Arithmetic

Time, Speed & Distance

Medium

Two ships meet mid-ocean, and then, one ship goes south and the other ship goes west, both travelling at constant speeds. Two hours later, they are $60 \mathrm{~km}$ apart. If the speed of one of the ships is $6 \mathrm{~km}$ per hour more than the other one, then the speed, in km per hour, of the slower ship is

$24$

$18$

$12$

$20$

✅ Correct Option: 2

Two ships meet at a point and then travel in perpendicular directions - one goes south and the other goes west. After 2 hours, they are 60 km apart. One ship is 6 km/hr faster than the other.

Since the ships travel at right angles to each other, we can use the Pythagorean theorem to solve this problem.

Let's define our variables:

Speed of slower ship = $s$ km/hr

Speed of faster ship = $(s + 6)$ km/hr

We don't know which ship (south or west) is faster initially, but we'll call the slower speed 's' and work from there.

Using Distance = Speed × Time:

Distance traveled by slower ship = $s × 2 = 2s$ km

Distance traveled by faster ship = $(s + 6) × 2 = 2s + 12$ km

Since the ships travel at right angles (south and west), they form a right triangle.

Using $a^2 + b^2 = c^2$ :

$(2s)^2 + (2s + 12)^2 = 60^2$

Expanding $(2s + 12)^2$ :

$(2s + 12)^2 = (2s)^2 + 2(2s)(12) + 12^2$

$= 4s^2 + 48s + 144$

Substituting back:

$4s^2 + (4s^2 + 48s + 144) = 3600$

$8s^2 + 48s + 144 = 3600$

$8s^2 + 48s - 3456 = 0$

Dividing the entire equation by 8:

$s^2 + 6s - 432 = 0$

We need two numbers that multiply to -432 and add to +6. These numbers are +18 and -24.

$(s + 24)(s - 18) = 0$

From $(s + 24)(s - 18) = 0$ :

$s = -24$ (rejected because speed cannot be negative)

$s = 18$

Therefore, the speed of the slower ship = 18 km/hr

Answer: 18 km/hr

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