2 Algebra questions? (distance type)?

Question

Help me with the steps to solve either/or both of these two please! (if you know about RESA, please do it that way. if you can't just steps to how you would solve it)

1. With a tail wind, a plane flew 240 km in 45 min. With no change in the wind, the return trip took 48 minutes. Find the wind speed and the air speed of the plane.

2. Rose took a half hour to row 3 km w/ the current. When she returned, she took 90 minutes. Find her rowing rate and the speec of the current.

Answer 1

1.Let the speed of the plane be x kmph and that of the wind be y km ph
Therefore by the problem
240/(x+y) =3/4 [45 minutes converted to 3/4 hours]
and,240/(x-y)=4/5 [48 minutes=4/5 hours)
by cross-multiplication ,we get from the first eqn
3(x+y)=4*240
=> x+y=960/3=320..(eqn.3)
From the second eqn. by cross-multiplication we get
4(x-y)=240X5
=>x-y=1200/4=300....(eqn4)
now adding eqn 3 and 4,we get
2x=620
=>x=310Kmph
Putting the value of x in eqn 3 we get y=10kmph
Therefore the speed of the plane is 310 Kmph and that of the wind is 10 Kmph
2.Let the speed of the boat be x kmph and that of the current be y kmph
By the problem
3/(x+y)=1/2.....(eqn.1),and
3/(x-y)=3/2 [9o minutes converted into hours] (eqn no 2)
.by cross multiplying,we get from eqn no 1
x+y=6
x-y=(3/2)/3=2
.by adding we get x=4kmph and putting this value of x ,we get y=2 KMph
The rowing rate is 4 Kmph and the rate of the current is 2 Kmph....

Answer 2

ok so we are assuming the plane is moving at some constant speed on its own and then there is another force that causes either an increased or decreased speed (the wind). knowing this we can write two equations:
240x + 240W = 45
240x - 240W = 48
(x is the speed of the plane, and W is the wind speed)

So now we have 2 equations and 2 unknowns
Lets use the first equation to solve for x:
240x + 240W = 45
240x = 45 - 240W
x = (45/240) - W

Now use the second equation
240x - 240W = 48
and sustitute x for what we just found out x equals:
240( 45/240 - W ) - 240W = 48
45 - 240W - 240W = 48
45 -480W = 48
-480W = 3
W = (-1/160) km/min (the negative sign means its blowing int he order direction)
Now that we have a real number for W, lets find the real # for x:
x = (45/240) - W
x = (45/240) - (-1/160)
x = 0.19375 km/min
---
The second problem is the same
Let x be the rowing rate and C be the speed of the current
Then
3x + 3C = 30
3x - 3C = 90

Ok in the example above I solved the two equations by substituting one inside the other... you can also do it by adding equations together. So if we add the top equation to the bottom equation we get:

3x + 3C = 30
+ 3x - 3C = 90
--------------------------------
6x + 0C = 120
(notice how the 3C and -3C cancelled each other out)
so we're left with
6x = 120
x = 20

So now you can just put x in one of the equations to find C:

3x + 3C = 30
3( 20 ) + 3C = 30
60 + 3C = 30
3C = 30 - 60
3C = -30
C = -10

Answer 3

First plane is going 320 km/ h

the plane is going 300 km/h

You take the average to find the speed of the plane so that is 310 km/h and then that means that the wind must be going 10 km/h


2. Find the speed with the current, which is 6 km/h

Find speed against current, which is 2 km/ h

Once again we take the average, which is 4 km/h and that means that the current must be going 2 km/h.


Hope that this helps you to do your questions.

To find the speed of the boat you set up fractions.

For example...

240 / 45 = x / 60

45x = 60(240)

45x = 14400

x = 14400 / 45

x = 320