Xavier Aptitude Test 2015 Paper

(d)At 5:00 PM, initial distance = AB = 1800 m
Minutes hand height = CD = 200 √3 m,
∠CAB = 30°

CB

AB

= tan 30° =

1

√3

⇒CB = 600 √3
Now, DB =600√3−200√3=400√3
At 5:10 PM: the minutes hand would move from CD to C1D(i.e. 60°) as follows : D and D1 are at same height.
And ∠C1DD1=30°
Now,

DD1

DC1

=cos‌30°
⇒

DD1

DC

=

√3

2

⇒ DD1200√3=

√3

2

⇒DD1=300 m
Also

C1D1

DD1

=tan‌30°
⇒

C1 D1

300

=

1

√3

⇒C1 D1=100√3 mAt 5:10 PM, the person moves from A to E , where∠C1 EB1=60°.
Now, C1 B1=B1D1+D1C1=BD+ D1C1
⇒C1B1=400√3+100√3=500√3 m
Now,

EB1

C1B1

=cot‌60°⇒

EB1

500√3

=

1

√3

⇒EB1=500 mts
The horizontal plane EBB ‌1 can be presented as below:Now, DD1= BB1=300 mts . and EB1=500 m
Now, EB=√(500)2−(300)2=400 m
In 10 minutes, distance travelled
= AE= AB− EB
=1800−400=1400 m
Speed=f1400×6010×1000=8.4 km/hr