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iVinArrow [24]
3 years ago
8

A cruise left port A and traveled towards port B 225 km away. After 1.5 hours of travel, the cruise was stopped for a half an ho

ur and then it had to increase its speed by 10 km/hour in order to reach port B on time. Find the original speed of the cruise.

Mathematics
1 answer:
Softa [21]3 years ago
7 0

Answer:

50

Step-by-step explanation:

v = the original speed of the cruise

d = 225 km the distance passed

t = d/v = 225/v the total time of the travel

v*1.5 + (v + 10)(d/v - 1.5 - 0.5) = d (using the formula distance = speed*time)

1.5v + (v + 10)(225/v - 2) = 225

Look at the images to see the solved equation

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You purchased concentrated cleaner that must be diluted (watered down)
abruzzese [7]

Answer:

The volume of the concentrated 80 gallons mixture is 20 gallons

The volume of water in the 80 gallons mixture is 60 gallons

Step-by-step explanation:

The given parameters are;

The content of Container A = The cleaner

The concentration of the cleaner = 20% solution

The content of Container B = Pure water

The concentration of the desired solution = 5%

The volume of the required solution = 80 gallons

Let x represent the volume of the concentrated solution and y represent the volume of water in the 80 gallons mixture

Therefore, we have;

20/100 × x + y×0 = 5/100×80

x + y = 80

y = 80 - x

20/100 × x + (80 - x)×0 = 5/100×80

0.2·x = 4

x = 4/0.2 = 20

x = 20 gallons

y = 80 - x = 80 - 20 = 60

y = 60 gallons

The volume of the concentrated solution (80 gallons mixture) = x = 20 gallons

The volume of water in the 80 gallons mixture = y = 60 gallons.

8 0
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