Algebra III
Lesson 25
Age Problems – Rate Problems
Age Problems
Example 25.1
Ten years ago Thomas was twice as old as Patricia. Five years from now,
Thomas will be 10 years older than Patricia. How old are both people now?
P = Patricia now
T = Thomas now
(T – 10) = 2(P – 10)
(T + 5) = (P + 5) + 10
T – 10 = 2P – 20
T + 5 = P + 15
T – 2P = -10
T – P = 10
T – P = 10
- (T – 2P = -10)
P = 20
T – 20 = 10
T = 30
Thomas is 30 & Patricia is 20.
Rate Problems
Two types
1st – Individuals working together
2nd – Groups working as one
1st - Individuals working together
Each person is working at their own rate.
Rate = Something/Time
R1T1 = stuff done by 1
($/hr, mi/hr, cm/yr, cars/min, etc)
R2T2 = stuff done by 2
R1T1 + R2T2 = stuff done by both
Example 25.2
Joe Frank can wash 3 cars in 5 hours, and Myrl can wash 4 cars in 3 hours.
Fifteen cars need to be washed. If Joe Frank works for 2 hours and is then
joined by Myrl, how long will Myrl and Joe Frank have to work together to finish
washing the fifteen cars?
Joe Frank
RJF
3 cars
=
5 hour
TJF = T + 2
Note: Always add the time
working alone.
RJFTJF + RMTM = stuff done by both
 3
 4
 (T + 2 ) +  (T ) = 15
 5
 3
(9)(T + 2) + (20)(T ) = 225
9T + 18 + 20T = 225
Myrl
RM =
4 cars
3 hour
TM = T
29T = 207
T = 207/29 hours
(times 15)
Example 25.3
Mayha could empty 2 bottles in 3 hours. Working with Wilbur, she found they
could empty 4 bottles in 3 hours. What was Wilbur’s rate of emptying bottles?
Mayha
Rm = 2/3
Wilbur
Rw = ?
For time use 3 hours,
since in that time
they both emptied 4
bottles.
RmTm + RwTw = bottles emptied
 2
 (3) + (Rw )(3) = 4
 3
2 + 3Rw = 4
3Rw = 2
Rw = 2/3 bottles/hr
2nd groups working together
Rate • Workers • Time = Jobs
RWT = J
In these problems, there will be information for two situations.
In one situation there will be enough information to find the
one part of the formula not given.
This is used with the rest of the information for the second
situation to find the answer.
Example 25.4
If 5 workers can do 1 job in 8 days, how many workers would it take to
do 3 jobs in 7 days?
RWT = J
1st situation
2nd situation
W = 5; T = 8; J = 1; R = ?
R = 1/40; T = 7; J = 3; W = ?
R•5•8=1
 1 
 (W )(7 ) = 3
 40 
R = 1/40
7W = 120
W = 120/7 = 17.14
Reality: 18 workers.
Example 25.5
Mary had 252 gallons (gal) of milk. This would feed 42 infants for 12 days.
Then 14 more infants were placed in the nursery and 48 more gallons were
provided. How long would the milk last now?
RWT = J
1st situation
2nd situation
W = 42; T = 12; J = 252; R = ?
R = ½; T = ?; J = 300; W = 56
R • 42 • 12 = 252
1
 (56 )(T ) = 300
 2
R = 252/504
28W = 300
R=½
W = 300/28
= 150/14 = 75/7 = 10 5/7 days
Practice
a) Five years ago, Orville was twice as old as Wilbur. Five years from now,
Orville will be 1 ½ times as old as Wilbur. How old are they both now?
R = Orville
W = Wilbur
(R – 5) = 2(W – 5)
(R + 5) =  3 (W + 5)
R – 5 = 2W - 10
2R + 10 = 3W + 15
R – 2W = -5
2R – 3W = 5
 2
(times -2)
-2R + 4W = 10
2R – 3W = 5
W = 15
R – 2(15) = -5
R = 25
times 2
b) Emerson can dig 2 holes in 3 hours and Johnna can dig 3 holes in 2 hours.
Six holes need to be dug. If Emerson digs of 1 hour and then is joined by
Johnna, how long will they work together to finish digging the 6 holes?
RETE + RJTJ = holes dug
RE = 2/3
RJ = 3/2
TE =T + 1
TJ = T
 3
 2
 (T + 1) +  (T ) = 6
 2
 3
4T + 4 +9T = 36
13T = 32
T = 32/13 hours
(times 6)
c) Sandra could build 3 sandcastles in 5 hours. Working with Frederick,
she discovered they could build 3 sandcastles in merely 3 hours. What was
Frederick’s rate of building sandcastles?
RSTS + RFTF = sandcastles built
RS = 3/5
RF = ?
3 sandcastles in 3 hours
 3
 (3) + (RF )(3) = 3
 5
(times 5)
9 + 15RF = 15
15RF = 6
RF = 6/15 = 2/5 sandcastles/hr
d) Six men can do 1 job in 9 days. How many men would it take to do 15
jobs in 5 days?
RWT = J
1st situation
2nd situation
W = 6; T = 9; J = 1; R = ?
R = 1/54; T = 5; J = 15; W = ?
R•6•9=1
 1
 (W )(5) = 15
 54 
R = 1/54
5W = 810
W = 162 men
e) Let f(x) = 2x and g(x) = 1 – x2. Evaluate:
1) (fg)(2)
2) (f/g)(2)
(fg)(2) = f(2) • g(2)
(f/g)(2) = f(2)/g(2)
= (2•2) •(1 – 22)
= (4)/(-3)
= (4) •(-3)
= - 4/3
= -12
3) (g ◦ f)(2)
(g ◦ f)(2) = g(f(2))
= g(4)
= 1 - 42
= -15