Question Bank  100 CAT level questions on Time, Speed and Distance topic

@mbatious total time for second case is 13 hrs, Am i right ??

@mbatious time is 6:24 am

t + t/12 = 9 (time taken in the return journey is 1/12th the time taken for the onward journey)
t = 108/13 = 8.30 hours
t + t/6 = 9.68 hours?

@mbatious
No of steps S= 30x+26 =18x+34 (where x=speed of escalator in steps/sec)
solving above eqn we get x=2/3 step/sec
hence number of steps = 20*2/3+26 =46
so ans is B

@mbatious
X= distance from Baghbazar Ghat to Ahiritola Ghat
S is swimmer speed and R is river speed
20/(SR)=3*20/(S+R) solving this we get S=2R
Now time taken by log to travel distance BA is equal to time taken by swimmer to travel from ahirtola to Howrah and back to Bagh bazar. So,
X/R=20/(SR)+(20+X)/(S+R)
X/R=20/R+(20+X)/3R (equate S=2R)
3X= 80+X (multiply by 3R)
so X=40 is the answer
i.e Option d

@mbatious
A(Arpit Speed)=2.5P (prakash speed)
d= distance from P to point of return
d=A*2=5P
2 +5(d10)/A =10/P (time taken by prakash and Arpit are same)
2+ 5(5P10)/2.5P=10/P
2+2(5P10)=10
12P=30
P=2.5
i.e ans is a

@mbatious
x = distance neha walk at g speed to pickup point
y= distance nitin drives at b speed to pickup point
x/g = y/b =30/2 =15
2(x+y)/b=40
x/b+y/b=20
x/b+15=20
x/b=5
(x/b)/(x/g)=5/15
g/b=1/3
so ratio is 1:3
hence C is ans

@mbatious
T=total steps
E= Escalator speed
B= speed of B
so speed of A=2B
time taken by A = 60Steps/2B = (T60)/E
time taken by B = 40Steps/B =(T40)/E
30E=(T60)B;40E=(T40)B
30E/40E=(T60)B/((T40)B)
3/4=(T60)/(T40)
Solving above you get T=120

a= speed of train starting from P
b= speed of train starting from Q
PQ/QR=a/b
(PQ+QR)/QR =(a+b)/b
QR=PR.b/(a+b)
QR/a=16
PR.b/((a+b).a)=16 ....(1)
Similarly PR.b/((a+b).a)=25 ....(2)
dividing 1 by 2
16a/b=25b/a
(a.a)/(b.b)=25/16
a/b=5/4
(a+b)/3=675
a+b=225
b=100
a=125
from 1
PR/a=16*(a+b)/b
=16*225/100
=36hr
so ans is b
