Age, Work, Mixture, Digit, Motion and Coin Problems
Submitted by: Ong, Mark M ark Lorenz P. Submitted to: Eng’r Rex H. Agustin
!. Mary is "! years o#d. Mary is t$i%e as o#d as Ann $as $&en Mary $as as o#d as Ann is no$. Ho$ o#d is Ann no$'
A. B. ). *.
( 18 " +
ast Mary x Anny "
Present "! x
24 – x 2 x
= x – 12
=36
X =18
!". -&e sum o /im’s and /e0in’s ages is 1. 2n 3 years, /im $i## be t$i%e as o#d as /e0in. 4&at are t&eir ages no$' A. B. ). *.
!, ! 5, 5 , 13 5, ( , "
Present uture /e0in x x63 /im y y63 7 6 y 8 1 9 8 1 x ;< e=uation > y 6 3 ? 8 " >x 6 3 ? ;< e=uation " Substitute y in e=uation ": ( 18 – x )+ 3 =2 x + 6 21 –
x =2 x + 6
X =5
Y =18 – 5 Y =13
!3. Robert is + ears o#der t&an &is brot&er Stan. Ho$e0er @y years ago, Robert $as t$i%e as o#d as Stan. 2 Stan is no$ @b years o#d and bb y?
A. C. ). *.
15 ( 5 1
Past
Present Robert b 6 + y b 6 + Stan by b ( b + 15 – y )= 2 (b – y )
b + 15 – y =2 b – 2 y 2b – b –2 y
+ y = 15
b – y =15
!!. DD is t&ree times as o#d as Dan;Dan. -&ree years ago, DD $as our times as o#d as Dan;Dan. -&e sum o t&eir ages is A. C. ). D.
" "! "1 36
Past Present DD 3x 3 3x Dan;Dan x3 x 3x 3 8 ! > x 3 ? 3x 3 8 !x " x " 8 3 x8F 3x 8 3 > F ? 3x 8 "5 Sum o t&e ages 8 F 6 "5 8 3( !+. A gir#s is one;t&ird as o#d as &er brot&er and 1 years younger t&an &er sister. -&e sum o t&eir ages is 31 years. Ho$ o#d is t&e gir#' A. C. C. *.
! + 6 5
Let: x 8 age o t&e gir# y 8 age o &er brot&er z 8 age o &er sister y 8 3x ;< E=. z8761 ;< E=. " x 6 y 6 z 8 31 ;< E=. 3 Substitute e=uations G " in e=uation 3: x 6 3x 6 >x 6 1? 8 31 x8(
!(. Pau#a is no$ 1 years o#d and &is %o##eague Moni%a is ! years o#d. Ho$ many years ago $as Pau#a t$i%e as o#d as Moni%a' A. C. ). D.
+ 5 1 1
Past Present Pau#a 1 x 1 Moni%a ! x ! ( 18 – x )= 2 (14 – x ) 18 –
x =28 – 2 x
x =10
!5. A at&er te##s &is son, @2 $as your age no$ $&en you $ere born. 2 t&e at&er is no$ 31 years o#d, &o$ o#d $as &is son " years ago' A. B. ). *.
+ 1! F "
Past Present at&er x 31 Son x -$o years ago, t&e son $as ( 19 – 2 )=17 years o#d
!1. Six years ago, i#da $as B0e times as o#d as Riza. 2n B0e years, i#da $i## be t&ree times as o#d as Riza. 4&at is t&e Iresent age o Riza'
A. C. ). *. Past i#da Riza
1! ( + ! Present uture +>x (? ' 3>x 6 +? x( x x6+ 3 ( x + 5 ) – 5 ( x + 6 )= x + 5 – ( x + 6 )
+
3 x 15 – 5 x
+ 30 = x + 5 – x + 6
−2 x + 45 =11 x =17
!F. At Iresent, t&e sum o t&e Iarents’ ages is t$i%e t&e sum o t&e %&i#dren’s ages. i0e years ago, t&e sum o t&e Iarent’s ages $as ! times t&e sum o t&e %&i#dren’s ages. iteen years &en%e, t&e sum o t&e Iarents’ ages $i## be e=ua# to t&e sum o t&e %&i#dren’s ages. Ho$ many %&i#dren are t&ere' A. C. C. *.
