Anonymous

Current word problem (algebra), need help?

1. A plane took 4h to fly 2200km from Saskatoon to Toronto with a tail wind. The return trip, with head wind, took 5h. Find the speed of the plane in still air and the wind speed.

2. A riverboat took 2h to travel 24km down a river with the current and 3h to make the return trip against the current. Find the speed of the boat in still water and the speed of the current.

Can't figure these out, help?

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• Lucy
Lv 7

PROBLEM #1: A plane took 4h to fly 2200km from Saskatoon to Toronto with a tail wind. The return trip, with head wind, took 5h. Find the speed of the plane in still air and the wind speed.

HINT: Write what you know in a table.

Variables:

d = distance

r = rate

t = time

w = speed of wind

......................... distance ............ rate ................ time

to Toronto .......... 2200 ….............. r + w ............… 4

from Toronto …... 2200 ..............… r - w .............… 5

Remember that the distance formula is d = r * t. Apply this to the table.

......................... distance .................... rate ................ time

to Toronto .......... 2200 = (r + w) * 4 ....... r + w ............… 4

from Toronto …... 2200 = (r - w) * 5 .....… r - w .............… 5

......................... distance ................ rate ................ time

to Toronto .......... 2200 = 4(r + w) ....... r + w ............… 4

from Toronto …... 2200 = 5(r - w) .....… r - w .............… 5

You have 2 equations.

2200 = 4(r + w)

2200 = 5(r - w)

Solve the first equation for r.

2200 = 4(r + w)

2200 = 4(r) + 4(w)

2200 = 4r + 4w

2200 - 4w = 4r

(2200 - 4w) / 4 = r

550 - w = r

Plug this into the second equation.

2200 = 5(r - w)

2200 = 5(550 - w - w)

2200 = 5(550 - 2w)

2200 = 5(550) + 5(-2w)

2200 = 2750 - 10w

10w = 2750 - 2200

10w = 550

w = 550 / 10

w = 55

Plug this solved w value into one of the equations to find r.

550 - w = r

550 - 55 = r

495 = r

ANSWER: The speed of the plane in still air is 495 km/h; the wind speed is 55 km/h.

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PROBLEM: A riverboat took 2h to travel 24km down a river with the current and 3h to make the return trip against the current. Find the speed of the boat in still water and the speed of the current.

HINT: Write chat you knoc in a table.

Variables:

d = distance

r = rate

t = time

c = speed of current

......................... distance ............ rate ................ time

to .......... 24 ….............. r + c ............… 2

from …... 24 ..............… r - c .............… 3

Remember that the distance formula is d = r * t. Apply this to the table.

......................... distance .................... rate ................ time

to .......... 24 = (r + c) * 2 ....... r + c ............… 2

from …... 24 = (r - c) * 3 .....… r - c .............… 3

......................... distance ................ rate ................ time

to .......... 24 = 2(r + c) ....... r + c ............… 2

from …... 24 = 3(r - c) .....… r - c .............… 3

You have 2 equations.

24 = 2(r + c)

24 = 3(r - c)

Solve the first equation for r.

24 = 2(r + c)

24 = 2(r) + 2(c)

24 = 2r + 2c

24 - 2c = 2r

(24 - 2c) / 2 = r

12 - c = r

Plug this into the second equation.

24 = 3(r - c)

24 = 3(12 - c - c)

24 = 3(12 - 2c)

24 = 3(12) + 3(-2c)

24 = 36 - 6c

6c = 36 - 24

6c = 12

c = 12 / 6

c = 2

Plug this solved c value into one of the equations to find r.

12 - c = r

12 - 2 = r

10 = r

ANSWER: The speed of the riverboat in still water is 10 km/h; the current speed is 2 km/h.

Source(s): For more help on distance word problems: http://www.purplemath.com/modules/distance.htm
• mecham
Lv 4
4 years ago

hi, a million. permit p = velocity of airplane permit w = velocity of wind then velocity with wind = p + w (4 hour trip) velocity against wind = p - w (5 hour trip) Now d = rt so we've. a million) 2200 = (p+w)*4 2) 2200 = (p-w)*5 Divide a million) via 4 3) 550 = p + w divide 2 via 5 4) 440 =p - w upload 3) and four) 990 = 2p so p = 495 km/hr Then from a million) 2200 = (p+w)4 putting p = 495 provides us: 2200 = (495 +w)*4 divide the two components via 4 and we've 550 = 495 + w now subtract 495 from the two components giving us w = fifty 5 km/hr. permit's examine a million) 2200 = (495 + fifty 5)*4 2200 = (550)*4 2200 = 2200 This one exams. Now for 2) 2) 2200 = (p-w)*5 2200 = (495 - fifty 5)*5 2200 = 440*5 2200 = 2200 This exams. wish This facilitates!!

speed = distance/time

#1

s+w=2200/4 or s=550-w

s-w=2200/5 or s=440+w

therefore

550-w=440+w

110=2w

55=wind

then plug 55 in for w:

s=440+55=495

Speed in still air = 495km/h, wind during the flights was 55km/h.

#2

s+c=24/2 or s=12-c

s-c=24/3 or s=8+c

12-c=8+c

20=2c

10=current

Then plug 10 in for c:

s+10=24/2

s+10=12

s=2

Speed of boat=2km/h, speed of current=10km/h