NIMCET 2015 — Mathematics PYQ

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Question

NIMCET 2015 — Mathematics PYQ

NIMCET | Mathematics | 2015

A harbour lies in a direction $60^\circ$ South of West from a fort and at a distance $30\text{ km}$ from it. A ship sets out from the harbour at noon and sails due East at $10\text{ km}$ an hour. The time at which the ship will be $70\text{ km}$ from the fort is:

Choose the correct answer:

Explanation

Let the Fort be at the origin $O(0, 0)$ .

The Harbour $H$ is at $30\text{ km}$ in direction $60^\circ$ South of West.

Coordinates of $H$ :

$x_H = -30 \cos 60^\circ = -30 \times \frac{1}{2} = -15$

$y_H = -30 \sin 60^\circ = -30 \times \frac{\sqrt{3}}{2} = -15\sqrt{3}$

The ship starts from $H$ at noon and moves East at $10\text{ km/h}$ .

After $t$ hours, the position of the ship $S$ is:

$x_S = -15 + 10t$

$y_S = -15\sqrt{3}$

Distance of ship from fort ( $OS$ ) is $70\text{ km}$ :

$OS^2 = 70^2$

$(-15 + 10t)^2 + (-15\sqrt{3})^2 = 4900$

$(10t - 15)^2 + 675 = 4900$

$(10t - 15)^2 = 4225$

Taking square root:

$10t - 15 = \sqrt{4225}$

$10t - 15 = 65$

$10t = 80$

$t = 8\text{ hours}$

Since the ship started at noon (12:00 PM), 8 hours later the time will be 8:00 PM.

Correct Option: (b)