Latest TCS Aptitude Question SOLUTION: A man walked a certain distance south and then the same distance plus 7km due west.He is now 13km from his starting point.What are the distances south and west that he walked?".

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A man walked a certain distance south and then the same distance plus 7km due west.He is now 13km from his starting point.What are the distances south and west that he walked?".

Asked In TCS Sohini Neogi (10 years ago)

Solved Devendra Marghade Read Solution (11)

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Read Solution (11) :

Devendra Marghade  (10 years ago)

5 Km & 12 Km

If a man walked distance(in km) 'x' towards south & then 'x+7'towards west.
Distance from starting point=Hypotenuse of a right triangle with sides x & (x+7)
So 13^2=x^2+(x+7)^2
x^2+7x-60=0 or (x+12)(x-5)=0, so x=5
Distance towards west=x+7=5+7=12 km.

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

Subrata Patra  (10 years ago)

5 km and 12 km
suppose he walked x km in south and x+7 km in west
then x^2 +(x+7)^2=169
=>2x^2+14x-120=0
=>x^2 +7x-60=0
=>(x+12)(x-5)=0
then x=-12,or 5
but distance can't be negative so x=5

Is this solution Helpfull? Yes (2) | No   

B. SIREESHA  (10 years ago)

Ans: 5km to south and 12 km towards west.
Let the distance covered as he goes south be x km
Since he covers the same distance and an additional distance of 7 km in west, therefore distance covered is (x+7)km
According to Pythagoras Theorem,
x^2+(x+7)^2= 13^2
Solving for x, we get x=5km (south)
so 5+7=12km (west)

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Mohit Thakral  (10 years ago)

5 12
x^2+(x+7)^2=13^2
x=5
12

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AAYUSH MITTAL  (10 years ago)

5km to south and 12 km to west
12^2+5^2=13^2

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Gourab Das  (9 years ago)

ans-5 km towards south and 12km toward west.
let the man walked x km south.therefore he walked x+7 west.
so we can write x^2+(x+7)^2=169
from here we can find x=5

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srishti talreja  (9 years ago)

distance travelled towards south is 5 km and towards west is 12 km.By phythagoras theorem.

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mounika mandapalli  (9 years ago)

south-5km,west=12km

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akash shrivastava  (9 years ago)

let he walked x km south.
and then x+7 km towards west.
Now the total distance from starting point is 13.
This forms a right angled triangle.
so
x^2+(x+7)^2=169
=> x^2+x^2+14x+49=169
=> 2x^2+14x-120=0 -----------eqn 1
dividing eqn 1 by 2
we get
x^2+7x-60=0
=> (x+12)(x-5)=0
=> x=-12 or x=5
Since distance cant be negative so x=5
Hence he walked 5 km south and then (5+7)=12 km west.

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savita kumari  (9 years ago)

(x^2)+(x+7)^2=(13^2)
by solving we get x = 5
so, south distance is 5 and west distance is 5+7=12

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AFSHAH NASIM  (9 years ago)

apply Pythagoras
2x^2+49+14x=169
solve
ans x=5 x+7=12

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