Time = \text{LCM}(600, 360, 200) Prime factorization of 600, 360, and 200:
600 = 2^3 \times 3 \times 5^2
360 = 2^3 \times 3^2 \times 5
200 = 2^3 \times 5^2
Taking the highest power of each prime factor:
\text{LCM} = 2^3 \times 3^2 \times 5^2 = 8 \times 9 \times 25 = 1800 \text{ seconds} Convert the time from seconds to minutes by dividing by 60:
\frac{1800}{60} = 30 \text{ minutes} Thus, the correct option is B.
Let the normal speed of the train be S. When the speed is reduced to \frac{3}{4}S, the time taken becomes \frac{4}{3} of the normal time.
The difference in delay between the two scenarios is:
40 \text{ minutes} - 30 \text{ minutes} = 10 \text{ minutes} This 10-minute difference in delay is caused over the 50 km distance where the train runs at the reduced speed instead of its normal speed.
Calculate the time taken to cover this 50 km stretch at normal speed:
Time\ at\ normal\ speed = \frac{50}{S} Calculate the time taken to cover the same 50 km stretch at the reduced speed of \frac{3}{4}S:
Time\ at\ reduced\ speed = \frac{50}{\frac{3}{4}S} = \frac{200}{3S} The difference between these two times equals the reduction in delay (10 minutes, which is \frac{1}{6} of an hour):
\frac{200}{3S} - \frac{50}{S} = \frac{1}{6}
Combine the terms on the left side:
\frac{200 - 150}{3S} = \frac{1}{6} \frac{50}{3S} = \frac{1}{6}
Cross-multiply and solve for S:
3S = 50 \times 6 3S = 300 S = \frac{300}{3} = 100 \text{ km/h} Thus, the correct option is C.
Let the downstream speed be v_d and the upstream speed be v_u. According to the problem, the time taken to row 5 km downstream is equal to the time taken to row 3 km upstream:
\frac{5}{v_d} = \frac{3}{v_u} From this, we can write the ratio of downstream speed to upstream speed:
\frac{v_d}{v_u} = \frac{5}{3}
Let v_d = 5k and v_u = 3k for some positive constant k. The total time taken to row 15 km downstream and 15 km upstream is given as 4 hours:
\frac{15}{v_d} + \frac{15}{v_u} = 4 Substitute 5k and 3k into the equation:
\frac{15}{5k} + \frac{15}{3k} = 4 \frac{3}{k} + \frac{5}{k} = 4 \frac{8}{k} = 4
Solve for k:
4k = 8 k = 2 Now, substitute k back to find the downstream and upstream speeds:
v_d = 5 \times 2 = 10 \text{ km/h} v_u = 3 \times 2 = 6 \text{ km/h}
The speed of the stream (s) is given by half the difference between the downstream and upstream speeds:
s = \frac{v_d - v_u}{2} = \frac{10 - 6}{2} = \frac{4}{2} = 2 \text{ km/h} Thus, the correct option is C.
Let the speed of the car be v m/s. The distance between two consecutive sound waves emitted by the gun is equal to the distance traveled by sound in the interval between the two shots (10 minutes):
Distance\ between\ shots = Speed\ of\ sound \times Time\ interval\ of\ shots Distance = 330 \times (10 \times 60) \text{ meters}
When the person is moving away from the source, the person and the sound are moving in the same direction. The distance between consecutive shots is also covered by the sound minus the distance the person moves during the interval between hearing the shots (11 minutes):
Distance = (Speed\ of\ sound \times Heard\ interval) - (Speed\ of\ car \times Heard\ interval) 330 \times (10 \times 60) = (330 \times (11 \times 60)) - (v \times (11 \times 60))
Divide the entire equation by 60 to simplify the time units:
330 \times 10 = (330 \times 11) - (v \times 11) 3300 = 3630 - 11v
Solve for v:
11v = 3630 - 3300 11v = 330 v = \frac{330}{11} = 30 \text{ m/s}
Convert the speed from meters per second (m/s) to kilometers per hour (km/h) by multiplying by \frac{18}{5}:
Speed\ in\ km/h = 30 \times \frac{18}{5} = 6 \times 18 = 108 \text{ km/h} Thus, the correct option is A.
