Two cars (A and B) start at the starting line at the same time on a 3-mile long circular track, going in opposite directions. As they drive around the course, they pass each other many times. Exactly one hour after starting, they pass each other for the 33rd time. At this point car A has completed exactly 20 laps. What is the average speed of Car B?
A man can cover a certain distance in 3 hr36 min if he walk at the rate of 5km/hr. If he cover the same distance on bike at the rate of 24km/hr, then what will be time taken by him?
Hello Nancy!
Distance travelled by the man in 3hr 36 min ( i.e 18/5 hours ) = (18/5) x 5 = 18 km .
Hence , to travel 18km with the speed of 24km/hr , it takes 18/24 = 3/4 = 45 minutes .
Alternate Approach : The distance is constant in the case . Now the speed of 24km/hr is 24/5 times of his walking speed. Therefore , the time taken will be 5/24 times of 3 hr 36 min i.e = (5/24) x (18/5) = 45 minutes
If a boy walks from his house to school at the rate of 4 km/hr, he reaches school 10 min earlier. However if he walks at the rate of 3km/hr he reach 10 min late . find the distance of his school from his house?
Let the time taken to travel the distance at the speed of 4km/hr be t hr .
Now the speed of 3km is 3/4 of its previous speed , therefore the time taken to travel the distance with the speed of 3km/ hr will be 4/3 of t .
Now , 4t/3 - t = 20min
t/3 = 1/3 hr
t = 1 hr .
Hence, the distance between his house to the school = 1 x 4 = 4km.
Alternate Approach :
House __________________D __________________School
Let the original time taken be t then ,
D/4 = ( t - 10)/ 60 or t - 10 = 15 D .... (1)
and (t + 10)/60 = D/3 or t + 10 = 20D .... (2)
Solving (1) and (2)
We get D = 4km .
At the end of each completed minute between 12:00 and 23:59 we measured in degrees the angle between the hour and the minutes hand of a regular wall clock. What is the measurement of the smallest recorded angle.
A. 0.25
B. 0.5
C. 2.5
D. 3
Sir please solve this problem
Hello Samyak ,
A __2x__ B ______5x_____C
A __2y__ B __y__D________C
2y = 2x => x = y
Now , DC = 4x
Distance traveled by Xavier => AC + CD = 11x
Distance traveled by Yohana => BC + CD = 5x + 4x = 9x
Time taken by Xavier and Yohana is same here Hence ratio of speeds = Ratio of distance travelled
i.e 11x :9x = 11 : 9
A certain distance when is travelled at 3km/hr more than the usual speed ,the time taken to cover the distance would have been reduced by 2 hours than the usual time.Also,if the same distance is travelled at 4km/hr less than the usual speed,the time taken would have been increased by 5 hours more than the usual time.What is the distance?
A certain distance when is travelled at 3km/hr more than the usual speed ,the time taken to cover the distance would have been reduced by 2 hours than the usual time.Also,if the same distance is travelled at 4km/hr less than the usual speed,the time taken would have been increased by 5 hours more than the usual time.What is the distance?
Let the original speed be V km/hr an original time taken => T hr .
Now, VT = ( V +3) ( T -2)
=> VT = VT - 2V + 3T - 6 = VT
3T - 2V = 6..... (1)
also , VT = ( V - 4) ( T +5)
VT + 5V - 4T - 20 = VT
-4T + 5V = 20 .... ( 2)
From equation (1) and (2) V = 12 , T = 10
Distance traveled : 12 × 10 = 120 km .
1.
Let the escalator have a total of n steps and let it have moved through m steps by the time Bob took 20 steps.
therefore since Sam runs twice as fast as Bob, the time Sam takes 30 steps is equal to time in which Bob takes 15 steps .
The escalator moved m x 15/20 = 3m/4 steps .
So m + 20 = n and 3m/4 + 30 = n
m/4 = 10
m = 40 steps , n = 60 steps.
If Nick walk half as fast as Bob then let him take t times the time Bob takes => escalator would move through 40t steps and he would take 20t/2 steps
40t + 10t = 60
t = 1.2 times
So Nick would take 12 steps.
A thief is spotted by a policemen from a distance of 100 metres . When the policeman starts the chase , the thief also starts running . If the speed of the thief be 8 km/hr , how far the thief will have run before he is overtaken ?
Hi Richa,
The question seems to be incomplete... the speed of the policemen is not known.
What is the correct answer ?
is it 200?
yes
Keshav goes one third distance upstream while going from point A to point B in a river in one hour, and then he drops his cloth in the river which travels now at the speed of the stream. While Keshav continues his journey towards point B and turns back immediately twards point A, and to his surprise he reaches point A at the same time when the cloth reaches there. Find the ratio of speed of stream to the speed of Keshav in still water.
Answer is 1: 1+√2 ?
Let total distance be 3x and Keshav removes his cloth after x distance
Now we have this equation for remaining journey
Also, let v be Keshav's and a be speed of stream then
2x/(v-a) + 3x/(v+a)= x/a
It means he goes 2x upstream with v-a speed and returns 3x downstream with v+a speed and in the same time clothes travel with the speed of stream i.e a
Solving this equation
(2v+2a +3v-3a)*a= (v^2 - a^2)
5va - a^2= v^2 - a^2
5a=v
a/v =5
Hence required ratio is 5:1
Sorry for the last line it will be a/v=1/5
So the ratio will be 1:5
A cyclist travels on a road parallel to railway track at a constant speed of 10 km/hour. He meets the train daily, at crossing, which also travelled in same direction as of him. One day, he was late by half an hour and meets the train 6 km before crossing. Find the speed of train.
Options ?
10
20
30
60
Let x be time taken by the cyclist while traveling from his home to crossing
Earlier boy travels 10x distance, in same time train travels vx (assume speed of train be v)
Now boy travels 10(x+1/2)-6 and train travels vx-6 distance in same time
Ratio of velocity would be same in both cases so
10x/vx = {10(x+1/2)-6}/(vx-6)
Solving this we will get v =60
solution please
16/3 min
|____d_____|______________2d___________|
d/2v + 2d/v = 20
5d/2v = 20
d/v = 8.
|____d_____|______________2d___________|
d/v + 2d/2v = T
T = 2d/v = 2 x 8 = 16.
Aniket, 16/3 is not possible
as 20/(16/3) = 3.xyz
means his average speed is more than thrice while coming back which is not possible
solution please
let distance be 'd' and speed of train be x and boy be y
d/(x+y)=20
d/(x-y)=30
x/d+y/d=1/20
x/d-y/d=1/30
adding both we get 2x/d=1/12
d/x=24 which is the time taken for train to cover distance bw two stations
hence 24 is the answer
Alternative Approach :
speed of a train => x
and the boy's speed => y.
Now when the train goes in the same direction as the boy relative speed is x-y similarly it is x+y for opposite direction.
Now imagine the boy stops moving, relative speed of the train to the boy would just be x right ?
Hence, x-y, x and x+y form an AP. ( and distance is constant) time taken should form a HP.
Hence (2 x 20 x 30)/(20+30) = 24 minutes .
