Guess-and-CheckExplanations
1.Joehas𝑥dollars.Hespendshalf,getsanother$10,spendshalfofwhathehasnow,and
endsupwith$8.Find𝑥.
(A)8
(B)10
(C) 12
(D) 18
Guess-and-Check:
Startwitheither(B)or(C).Ifwestartwith(B),thecalculationis:Joestartswith$10;he
spendshalf,sohe’sdownto$5;getsanother$10,puttinghimat$15;andspendshalfof
thatamount,endingupat$7.50.Sinceheneedstoendupat$8,thisistoolow.Wecan
crossout(B),aswellas(A),whichisevenlower.Goingthroughthesameprocesswith
answerchoice(C)leavesJoewith$8,thecorrectanswer.
Algebra:
Theeasiestmethodistoworkbackwards.IfJoespendshalfofanamounttoendupwith
$8,hemusthavehad$16before.Hehad$16aftergetting$10additional,sohehad$6
beforethat.Hethereforehad$6afterthefirsttimehespenthalfofhismoney,sohemust
havestartedwith$12.
!
!!"
!
Anequationcouldalsowork: ! = 8.
Solvingthisequationwillyield𝑥 = 12
2.Iftheproductoftwice𝑥andthreelessthan𝑥is56,whichofthefollowingcouldbe𝑥?
(A)7
(B)8
(C)9
(D)10
Guess-and-Check:
ThisisatextbookexampleofwhenyoucanusetheGuess-and-Checkstrategy:whena
questionasksforthevalueofx,givesyounumbersasanswerchoices,andlookslikeit
mighttakeawhiletosolveusingequationsandalgebra.
EverytimeweusetheGuess-and-Checkstrategy,westartwitheither(B)or(C).Ifweplug
inanswerchoice(B)first,thequestionreads:“Iftheproductoftwice8andthreelessthan
8is56….”
Productmeansmultiply,sowe’llmultiply“twice8”and“threelessthan8”-inotherwords,
we’llmultiply16and5.Thatgetsus80;we’relookingfor56,soclearlyanswerchoice(B)
istoohigh.[Ifwehadstartedwithchoice(D),thatwouldhavegivenus140,muchtoohigh,
andwewouldhavethentriedanswerchoice(B).]
Since(B)istoohighandtheanswermustbetheonlyloweranswer,(A).Wedon’teven
needtocheckit.
Algebra:
Translatethequestionintoanequation:2𝑥(𝑥 − 3) = 56;thendistributetoget2𝑥 ! −
6𝑥 = 56.Sincethisisaquadratic(𝑥 ! )equation,wemustmakeonesideequalzeroand
thensolve:
2𝑥 ! − 6𝑥 − 56 = 0
2 𝑥 ! − 3𝑥 − 28 = 0
𝑥 ! − 3𝑥 − 28 = 0
𝑥 − 7 𝑥 + 4 = 0
𝑥 = 7 𝑎𝑛𝑑 𝑥 = −4
3.Giventhediagramabove,whatis𝑥?
(A)15
(B)20
(C)25
(D)35 Guess-and-Check:
Westartwitheither(B)or(C);let’sstartwith(B).If𝑥is20,then∠𝐶is60°,sincethesmall
trianglemustaddupto180°.Butif∠𝐶is60°,thenthelargertriangleaddsupto195°,
whichisimpossible.
Soweknowthat20iswrong,butwedon’tknowwhetherourrealanswermustbehigher
orlower.That’sOK;let’stry(C)andseewhathappens.Usingthesameseriesof
calculations,if𝑥is25thenthelargertrianglewouldaddupto190°.Ouranswerchoices
looklikethis:
(A)15 (B)20195
(C)25 190
(D)35
Avalueof180°canonlybeattheverybottomofthislist,sotheanswermustbe(D).
Algebra:
Thelargertrianglemustaddupto180°,anditalreadyhasanglesof50°and85°.Therefore
thethirdangle(inthelower-rightcorner)is180° − 50° − 85° = 45°.Thesmalltriangle
alsoaddsupto180°,andithasanglesof100°,45°,and𝑥°.Wesetup100° + 45° + 𝑥° =
180andsolvefor𝑥 = 35.
Ifyourquestion#4isaboutgolf,seethenextpage.
4.Fredhas$1’s,$5’s,and$20’s.Hehasfifteenbillsandatotalof$119.Ifhehasequal
numbersof$1’sand$20’s,howmany$5’sdoeshehave?
(A)7
(B)9
(C)11
(D)13
Guess-and-Check
Asalways,westartGuess-and-Checkwitheither(B)or(C).Ifwestartwith(B),weassume
thatFredhas9five-dollarbills.Sincehehas15totalbills,hehas15 − 9 = 6otherbills
besidesthefives.Thequestionsaysthatthesearesplitevenlybetweenonesandtwenties,
sotheremustbe6 ÷ 2 = 3ofeach.HowmuchmoneywouldFredhaveinthisscenario?He
wouldhave9 $5 + 3 $1 + 3 $20 = $108.Thisisnottherightanswer,sowewrite
downthisnumbernexttoanswerchoice(B).