3 ! 5 ( Past
Present Parents "x )&i#dren x +n x 2 x – 10 =4 ( x – 5 n ) 2 x – 10
2 x
"x x 6 +n
"x 6 3
=4 x – 20 n
=20 n – 10
x =10 n – 5 2 x
uture
;< E=.
+30 = x + 15 n
x =15 n – 30
;< E=. "
E=uate e=uations and ": 15 n – 30= 10 n – 5
N =5
+. *ebbie is no$ t$i%e as o#d as Derry. our years ago, *ebbie $as t&ree times as o#d as Derry t&en. Ho$ o#d is *ebbie' A. B. ). *.
! 16 1 "!
Past *ebbie Derry
Present 3>x !? x! x 2 x – 3 ( x – 4 )= x – ( x – 4 )
"x
2 x – 3 x
+ 12= x – x + 4
− x + 12= 4 x =8 2 x
=16
+. A IumI %an IumI out $ater orm a tank in &ours. Anot&er IumI %an IumI out $ater rom t&e same tank in " &ours. Ho$ #ong $i## it take bot& IumIs to IumI out t&e $ater in t&e tank'
A. C. ). *.
! "ours ( &ours 5 J &ours ( J &ours
Let: x 8 time neede to %omI#ete t&e $ork 1 20
+
1 11
=
1
x
X =7.096 hours
+". A !;mm IiIe %an B## t&e tank a#one in + &ours and anot&er (;mm IiIe %an B## t&e tank a#one in ! &ours. A drain IiIe 3;mm %an emIty t&e tank in " &ours. 4it& a## t&e t&ree IiIes oIen, &o$ #ong $i## it take to B## t&e tank' A. B. ). *.
". &ours #.5 "ours "."+ &ours ".5+ &ours 1 5
1
1
4
20
+ −
=
1
x
X =2.5 hours
+3. A tank is B##ed $it& an intake IiIe in " &ours and emItied by an out#et IiIe in ! &ours. 2 bot& IiIes are oIened, &o$ #ong $i## it take to B## t&e emIty tank' A. B. ). *.
3 &ours $ "ours + &ours ( &ours
1
1
1
4
x
− =
2
x =4 hours
+!. A tank %an be B##ed in F &ours by one IiIe, " &ours by a se%ond IiIe and %an be drained $&en u## by a t&ird IiIe in + &ours. Ho$ #ong $i## it take to B## an emIty tank $it& a## IiIes in oIeration' A. C. ). D.
5 5 5 ! 1 9
&ours and " minutes &ours and 3" minutes &ours and !" minutes "ours and 5 minutes
+
1 12
–
1
1
15
= x x
x 8 5.1"( &ours K 5 &ours and + minutes ++. 2 A %an do t&e $ork in @x days and C in @y days, &o$ #ong $i## t&ey Bnis& t&e ob $orking toget&er' A. C. C. *.
>769?K>xy? >769?K" %x&'(%)*+'
√xy
Let: n 8 number o days neede to %omI#ete t&e $ork 1
1
1
x y
n
+ =
x + y n xy 1
=
xy n= x + y +(. Pedro %an Iaint a en%e + aster t&an Duan and "+ aster t&an Pi#ar, and toget&er t&ey %an Iaint a gi0en en%e in ! &ours. Ho$ #ong $i## it take Pedro to Iaint t&e same en%e i &e &ad to $ork a#one' A. C. C. *.
( 1 1 "
Let: A 8 number o &ours, Pedro %an Iaint t&e &ouse C 8 number o &ours, Duan %an Iaint t&e &ouse ) 8 number o &ours, Pi#ar %an Iaint t&e &ouse 1
1
1
B
C
+ + = ¼
A 1
;< E=.
