Latest TCS Aptitude Question SOLUTION: Jake is faster than Paul. Jake and Paul each walk 24 km. The sum of their speeds is 7 km/h and the sum of time taken by them is 14 hours. Then Jake's speed is equal to :

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Jake is faster than Paul. Jake and Paul each walk 24 km. The sum of their speeds is 7 km/h and the sum of time taken by them is 14 hours. Then Jake's speed is equal to :

Asked In TCS Dipty Sharma (10 years ago)

Unsolved Read Solution (15)

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Read Solution (15) :

rohit kumar  (10 years ago)

Given that speed of Jake is greater than Paul
Distance = 24 km
Sum of their speed is 7 km/h = J+P
So possible speed ratio between J & P is
Go by Option
6:1 Not in option
5:2 = (24/5)+(24/2) ≠ 14 Hours
4:3 = (24/4)+(24/3) = 14 Hours
So Jake’s speed is 4km/h

Is this solution Helpfull? Yes (112) | No (3)  

kumar amit  (10 years ago)

since distance=24 km
let jake speed=x kmph
paul speed=y kmph
a/c to question
x+y=7
or
y=7-x
now,
24/x+24/y=14
or 24/x+24/7-x=14
or
1/x+1/7-x=14/24=7/12
x+7-x/7x-x*x=7/12
by solving
x=3,4
or
y=4,3
by using option
4km/h

Is this solution Helpfull? Yes (30) | No (1)  

Bholanath   (10 years ago)

V(p)+V(j)=7................(a)
distance traveled by each= 24 Km
T(j)+T(p)=14
24/V(j)+24/V(p)=14
1/V(j)+1/V(p)=14/24
(V(p)+V(j))/V(p)V(j)=7/12;
7/V(p)V(j)=7/12, from ..(a)
V(p)V(j)=12................(b)

square of (V(j)-V(p))=square of (V(j)+V(p))-4*V(p)V(j)
by putting values from a and b
we get,
V(j)-V(p)=1.............(c)
on solving a and c
V(j)=4 Km/h
and V(p)=3 Km/h

Is this solution Helpfull? Yes (7) | No (10)  

shrey  (9 years ago)

jake's speed=x, paul's speed=y
average speed= 2xy/x+y
(24+24)/14=2xy/7
xy=12
y=12/x....1

x+y=7.....2

from 1 and 2 we get x=4
x=

Is this solution Helpfull? Yes (5) | No   

SURABHI SRIVASTAVA  (8 years ago)

given sum is 7 so one is x and other is 7-x
so 24/x+24/7-x = 14
check by option as
put 4 we get 6+8 =14
where it satisfy equation 14=14 so 4km/h ans

Is this solution Helpfull? Yes (4) | No   

kunwar   (8 years ago)

SIMPLE LET JAKE SPEED BE 4 AND PAUL BE 3 AS SUM IS 7 AND JAKE IS FASTER THAN PAUL
AND TIME TAKEN BY JAKE=24/4=6 AND PAUL,S TIME=24/3=8
THEN 8+6=14

Is this solution Helpfull? Yes (2) | No   

D venkatesh babu  (10 years ago)

Here it is given that
sum of speeds V1+V2=7;------------(i)
sum of time taken T1+T2=14--------(ii)
Here they given jack speed is more than paul
hence 3 cases arises it may be (v1,v2)=(6,1) or (5,2) or (4,3)
by solving them in the equation (i) we will get v1,v2=4,3
also it also satisfies the (ii) equation T1,T2
v1+v2=4+3=7;
T1+T2=6+8=14;
hence jack speed =speed of paul+1 (V1=V2+1)

Is this solution Helpfull? Yes | No (4)  

rakesh  (10 years ago)

given speed(j) + speed(paul) = 7km/hr

consider speeds:- 6(4hrs) + 1(24 hrs) [28>14]
5(almost 5hrs) + 2(12 hrs) [17>14]
4(6 hrs) + 3(8 hrs) [14=14]
Therefore speed of jake is equals to 4km/h

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DIVYA KUKRETI  (10 years ago)

let speed and time of Jake=x and t1 respectively
Speed and time of Paul=y and t2 respectively(x>y.. given)so t1

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Aritra Chakraborty  (8 years ago)

Jake = j kmph, paul = p kmph, total= j+p kmph, total time= 24/j + 24/p.. thus j+p = 7, 24/j+24/p=14.. solving we get j=4 kmph

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Tushar Bedekar  (8 years ago)

let the speed of jake is s1 and that of paul is s2.also time taken will be t1 and t2 respectively.
so s1+s2=7 and t1+t2=14 also (s1>s2) solve the above conditions to find the value of s1 which is 4km/h(ans)

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Manikandan. P  (7 years ago)

Speed=j+p=7kmph
Time =j+p=14hr
Totally 4:3. p-4 j-3 so the ans 4kmph

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nitya kumari  (7 years ago)

let speed of jake =x
speed o fpaul =7-x
total distance is 24km
24/x +24/(7-x) =14
solving above equation we get
x^2-7x+12=0
x=4,3
jake is faster so his speed will be 4km/hr

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Rajaboina Radhika  (6 years ago)

4kmph

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Pooja Kumari  (4 years ago)

Speed of joke =x
Speed of paul=7-x
Given that each walk 24km
Tj+tp=14
24/x + 24/(7-x) =14 after solving this eqn
X=3,4 since joke is faster than Paul
Speed of joke =4

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