rate problems. introduction several types of problems fall into the category known as “rate...
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RATE PROBLEMS
INTRODUCTION
Several types of problems fall into the category known as “rate problems”:– Distance– Work– Percent problems– Mixture problems
Distance Problems
Rate x Time = Distance
Within the category of distance problems, we encounter 3 types of questions:– Opposite direction– Same direction– Round-trip
Opposite Direction
Key Words: opposite direction, towards each other, away from each other
Example: Two trains start from the same point and travel in opposite directions. The northbound train averages 45 mph and starts 2 hours before the southbound train, which averages 50 mph. How long after the northbound train starts will the trains be 470 miles apart?
Opposite Direction (Cont.)
What do we know?
Northbound– 45 mph
Southbound– 50 mph
Northbound train starts 2 hours earlier.
What do we want to find out?
“How long…” indicates time.
Organize the information:
RATE TIME DISTANCE
N 45 X 45x
S 50 X-2 50(x – 2)
470)2(5045 xx
Opposite Direction Practice
Two boats leave at 7:00 AM from ports that are 252 miles apart and cruise toward each other at speeds of 30 mph and 26 mph. At what time will they pass each other?Answer: 11:30
Same Direction
Key words: same direction, pass, overtake, catch up to
Example: In a cross country race, Adam left the starting line at 7:00 AM and drove an average speed of 75 km/h. At 8:30, Manuel left the starting point and drove the same route, averaging 90 km/h. At what time did Manuel overtake Adam?
Same Direction (Cont.)
What do we know?
Adam– 75 km/h
Manuel– 90 km/h
Manuel left 1.5 hours later.
What do we want to find out?
“What time…” indicates time.
Organize the information: RATE TIME DISTANCE
A 75 x+1.5 75(x+1.5)
M 90 X 90x
xx 90)5.1(75
Same Direction Practice
In a cycling race, Mary left the starting line at 6:00 AM averaging 25 mph. At 6:30 AM, Peter left the starting point and traveled the same route at 30 mph. At what time did Peter pass Mary?Answer: 9:00
Round Trip
Key words: round trip, to and from, there and back
Example: A ski lift carried Maria up a slope at the rate of 6 km/h and she skied back down parallel to the lift at 34 km/h. The round trip took 30 min. How far did she ski?
Round Trip (Cont.)
What do we know?
Uphill– 6 km/h
Downhill (ski) – 34 km/h
Round trip took 30 min.
What do we want to find out?
“How far…” indicates distance.
“How long…” indicates time
Organize the information:
RATE TIME DISTANCE
U 6 ½ - x 6(1/2 – x)
D 34 x 34x
)2
1(634 xx
Practice
1) At noon, a plane leaves Hawaii for California at 555 km/h. At the same time, a plane leaves California for Hawaii at 400 km/h. The distance between the two cities is about 3800 km. At what time do the planes meet?
SOLUTION:
3.97t
3800955t
3800400t555t
planesboth for time
t
Practice
2) Maria and Tom start at the park. They drive in opposite directions for 3 hours. The are then 510 km apart. Maria’s speed is 80 km/h. What is Tom’s speed?
SOLUTION:
km/h 90r
2703r
5102403r
speed sTom'
r
Practice
3) Kara floats from A to B at 6 km/h. She returns by motorboat at 18 km/h. The float trip is 4 hours longer than the motorboat trip. How far is it from A to B?
SOLUTION:
hours 2
2412
24618
)4(618
float tripfor time 4t
motorboatby time
t
t
tt
tt
t
The question asks us to find distance. Since distance = rate x time,
D = 18(2) = 36 km
Head Wind- Tail Wind
Some distance problems add the variable of wind speed or water current.
Tail Wind– Wind blowing in the same direction as the vehicle (increases speed)
Head Wind– Wind blowing in the opposite direction (decreases speed)
Upstream– Craft is traveling against the current (decreases speed)
Downstream– Craft is traveling with the current (increases speed)
Example # 1
A boat traveled downstream 160 km in 8 hours. It took the boat 10 hours to travel the same distance upstream. Find the speed of the boat and the current.
Let x = speed of the boat
y = speed of the current
Rate Time Distance
Up x – y 10 10(x-y)
Down x + y 8 8(x+ y)
160)(8
160)(10
yx
yx
2
18
362
20
16
y
x
x
yx
yx
Example # 2
When flying with a tailwind, a plane traveled 1200 miles in 3 hours. Flying with a headwind, the plane traveled the same distance in 4 hours. Find the plane’s speed and the speed of the wind.