1
1
1
A
B
B
A
1
1
1
1
C
C
A
=1.5 ( ); = 0.666 ( )
=1.2 ( ) ; =0.833 ( )
A
;
;
Substitute e=uations " G 3 in e=uation : 1
1
1
A
A
+0.666 ( )+ 0.833 ( )=¼
A
A =10 hours
+5. N#enn %an Iaint a &ouse in F &ours $&i#e Ste$art %an Iaint t&e same &ouse in ( &ours. -&ey $ork toget&er or ! &ours. Ater ! &ours, Ste$art #et and N#enn Bnis&ed t&e ob a#one. Ho$ many more days did it take N#enn to Bnis& t&e ob'
A. C. ). *. ote:
#.!5 ".+ "."+ 3.
(rate )( time)= 1 (complete job ) 1
1
9
16
( +
0.6944
1
) 4 + ( x )= 1 9
+ 0.111 x =1
x =2.75 hours
+1. 2t takes Cut%& t$i%e as #ong as it takes *an to do a %ertain Iie%e o $ork. 4orking toget&er t&ey %an do t&e $ork in ( days. Ho$ #ong it take *an to do it a#one'
A. C. ). *.
"
Let: x 8 time or Cut%& to Bnis& a %ertain ob $orking a#one y 8 time or *an to Bnis& a %ertain ob $orking a#one
1
1
1
x y
6
+ =
;
x =2 y
;
Substitute e=uation " in e=uation : 1 2 y 1
1
1
y
6
+ =
+2
2 y
=
1 6
y =9 days +F. A and C $orking toget&er %an Bnigs& Iainting a &ouse in ( days. A $orking a#one %an Bnis& it in + days #ess t&an C. &o$ #ong $i## it take ea%& o t&em to Bnis& t&e $ork a#one' A. B. ). *. 1
1, 3 1, 15 (, 5, "
1
1
B
6
+ =
A
;
A = B – 5
;
Substitute e=uation " in e=uation : 1
B −5
1
1
B
6
+ =
B +( B + 5) 1 = B ( B −5 ) 6
−5 1 = B ² −5 B 6 2B
12 B – 30
=B ² −5 B
B ² −17 B + 30= 0
( B – 15 )( B – 2)= 0 B =15 B =2 ( absurd )
Substitute B = 15 in equation 2
A =15 – 5=10 days
(. A and C %an do a Iie%e o $ork in !" days, C and ) in 3 days and ) and A in " days. 2n &o$ many days %an a## o t&em do t&e $ork toget&er'
A. C. ). *.
1 5 " +
Let: x 8 number o days neede by A, C and ) to Bnis&ed t&e $ork $orking toget&er. 1 / A + 1 / B + 1 / C = 1 / X 1
/ A + 1 / B =1 / 42
;
1
/ B + 1 /C =1 / 31
;
1
/ C + 1 / A =1 / 20
;
Add t&e t&ree e=uations: ( 1 / A + 1 / B )+(1 / B + 1 /C )+(1 / C + 1 / A )=1 / 42 + 1 / 31+ 1 /20 2
/ A + 2 / B + 2 / C =0.106
1
/ A + 1 / B + 1 / C = 0.053=1 / X
x =18.37 days ∨19 days
(. 2t takes My#ine t$i%e as #ong as Deana to do a %ertain Iie%e o $ork. 4orking toget&er, t&ey %an Bnis& t&e $ork in ( &ours. Ho$ #ong $ou#d it take Deana to do it a#one' A. B. ). *.