A clock face is divided into 60 minute spaces. The minute hand completes a full circle of 60 minute spaces in 60 minutes, meaning its speed is 1 minute space per minute.
The hour hand completes 5 minute spaces (from 12 to 1) in 60 minutes, meaning its speed is \frac{5}{60} = \frac{1}{12} of a minute space per minute.
When the time is exactly 4:00, the hour hand is at the 4 position, which is 20 minute spaces ahead of the 12 position where the minute hand starts. Since both hands move in the same direction, the relative speed of the minute hand with respect to the hour hand is the difference between their individual speeds:
Relative\ Speed = 1 - \frac{1}{12} = \frac{11}{12} \text{ minute spaces per minute}
To coincide, the minute hand must cover the initial gap of 20 minute spaces. The time taken to cover this gap is:
Time = \frac{Distance\ gap}{Relative\ Speed} = \frac{20}{\frac{11}{12}} = \frac{20 \times 12}{11} = \frac{240}{11} \text{ minutes}
Convert the improper fraction into a mixed fraction:
\frac{240}{11} = 21\frac{9}{11} \text{ minutes} Thus, the hands coincide at 21 \frac{9}{11} minutes past 4. Therefore, the correct option is C.
When a train passes a bridge or a platform, the total distance to be covered is the sum of the length of the train and the length of the bridge:
Total\ Distance = Length\ of\ train + Length\ of\ bridge = 150 + 250 = 400 \text{ meters}
Given:
Speed of the train = 60 \text{ km/h}
Convert the speed from kilometers per hour (km/h) to meters per second (m/s) by multiplying by \frac{5}{18}:
Speed\ in\ m/s = 60 \times \frac{5}{18} = \frac{300}{18} = \frac{50}{3} \text{ m/s}
Now, calculate the time taken using the formula:
Time = \frac{Total\ Distance}{Speed} = \frac{400}{\frac{50}{3}} = 400 \times \frac{3}{50} = 8 \times 3 = 24 \text{ seconds} Thus, the correct option is B.
First, calculate the distance traveled by the thief before the owner starts the pursuit. The thief drives for 30 minutes (from 2:30 PM to 3:00 PM) at a speed of 60 km/h:
Distance\ of\ thief = Speed \times Time = 60 \times \frac{30}{60} = 60 \times 0.5 = 30 \text{ km}
When the owner starts chasing, both vehicles are moving in the same direction. The relative speed of the owner with respect to the thief is the difference between their speeds:
Relative\ Speed = 75 - 60 = 15 \text{ km/h} The time taken by the owner to cover the 30 km gap at this relative speed is:
Time = \frac{Distance}{Relative\ Speed} = \frac{30}{15} = 2 \text{ hours} Since the owner started at 3:00 PM, the owner will catch the thief 2 hours later:
3:00\text{ PM} + 2\text{ hours} = 5:00\text{ PM} Thus, the correct option is B.
When A runs the full 100-meter race, B is given a start of 20 meters, meaning B runs:
100 - 20 = 80 \text{ meters} Similarly, when A runs 100 meters, C is given a start of 28 meters, meaning C runs:
100 - 28 = 72 \text{ meters} This implies that when B covers 80 meters, C covers 72 meters.
To find how many meters C covers when B runs a full 100-meter race, we set up a proportion:
Distance\ of\ C = \frac{72}{80} \times 100 Distance\ of\ C = \frac{9}{10} \times 100 = 90 \text{ meters} Therefore, in a 100-meter race between B and C, B beats C by:
100 - 90 = 10 \text{ meters} Thus, the correct option is B.
