Latest Infosys Aptitude Question SOLUTION: A moves 3km. east from his starting point,he then travels 5km north,from that point he moves 8kms to the east. how far is A from his starting point?

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A moves 3km. east from his starting point,he then travels 5km north,from that point he moves 8kms to the east. how far is A from his starting point?

Asked In Infosys (13 years ago)

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JOHNSI GEORGE  (12 years ago)

Hope the ans is 13. A moves 3 km east and then 5 km towards north it doesn't mean from the point he left.(when we give a diagrammatic representation it ll be like 3'0 clk) .It is given in the ques that FROM THAT point so from 5 goes 8km E. so 3+5+(8-3)=13

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vinoth  (8 years ago)

TOTAL KMS TRAVELLED TOWARDS EAST =3+8=11KMS
TOTAL KMS TRAVELLED TOWARDS NORTH =5KMS
ACCORDING TO HYPOTENUSE RULE,
(HYPOTENUSE)2=SUM OF THE SQUARES OF TWO SIDES
SQRT(121+25)=12.12KMS

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Unknown  (6 years ago)

11 kms.......

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brundha sholaganga  (10 years ago)

(sqrt(3^2+(2.5)^2))+sqrt(8^2+(2.5)^2))

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vinoth  (8 years ago)

TOTAL KMS TRAVELLED TOWARDS EAST =3+8=11KMS
TOTAL KMS TRAVELLED TOWARDS NORTH =5KMS
ACCORDING TO HYPOTENUSE RULE,
(HYPOTENUSE)^2=SUM OF THE SQUARES OF TWO SIDES
SQRT(121+25)=12.12KMS

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ANUSHKA BAJPAI  (8 years ago)

displacement from original position is 13. ((11)^2+(5^2)sq root)

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Sindhu R  (5 years ago)

N
W E
S
3 kms east,5kms north,8kms east


fig. B;----------------- A
| 8 :
5 | : 5
| :
C----------;.................... E
3 D 8
now in fig we can see we can calculae the distance from A to C by applying pithagoras
theorem.
coz AB = 8 km
so DE =8 km
and BD = 5 km
so AE = 5 km
now we can calculate CE by adding CD+DE
now AC^2 = AE^2+CE^2 CE=CD+DE
AC^2=5^2+11^2
AC^2=25+121
AC^=146
AC~12.08

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krishna  (5 years ago)

A is 13 km away from his starting point
diagramatically:
8
------------
| |
5 | |5
------ ----------
3 8

12

therefore,
by pythogrous theorem
144+25=169
which is sqrt(169)=13

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