Let x = speed of the plane
y = speed of the wind
Rate Time Distance
Headwind x – y 4 4(x-y)Tailwind x + y 3 3(x+y)
1200)(3
1200)(4
yx
yx
50
350
7002
400
300
y
x
x
yx
yx
Work Rate
Work Rate problems are directly related to distance problems:– Rate x Time = Work Done
When setting up the table, – Rate = amount of work
completed per unit of time– Time = amount of time working
together
Example #1
An office has two envelope stuffing machines. Machine A can stuff a batch of envelopes in 5 hours, while Machine B can stuff a batch of envelopes in 3 hours. How long would it take the two machines working together to stuff a batch of envelopes?
Example # 1 Cont.
What do we know?
Machine A takes 5 hours.
Machine B takes 3 hours.
Rate
(Per hr.)
Time (together)
Work Done
A t
B t
5
1
3
15
t
3
t
135
tt
hrs. 875.1
158
1553
)1(153
155
15
15
t
t
tt
tt
LCD
Example #2
Mary can clean an office complex in 5 hours. Working together John and Mary can clean the office complex in 3.5 hours. How long would it take John to clean the office complex by himself?
Example # 2 Cont.
What do we know?
Mary can clean the office in 5 hours.
Working together, it takes 3.5 hours.
Rate
(Per hr.)
Time (together)
Work Done
Mary 3.5
John 3.5
5
1
t
1
7.
t
5.3
15.3
7.0 t
hours 67.11
3.5.3
5.37.
)1(5.3
)(7.
t
t
tt
tt
tt
tLCD
Mixture Problems
Mixture problems fall into 2 different categories:– Concentration/ percent– Money/ Cost
There are 2 different strategies for solving mixture problems:– Table– Diagram
Example #1
CONCENTRATION
A solution is 10% acid. How much pure acid must be added to 8 L of solution to get a solution that is 20% acid?
Example # 1 (Cont.)Strategy: TABLE
Amount of solution
% of acid
Amount of acid
Original solution
8 .10 .8
Acid added
X 1 x
New solution
8+x .20 .20(8+x)
What is the resultingConcentration?
Use the last column in the table to write an equation.
)8(20.8.0 xx
1
8.8.
6.18.8.
2.6.18.
x
x
x
xx
You need to add 1 L of acid.
Example # 1 (Cont.)
Strategy: DIAGRAM
8L + x = 8+x
Start Add Result
.10 1 .20
To write the equation,Multiply the percent on
The outside with theAmount on the inside!
)8(20.8.0 xx
1
8.8.
6.18.8.
2.6.18.
x
x
x
xx
Example #2
Cost/ Money
A mixture of raisins and peanuts sells for $4 a kilogram. The raisins sell for $3.60 a kilogram. The peanuts sell for $4.20 a kilogram. How many kilograms of each are used in 12 kilograms of the mixture?
Example # 2 (Cont.)Strategy: TABLE
Amount of ingredient
Price Cost
Raisins R 3.60 3.60R
Peanuts 12 – R 4.20 4.20(12-R)
Mixture 12 4 48
Use the last column in the table to write an equation.
48)12(20.460.3 RR
4
4.26.0
486.04.50
482.44.506.3
R
R
R
RR
You need 4 kg of raisins, and 8 kg of peanuts.
Example # 2 (Cont.)
Strategy: DIAGRAM
R + 12-R = 12
Raisins Peanuts Mixture
3.60 4.20 4
To write the equation,Multiply the cost onThe outside with the
Amount on the inside!
48)12(20.460.3 RR
4
4.26.0
486.04.50
482.44.506.3
R
R
R
RR
Example #3
CONCENTRATION
You have 40g of a 50% solution of acid in water. How much water must you add to make a 10% acid solution?
Example # 3 (Cont.)
Strategy: DIAGRAM
40g + x = 40+x
Start Add Result
.50 0 .10
To write the equation,Multiply the percent on
The outside with theAmount on the inside!
)40(1.)(0)40(5. xx
x
x
x
160
1.16
1.420We need 160g of
water!
Example 4-- Investment
Mervin Invested an amount of money at 5% interest and another amount at 8.25% interest. The amount invested at 5% was $1000 more than the amount at 8.25%. If his total income from simple interest for 1 year was $315, how much did he invest at each rate?
Example 4 Cont.
Amount Invested
% Invested
Income
5% X .05 .05x
8.25% X-1000 .0825 .0825(x-1000)
315)1000(0825.05. xx
STRATEGY: TABLE
3000
5.3971325.
3155.821325.
3155.820825.05.
x
x
x
xx
$3000 invested at 5%.
$2000 invested at 8.25%