Verbal Reasoning - Direction Sense Test - Discussion
Discussion Forum : Direction Sense Test - Direction Sense 1 (Q.No. 12)
12.
A man walks 2 km towards North. Then he turns to East and walks 10 km. After this he turns to North and walks 3 km. Again he turns towards East and walks 2 km. How far is he from the starting point?
Answer: Option
Explanation:
Discussion:
37 comments Page 4 of 4.
Haritha said:
1 decade ago
I agree with @Kiran's point. Can anybody explain it more clearly?
Nidhi said:
1 decade ago
Cannot understand the solution! why are we applying Pythagoras theorem ?
Murali said:
1 decade ago
12km
.......................end
.
. 5km
.
.
.
Start
Now we want to find distance from startin g point i.e start to end.
We know that pythagarus theorem,
H^2 = side^2+side^2.
Therefore,
H = sqrt(25+144).
H = sqrt169.
H = 13km.
12km comes from adding north distance and similarly 5km.
.......................end
.
. 5km
.
.
.
Start
Now we want to find distance from startin g point i.e start to end.
We know that pythagarus theorem,
H^2 = side^2+side^2.
Therefore,
H = sqrt(25+144).
H = sqrt169.
H = 13km.
12km comes from adding north distance and similarly 5km.
Sudhakar said:
1 decade ago
Hi this is calculated using Pythogoras theorem,
Traveled towards east = 10+2 = 12Km
Traveled towards north = 2+3 = 5 Km
We need to substitute in the formula.
= 5^2 + 12^2
= root of 25 + 144
= root of 169
= 13.
Traveled towards east = 10+2 = 12Km
Traveled towards north = 2+3 = 5 Km
We need to substitute in the formula.
= 5^2 + 12^2
= root of 25 + 144
= root of 169
= 13.
Kiran4jalsa said:
1 decade ago
@ Nagendra..
can you explain it in brief...
i think you are wrong
according to the figure given above there are two right angled triangles rite?
so we have find both AC and CE by using formula (hypotys^2=side^2+side^2) and then we have to calculate AC+CE which gives AE which is required solution...
can you explain it in brief...
i think you are wrong
according to the figure given above there are two right angled triangles rite?
so we have find both AC and CE by using formula (hypotys^2=side^2+side^2) and then we have to calculate AC+CE which gives AE which is required solution...
Vijay said:
1 decade ago
Nice nagendra muruthy.
Nagendramurthy said:
1 decade ago
No. Here we have to find the shortest distance from a to e. so that can be find out like this..
Total distance travel towards north 2+3= 5km
Total distance travel towards east 10+2= 12km
Therefore the shortest distance will sqrt of (5*5+12*12) = sqrt(169) = 13
I think u got my point. Have a nice day.
Total distance travel towards north 2+3= 5km
Total distance travel towards east 10+2= 12km
Therefore the shortest distance will sqrt of (5*5+12*12) = sqrt(169) = 13
I think u got my point. Have a nice day.
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