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 = 4X-3X = X
Distance from P at the end of 9th minute = X+2X = 3X
Distance from P at the end of 10th minute = 3X-X = 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/(S-R)=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/(S-R)+(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(d-10)/A =10/P (time taken by prakash and Arpit are same)
2+ 5(5P-10)/2.5P=10/P
2+2(5P-10)=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 = (T-60)/E
time taken by B = 40Steps/B =(T-40)/E
30E=(T-60)B;40E=(T-40)B
30E/40E=(T-60)B/((T-40)B)
3/4=(T-60)/(T-40)
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+D-0.2 total distance
Slower one = 0.2 total distance
Time is same so
2D-0.2/3 =0.2/2
4D-0.4=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 @Prem-Mora
-
@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 minute-hand and hour-hand 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 .
.
