Latest TCS Aptitude Question SOLUTION: Jake and paul start together to walk 10kms jake walks 1.5kmph faster than paul and arrives at his destination 1.5 hours earlier than paul jakes speed walking is a)6kmph b)2kmph c

• Placement Papers
• /
• TCS
• /
• Numerical Ability
• /
• Time Distance and Speed

TCS Question Papers: TCS Interview Experience(90) TCS Company Profile

(#M40154847) TCS Numerical Ability question Apti Keep an EYE

Jake and paul start together to walk 10kms jake walks 1.5kmph faster than paul and arrives at his destination 1.5 hours earlier than paul jakes speed walking is a)6kmph b)2kmph c)4 kmph d)8 kmph

Asked In TCS dhritiman (8 years ago)

Unsolved Read Solution (3)

Is this Puzzle helpful?   (4)   (2) Submit Your Solution Time Distance and Speed

Read Solution (3) :

ROHIT KUMAR DUBEY  (8 years ago)

Let speed of jake(Sj)= x kmph. then speed of paul (Sp)=( x-1.5 )kmph.
time difference =1.5hours
total distance =10km
10/Sj -10/Sp=1.5
10/x-1.5 -10/x=1.5
solving above equation
x=4kmph
ans=C

Is this solution Helpfull? Yes (13) | No   

gopal  (8 years ago)

ans-3.175
ts=(t-1.5)(s+1.5)&st=10

Is this solution Helpfull? Yes | No (3)  

Parthiban  (8 years ago)

[1.5hr * 60 min =90 min earlier. hence 10km/90=0.1 km/ min walk faster by jake]
for an hour, 6km/hr faster. now 10-6 = 4 kms for 90 min taken by paul .
4km=90 min
hence 1km = 22.5 min
then calc for 1 hr, 60/22.5= 2.66 kmph taken by paul.
Jake 1.5kmph faster than paul
Finally 2.66*1.5= 4 kmph taken by Jake

Is this solution Helpfull? Yes | No   

Please Login to Read Solution.