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

@mbatious Let the starting point be P
let the distance travelled in 1min be X km.
Distance from P at the end of 4th minute = 4X
Distance from P at the end of 7th minute = 4X3X = X
Distance from P at the end of 9th minute = X+2X = 3X
Distance from P at the end of 10th minute = 3XX = 2X
2X = 2 x 4 x 1/60 km = 2/15km

@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


@mbatious !
https://m.facebook.com/events/694814237601263?view=permalink&id=694820154267338Total 40 min adding 5 min lead

@mbatious relative speed = 40+60=100
0.2/100*10 = 0.02 or 20mtrs


@mbatious H............D(distance).........S
Fastest one = D+D0.2 total distance
Slower one = 0.2 total distance
Time is same so
2D0.2/3 =0.2/2
4D0.4=0.6
D=1/4 or 250 mtr

@mbatious
Shikha .........48min drive......rohan...................concert
...........shikha/rohan..........................................concert
meet somewhere in between and saves 8 min drive either side
so rohan covers distance worth 40min drive in the same time shikha covers distance worth 8min drive
therefore ratio of speed = 5:1 ( rohan:shikha)
hence auto rikshaw speed = 60/5 = 12kmph

@mbatious C?
Where will i find the answers?
Please tell @mbatious @arvindkjha @Aparajita @PremMora

@mbatious said in Question Bank  100 CAT level questions on Time, Speed and Distance topic:
A faulty clock gains 10 minutes every hour. If the time is set correctly at 12 Noon on 1st Jan 2010, then how many times will its minutehand and hourhand meet in the next 24 hours
24:28 is the ratio of correct:faulty clock
The new speed of minute hand will be 70 div/60 min.
The new speed of hour hand will be 5.8333 div/60 min.so there are 28 * 60 divisions.
And the relative speed is 64.167.
so,
28 * 60/64.167=26.18 ==> 26 meetings.

@mbatious 1296 km . . .

@mbatious 17.5 km .
.