Latest written test Aptitude Question SOLUTION: Jake and Paul each walk 10 km. The speed of jack is 1.5 faster than paul speed . Jack reaches the destination 1.5 hrs before paul .Then Jake's speed is equal to

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(#M40039369) WRITTEN TEST Logical Reasoning question answer Keep an EYE

Jake and Paul each walk 10 km. The speed of jack is 1.5 faster than paul speed . Jack reaches the destination 1.5 hrs before paul .Then Jake's speed is equal to

Asked In written test sudhakaravsd (9 years ago)

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vijayalakshmi  (9 years ago)

Let pouls speed be xkmph
Then jacks speed is (x+1.5)kmph
(10/x)-(10/1.5+x)=1.5
X=2.5
Jacks speed=2.5+1.5=4kmph

Is this solution Helpfull? Yes (17) | No (5)  

SOMESH SANJAY GOLAIT  (6 years ago)

Let speed of jake = x kmph.
So, speed of paul = ( x - 1.5 ) kmph.
Time difference =1.5hours
Total Distance = 10km
10/(x-1.5) -10/x = 1.5
By solving the above eq.
x = 4 kmph

Is this solution Helpfull? Yes (3) | No   

jagadeesh  (8 years ago)

(10/x)-(10/1.5x)=1.5
=4.44

Is this solution Helpfull? Yes (2) | No (4)  

Anusha  (4 years ago)

Let speed of Paul = x km/h
then , speed of Jack = (x + 1.5 ) km/h

a/c to question, Jack reaches the destination 1.5 hrs before Paul.
e.g., time taken by Jack + 1.5hrs = time taken by Paul

time taken by Jack = distance covered by Jack/speed of Jack
= 10km/(x + 1.5) hrs

time taken by Paul = distance covered by Paul/speed of Paul
= 10km/x hrs

now, 10/(x + 1.5) + 1.5 = 10/x

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

=> 1.5 = 10{(x + 1.5 - x)}/x(x + 1.5)

=> 1.5 = 10 × 1.5/(x² + 1.5x)

=> x² + 1.5x = 10

=> x² + (3/2)x - 10 = 0

=> 2x² + 3x - 20 = 0

=> 2x² + 8x - 5x - 20 = 0

=> 2x(x + 4) -5(x + 4) = 0

=> (x + 4)(2x - 5) = 0

x = -4 , 5/2

hence, speed of Paul = 2.5 km/h
so, speed of Jack = x + 1.5 = 2.5 + 1.5 = 4 kmph

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

sivasankari  (6 years ago)

(10/x)-(10/1.5+x)=1.5 derive this line

Is this solution Helpfull? Yes | No (2)  

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