F &ours 18 "ours " &ours ! &ours
Let: x 8 time or My#ine to Bnis& t&e ob y 8 time or Deana to Bnis& t&e ob 1 / x + 1 / y =1 / 6 ;Ky? Substitute e=uation " in e=uation : 2 / y + 1 / y =1 / 6 3
/ y =1 /6
;
y =18 hours
(". Mike, Loui and Doy %an mo$ t&e #a$n in !, ( and 5 &ours resIe%ti0e#y. 4&at ra%tion o t&e yard %an t&ey mo$ in &our i t&ey $ork toget&er'
A. C. ). *.
$!(8$ !+K1! 1!K!5 3FK(
Let: x 8 ra%tion on t&e #a$n t&at %an mo$ed ater one &our x =(1 / 4 + 1 / 6 + 1 / 7 )( 1 )
x =( 42 + 28 + 24 )/ 168 x =94 / 168∨ 47 / 84
(3. A armer %an I#o$ t&e Be#d in 1 days. Ater $orking or 3 days, &is son oins &im and toget&er t&ey I#o$ t&e Be#d in 3 more days. Ho$ many days $i## it re=uire or t&e son to I#o$ t&e Be#d a#one' A. C. C. *.
1# 3
Let: x 8 time or t&e armer to o$ t&e Be#d y 8 time or t&e son to o$ t&e Be#d ( 3 )( 1 / x )+(1 / x + 1 / y )( 3)= 1
;
Substitute x 8 1 in e=uation ( 1 / 8 (3 )+( 1 / 8 +1 / y )( 3 )=1 ) 1 /3
/ + / + / y =1 /3
1 8 1 8 1
y =12 days
(!. )re$ o. %an Bnis& insta##ation o an antenna to$er in " man;&our $&i#e )re$ o. " %an Bnis& t&e same ob in 3 man;&our. Ho$ #ong $i## it take bot& %re$s to Bnis& t&e same ob, $orking toget&er' A. B. ). *.
man;&our 1# man-"our ! man;&our ( man;&our
Let: x 8 number o man;&ours neede by %re$ number and number " to Bnis& t&e ob 1 / 200 + 1 / 300=1 / x
x =120 man −hours
(+. On one ob, t$o Io$er s&o0e#s ex%a0ate ", %ubi% meters o eart&, t&e #arger s&o0e# $orking ! &ours and t&e sma##er or 3+ &ours. On anot&er ob, t&ey remo0ed !, %ubi% meters $it& t&e #arger s&o0e# $orking 5 &ours and t&e sma##er $orking F &ours, &o$ mu%& eart& %an ea%& remo0e in &our $orking a#one' A. C. C. *. Let:
(F.", "15.3 51.3, "F!. 1!3., 3$!.8 "., 3".! x 8 %aIa%ity o t&e #arger s&o0e# in m³/hr
y = capacity of the smaller shovel in m³/hr 40 x
+ 35 y =20000
x =500 – 0.875 y 70 x
-!q"1
+ 90 y =40000
-!q"2
Substitute equation 1 in equation 2#
(
70 500 – 0.875 y
35000 – 61.25 y
)+ 90 y = 40000
+ 90 y = 40000
y =173.9 m ³ / hr
x =500 – 0.875 ( 173.9 ) x =347.8 m ³. hr
((. -en #iters o "+ sa#t so#ution and + #iters o 3+ sa#t so#ution are Ioured into a drum origina##y %ontaining 3 #iters o sa#t so#ution. 4&at is t&e Ier%ent %on%entration o sa#t in t&e mixture'
A. C. ). *. "+
1.55 "".+ "5.+ "+.5" 6
3+ +
6
3
8
x ++
( )+ 0.35 (15 )+ 0.10 (30 )= x (55 )
0.25 10
x =19.55
(5. A )&emist o a disti##ery exIerimented on t$o a#%o&o# so#utions o dierent strengt&, 3+ a#%o&o# and + a#%o&o#, resIe%ti0e#y. Ho$ many %ubi% meters o ea%& strengt& must &e use in order to Irodu%e a mixture o ( %ubi% meters t&at %ontain ! a#%o&o#' A. C. ). D.
