Latest TCS Aptitude Question SOLUTION: Jack and Paul covers 10 km. Jack has speed 1.5 km/h more than Paul and reaches the destination 1.5 Min before Paul. So what is the speed of Jack.

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

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

(#M40039352) TCS Numerical Ability question Today on 9th Sep in KGEC Keep an EYE

Jack and Paul covers 10 km. Jack has speed 1.5 km/h more than Paul and reaches the destination 1.5 Min before Paul. So what is the speed of Jack.

Asked In TCS Zishan Khan (9 years ago)

Unsolved Read Solution (14)

Is this Puzzle helpful?   (8)   (3) Submit Your Solution Time Distance and Speed

Read Solution (14) :

dibyendu  (9 years ago)

let speed of paul =x kmph
speed of jack=(x+1.5) kmph
(10/x)-(10/x+1.5)=(1.5/60) hr
solving we get
x=23.75 kmph
jack speed should be=25.25 kmph

Is this solution Helpfull? Yes (33) | No (1)  

Kethineni Vinodh  (9 years ago)

Let speed of paul =x kmph
then speed of jack=x+1.5
now the equation becomes
(10/x)-(10/x+1.5)=1.5 min
solving the equation we get
x^2+1.5x-10=0
x=2.5 and -4
so paul speed=2.5kmph
then Jack Speed=4kmph

Is this solution Helpfull? Yes (23) | No (17)  

Parshant garg  (9 years ago)

statement wrong it should be 1.5 hours
then
10/x-10/(x+1.5)=1.5
we get x=2.5
and jack speed is 4

Is this solution Helpfull? Yes (7) | No (3)  

Amandeep Singh  (9 years ago)

let speed of paul=x
then speed of jack=x+1.5
now the equation becomes-
(10/x)-(10/x+1.5)=1.5
solving the equation u get
x^2+1.5x-10=0

now in TCS paper options were given-

a)2
b)4
c)8
d)6

by hit and trial subtract 1.5 from all the above and put in the above equation to get result.

Is this solution Helpfull? Yes (5) | No (2)  

Arth Malviya  (9 years ago)

let speed of paul =x kmph

speed of jack=(x+1.5) kmph

(10/x)-(10/x+1.5)=(1.5/60)

solving above equ.
we get

x=23.75 kmph

paul speed = 23.75

jack speed should be=25.25 kmph

Is this solution Helpfull? Yes (3) | No   

asd  (9 years ago)

j=1.5 m/h let time taken by jack(x-1.5/60)
----------------------------
p let speed s 10 km Let paul take x h to reach

for paul from formula speed=d/t
1.5=10/(x-1.5/60)
we get x=803/120

for paul s=10/x=10*120/803 simply putting value of x
s=1.494 (ANSWER) ie paul speed less than jack

Is this solution Helpfull? Yes (2) | No (9)  

Subhankar Laha  (9 years ago)

I think it's 25.25kmph

Is this solution Helpfull? Yes (2) | No (8)  

Ritu Jha  (9 years ago)

sorry for the wrong answer. the equation i m getting is:
p^2 +1.5p -600=0
where p=pauls speed
and jake speed is j=1.5+p

Is this solution Helpfull? Yes (2) | No (3)  

KRISHNA KUMAR  (9 years ago)

Let speed of Paul=x km/hr
then
speed of jack=(x+1.5)km/hr
So a/q
(10/x)-(10/(x+1.5))=(1.5/60)
solve for x
we get x=23.75 and x=-25.25(not possible)
so speed of jack is 23.75+1.5=25.25 km/hr

Is this solution Helpfull? Yes (1) | No   

sapna agarwal  (9 years ago)

I THINK JACK SPEED IS 4 AND PAUL SPEED IS 2.5

Is this solution Helpfull? Yes | No (8)  

Ritu Jha  (9 years ago)

i am getting Jacks speed as 200 kmph.

Is this solution Helpfull? Yes | No (6)  

sapan kumar paul  (9 years ago)

ANS=10KM/Hr

Is this solution Helpfull? Yes | No (1)  

saravanakumar  (9 years ago)

the answer is 4

Is this solution Helpfull? Yes | No (1)  

satyam  (9 years ago)

If YOU CONVERT 1.5 MINUTE IN HOUR THEN ANSWER WILL BE 22.5 OTHERWISE 4

Is this solution Helpfull? Yes | No (1)  

Please Login to Read Solution.