Distance = Rate x Time

[Pages:6]Math 233A

Chapter 2.4: More Application Problems

Objectives: ? Finding distance, rate, and time for going in the same direction ? Finding distance, rate, and time for going in towards each other ? Finding distance rate, and time for going in opposite directions ? Finding distance, rate, and time for going slow, then going faster ? Solve mixture problems

Chapter 2.4

Steps for Solving Application Problems: 1. Read, throw out nonsense numbers 2. Assign a variable (What is it asking for?) 3. Write an equation 4. Solve the equation 5. Answer the question! 6. Check, does it make sense?

Distance = Rate x Time

Finding Distance, Rate, and Time For Going in the Same Direction

Ex: Two people start walking at the same time in the same direction. One person walks 4 mph and the other person walks 6 mph. How long until they are 0.5 miles apart.

Let

Rate Slower Faster

Time Distance

Equation:

faster's dislance

slower's distance

total distance apart

Answer: 0.25 hours = time it takes to be 0.5 miles apart

Finding Distance, Rate, and Time For Going in the Same Direction and Catching Up

Ex: Say you and your friend are hanging out. Your friend takes off going 55 mph down the freeway. Five minutes later, you decide to catch up with her and take off going 60 mph down the same freeway. How long does it take you to catch up to her, and how long has she been driving?

Let Then

Your friend

Rate

You

Time

Distance

Equation:

friend's distance

your distance

Answer: 55 minutes = time it takes took you to catch up 1 hour = time she has been driving

Finding Distance, Rate, and Time For Going Towards Each Other

Ex: The distance between Gilroy and Watsonville is 24 miles. If you left Gilroy going 40 mph at the same time as your friend leaves Watsonville, going 45 mph, how long will it take to meet up?

Let

you friend

Rate

Time Distance

Equation:

your distance

friend's distance

total distance traveled

Answer: 0.28 hours (about 17 minutes)? time it takes for both to travel 24 miles

You try:

1. Two cars are traveling toward each other from two cities 270 miles apart. One car is going 5 mph slower than the other. They meet up after 2 hours. How fast are they each going?

Let

=

Then

=

Faster Slower

Rate

Time Distance

faster's distance slower's distance total distance traveled

Answer: 70 mph = faster car's rate 65 mph = slower car's rate

Finding Distance, Rate, and Time For Going In Opposite Directions

Ex: Two cars are going in opposite directions and one car is going 10 mph slower than the other. After 1.5 hours they are 150 miles apart. How fast are they each going?

Let_^ Then

Faster Slower

Rate

Time Distance

Equation:

+

faster's distance slower's distance

total distance traveled

Answer: 55 mph = faster car's rate 45 mph = slower car's rate

You try: 1. A freight train and a passenger train start from the same place and travel in opposite directions.

The passenger train travels 15 mph faster than the freight train. After 2 hours, they are 278 miles apart. How fast are they each going?

Let

=

Then

=

Freight Passenger

Rate

Time Distance

Equation:

freight's distance passenger's distance total distance traveled

Answer: 62 mph = freight train's rate 77 mph = passenger train's rate

Solving Mixture Problems

Ex: Three pounds of type A sunflower seed, which sells for $3.60 per pound, are mixed with 2 pounds of type B sunflower seeds, which sells for $4.80 per pound. What should the mixture cost per pound?

Let

Equation:

$3.60 + $4.80 = $x

3 pounds

2 pounds

5 pounds

+

--

3 pounds for S3.60 each 2 pounds for $4.80 each 5 pounds for Sx each

Answer: $4.08 per pound = cost of mixture

You try:

1. A nurse mixes 3.3 quarts of a 12% saline solution and 6.7 quarts of a 9% saline solution. What percent of saline solution will the mixture be? (round to the nearest percent)

Let

Equation:

+

=

quarts

quarts

quarts

V

quarts of

+

% solution

V

quarts of

=

% solution

V

quarts of % solution

Answers: 10% = percent of saline solution in the 10 quart mixture

Ex: How many ounces of 16% and 30% iodine solution should be mixed together to get 20 ounces of an iodine solution that is 25%?

Let_

Then

Equation:

16% + 30% = 25%

X ounces 20-.x ounces 20 ounces

+

X ounces of 16% solution 20-x ounces of 30% solution

20 ounces of 25% solution

Answer:

7.14 ounces = number of ounces of 16% iodine solution

12.86 ounces = number of ounces of 30% iodine solution

You try:

Grass seed A sells for $2.65 per pound and grass seed B sells for $2.80 per pound. How many

pounds of each type should be used to get a 12-pound mixture that sells for $2.70 per pound?

Let Then Equation:

$

$

+

$ =

pounds

pounds

pounds

pounds for

+

each

pounds for

each

pounds for

each

Answer: 8 pounds = the number of pounds of grass seed A, that sells for $2.65/lb 4 pounds = the number of pounds of grass seed B, that sells for $2.80/lb

Ex: How many liter of an 8% sulfuric acid solution must be mixed with 6 liters of a 15 % solution to get a solution that is 12% sulfuric acid?

Let Then Equation:

8% + 15% = 12%

X liters

6 liters

(A:+6) liters

X liters of 8% solution 6 liters of 15%solulion x+6 liters of 12% solution

Answer:

4.5 = number of liters of 8% sulfuric acid solution

10.5 = number of liters of 12% sulfuric acid solution

You try:

1. How many liter of a 12% sulfuric acid solution must be mixed with 2 liters of a 9% solution to

get a solution that is 11% sulfuric acid?

Let

=

Then

+

==

liters

liters

liters

+

^

V

'

>

V

'

>

V

'

liters of % solution

liters of % solution

liters of % solution

Answer: 4 liters = number of liters of 12% sulfuric acid solution 6 liters = number of liters of 11% sulfuric acid solution

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