" m³ of solution $ith %5& alcohol' () m³ of solution $ith 5)& alcohol
5) m³ of solutoin $ith %5& alcohol' 2) m³ of solution $ith 5)& alcohol 2) m³ of solution $ith %5& alcohol' 5) m³ of solution $ith 5)& alcohol 40 m³ of solution with 35% alcohol, 20 m³ of solution with 50% alcohol
3+ 6 + 8 ! x (;x ( or t&e 3+ so#ution: 0.35 ( x )+ 0.50 ( 60 − x )= 0.40 ( 60 ) 0.35 x
+ 30 −0.50 x =24
x =40 m ³ or + so#ution: 60 – x =20 −¿ m ³ (1. A go#dsmit& &as t$o a##oys o go#d, t&e Brst being 5 Iure and t&e se%ond being ( Iure. Ho$ many oun%es o t&e 3 Iure go#d must be used to make oun%es o an a##oy $&i%& $i## be (( go#d'
A. C. ). *.
$ 3+ !+ 31
5 6 ;x
3( 0.70 ( 100 − x )+ 0.60 ( x )=0.66 ( 100 ) 70
( x
8
− 0.7 x + 0.6 x = 66
x =40 ounces
(F. -$o t&ousand >"? kg o stee# %ontaining 1 ni%ke# is to be made by mixing a stee# %ontaining ! ni%ke# $it& anot&er %ontaining ( ni%ke#. Ho$ mu%& o ea%& is neede'
A. C. C. *.
+ kg o stee# $it& ! ni%ke#, + kg o stee# $it& ( ni%ke# 5+ kg o stee# $it& ! ni%ke#, "+ kg o stee# $it& ( ni%ke# 5 kg o/ steel 0it" 1$ nikel, 15 kg o/ steel 0it" 6 nikel "+ kg o stee# $it& ! ni%ke#, 5+ kg o stee# $it& ( ni%ke#
! 6 ( 8 1 7 ";x " or ! substan%es 0.14 ( x )+ 0.06 ( 2000 − x )= 0.08 ( 2000 ) 0.14 x
+ 120 – 0.06 x =160
x =500 kg or ( substan%e: 2000 – x =1500 kg
5. Ho$ mu%& $ater must be e0aIorated rom kg so#ution $i%& &as ! sa#t to make a so#ution o sa#t' A. C. C. *. !
! kg + kg 6 kg 5 kg ;
8 x ;x 0.04 ( 10) – 0 ( x )= 0.10 ( 10 – x ) 0.4
=1 – 0.1 x
x =6 kg
5. 2 a t$o digit number &as x or its unit’s digit and y or its ten’s digit, reIresent t&e number. A. B. ). *.
x 6 y 1&* x yx xy
Let: x 8 ten’s digit o t&e number y 8 units’ digit o t&e number -&e to$ digit number is reIresented by: 10 y + x
5". One number is + #ess t&an t&e ot&er. 2 t&eir sum is 3+, $&at are t&e numbers' A. C. C. *. Let:
1+, + 1, ++ !, 65 5+, ( x 8 Brst number x + 8 se%ond number x +( x – 5 )=135 2 x
=140
X =70 X – 5 =65
53. -en #ess t&an our times a %ertain number is !. *etermine t&e number.
A. C. ). *.
6 5 1 F
Let: x 8 t&e number 4 x – 1 c =14
x =6
5!. -&e sum o t$o number is " and one number is t$i%e t&e ot&er. ind t&e numbers. A. B. ). *. Let:
(, + !, 1$ 1, 3 F, " x 8 t&e Brst number "x 8 t&e se%ond number x + 2 x =21
x =7 2 x
=14
5+. 2 eig&t is added to t&e Irodu%t o nine and t&e numeri%a# number, t&e sum is se0enty;one. ind t&e unkno$n number. A. C. C. *.
+ ( ! 1
Let: x 8 t&e number 9 x + 8=71
x =7
5(. ind t&e ra%tion su%& t&at i " is subtra%ted rom its terms its be%omes Q, but i ! is added to its terms it be%omes J. A. C. C. *.
