Help?

Prep School Admissions
1

<p>I know this is probably the wrong place, but I really don't get this and I figured a lot of you on CC would know how to do this. And I have a test tomorrow morning and you guys have been so helpful with like everything:
5. A helicopter flew toward a ship that was heading for shore. The average speed of the ship was 40 km/h and the average speed of the helicopter was 300 km/h. The helicopter started from shore when the ship was 510 km away. How long did it take them to meet? </p>

<p>Does anyone know? If not, thanks anyways, and if you do could you please help? Thanks so much,
Kate</p>

2

<p>Haha I remember doing this. I think the formula looks something like this.</p>

<p>Distance Apart
________________ = Time it will take to meet</p>

<p>Speed 1 + Speed 2</p>

<p>So 510 divided by 340 (300+40) = 1.5</p>

<p>So the answer is 1.5 hours.</p>

<p>I hoped this help.</p>

<p>These problems only get harder so I suggest you try to get on board =)</p>

3

<p>Looking back on this again, I don't think this is the proper formula, but I am pretty sure it works.</p>

4

<p>Looking at this problem and answer might be helpful for potential Jan SSAT takers out there because it is a thinking problem, all about setting it up, then a quick computation.</p>

<p>Hockey Kid is right. More generally, remember that rate x time = distance. In this problem it doesn't matter how much of the 510 km is covered by each vehicle, so you can think of the rate as 300+40=340 km/hr. Solve for time.510/340 is 1 1/2.</p>