RRB Junior Engineer Stage 1 Model Papers 10

Question : 23 of 98

Marks: +1, -0

A ship after sailing 12km towards south from a particular place covered 5 km more towards east. Then the straightway distance of the ship from that place is

18 km
15 km
13 km
11km

Validate

Solution:

Given, AB = 12 km, BC = 5 km
Straight way distance of ship

AC=√{AB^2+BC^2}=√{12^2+5^2}=√169=13

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