Algebra Age Problems When solving age problems, you need to represent the following in terms of a variable: - the
present ages of the people or things involved - the age, at the other specified time, of the people or things involved Then, form an equation based on these representations. Cary is 9 years older than an. !n " years, the sum of their ages will equal 9#. $ind both of their ages now. % & an's age now % ( 9 & Cary's age now )Cary is 9 yrs older than % ( " & an's age in " years % ( +, & Cary's age in " years % ( " ( % ( +, & 9# )in seven years the sum of their ages will be 9#* -% ( -# & 9# )combined li.e terms* -% & "/ )subtracted -# from both sides* % & #0 )divided both sides by #0* % ( 9 & 11 )substituted #0, in for %, into % ( 9* an is #0 Cary is 11 -------------------------------------------------------------------------------$red is 1 times as old as his niece, 2elma. Ten years from now, he will be twice as old as she will be. 3ow old is each now4 % & 2elma's age now 1% & $red's age now )$red is 1 times as old as 2elma* % ( +/ & 2elma's age in +/ years 1% ( +/ & $red's age in +/ years 1% ( +/ & -5% ( +/6 )his age in +/ yrs is twice her age in +/ years* 1% ( +/ & -% ( -/ )used distributive property* -% ( +/ & -/ )subtracted -% from both sides* -% & +/ )subtracted +/ from both sides* % & 0 )divided both sides by -* 1% & -/ )substituted 0, in for %, into 1%* 2elma is 0 now $red is -/ now -------------------------------------------------------------------------------7n eagle is 1 times as old as a falcon. Three years ago, the eagle was " times as old as the falcon. $ind the present age of each bird. % & falcon's age now 1% & eagle's age now )the eagle is 1 times as old as falcon* % - # & falcon's age # years ago 1% - # & eagle's age # years ago 1% 8 # & "5% 8 #6 )three years ago, eagle was " times the falcon* 1% 8 # & "% 8 -+ )used distributive property* 1% & "% -+9 )added # to both sides* -#% & -+9 )subtracted "% from both sides* % & , )divided both sides by -#* 1% & -1 )substituted ,, in for %, into 1%* falcon is , now eagle is -1 now -------------------------------------------------------------------------------:renda is 1 years older than Walter, and Carol is twice as old as :renda. Three years ago, the sum of their ages was #0. 3ow old is each now4 % & Walter's age now % ( 1 & :renda's age now ):renda is 1 yrs older than Walter* -5% ( 16 & -% ( 9 & Carol's age now )Carol is twice as old as :renda, used distributive property* % - # & Walter's age # years ago )subtracted # from %* % ( + & :renda's age # years ago )subtracted # from % ( 1* an*
-% ( 0 & Carol's age # years ago )subtracted # from -% ( 9* 5% - #6 ( 5% ( +6 ( 5-% ( 06 & #0 )sum of ages, # years ago, was #0* % - # ( % ( + ( -% ( 0 & #0 )too. out parentheses* 1% ( # & #0 )combined li.e terms* 1% & #- )subtracted # from both sides* % & 9 & Walter now )divided both sides by 1* % ( 1 & +- & :renda now )substituted 9, in for %, into % ( 1* -5% ( 16 & -1 & Carol now )substituted 9, in for %, into -5% ( 16* Walter is 9 now :renda is +- now Carol is -1 now NUMBER AND DIGIT PROBLEM
WORK PROBLEMS
;ne pipe can fill a pool +.-0 times faster than a second pipe. When both pipes are opened, they fill the pool in five hours. 3ow long would it ta.e to fill the pool if only the slower pipe is used4
Convert to rates: hours to complete <ob: fast pipe: f slow pipe: 1.25f together: 5 completed per hour: fast pipe: 1/f slow pipe: 1/1.25f together: 1/5 adding their labor:
1/ f
+ 1/1.25f = 1/5
multiplying through by 5f: 5 + 5/1.25 = f 5+4=f=9 Then 1.25f = 11.25, so the slower p pe ta!es 11.25 ho"rs.
Mot o# $D sta#%e& 'orm"la
Motion problems are based on the formula d = rt where d = distance, r = rate and t = time. When sol in! motion problems, a s"etch is often helpful and a table can be used for or!ani#in! the information. $ohn and %hilip who li e 14 miles apart start at noon to wal" toward each other at rates of & mph and 4 mph respecti el'. (n how man' hours will the' meet) Solution: *et x = time wal"ed.
r $ohn %hilip
&x + 4x = 14 +x = 14 x=2
t x x
d &x 4x
& 4
,he' will meet in 2 hours. Example: (n still water, %eter-s boat !oes 4 times as fast as the current in the ri er. .e ta"es a 15/mile trip up the ri er and returns in 4 hours. 0ind the rate of the current. Solution:
*et x = rate of the current.
r down ri er up ri er 4x + x 4x / x t 15 / 5x 15 / &x d 15 15
