act 2 explanatory answers - math

8/12/2019 ACT 2 Explanatory Answers - Math

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M THEM TICS PR CTICE TEST 2 EXPL N TORY NSWERS

Question 1. The correct answer is C. The restaurant is rotat ing 180 0 in 45 minutes. Thntis 60 0 each 15 minutes. That is 240 0 every 60 minutes. You might want to sketc h so methin g likethe draw ing below.

f yn n like an algebraic= ( : ~ )= 4 . 60 = 240.

45 min1800

15 min

30 m in

solut ion, you can se t up a pwportion .. --L.- and solve it :4S min 60 mill

Most peop le w ho do not get th is qu estion right choo se D, which gives th e number of degrees therestau rant would rotate in 45 minutes i f it mad e 1 complete rotat ion each hour.

Question 2. The 'correct ~ s w e ris J. The 12 vases cost 18, so eacp vase costs ~ ~ 8= 1.50.If you c hose F you probably divided 12 by 18 rath er than 18 by 12 . If vases cost 0.67 each, then12 vases wou ld cost less than 12 .

Question 3. The correct answer is B. Th e lo nger side of the apartment is 30 feet long , andit is 6 inche s lon g on t he scale drawin g. So the lengt h of the room, in feet, is 5 times the lengt h o nthe drawing , in inc hes. Using this relationsh ip, the length of the short er side of the apartment is 5tim es the 4 inches from the scale d rawing. Th is is 20 feet.

Alternately, you co uld not ice that the length of the sho rter side is j th e length of the long er side

on the drawing, and so the length of the shorter s ide of the room is j of 30 feet, which is 20 feet.

These so lution s are equivale nt to using a proporrion suc h as O i ~= ~ and solv ing: = \3 = = 2 0feet

Question 4. The correct answer is H. The total pro fit for the 5 vea rs was, in million s,8 8 8 9 9 = 42. T hen the average profit, in million s, was ~ 2 = 8.4.

The most common wrong answer was j , which is the average of 8 a nd 9. Because there were moreyea rs with a profit of 8 million than with 9 mi llion, the average for the 5 years must be clo serto 8 million than to 9 million.


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uestion 5. T he correct answer is A. If ~ van were dri ve n for 20 mi les, the cos t for thosemile s would be $ 0.30 . 20 = $6. Th en the dai ly charge of $25 wou ld have to be a dded in, for atota l of $31. Sim ila rl y, if the van were d riven for n m iles, rhe cost fo r tho se miles would be a 30mdoll ars, and th e daily charge would mak e the total a. 3 0m + 25 dollars.

C co mes from treating th e 30 ce nt s like it was 30 doll ars.

uestion 6 . Th e correct answer is F The sum of the measures of all fo ur inter ior angles in anyquadrilate ral is always 360, The given three angle measures add up to 65 + 100 + 75 = 240 ,so th e missing ang le measure is 360 - 240 =120,lf AB we re par allel to CD then the measure of L B and th e meas ur e of L C would add up to 180 0and the answer would be G. Bu t th e probl em does not say t hat th ose s ides a re parallel , an d it turn sout th at they are not parallel.

If you cho se K; yo u ma y have calc ula ted the average of th e th ree given angle meas ur es or take nrhe supplement of L A

uestion 7 The correct answer is C . Th e shorte r two sides are the same lengt h in Cso the sa me thin g has to happ en in any triang le sim ila r to it. That means that DE is th e sa melengt h as EF w hich is 3 met ers. Then, the per imeter of 6.DE F is 3 + 3 + 5 = 11 met ers.

Th e mo st co mmo n incorrect an swe r is A, correc tl y fin din g the leng th of DE and stopping th ere

uestion 8 . The correct answer ,is K. T his problem can be solved by substit uting the

Ce lsius temp era tur e C = 38 into the form ula and so lving fo r F The sub stitut ion step gives

F = t 3 8) + 32, which can be so lved as follow s:' F = 68.4 + 32 > F =100.4. It is ap p rop ria te toround this to th e nearest deg ree Fahrenheit beca use the precision o f the Celsius tempe ratur e wasonl y to the nea rest degree Cel sius.