When a distance is divided into three equal parts and covered at three different speeds (v_1, v_2, v_3), the average speed is given by the harmonic mean formula:
Average\ Speed = \frac{3}{\frac{1}{v_1} + \frac{1}{v_2} + \frac{1}{v_3}} Given:
v_1 = 10 \text{ km/h} v_2 = 20 \text{ km/h} v_3 = 30 \text{ km/h}
Substitute the values into the formula:
Average\ Speed = \frac{3}{\frac{1}{10} + \frac{1}{20} + \frac{1}{30}} Find a common denominator for the fractions in the denominator (which is 60):
\frac{1}{10} + \frac{1}{20} + \frac{1}{30} = \frac{6}{60} + \frac{3}{60} + \frac{2}{60} = \frac{11}{60}
Now, divide 3 by this fraction:
Average\ Speed = \frac{3}{\frac{11}{60}} = \frac{3 \times 60}{11} = \frac{180}{11} \approx 16.36 \text{ km/h} Thus, the correct option is C.
Let the time they travel until they meet be t hours. Since both trains start at the same time and travel for the same duration until they meet, the distance traveled by each train is:
Distance traveled by the first train (d_1) = 60t
Distance traveled by the second train (d_2) = 40t
According to the problem, one train has traveled 20 km more than the other:
60t - 40t = 20 20t = 20 t = 1 \text{ hour}
The total distance between stations A and B is the sum of the distances traveled by both trains:
Total\ Distance = 60t + 40t = 100t Substitute t = 1:
Total\ Distance = 100 \times 1 = 100 \text{ km} Thus, the correct option is B.
Let the downstream speed be v_d and the upstream speed be v_u.
Based on the given information, we can set up two equations for the time taken:
\frac{20}{v_d} + \frac{15}{v_u} = 5 \frac{30}{v_d} + \frac{20}{v_u} = 7
To solve these simultaneous equations, let x = \frac{1}{v_d} and y = \frac{1}{v_u}:
20x + 15y = 5 30x + 20y = 7 Multiply the first equation by 2 and the second equation by 1.5 to align the coefficients of x:
40x + 30y = 10 45x + 30y = 10.5
Subtract the first modified equation from the second:
(45x + 30y) - (40x + 30y) = 10.5 - 10 5x = 0.5 x = \frac{0.5}{5} = 0.1
Since x = \frac{1}{v_d} = 0.1, we get:
v_d = \frac{1}{0.1} = 10 \text{ km/h} Substitute x = 0.1 back into the first equation 20(0.1) + 15y = 5:
2 + 15y = 5 15y = 3 y = \frac{3}{15} = \frac{1}{5} = 0.2
Since y = \frac{1}{v_u} = 0.2, we get:
v_u = \frac{1}{0.2} = 5 \text{ km/h} The speed of the stream (s) is given by half the difference between the downstream and upstream speeds:
s = \frac{v_d - v_u}{2} = \frac{10 - 5}{2} = \frac{5}{2} = 2.5 \text{ km/h} Thus, the correct option is B.
When a train passes a man running in the same direction, the relative speed of the train with respect to the man is the difference between their speeds:
Relative\ Speed = Speed\ of\ train\ (v_t) - Speed\ of\ man\ (v_m)
First, calculate the relative speed in meters per second (m/s) using the length of the train and the time taken to pass the man:
Relative\ Speed\ in\ m/s = \frac{Length\ of\ train}{Time} = \frac{150 \text{ meters}}{10 \text{ seconds}} = 15 \text{ m/s}
Convert this relative speed from meters per second to kilometers per hour (km/h) by multiplying by \frac{18}{5}:
Relative\ Speed\ in\ km/h = 15 \times \frac{18}{5} = 3 \times 18 = 54 \text{ km/h}
Since the relative speed is equal to the speed of the train minus the speed of the man, we can set up the equation:
54 = v_t - 6 v_t = 54 + 6 = 60 \text{ km/h} Thus, the correct option is B.
When A runs the full 1000-meter race, B runs 1000 - 100 = 900 meters, and C runs 1000 - 190 = 810 meters. Therefore, the ratio of the distances covered by B and C in the same time is:
\frac{B}{C} = \frac{900}{810} = \frac{10}{9}
This means that for every 10 meters B runs, C runs 9 meters.
In a 500-meter race between B and C, B runs the full 500 meters. The distance covered by C in that same time is:
Distance\ of\ C = \frac{9}{10} \times 500 = 450 \text{ meters} Thus, B beats C by:
500 - 450 = 50 \text{ meters} Thus, the correct option is B.
