Xavier Aptitude Test 2015 Paper

Question : 74 of 113

Marks: +1, -0

A person is standing at a distance of 1800 meters facing a giant clock at the top of a tower. At 5.00 P.M., he can see the tip of the minute hand of the clock at 30 degree elevation from his eye-level. Immediately, the person starts walking towards thetower. At 5.10 P.M., the person noticed that the tip of the minute hand made an angle of 60 degree with respect to his eye-level. Using three–dimensional vision, find the speed at which the person is walking. The length of the minutes hand is 200

√3

meters (

√3

= 1.732).

7.2 km/hour
7.5 km/hour
7.8 km/hour
8.4 km/hour
None of the above

Validate

Solution:

(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

C_1

D(i.e. 60°) as follows :

\ {D}

and

\ {D}_{1}

are at same height.
And

\∠{C}_{1 }{DD}_{1}=30^{\°}

Now,

\ {\ {DD}_{1}}/{ {DC}_{1}}=\cos 30^{°}

\⇒{\ {DD}_{1}}/{ {DC}}=\ {\√{3}}/{2}

\⇒{\ {DD}_{1}}{200 \√{3}}={\√{3}}/{2}

\⇒{DD}_{1}=300 \ {m}

Also

\ {\ {C}_{1} {D}_{1}}/{\ {D} {D}_{1}}=\tan 30^{\°}

\⇒{ {C}_{1} \ {D}_{1}}/{300}= {1}/{\√{3}}

⇒{C}_{1} \ {D}_{1}=100 \√ {3} \ {m}

5 : 10 \ {PM},

the person moves from A to

\ {E}

, where

\∠{C}_{1} \ {EB}_{1}=60^{\°} .

Now,

\ {C}_{1} \ {B}_{1}= {B}_{1 } {D}_{1}+ {D}_{1} {C}_{1}= {BD}+\ {D}_{1} {C}_{1}

\⇒ C_{1} B_{1}=400\√{3}+100 \√{3}=500 \√{3} \ {m}

Now,

{E B_{1}}/{C_{1} B_{1}}=\cot 60^{\°} ⇒{E B_{1}}/{500 √{3}}=\ {1}/{\√{3}}

\⇒E B_{1}=500 \ {mts}

The horizontal plane EBB

_{1}

can be presented as below:

Now,

\ {DD}_{1}=\ {BB}_{1}=300 \ {mts}

. and

\ {EB}_{1}=500 \ {m}

Now,

\ {EB}=\√{(500)^{2}-(300)^{2}}=400 \ {m}

In 10 minutes, distance travelled

=\ {AE}=\ {AB}-\ {EB}

=1800-400=1400 \ {m}

Speed

=\f {1400 \× 60}{10 \× 1000}=8.4

km/hr

Go to Question:

Prev Question

Next Question