If yo u cho se J, yo u may have ad d ed 38 + 32 and mi sse d th e t . An a nswer of H might co mefrom calculating t 38) cor r ec tly and for gettin g to add in 32.

uestion 9. The correct answer is B. N ick can only order who le cases, w hich co ntain24 boxes of pens with 10 pens per box, for a tota l of 24 . 10 =240 pens pe r case. An order of2 cases would be 480 pen s, which falls short of the desi red 500 pens. To get 500 pe,ns from hissup p lier, Ni c k n eeds to order 3 cases, and h e will get 720 pens.

If you got answer A, you may ha ve correctly divided 500 by 240 to get approximately 2.08 cases,but you may ha ve rounded tha t to the near est integer, which do es not give the cor r ect answer inthis con text . Ans,wer E repre sents the n umber of boxes (not cases) of pens needed if N ick co uldorder any wh o le number of boxes.

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Question 10. The correct answer is K. When a + b = 6, then 2{a + b +becomes 2 6) + 6)2 - 2, which simplifies [ 12 + 1 + 36 - 2 = 47.

0+6 +( a +b ) - 2

Question 11. The correct answer is C. If you bought 1 hamburge r and 1 soft drink, itwould cost 2.10. If you bought 1 hamburger mo re, your order would cost 3.50. So, t he cost ofthe additi ona l hamburger was 3.50 - 2.10 = 1.40. Because 1 hambu rger and 1 soft drink cost

2.10, a soft dr ink must cost 2 .10 - 1.4 0 = 0.70.

Alternat ively, y o ~cou ld set up two eq uat ions w ith two unknown s. Let h dollars be the cost of eac hhamburger and 5 do llars be the cost of each soft drink. T hen h + s =2.10 and 2h + s = 3.5 0.Subtract ion g ~ v e s

2h + 5 = 3.50-(h + 5 = 2.10)

h = 1.40

And then, su bstit utin g 1.40 for h iil h + s =2.10 gives s =2.10 -1 .40 =0.70.The most common wrong answe r is E, which is the correct cos t of a hambu rger. However, thequestion asks for the cost of a soft drink. Answer cho ice D is half of 2.10, whi,ch would only becorrect if a soft drink cost the same as a hambur ger.

Question 12. The correct answer is K. The re are many ways [ solve this equation. Onesolu tion is the following.

start with 1 x = -8 10 - x)

div ide both sides by -8 > - ~ x = 1 x

a dd x to both sides > - x = 10multiply both sides by -2 => x = . -20

Another solut ion method would be to graph y = 12x and y = -8 10 - x and see where the twogra phs intersect. A calc ulator would p rod uce these grap hs, and you cou ld find an approx imatesol ution . That is good e nou gh fo r this problem, because the answer choices are spread apart.

This is a problem where checking yOUL answer is easy and ca n payoff. When x = -20, the left sideof the original equation is 12(-20), which is -240. Th e right side is -8(10 - (-20)), wh ich simpl ifies [ -8{10 + 20), then to -8(30), then to -2 40. Thi s so lution checks. No ot her answer choicewould sat isfy .the equatio n.

If you chose H or j , you may have multipli ed out -8 10 - x) to ge t -80 + x or -80 - x and donethe rest of the steps correctly. You co uld hav e ca ught this by chec king you r answer. 1 you madean erro r wit h minu s signs, you may ha ve gotten one of the other 'answer choices .

e

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uestion 13. The correct answer is C. T here would be 6 t iles a lo ng the 24 s ide(6 . 4 = 24 ). There would be 16 t iles a long the 64 side (16 4 = 64 ). Then, 6 . 16 = 96 tilesare needed to completely cover t he recta ngu lar counrertop:

1 ~ ~ 6 4 ~2 4 H - + + t - 1 H - + + H - + + + - H -I

1- - t-t -H- H- -I

An alteniate solut ion is to figu re the ar ea of the co un ter top in squar e inches, 2 4 . 64, w h ich is

1,536 sq uare inches. Then, div ide that by the area of a ti le, which is 16 sq uare inches. The resultis 96, whic h is the number of t iles needed. T he tiles cover the 'area without bei ng cut because theside lengths of the cou nt ertop are divisible by th e side length of a tile.