When two people move in opposite directions around a circular track, they cover the entire circumference of the track together to meet each other anywhere on the track. The relative speed of the two runners moving in opposite directions is the sum of their individual speeds:
Relative\ Speed = Speed\ of\ P + Speed\ of\ Q = 6 + 4 = 10 \text{ m/s}
The time taken to meet for the first time is the total distance of the track divided by their relative speed:
Time = \frac{Circumference}{Relative\ Speed} = \frac{400 \text{ meters}}{10 \text{ m/s}} = 40 \text{ seconds} Thus, the correct option is C.
Let the total number of steps of the escalator be N. When the person walks up the moving escalator, his effective speed is the sum of his walking speed (v_m) and the escalator’s speed (v_e). The time taken is 20 seconds:
v_m + v_e = \frac{N}{20}
When the person stands still on the moving escalator, he is carried up solely by the escalator’s speed in 60 seconds:
v_e = \frac{N}{60}
Substitute v_e into the first equation to find the person’s walking speed (v_m):
v_m + \frac{N}{60} = \frac{N}{20} v_m = \frac{N}{20} - \frac{N}{60} = \frac{3N - N}{60} = \frac{2N}{60} = \frac{N}{30}
When the person walks down the escalator while it is still moving upwards, his direction is opposite to the escalator’s movement. Therefore, his net downward speed is the difference between his walking speed and the escalator’s speed:
Net\ Speed = v_m - v_e = \frac{N}{30} - \frac{N}{60} = \frac{2N - N}{60} = \frac{N}{60}
Now, calculate the time taken to walk down the escalator:
Time = \frac{Distance}{Net\ Speed} = \frac{N}{\frac{N}{60}} = 60 \text{ seconds} Thus, the correct option is C.
When two trains move in opposite directions, the total distance to be covered to completely pass each other is the sum of their lengths:
Total\ Distance = Length\ of\ train\ 1 + Length\ of\ train\ 2 = 120 + 180 = 300 \text{ meters}
Since they are moving in opposite directions, their relative speed is the sum of their individual speeds:
Relative\ Speed = 68 + 52 = 120 \text{ km/h} Convert the relative speed from kilometers per hour (km/h) to meters per second (m/s) by multiplying by \frac{5}{18}:
Relative\ Speed\ in\ m/s = 120 \times \frac{5}{18} = \frac{600}{18} = \frac{100}{3} \text{ m/s}
Now, calculate the time taken to cross each other using the formula:
Time = \frac{Total\ Distance}{Relative\ Speed} = \frac{300}{\frac{100}{3}} = 300 \times \frac{3}{100} = 3 \times 3 = 9 \text{ seconds} Thus, the correct option is C.
Let the length of the train be L meters and the uniform speed of the train be S m/s.
When the train passes the first platform (100 meters long), the total distance covered is the sum of the train’s length and the platform’s length:
L + 100 = S \times 15 When the train passes the second platform (300 meters long), the total distance covered is:
L + 300 = S \times 25
Subtract the first equation from the second equation to find the speed S:
(L + 300) - (L + 100) = 25S - 15S 200 = 10S S = \frac{200}{10} = 20 \text{ m/s}
Now, substitute the value of S back into the first equation to find the length of the train L:
L + 100 = 20 \times 15 L + 100 = 300 L = 300 - 100 = 200 \text{ meters} Thus, the correct option is B.
When two people move in the same direction around a circular track, the faster runner must gain a full lap (one complete circumference) over the slower runner to meet them anywhere on the track for the first time.
The relative speed of the two runners moving in the same direction is the difference between their individual speeds:
Relative\ Speed = Speed\ of\ P - Speed\ of\ Q = 5 - 3 = 2 \text{ m/s}
The time taken to meet for the first time is the total circumference of the track divided by their relative speed:
Time = \frac{Circumference}{Relative\ Speed} = \frac{500 \text{ meters}}{2 \text{ m/s}} = 250 \text{ seconds} Thus, the correct option is B.
