Part 1 1. Tony and Belinda have a combined age of 56. Belinda is 8 more than twice Tony’s age. How old is each?
Tony’s age = t years, Belinda’s age = 56 - t years
56 - t = 2t + 8
56 - 8 = 2t + t ==> 3t = 48 ==> t = 16 years
THUS Tony = 16 years, and Belinda = 40 years 2. Salisbury High School decided to take their students on a field trip to a theme park. A total of 150 people went on the trip. Adults pay $45.00 for a ticket and students pay $28.50 for a ticket. How many students and how many adults went to the park if they paid a total of $4770?
Adults = a, and Students = 150 - a
45a + 28.5(150 - a) = 4770
45a + 4275 - 28.5a = 4770 ==> 16.5a = 4770 - 4275 = 495 a = 495 / 16.5 = 30,
Adult = 30, Students = 120 3. Your piggy bank has a total of 46 coins in it; some are dimes and some are quarters. If you have a total of $7.00, how many quarters and how many dimes do you have?
Dimes = d, and Quarter = 46 - d d/10 + (46 - d) /4 = 7
Multiply both sides by 20, which gives
2d + 5(46 - d) = 140 ==> 2d + 230 - 5d = 140
-3d = 140 - 230 = -90 ==> d = 30
Dimes = 30, and Quarters = 16
Part 2 c = current of river b = rate of boat d = s(t) will represent (distance = speed X time)
Upstream: 60 = 6(b-c)
Downstream: 60 = 3(b+c)
There are now two separate equations: 60 = 6b - 6c and 60 = 3b + 3c
Solve both equations for b: b = 10 + c b = 10 – c
Now make both equations equal each other and solve for c:
10 + c = 20 – c
2c = 10 c = 5
The speed of the current was 5 mph
Now, plug the numbers into one of either the original equations to find the speed of the boat in still water.
I chose the first equation: b = 10 + c or b = 10 + 5 b = 15 The speed of the boat in still water must remain a consistent 15 mph or more in order for Wayne and his daughter to make it home in time or