Th e most O mm on wrong answer is E, which comes from correcrly ca lculatin g the are a of thecounte rtop (1,536 square inches), but dividin g by th e length of the side of th e sq uare (4 inches)rather than by 4 '4 sq uare inches. If you cho se A o r B, you ma y have con fu sed per imeter a nd area.

uestion 14. The correct answer is H. If the answer c hoices g ive 2 of h 3 interior a nglemea Sll1'es in a triangle, the n the thi rd a ngle measure is 180 0 mi nus th e sum of the given a ngle measur es. Th e c hart below shows this calc ula tion.

1st angle 2nd anglc Sum of 3rd anglc

1st 2.nd an les . (1800

- sum of lst 2nd an I 'es)F. 20' 40' 60' 120'G. 30' 60 ' 90' 90'R. 40' 100 140' 40 'J. 45' 120 165' 15 'K. 50' 60' 0 70'

The on ly place w here the thi rd angle has measure equal to on e of t he first two angles is in H, wherethere are two 40 angles. .

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Question 15 . The correct answer is C. The perimeter of the trian g le 66 'inches, is thelength of th e th ree sides added together. Because one side is 16 inches long, the lengths of the othertwo sides added together must be 66 - 16 =50 inc hes. The ratio of the lengths 'of these two sidesis 2:3. This ratio denot es 2 parts for the first side, 3 parts for the second side, a nd therefore 5 partsaltogether. Because there are 50 inches altogether, and this must make up the 5 parts, each part is10 inches long. That ma kes one side 2 parts, or 20 inches long, a nd the other s ide 3 parts, or30 inches long. So the ongest side of the triangle is 30 inches long.

B is the length of another side of the triangle, but not the longe st side.

Question 16. The correct answer is H . Sketc hin g a picture can be helpfu l.

y

-3,6)

3,2)

l t ; ; o H t t ' x

The po ints (-3,6) and (3,2) are on opp o site sides of the y-ax is, and an equal distance away fromthe y-axis, because the x-coord inates are 3 and -3 . 0, the midpoint vill be ,the y- int ercept of the

line . The y-coordinate of the midpoin t is 6;2 which is 4.

If the poims were not so nicely spaced with the y-axis, you cou ld find the equation of the line

through the two points an d use it to identify the y-inte rcept . The slope of the line would be the. ,

change in y divided by the change in x ~ 3 _ 3= - ~ T he point-slope form of the equation is then

y - 2 = - ~ ( x- 3). To find the y-intercept, set = 0 and solve for y so y - 2 = 2 y = 4. Th is shows

th at 4 is the y-inte rcept.

From t he picture alone, yo u cou ld deduce that the lin e had to cross the y-axis between 2 and 6.This observa tion eliminates all t he answer c hoices except th e correct on e .

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MATHEMATICS PRACTICE TEST 2 EXPLANATORY ANSWERS

Question 17 . The correct answer is C In the figure bel ow, the angles with measure a and100 form a straight angle along line m This means a 100 = 180, or a =80. Now, you knowtwo of the th r ee angles in the larger triangle. Th e sum of a ll th ree must be 180. So,80 + bO + 65 = 180, which means that bO = 3SO The angle measure bO is equal to angle measure XO because vertical angles ha ve the same measure. SO XO = 35.