Wedon'tknowimmediatelywhetherweneedmoreorfewerfive-dollarbills,sowenow
try(C).WeassumethatFredhas11five-dollarbills.Sincehehas15totalbills,hehas
15 − 11 = 4otherbillsbesidesthefives.Thequestionsaysthatthesearesplitevenly
betweenonesandtwenties,sotheremustbe4 ÷ 2 = 2ofeach.Howmuchmoneywould
Fredhavenow?Hewouldhave11 $5 + 2 $1 + 2 $20 = $97.Wewritethisnextto
answerchoice(C).Here'showouranswerchoiceslooknow:
(A)7 (B)9$108
(C)11 $97
(D)13
Ifwe'relookingfor$119,itcanonlybeatthethetopofthislist,sotheansweris(A).
Algebra:
Moreannoyingtosetupthanisworthit.Where𝑥isthenumberoffive-dollarbills,the
correctequationwouldbe:
15 − 𝑥
15 − 𝑥
5𝑥 + 1
+ 20
= 119
2
2
Ifyourquestion#4isaboutdollarbills,seethepreviouspage.
4.Inacertain18-holeroundofgolftotaling75shots,agolferonlymakesscoresof3,4,and
5onthevariousholes.Ifthegolferhadequalnumbersof4'sand5's,onhowmanyholes
didthegolferscorea3?
(A)4
(B)6
(C)8
(D)10
Guess-and-Check:
Westartwith(B)byassumingthatthegolferscoreda3onsixoftheholes.Thisleaves12
otherholes.Wearetoldthatthegolfermakesequalnumbersof4'sand5's,soshemust
havemadesix4'sandsix5's.Doesthisequal75?No—six3's,six4's,andsix5'saddsupto
72,so(B)isincorrect.
Ifwewanttobeclever,wecanseethatthegolfer'sscoreistoolow,soshemustneedmore
highscores(4'sand5's)andfewerlowscores(3's).Theonlyanswerchoicewherethe
golferscoresfewer3'sisanswerchoice(A).
Ifwewanttobesafeandthorough,wecantryanswerchoice(C)andassumethatthe
golfermakeseight3's.Thereforetheothereightholesaresplitwithequalnumbersof4's
and5's—fiveofeach.Eight3's,five4's,andfive5'saddsupto69.
Whetherwestartedwith(B)or(D),ouranswerchoiceswouldlooklikethis:
(A)4
(B)672
(C)8 69
(D)10
Followingthepattern,avalueof75wouldhavetobeatthetopofthislist,sotheanswer
mustbe(A).
Algebra:
Moreannoyingtosetupthanisworthit.Where𝑥isthenumberof3'sthegolferscores,the
correctequationwouldbe:
18 − 𝑥
18 − 𝑥
3𝑥 + 4
+5
= 75
2
2
5.Whatis𝑥iftheaverage(arithmeticmean)of3,17,and𝑥is19?
(A)34
(B)35
(C)36
(D)37
Guess-and-Check:
Westartbyassuming𝑥 =either35or36(BorC).Ifwestartwith(B)35,theaverageof3,
!
17,and35is18 !.Thisistoolow,sowecrossout(B)and(A).Ifwestartwithorthen
!
attempt(D)36,theaverageof3,17,and37is18 !.Thisisstilltoolow,butgettingcloser,so
theansweris(D)37.
Algebra:
!"#
Usetheaverageformula,𝐴 = !"#$%& !" !"#$%:
19 =
Usealgebratosolvefor𝑥 = 37.
3 + 17 + 𝑥
3
6.If𝑓(𝑥) = 𝑥 ! + 3𝑥 − 8,and𝑓(𝑟) = 62,whatisr?
(A)4
(B)5
(C)6
(D)7
Guess-and-Check:
Startwitheither(B)or(C);let’stry(C)thistimejustforvariety.6! + 3(6) − 8is46.It's
toolow,andthere'sonlyonehighervalue,sothatisthatonlypossibleanswer.That'sallwe
havetodo;theansweris(D).
Algebra:
Thisquestionwillbeverydifficulttosolveifyoudon’thaveabasicideaofwhatafunction
is,nomatterwhattechniqueyouuse.Hopefullyyouunderstandthatinthiscase,the
questionissayingthat𝑟 ! + 3𝑟 − 8 = 62.Sinceit’saquadratic(𝑥 ! )equation,wemust
makeonesideequalzero,factor,andsolveasfollows:
𝑟 ! + 3𝑟 − 8 = 62
𝑟 ! + 3𝑟 − 70 = 0
𝑟 – 7 𝑟 + 10 = 0
𝑟 − 7 = 0 𝑎𝑛𝑑 𝑟 + 10 = 0
𝑟 = 7 𝑎𝑛𝑑 𝑟 = −10
Only7isoneoftheanswerchoices,sotheansweris(D).
