XAT 2012 Question 76

Question 76

Shyam, a fertilizer salesman, sells directly to farmers. He visits two villages A and B. Shyam starts from A, and travels 50 meters to the East, then 50 meters North-East at exactly 45° to his earlier direction, and then another 50 meters East to reach village B. If the shortest distance between villages A and B is in the form of $$a\sqrt{b+\sqrt{c}}$$ meters, Find the value of a+b+c.

Solution
Shortest distance between A and B = $$d = \sqrt{(AM)^2 + (BM)^2}$$ ----------Eqn(1)

In $$\triangle$$ OPN

=> $$sin 45 = \frac{PN}{PO}$$

=> $$\frac{1}{\sqrt{2}} = \frac{PN}{50}$$

=> $$PN = \frac{50}{\sqrt{2}} = 25 \sqrt{2}$$

=> $$BM = PN = 25 \sqrt{2}$$

Again, $$tan 45 = \frac{PN}{OP}$$

=> $$OP = 25 \sqrt{2}$$

=> $$AM = 50 + 25 \sqrt{2} + 50 = 100 + 25 \sqrt{2}$$

Using eqn(I), we get :

=> $$d^2 = (100 + 25 \sqrt{2})^2 + (25 \sqrt{2})^2$$

=> $$d^2 = 10000 + 5000 \sqrt{2} + 1250 + 1250 $$ = $$12500 + 5000 \sqrt{2}$$

=> $$d^2 = 2500 (5 + 2 \sqrt{2}) = 2500 (5 + \sqrt{8})$$ -----------Eqn(II)

Also, it is given that : $$d = a\sqrt{b+\sqrt{c}}$$

=> $$d^2 = a^2 (b + \sqrt{c})$$ -----------Eqn(III)

Comparing, eqn(II) & (III), we get :

=> $$a^2 (b + \sqrt{c}) = 2500 (5 + \sqrt{8})$$

=> $$a = 50 , b = 5 , c = 8$$

$$\therefore a + b + c = 50 + 5 + 8 = 63$$



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