3K+ +K" 5(1$ (K3
Let xKy 8 t&e ra%tion ( x – 2)/ ( y – 2)= ¼ 4 x – 8
= y – 2
y = 4 x – 6−¿ Eq .1
( x + 4 )/ ( y + 4 )= ½ 2 x
+8 = y + 4 −¿ Eq .2
Substitute e=uation in e=uation ": 2 x + 8 =( 4 x – 6 )+ 4 10
=2 x
x =5
y = 4 ( 5 ) – 6 y =14
55. -&e Irodu%t o Q and K+ o a number is +. 4&at is t&e number' A. + C. 5+ C. 1
*. "+ Let: x 8 t&e number ( 1 / 4 ( x ))(1 / 5 ( x ))=500
x ² / 20=500
x ² =10000 x =100
51. 2 3 are subtra%ted rom t&e numerator o a %ertain ra%tion, t&e 0a#ue o t&e ra%tion be%omes 3K+. 2 is subtra%ted rom t&e denominator o t&e same ra%tion it be%omes "K3. ind t&e origina# ra%tion. A. B. ). *.
3+K++ 36(55 3K5 3"K!
Let xKy 8 t&e ra%tion ( x −3 )/ y =3 / 5 3 y
=5 x – 15
Y =5 / 3 x – 5−¿ Eq .1 x /( y – 1 )=2 / 3 3 x
=2 y – 2−¿ Eq .2
Substitute e=uation in e=uation " 3 x =2 ( 5 / 3 x −5 )−2 3 x
=10 / 3 x – 10 – 2
0.333 x
=12
X =36 Y =5 / 3 ( 36 ) – 5=55
5F. -&e denominator o a %ertain ra%tion is t&ree more t&an t$i%e t&e numerator. 2 5 is added to bot& terms o t&e ra%tion, t&e resu#ting ra%tion is 3K+. ind t&e origina# ra%tion. A. 1K+ C. 3K+
C. 5(13 *. 3K+ Let:
x 8 numerator o t&e ra%tion y 8 denominator o t&e ra%tion y =2 x + 3−¿ Eq .1
( x + 7 )/( y + 7 )=3 / 5 +
5 x 35
=3 y + 21−¿ Eq .2
Substitue e=uation in e=uation ": 5 x + 35 =3 ( 2 x + 3)+ 21
+ =6 x + 9 + 21
5 x 35
x =5
y =2 ( 5 )+ 3 y =13
1. ind t&e Irodu%t o t$o numbers su%& t&at t$i%e t&e Brst added to t&e se%ond e=ua#s F and t&ree times t&e Brst is " more t&an t&e se%ond.
A. C. ). *. Let:
#$ 3" 1 " x 8 t&e Brst number y 8 t&e se%ond number 2 x + y =19
y =19 – 2 x −¿ Eq .1 3 x
= y +21−¿ Eq .2
Substitute e=uation in e=uation ": 3 x =( 19 – 2 x )+ 21 5 x
=40
X =8
Y =19 – 2 ( 8 ) Y =3 -&e Irodu%t o t&e numbers is 1>3? 8 "!
1. -&e tens’ digit o a number is 3 #ess t&an t&e units’ digit. 2 t&e number is di0ided by t&e sum o t&e digits, t&e =uotient is ! and t&e remainder is 3. 4&at is t&e origina# number' A. B. ). *.