First, calculate the distance covered by Train X during the 1 hour it travels alone before Train Y starts (from 8:00 AM to 9:00 AM):
Distance = Speed \times Time = 40 \text{ km/h} \times 1 \text{ hour} = 40 \text{ km}
Find the remaining distance between the two trains at 9:00 AM when both are moving:
Remaining\ Distance = 340 \text{ km} - 40 \text{ km} = 300 \text{ km}
Since both trains are now traveling towards each other, their relative speed is the sum of their individual speeds:
Relative\ Speed = 40 + 60 = 100 \text{ km/h} Calculate the time taken by the trains to meet after 9:00 AM:
Time = \frac{Remaining\ Distance}{Relative\ Speed} = \frac{300}{100} = 3 \text{ hours}
Add this time to 9:00 AM to find the exact meeting time:
9:00\text{ AM} + 3\text{ hours} = 12:00\text{ PM} Thus, the correct option is B.
Let the speed of the train be S_t and the speed of the car be S_c.
Case 1:
The person travels 240 km by train and the remaining 480 - 240 = 240 km by car in 7 hours:
\frac{240}{S_t} + \frac{240}{S_c} = 7
Case 2:
The person travels 300 km by train and the remaining 480 - 300 = 180 km by car in 7 hours and 15 minutes (which is 7\frac{1}{4} = \frac{29}{4} hours):
\frac{300}{S_t} + \frac{180}{S_c} = \frac{29}{4}
To solve these simultaneous equations, let x = \frac{1}{S_t} and y = \frac{1}{S_c}:
240x + 240y = 7 300x + 180y = \frac{29}{4}
Multiply the first equation by 3 and the second equation by 4 to align the coefficients:
720x + 720y = 21 1200x + 720y = 29
Subtract the first modified equation from the second:
(1200x + 720y) - (720x + 720y) = 29 - 21 480x = 8 x = \frac{8}{480} = \frac{1}{60}
Since x = \frac{1}{S_t} = \frac{1}{60}, we get:
S_t = 60 \text{ km/h} Substitute x = \frac{1}{60} back into the first equation 240\left(\frac{1}{60}\right) + 240y = 7:
4 + 240y = 7 240y = 3 y = \frac{3}{240} = \frac{1}{80}
Since y = \frac{1}{S_c} = \frac{1}{80}, we get:
S_c = 80 \text{ km/h} Thus, the speeds of the train and the car are 60 km/h and 80 km/h respectively. Therefore, the correct option is B.
When a swimmer wants to reach a point directly opposite on the other bank of a flowing river (shortest path perpendicular to the flow), they must swim upstream at an angle so that the lateral component of their velocity cancels out the stream’s speed.
Let:
Speed of the man in still water (v) = 5 \text{ km/h}
Speed of the river stream (u) = 3 \text{ km/h}
To reach the exact opposite point, the effective speed across the river perpendicular to the banks (v_{eff}) forms a right-angled triangle where the man’s swimming speed is the hypotenuse and the stream speed is one of the legs:
v_{eff} = \sqrt{v^2 - u^2}
Substitute the given values:
v_{eff} = \sqrt{5^2 - 3^2} = \sqrt{25 - 9} = \sqrt{16} = 4 \text{ km/h} Now, calculate the time taken to cross the river of width 1 km at this effective speed:
Time = \frac{Distance}{v_{eff}} = \frac{1 \text{ km}}{4 \text{ km/h}} = \frac{1}{4} \text{ hour}
Convert the time into minutes by multiplying by 60:
\frac{1}{4} \times 60 = 15 \text{ minutes} Thus, his effective speed is 4 km/h and it takes him 15 minutes to cross. Therefore, the correct option is A.
The total distance of the journey is given as 240 km, and it is divided into two equal halves:
First\ Half\ Distance = Second\ Half\ Distance = \frac{240}{2} = 120 \text{ km}
Step 1: Calculate the time taken for the first half of the journey
Time_1 = \frac{Distance_1}{Speed_1} = \frac{120 \text{ km}}{40 \text{ km/h}} = 3 \text{ hours}
Step 2: Calculate the time taken for the second half of the journey
Time_2 = \frac{Distance_2}{Speed_2} = \frac{120 \text{ km}}{60 \text{ km/h}} = 2 \text{ hours}
Step 3: Calculate the total time taken
Total\ Time = Time_1 + Time_2 = 3 + 2 = 5 \text{ hours} Thus, the total time taken for the entire journey is 5 hours.
