1950 AMC 12 Problem 50

Question

A privateer discovers a merchantman

10

miles to leeward at

11{:}45

a.m. and with a good breeze bears down upon her at

11

mph, while the merchantman can only make

8

mph in her attempt to escape. After a two hour chase, the top sail of the privateer is carried away: she can now make only

17

miles while the merchantman makes

15.

The privateer will overtake the merchantman at:

3{:}45

p.m.

3{:}30

p.m.

5{:}00

p.m.

2{:}45

p.m.

5{:}30

p.m.

Find the remaining gap after the first two hours

After the sail is lost, the speed ratio is

17:15

while the merchantman still travels at

8

mph

During the first two hours, the number of miles the privateer gains is

2(11-8)=6,

reducing the gap from

10

miles to

4

miles at

1{:}45

p.m.

After the damage, the ships’ speeds are in the ratio

17:15.

Since the merchantman still travels at

8

mph, the privateer’s new speed is

8\cdot\tfrac{17}{15}=\tfrac{136}{15}

mph. The closing speed is therefore

\tfrac{136}{15}-8=\tfrac{16}{15}

mph, so the number of hours needed to close the remaining gap is

\frac{4}{\frac{16}{15}}=\frac{15}{4}=3\frac34.

Adding

3

hours

45

minutes to

1{:}45

p.m. gives

5{:}30

p.m.

Thus, the correct answer is E.

Problem 50 in other years