Xavier Aptitude Test 2012 Paper

Question : 75 of 85

Marks: +1, -0

Shyam, a fertilizer salesman, sells directly to farmers. He visits two villages A and B. Shyam starts from A, and travels 50 metres to the East, then 50 metres North-East at exactly 45° to his earlier direction, and then another 50 metres East to reach village B. If the shortest distance

between villages A and B is in the form of

a \√{b+\√{c}}

metres, find the value of

a + b + c.

52
54
58
59 .
None of the above

Validate

Solution:

(e) The shortest distance between A and B will be

d=\√{\ {AM}^{2}+\ {BM}^{2}}

\ {BM}= {50}/{\√{2}}=25 \√{2}

\ {AM}=50+25 √{2}+50=100+25 \√{2}

d^{2}=(100+25 \√{2})^{2}+(25 \√{2})^{2}

=a^{2}(b+\√{c})

d^{2}=(100+25 \√{2})^{2}+(25 \√{2})^{2}

=a^{2}(b+\√{c})

\⇒ d=a \√{(b+\√{c})}

2500(5+2 √{2})=a^{2}(b+\√{c})

⇒50^{2}(5+\ √{8})

a^{2}=2500, b=5, c=8

\∴\ a+b+c=50+5+8=63

Hence correct option is (e)

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