Algebra level 1 unit 4 launch problem
name:
o cruise ships are leaving the same harbor and traveling to an exotic island. the sun
harbor at 6: 00 am and traveled at an average rate of 40 mph. the ship named tuna can
traveled on the same course as sea grappler at an average rate of 60 mph.
what time of day did tuna catcher catch up to sea grappler?
how far had each boat traveled at the moment that tuna catcher caught up to sea graget
t
if the exotic island the ships were headed to was 250 miles away from the harbor the bes
boat arrived there first?

At 12 o'clock at noon both the ships will meet.

The distance traveled till they meet is 240 miles.

The ship Tuna Catcher will reach the island first.

Step-by-step explanation:

Let after x hours past 6:00 A.M. the two ships meet each other on the same path moving in the same direction from the harbor to the island.

The ship named Sea Grappler starts at 6:00 A.M. from the harbor and moving at an average rate of 40 mph and the ship named Tuna Catcher starts at 8: 00 A.M. from the same harbor and moving at an average rate of 60 mph.

So, the distance traveled by Sea Grappler in x hours equals the distance traveled by Tuna Catcher in (x - 2) hours are the same.

Hence, 40x = 60(x - 2)  

⇒ 20x = 120

⇒ x = 6 hours.

So, at 12 o'clock at noon both the ships will meet. (Answer)

Now, the distance traveled by each ship till 12 o'clock at noon is  

(40 × 6) = 240 miles. (Answer)

If the exotic island is 250 miles away from the harbor and as the two ships meet each other before reaching the island, so, the ship that has greater speed i.e. ship Tuna catcher will reach the island first. (Answer)

answer: i believe its b. if you sound out the sentences that sounds wrong.

hi! i think the answer is 1/2 : )