Aricent Time, Speed and Distance Problems with Solutions

m*d*t/w=m1*d1*t1/w so 3/50=6*x/300 ans is x= 300*3/50*6 x=3

let the speed is a & b (b>a)
total distance =x
two equation will form
18.5/a=(x-18.5)/b (Deep end equation)
x=18.5(a+b)/a..................1
also,
x/b+45+10.5/b=x/a+45+(x-10.5)/a (swallow end equation)
x=10.5(a+b)/(2a-b).........2
equating 1 & 2
we get a/b=37/53
so, x=18.5*90/37=45

if two trains of lengths a and b are moving towards each other
with speeds xkmph and ykmph respectively, then time taken
by them to cross each other= a+b/u+v
here t=9secs.
9=x+270/120+80 (assuming second train length as x)
9=x+270/200*5/18 (converting kmph into m/s)
9=x+270/11.11*5
x=229.95=230

train and man ismoving in the same direction so the
relative speed will be= (63-3)km/hr
= 60km/hr
then 60km/hr * 5/18= 50/3 m/sec
therefore, time = distance/speed
= 500/ (50/3)
= 30 seconds
so the answer is 30 seconds

speep =distance/time
speed of first train
5x =350/2=175kmph
x=35
therefore speep of second train is=3x=3*35=105kmph.

speed1=48km/hr and time1=50/60hr
distance=speed*time
distance=48*50/60=40km
now it is given that the distance remains same but
she has to win the race in 40 min
therefore now:
time2=40/60 hr and distance=40km
speed2=40/(40/60)=40*60/40=60 km/hr

m1 * d1 = m2 * d2
3 *4 = 4 * x
where x = no. of days taken by 4 men to paint

Therefore, x=3

let from his home to station distance = x
so,
(x/40) + ((52-x)/80) = 1
solving this x = 28 km
so from office to station = 52-x = 52-28 = 24 km

If A takes x seconds and B takes y seconds to run 1km, then:
x+19=960y/1000
and
960/1000+30=y
⇒ y=150 sec and x=125 sec
Answer =150/1000×5000
= 750 sec

the person from p moves 6/11 km in 1 hr so 3-6/11=27/11
6/11 t+5/22 t=27/11
17/22 t=27/11
t=54/17 =3.17h
so time=3:17 pm