ব্যাখ্যা
Distance = Speed×Time
Distance = (6× 50)/60)km
Distance = 5 km
∴ Required time
= Distance/ Speed
= 5/10 hrs
= 1/2 hrs
= 30 min
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Distance = Speed×Time
Distance = (6× 50)/60)km
Distance = 5 km
∴ Required time
= Distance/ Speed
= 5/10 hrs
= 1/2 hrs
= 30 min
Let the speed of car be x km/hr
Distance= Speed × Time
Distance = 8x km
According to the question,
⇒ (x+4)×7.5= 8x
⇒ 7.5x+30= 8x
⇒ 8x−7.5x= 30
⇒ 0.5x= 30
⇒x= (30/0.5)= 60 km/hr
Required distance:
= 8 × 60
= 480 km
Required ratio:
= (65×8):(70×4)
= 520:280
= 13:7
According to the question,
∴ 50 m = 20 m
∴ 1m = 20/50 m
∴ 1000 m = (20/50)×1000 m
= 400 metres
Average speed= Total distance/Total time
Average speed= (10+12)/(10/12)+(12/10)
Average speed= 10.8 km/hr
Time taken by man if he did not stop
= 5 km/10 kmph
= 1/2 hr
= 30 min
∵Man takes rest for 5 minutes on each km
Total rest time= 5×4= 20 min
Total travelling time:
= 30 min+20 min
= 50 min
Cyclist:Jogger
Ratio of distance→ 2:1
Ratio of time→ 1:2
Ratio of their speed (Jogger:Cyclist)
= (1/2):(2/1)
= 1:4
Relative speed= 24-18= 6 km/hr
Time required by faster train to overtake slower train
= 27/6 hr
= 4(1/2) hr
∴ Distance between Q and R:
= 18×4(1/2)
= 81 km
If the required distance be x km, then
(x/3)−(x/4)= 1/2
⇒ (4x−3x)/12= 1/2
⇒ x/12= 1/2
⇒ x= 6 km
Time taken to ride one way = 3/2 = 1.5 hrs
Time taken to walk one way:
= 4.5 - 1.5
= 3 hrs
Time taken to walk both way :
= 3×2
= 6 hours
Let speed of C = X
Then Speed of B = 3X
Then Speed of A = 6X
Ratio of the speed of A and C = 1:6
So, Greater the speed less time taken in journey.
C’s speed is 6 times less than A So A will take
1/6 of the total time taken C to covered same distance.
So, Time taken by A
= 3/(2×6)
= 1/4 hours
= 15minutes
Let the time in which he traveled on foot = x hour
Time for travelling on bicycle = (9-x) hr
Distance = Speed×Time, and Total distance = 61 km
So,
4x + 9(9-x) = 61
=> 5x = 20
=> x = 4
So distance traveled on foot = 4x4 = 16 km
We know Time = Distance/speed
So total time taken = (160/64) + (160/80)
Time taken for 320 Km = 320× (2/9) = 71.11km/hr
In every two minutes he is able to ascend 1 m.
In this fashion he ascends up to 12 m because when he reaches at the top he does not slip down.
Thus, up to 12 m he takes 12×2=24 min. and for the last 2 m he takes 1 m.
Therefore, total time taken by him is 24+1 = 25 min to reach the top.
Let after T hours they meet
Then, 3T+4T=17.5
T=2.5
Time = 10:00 am + 2.5 hour = 12:30 pm
Here, 2xy/(x+y) = (2×40×60)/(40+60) = 48
A:B = 1000:900
B:C = 400:360 = 100:90 = 900:810
⇒ A:B:C = 1000:900:810
⇒ A:C = 1000:810
⇒ A:C = 500:405
⇒ In a 500 m race, A beats C by (500-405) m = 95 m
Number of games played = 40
Number of won games = 24
Percentage of games played = (24/40)× 100
= 60%
When P scores 80, Q scores 64.
When R scores 80, P scores 60
Hence, when R scores 150, Q scores (60×64×150)/(80×80) = 90
Therefore, in a game of 150, R can give 60 points to Q.
B runs 45/2 m in 6 sec.
B covers 300 m in (6x2)/45 x300 sec
= 80 sec
Distance to be covered = πD = 40π km
Relative speed of bikers = 16 + 14 = 30 kmph.
Now, time = distance/speed = 40π/30
= 4.18 hrs.
Distance covered by B in 9 sec.
=(100/45)x 9m= 20 m.
A beats B by 20 metres.
speed of A : speed of B = 7/3:1 = 7:3
It means, in a race of 7 metres, A gains (7−3 ) = 4 metres.
If A needs to gain 80 metres, race should be of (7/4)×80 = 140 metres.