. s

- - - , - - , - - - r l - ,- - m

\ ~ n

If you chose A, you may have calcu lated 18 0 - 65 + 100). These three given angle measures,xo 65, and 100, do not need to add to 180. They. are no t the measures of the three interiorangles in the s me triangle.

Question 18. The correct answer is H. Each week, both ponds ge t shallower . The tablebelow shows what happ ens for the first few weeks.

Now 1 week 2 weeks 3 weeks 4 weeks

First pond 180cm 179 Col 178 Col 177 em 176 cm

Seco nd pond 160 em 15 9 .5 em 159 em 158.5 em 158 em

Difference 20em 19. 5 em 19 em 18.5 cm 18cm

The ponds are getting closer and closer to the sa m e depth, at th e rate of 0.5 em per week. Becauseth e ponds sta rt ed out 20 em different, it w ill tak e 20 0.5 = 40 weeks to bring them to the samedepth.

You cou ld also check the answer c hoices. Answer F is no t correct because, :after 10 weeks, the f irstpond wo u ld be 180 - 1 0 = 170 em deep and the second pond would be 160 - 5 = 155 cm deep.The other incorrect answers can be e1iminated ,in the same way.

The most common incorrect answer is G. I you c ho se that answer, you may ha ve reasoned that itwou ld take 20 weeks for th e fir st pon d to get do.wn to the level of the second pond. And that iscorrect, except that the second pond has gotte n sha llower durin g th at 20 weeks, so the ponds arenot the sa m e depth.

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Question 19 The correct answer is E. A line wi ll always ha ve an equation of the formx = a or y = mx + b for suit able cons ta nt s a m, and b. And, i f a grap h has an equation that canbe put into one of these forms, the grap h is a line.

Equation A is already in th e first form, w her e a = 4.

Start ing with Equation B, divide both sides by 3 and yo u will get y = 2. This is the seco nd form,with m =0 and b =2.Equation C can be manipulated into the second form as follows: x - y = 1 > -y = 1 - x > y =-1 + x > y = x - 1, which has m = 1 and b = l .

Equation D is already in the seco nd form, with m =, t a nd b = -2 .Equation E can be written as y = 2 + 5. This is the equation of a parabola, shown below. It isthe only one of the eq uations that is no t a line.

y

onl - - - x

Question 20 T he correct answer is G . All the answer choices are in terms of LA. Manypeople remember the trigonome .tric functions in the contex,( of a right triangle, in terms of thelengths of the side o p p o s t ~the a ngle, the side a dj acent to the angle, and the hypotenuse. For thetriang le g iven in the probl em, t he side opposite L A has length 12 COl, the side adjacent to LA ha slength 5 COl. a nd th e hypotenuse has length 13 cm. The values of the trig functions are

A adjacent 1. . A _ opposite1 d A _ ' J..

cos = hypotenuse 13, sm - hypotenuse 13' a n tan - adjacent 12

The only answer c hoi ce tbat gives one of these is G.

I f you chose F, you may have been expecting the angle to be at one end of the horizontal base of the

triang le. The value of cos B is g.Triangles can be rotated to any position. T he terms oppos ite, adjace nt , and hypotenuse are chose n t o apply when the triangle is in an arbitrary position.

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MATHEMATICS PRACTICE TEST 2 EXPLANATORY ANSWERS

Question 21. The correct answer is ~ . Pa rallel lin es ha ve th e same slop e, so any line pa ra llel to this line has th e sa me sl op e as this line. To find t he s lo pe of this line, yOll cou ld put it i nt oslop e-int ercept form, and then the slop e is the coefficie nt of x 7x 9y = 6 => 9y = -7x 6 => y =- ; x + ~ ,an d so the slop e s ~ .

If yOll chose E, you pr obably knew that the slope is the coefficient of x but that only is tr u e whenthe equ ation is in slope-inrercept form.

If yo u c hose A, you may have got ten the eq uation into the f,:mn 9y = -?x 6 and then read of fthe coeff icie nt o f x You didn t q uit e have it in slope -inter ce pt fo rm.