Therefore, the correct option is C.
First, calculate the total number of hours that have elapsed from 12:00 noon on Monday to 2:00 PM on Wednesday:
Monday\ 12:00\ PM\ to\ Tuesday\ 12:00\ PM = 24\ hours Tuesday\ 12:00\ PM\ to\ Wednesday\ 12:00\ PM = 24\ hours Wednesday\ 12:00\ PM\ to\ Wednesday\ 2:00\ PM = 2\ hours Total\ Time\ Elapsed = 24 + 24 + 2 = 50\ hours
Given that the clock loses 2 minutes every hour, the total time lost by the clock in 50 hours is:
Total\ Loss = 50 \times 2 = 100\ minutes Convert 100 minutes into hours and minutes:
100\ minutes = 1\ hour\ and\ 40\ minutes
Subtract this lost time from the actual time (2:00 PM on Wednesday):
2:00\ PM - 1\ hour\ 40\ minutes = 12:20\ PM\ on\ Wednesday Thus, the clock will show 12:20 PM on Wednesday. Therefore, the correct option is B.
First, calculate the circumference of the wheel using its diameter (d = 70 \text{ cm}):
Circumference = \pi \times d = \frac{22}{7} \times 70 = 220 \text{ cm}
Convert the circumference into meters:
220 \text{ cm} = \frac{220}{100} = 2.2 \text{ meters} Convert the total distance from kilometers to meters:
Total\ Distance = 2.64 \text{ km} = 2.64 \times 1000 = 2640 \text{ meters}
Now, calculate the number of revolutions by dividing the total distance by the circumference of the wheel:
Number\ of\ Revolutions = \frac{Total\ Distance}{Circumference} = \frac{2640}{2.2} = \frac{26400}{22} = 1200 Thus, the wheel will make 1200 revolutions. Therefore, the correct option is B.
First, calculate the distance covered by the thief during the 15-minute head start (which is \frac{15}{60} = \frac{1}{4} of an hour):
Distance = Speed \times Time = 60 \text{ km/h} \times \frac{1}{4} \text{ hour} = 15 \text{ km}
When the owner starts chasing, both vehicles are moving in the same direction. Therefore, the relative speed of the owner with respect to the thief is the difference between their speeds:
Relative\ Speed = 80 \text{ km/h} - 60 \text{ km/h} = 20 \text{ km/h} The time taken by the owner to cover the 15 km gap and catch the thief is:
Time = \frac{Distance}{Relative\ Speed} = \frac{15 \text{ km}}{20 \text{ km/h}} = \frac{3}{4} \text{ hour}
Convert \frac{3}{4} of an hour into minutes:
\frac{3}{4} \times 60 = 45 \text{ minutes} Thus, the owner will catch the thief 45 minutes after he starts. Therefore, the correct option is B.
To find the angle between the hands of a clock at 4:20 PM, we can calculate the position of each hand relative to the 12 o’clock position (0°).
Step 1: Calculate the position of the minute hand
The minute hand completes a full $360^\circ$ circle in 60 minutes, which means it moves at a rate of 6^\circ per minute (\frac{360^\circ}{60}).
At 20 minutes past the hour:
Position\ of\ minute\ hand = 20 \text{ minutes} \times 6^\circ/\text{minute} = 120^\circ
Step 2: Calculate the position of the hour hand
The hour hand completes a full $360^\circ$ circle in 12 hours, moving at a rate of 30^\circ per hour (\frac{360^\circ}{12}), or 0.5^\circ per minute (\frac{30^\circ}{60}).
At 4 hours and 20 minutes:
* Movement due to the 4 hours: 4 \times 30^\circ = 120^\circ
* Movement due to the 20 minutes: 20 \times 0.5^\circ = 10^\circ
* Total position of the hour hand: 120^\circ + 10^\circ = 130^\circ
Step 3: Calculate the difference between the two hands
Angle = |Position\ of\ hour\ hand - Position\ of\ minute\ hand|
Angle = |130^\circ - 120^\circ| = 10^\circ Thus, the exact angle between the hands at 4:20 PM is 10°. Therefore, the correct option is B.