3( $! +1 (F
Let:
t 8 ten’s digit o t&e number u 8 unit’s digit o t&e number umber 8 t 6 u t =u – 3 −¿ Eq .1
( 10 t + u )/( t + u )= 4 + 3 /( t + u)
( 10 t + u – 3)/( t + u )= 4 + u – 3 = 4 t + 4 u
10 t
6 t – 3
=3 u−¿ Eq .2
Substitute e=uation in e=uation ": 6 ( u – 3) – 3 = 3 u 6 u – 18
−3=3 u
=21
3u
u= 7 t =7 – 3
t =4 Number=10 ( 4 )+ 7
Number= 47
1". -&e Se%ond o t&e our numbers is t&ree #ess t&an t&e Brst, t&e t&ird is our more t&an t&e Brst and ourt& is t$o more t&an t&e t&ird. ind t&e ourt& number i t&eir sum is 3+. A. C. ). D. Let:
" 13 $ 8 Brst number x 8 se%ond number
y 8 t&ird number z 8 ourt& number x = – 3−¿ Eq .1 a
y = + 4 −¿ Eq .2
! = y + 2=( + 4 )+ 2 ! = + 6−¿ Eq .3
+ x + y + ! =35 −¿ Eq .4 Substitue e=uation , ", and 3 in e=uation !: +( – 3 )+( + 4 )+( + 6 )=35 4
=28
=7 Substitute $ 8 5 ni e=uation 3 ! =7 + 6
! =13
13. A ogger starts a %ourse at a steady rate o 1 kI&. i0e minutes #ater, a se%ond ogger starts t&e same %ourse at kI&. Ho$ #ong $i## it take t&e se%ond ogger to %at%& t&e Brst'
A. C. ). *. Let:
# min " min "" min 1 min 8 0e#o%ity o t&e Brst ogger " 8 0e#o%ity o t&e se%ond ogger
" 1= " 2=
8 km
hr 10 60
x
1 hr 60 min
km / min
s 1=s 2 " 1 t 1 = " 2 t 2
=
8 60
km/ min
/
= 10 / 60 (t 1 −5)
8 60 t 1
=10 t 1 – 50
8 t 1
t 1=25 min t 2= 25 – 5
t 2= 20 min 1!. A boat man ro$s to a I#a%e !.1 mi#es $it& t&e stream and ba%k in ! &ours, but Bnds t&at &e %an ro$ ! mi#es $it& t&e stream in t&e same time as 3 mi#es against t&e stream. ind t&e rate o t&e stream. A. C. ). D.
.+ mi#es Ier &our mi#e Ier &our .1 mi#e Ier &our .6 mile 2er "our
4.8
" 1+ " 2
+
4.8
" 1− " 2
=14
4.8
( " 1+ " 2 )+ 4.8 ( " 1 + " 2) =14 ( " 1 + " 2 ) ( " 1 −" 2) − 4.8 " 2 + 4.8 " 1+ 4.8 " 2 8! " 1 ²− " 1 " 2 + " 1 " 2− " 2 ²
4.8 " 1
=14 " 1 ² −14 " 2 ²
9.6 " 1
14
" 1+ " 2
=
;
3
" 1− " 2
−14 " 2=3 " 1 + 3 " 2
14 " 1
" 1=1.545 " 2
;
Substitute e=uation " in e=uation 2 49.6 ( 1.545 " 2 )=14 ( 1.545 " 2 ) −14 " 2² 14.8432 " 2
=19.418 " 2 ²
# 2=0.76 mph 1+. A man ro$s do$nstream at t&e rate o + mI& and uIstream at t&e rate o " mI&. Ho$ ar do$nstream s&ou#d &e go i &e is to return 5K! &ours ater #ea0ing'
A. C. ). *. ote:
#.5 miles 3.3 mi#es 3. mi#es ".5 mi#es
time =
distance "elocity
t 1 + t 2 =ttotal s s
7
5
4
+ = 2
s =2.5 miles
1(. An airI#ane ying $it& t&e $ind, #ook " &ours to tra0e# km and ".+ &ours in ying ba%k. 4&at $as t&e $ind 0e#o%ity in kI&'
A. C. ). *.
5 ( 5 !
Let: 8 0e#o%ity o airI#ane " 8 0e#o%ity o $ind
" 1 + " 2=
1000 2
=500
*ire%tion o t&e $ind
" 1− " 2 =
1000 2.5
E=.
= 400
E=."
Subtra%t e=uation " rom e=uation ( " 1 + " 2 )−( " 1 −" 2 )=500 −400
=100
2" 2
" 2=50 kph
15. A boat tra0e#s do$nstsream in "K3 o t&e time as it goes going uIstream. 2 t&e 0e#o%ity o t&e ri0er’s %urrent is 1 kI&, determine t&e 0e#o%ity o t&e boat in sti## $ater.
