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Failing Math! Need help!!! Please!?

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Failing Math! Need help!!! Please!?

Postby muata2 » Sat Dec 17, 2011 9:22 pm

Solve each of the following problems using a system of equations. Show all of your work and state your solution in a complete sentence.

Tony and Belinda have a combined age of 56. Belinda is 8 more than twice Tony's age. How old is each? (2 points)
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? (2 points)
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? (2 points)
Follow the 5 steps below to complete this problem. (4 points)

Step 1:

Pick a friend or family member to be the character of your word problem. This friend or family member may do one of the following:
Drive a boat
Drive a jet ski

Step 2:

Select a current speed of the water in mph.

Step 3:

Select the number of hours (be reasonable please) that your friend or family member drove the boat or jets ski against the current speed you chose in step 2.

Step 4:

Select the number of hours that your friend or family member made the same trip with the current (this should be a smaller number, as your friend or family member will be traveling with the current).

Step 5:

Write out the word problem you created and calculate how fast your friend or family member was traveling in still water. Round your answer to the nearest mph.
muata2
 
Posts: 23
Joined: Sat Apr 02, 2011 7:52 am

Failing Math! Need help!!! Please!?

Postby ludano17 » Sat Dec 17, 2011 9:32 pm

Age of Tony = x
age of Belinda = y

x + y = 56 ----- (1)
y = 2x + 8 ----- (2)

substitute (2) in (1)
x + 2x + 8 = 56
3x = 48
x = 16
therefore tony's age = 16
substituting x= 16 in (1)
16 + y = 56
y = 40
belinda's age = 40



#2
no of students = x
no of adults = y

x + y = 150 ----- (1)
28.5x + 45y = 4770 ----- (2)

subject y in (1)
y = 150 - x ----- (3)
substituting (3) in (2)
28.5x + 45(150 - x) = 4770
28.5x + 6750 - 45x = 4770
-16.5x = -1980
x = 120
substituting x = 120 in (1)
120 + y = 150
y = 30
therefore no of students = 120
no of adults = 30
ludano17
 
Posts: 25
Joined: Thu Mar 31, 2011 12:21 pm

Failing Math! Need help!!! Please!?

Postby reuben29 » Sat Dec 17, 2011 9:35 pm

Age of Tony = x
age of Belinda = y

x + y = 56 ----- (1)
y = 2x + 8 ----- (2)

substitute (2) in (1)
x + 2x + 8 = 56
3x = 48
x = 16
therefore tony's age = 16
substituting x= 16 in (1)
16 + y = 56
y = 40
belinda's age = 40



#2
no of students = x
no of adults = y

x + y = 150 ----- (1)
28.5x + 45y = 4770 ----- (2)

subject y in (1)
y = 150 - x ----- (3)
substituting (3) in (2)
28.5x + 45(150 - x) = 4770
28.5x + 6750 - 45x = 4770
-16.5x = -1980
x = 120
substituting x = 120 in (1)
120 + y = 150
y = 30
therefore no of students = 120
no of adults = 30
You must be joking. DO YOUR OWN HOMEWORK!!!!


GET TUTORING IN YOUR SCHOOL, EVERY SCHOOL HAS IT FOR FREE
reuben29
 
Posts: 18
Joined: Thu Mar 31, 2011 5:13 am

Failing Math! Need help!!! Please!?

Postby lorimar » Sat Dec 17, 2011 9:40 pm

ok with any word problem you need 2 equatios 2 unknowns or more....in the first 3 you only have 2 unknowns....

first t is tony's age and b is belinda's age
combined age b+t=56 b=2t+8

now 2t+8+t=56 3t=48 t=16........plug in for b

2.
a is adult
s is student

a+s=150
45a+28.5s=4770
45*(150-s)+28.5s=4770
solve for s.....plug s into first equation to get a

3....

d is dimes
q is quarters

d+q=46
10d +25q = 700

I multiplied both sides by 100 to get pennies as whole numbers since students don't like decimals....

10(46-q) +25q=700
solve for q .....then plug into first to get d.....

Well the next one just asks you to step out with the questions......

1.aLet's assume you and friend were boating.....

b. water was moving at 15 mph
c. you were driving against water and it took 4 hours.....
d. friend drove back it took 1.5 hours....

e is the equations to solve for speed in still water.....

we know distance=rate*time
distances would be same.....

I am not getting this part right now....its late.....I will try to contemplate.....and edit my answer later.....
lorimar
 
Posts: 19
Joined: Thu Mar 31, 2011 5:29 am


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