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.

Answer: E

Concepts:distance rate and time relative speed ratio and proportion

Difficulty rating: 1900

Small Hint:

Find the remaining gap after the first two hours

Big Hint:

After the sail is lost, the speed ratio is $17:15$ while the merchantman still travels at $8$ mph

Solution:

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.

1950 AMC 12 Problem 50

Answer: E

Difficulty rating: 1900

Small Hint:

Big Hint:

Solution:

Problem 50 in Other Years