A. $ k2"
C. + kI& ). 3 kI& *. ( kI& Let: 8 0e#o%ity o t&e boat in sti## $ater S 8 distan%e tra0e#ed uIstream S" 8 distan%e tra0e#ed do$nstream s 1=s 2
( " −8 ) t =( " + 8 ) 2
16
3
3
" −8= " +
( ) 2 3
t
" =40 kph
11. -$o I#anes #ea0e Mani#a or a sout&ern %ity, a distan%e o F km. P#ane A tra0e#s at a ground sIeed o F kI& aster t&an t&e P#ane C. P#ane A arri0es in t&eir destination " &ours and + minutes a&ead o P#ane C. $&at is t&e ground sIeed o P#ane C' A. C. C. *.
"+ kI& 3+ kI& #$ k2" "15 kI&
Let: 0 8 ground sIeed o I#ane A 0" 8 ground sIeed o I#ane C s 1=s 2
#t = (# + 90 ) ( t −2.25 ) Cut
t =
900
"
−90
2.25 #
( )+ 900
"
202.5
=0
Mu#tiI#y bot& sides by : 2 2.25 # −81000 + 202.5 # = 0 2
2.25 #
−81000 + 202.5 # = 0
*i0ide a## by "."+ 2 # + 90 " −36000=0
( # −240 ) ( # + 150 )= 0 # =240 kph
# =−150 kph ( absurd )
1F. A train, an &our ater starting, meets $it& an a%%ident $&i%& detains it an &our, ater $&i%& it Iro%eeds at 3K+ o its ormer rate and arri0es t&ree &our ater time but &ad t&e a%%ident &aIIened + mi#es art&er on t&e #ine, it $ou#d &a0e arri0e one and one;&a# &our sooner. ind t&e #engt& o t&e ourney. A. B. ). *.
FKF mi#es 8( miles F"KF mi#es 1+KF mi#es
Let:
t 8 time needed to tra0e# and rea%& destination $it&out any de#ay 0 8 0e#o%ity o t&e train Nenera# E=uation: -ime %onsumed by t&e train tra0e#ing, beore t&e a%%ident 6 time during $&i%& t&e train $as detained 6 time needed to %ontinue t&e %ourse and rea%& t&e destination 8 time needed to tra0e# and rea%& t&e destination $it&out any de#ay 6 time o de#ay )ondition : 2 t&e a%%ident &aIIened &our ater, substitute 0a#ues to t&e genera# e=uations. $ −# =t + 3 1 + 1+ 3 # 5
$ −# 3 5
# t =
Substitute: 5 3 # 2 3
=t + 1
$ : #
( $ −# )= $ + 1 #
$ 8 = # 3
# =
$ 4
)ondition ":
;< E=.
2 t&e a%%ident &aIIened + mi#es art&er, substitute 0a#ues to t&e genera# e=uation: $ −( 50 + # ) 50 + # 3 +1 + =t + # 3 2 # 5
[
50
+ #
#
50
+
5 3 #
#
5
250
3
3
+ # + $−
]
( $ −50− # )= $ + 1 # 2
5
#
3
2
− # =$ +
Substitute e=uation in abo0e e=uation: $ 5 250 5 $ 1 $ 100 50 + + $ − − =$+ − 4
$=
3
800 9
3
()
3 4
2 4
3
miles
F. On a %ertain triI, Edgar dri0e "3 km in exa%t#y t&e same time as Er$in dri0e 31 km. 2 Er$in’s rate ex%eeded t&at o Edgar by 3 kI&, determine t&e rate o Er$in. A. C. ). D. Let:
3F kI& !! kI& !1 kI& 5# k"2 8 rated o Er$in ;3 8 rate o Edgar t 1=t 2 231
# −13 231 #
=
308
#
=308 # − 4004
# =52 kph