!
!
7.Giventhat𝑎 = ! + 1,𝑏 = ! + 2,and𝑎 = 5,whatis𝑐?
(A)16
(B)18
(C)20
(D)22
Guess-and-Check:
!
!"
Ifwestartwith(B)byassumingthat𝑐 = 18,then𝑏 = ! + 2.Therefore𝑏 = ! + 2,so
!
!!
!
𝑏 = 11.We'realsotoldthat𝑎 = ! + 1,so𝑎 = ! + 1,andwesolvefor𝑎 = 4 !.Butthe
questionsays𝑎 = 5;thismeansthatanswerchoice(B)isnotcorrect.Sinceourresultwas
toolow,we'llcrossoff(B)and(A)andnowtry(C).
!"
!
!"
Using(C),𝑐 = 20,then𝑏 = ! + 2,so𝑏 = 12.We'realsotoldthat𝑎 = ! + 1,so𝑎 = ! + 1,
andwesolvefor𝑎 = 5,aswehoped.Theansweris(C).
Algebra:
!
!
!
!
𝑎 = 5and𝑎 = ! + 1,so5 = ! + 1.Solvethisfor𝑏 = 12.Also,𝑏 = ! + 2,so12 = ! + 2.
Solvethisfor𝑐 = 20.
8.Charliepurchased7videogamesforatotalcostof$270.Ifallofthegamescosteither
$35or$40,howmany$40videogamesdidCharliepurchase?
(A)4
(B)5
(C)6
(D)7
Guess-and-Check:
Wecanstartwith(B)byassumingthatCharliepurchasedfive$40videogames.Sincehe
purchasedsevengamestotal,hewouldalsohavepurchasedtwo$35games.Five$40
gamesandtwo$35gameswouldaddupto$270,so(B)isthecorrectanswer.
Ifwehadstartedwithwithanswerchoice(C),six$40gamesandone$35gamewouldyield
atotalcostof$275.Thisistoohigh,sothecorrectanswerwouldincludefewerofthemore
expensivegames.Wewouldthentryanswerchoice(B).
Algebra:
Let𝑥 =thenumberof$40gamesand𝑦 =thenumberof$35games.Wehaveoneequation
forthenumberofgames,𝑥 + 𝑦 = 7.Wehaveasecondequationforthetotalcost,
40𝑥 + 35𝑦 = 270.
Usesubstitution:solvefor𝑦intermsof𝑥toget𝑦 = 7 − 𝑥.Substitutethisvalueintothe
secondequationtoget40𝑥 + 35 7 − 𝑥 = 270,andsolvefor𝑥 = 5.
9.Sarahis3inchestallerthanCamille.Marlynis6inchesshorterthanLouiseand5inches
shorterthanCamille.IfLouiseis68inchestall,howtallisSarah?
(A)69inches
(B)70inches
(C)71inches
(D)72inches
Guess-and-Check:
Westartwith(B)byassumingthatSarahis70inchestall.Sarahis3inchestallerthan
Camille,soCamillemustbe67inches.Wedonothaveenoughinformationtousethe
informationaboutMarlynandLouiseyet,butwearetoldthatMarlynis5inchesshorter
thanCamille.Camilleis67inches,soMarlynis62inches.Nowwecanusetheinformation
thatMarlynis6inchesshorterthanLouise.Marlynis62inches,soLouiseis68inches.The
questionsaysthatLouiseactuallyis68inchestall,soanswerchoice(B)iscorrect.
Ifwestartedwith(C),wewouldhavefoundthatLouiseis69inchestall.Thiswouldbetoo
highanumber,sowewouldhaveeliminated(D)andtriedanswerchoice(B)next.
Algebra:
Asystemoffoursimpleequations—easytosolvewithsubstitution,butnotaveryefficient
waytosolvethisproblem.
10.Theaverageageofagroupoffourpeopleis49years.Ifoneofthepeopleis40years
old,whatistheaverageageoftheotherthreepeople?
(A)52
(B)54
(C)56
(D)58
Guess-and-Check:
Westartbyassumingthattheaverageageoftheotherthreepeopleis(B)54years.That
meanstheaverageageofallfourpeopleis
40 + 54 + 54 + 54
𝐴=
4
𝐴 = 50.5
Butweknowtheactualaverageshouldbe49years,not50.5years.Since(B)istoohigh,
theanswercanonlybe(A)52.
Ifwehadstartedwith(C),wewouldhaveobtainedanaverageof52—toohigh—and
thentried(B)asdescribedabove.
Algebra:
Theaverageageofone40yearoldpersonand3peoplewithanunknownaverageageis
!"#
49.Wecanusetheaverageformula,𝐴 = !"#$%& !" !"#$%,andsetupanequation
representingatotaloffourpeople:
49 =
Solvingthisequationfor𝑥yields𝑥 = 52.
40 + 3𝑥
4