When sound travels towards a person who is moving in the opposite direction (approaching the source), the time interval between hearing successive sounds decreases because the person meets the sound waves sooner.
The relationship between the time intervals and speeds is given by the formula:
\frac{\text{Time interval between shots}}{\text{Time interval between reports}} = \frac{\text{Speed of sound } (S)}{\text{Speed of sound } (S) - \text{Speed of person } (v)}
Given:
* Time interval between shots = 10 minutes
* Time interval between reports (heard by the person) = 9 minutes
* Speed of sound (S) = 330 m/s
Substitute the given values into the formula:
\frac{10}{9} = \frac{330}{330 - v}
Cross-multiply and solve for v:
10 \times (330 - v) = 9 \times 330 3300 - 10v = 2970
10v = 3300 - 2970 10v = 330 v = 33 \text{ m/s} Thus, the speed of the person is 33 m/s. Therefore, the correct option is B.
For a faulty clock (whether it gains or loses time) to show the exact correct time again, it must accumulate a total error of a full 12-hour cycle, which is equivalent to 12 \text{ hours} \times 60 \text{ minutes/hour} = 720 \text{ minutes}.
Given:
* The clock loses 10 \text{ minutes} every 24 \text{ hours}.
To find the total time required for a 720-minute loss, we can set up a proportion or unitary method:
Time\ needed\ (in\ hours) = \frac{Total\ required\ error\ (720 \text{ minutes})}{Error\ rate\ per\ day\ (10 \text{ minutes/day})} \times 24 \text{ hours/day}
Substitute the values:
Time\ needed = \frac{720}{10} \times 24 = 72 \times 24 \text{ hours} Convert the total hours into days by dividing by 24:
Number\ of\ Days = \frac{72 \times 24}{24} = 72 \text{ days} Thus, the clock will show the correct time again after 72 days.
Therefore, the correct option is B.
When multiple runners run around a circular track in the same direction, they will all meet together at the starting point again at a time that is the Least Common Multiple (LCM) of the individual times each runner takes to complete one full lap.
Step 1: Calculate the time taken by each runner to complete one lap
* Time taken by A (T_A) = \frac{\text{Track Length}}{\text{Speed of A}} = \frac{1200 \text{ meters}}{4 \text{ m/s}} = 300 \text{ seconds}
* Time taken by B (T_B) = \frac{1200}{6} = 200 \text{ seconds}
* Time taken by C (T_C) = \frac{1200}{8} = 150 \text{ seconds}
Step 2: Find the LCM of the individual lap times
We need to find \text{LCM}(300, 200, 150):
* Prime factorization of 300 = 2^2 \times 3 \times 5^2
* Prime factorization of 200 = 2^3 \times 5^2
* Prime factorization of 150 = 2 \times 3 \times 5^2
Taking the highest power of each prime factor:
\text{LCM} = 2^3 \times 3 \times 5^2 = 8 \times 3 \times 25 = 600 \text{ seconds} Thus, all three runners will meet at the starting point for the first time after 600 seconds.
Therefore, the correct option is C.
Let the speed of the man in still water be u and the speed of the stream/current be v.
The downstream speed is the sum of the man’s speed in still water and the stream’s speed:
u + v = 15 \text{ km/h} \quad \text{--- (Equation 1)}
The upstream speed is the difference between the man’s speed in still water and the stream’s speed:
u - v = 9 \text{ km/h} \quad \text{--- (Equation 2)}
To find the man’s speed in still water (u), add Equation 1 and Equation 2:
(u + v) + (u - v) = 15 + 9 2u = 24 u = \frac{24}{2} = 12 \text{ km/h}
To find the speed of the current (v), subtract Equation 2 from Equation 1:
(u + v) - (u - v) = 15 - 9 2v = 6 v = \frac{6}{2} = 3 \text{ km/h} Thus, the speed of the man in still water is 12 km/h and the speed of the current is 3 km/h.
Therefore, the correct option is C.