If you cho se C, you may have made a mis take purri ng the equation in slope-i nt ercept form.

An swe r choic e D co uld be read ing off the cons tant 6 from th e o riginal equation.

Question 22 The correct answer is H While it may not be wo rth your time to ske tch agraph duri ng th e ACT, yo u sho uld at lea st have a general idea of the situatio n in your mind. Thisis the equati on of a parabola. Co mmon se nse w ill tell you tha t if the car is goi ng faster, it will takea longer distance to b rake to a stop. So, the parabola is opening upward.

y

o [0 20 30 40 5 60 xspeed mph)

The horizontal dashed line on the g raph is w here t he brak ing d istance is 15 feet. Th e re are twopoints w here th e parabola interse c ts th is line. One is to t he left of the y-axis, w hich re presents anega tive speed fo r the car. If yo u so lve an eq uation to find the intersec tion poi nts, you will haveto discar d th e int ersect ion po in t i t h a negat ive speed.

If ll took the time to create a graph, per hap s even on a graphi ng calculator , yOll might be ableto es tim ate closely enoug h to choose among t he answer c ho ices. T he des ired spee d is th e x-co ordinate of the intersect ion po int.

To get this speed alg ebra ica lly, the desired braking d istance is 150 feet, and the , eq ua tio n

Y = 3 V 4 ~ I O Xgives the relation betwee n speed x) an d braki ng dista nce y . So you can so lve the

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PRACTICETEST 2 EXPLANATORYANSWERS

, 150 3(. +10,) 0 I I b I I b h d b 40 dquation = 40 . ne so un on pa t 1 starts y mu tiP ymg ot Sl es y an con tll1ues

as fo Uows : 150 40 = 3 x + l Ox) :::::;> 6 ~ = x + lOx:::::;>X l + l Ox - 2,0 00 = 0 :::::;> x + 50 ){x - 40 )

= 0 :::::;>x = -50 or x = 40. You mu st disca rd th e fi rst so lut ion. Th e o nl y remaini ng so luti o n is x =

40, w hich represents a speed of 40 mil es pe r hour.

H yo u cho Je J, you may ha ve don e eve ryt hin g correctly excep t that w hen you so lvedx . 50) x ... 40 ) =0, yo u thought x = 50 was a so luti on. If yo u sub stitu te 50 for x in the equa tionx + 50) x - 40) = 0, you wi ll see tha t it is not a so lution.

If yo u cho se F yo u may h ave made a num erica l mi stake w hen yo u sub sti tu t ed x = 10 into t ~e brak-

, d T I I . 3 102

+10.10 ) 3 200) I h b dmg - Istance equatIOn. le va ue IS 40 t mi g t e temptlllg to re lice

4 ~ to 50 and then ge t 3(5 0) = 150, wh ich is w ha t you are lookin g fo r. But 4 ~ is 5, not 50.Sub stitutin g the a nswe r choi ces into th e e q ~ at io n is a reasonable strategy for this problem. Youca n u se th e res ult s from o ne of the a nswe r cho ices t o hetp you c ho ose t he next on e to substitu te,sinc e you probably know that it takes lon ger t o stop if you a re g oin g fas ter. T he results of thesesub stit ution s a re g iven below.

Speed C alculations .Brak ing distancemp h ) ft)

F. 10 3 20 0 ) = 3 5) 1540

G, 30 31l,200) = 3 3 0) 9040

H, 40 3 i,ooo) = 3 5 0) 15040

J 50 3 3,000) = 3 75) 22540

K. 60 3 4,200) = 3 105) 31540

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MATHEMATICS PRACTICE TEST 2

Question 23. The correct answer is A. Substitution gives g 4 ) = .J4 =2 and 1) = + 1h ~ l+ 5 = 7 . T en / 1) - 7

If yo u chose C, yo u ma y ha ve s t r r ~ su bstitut ing 4 into function g and rhen contin ued substit ut -

. ing 4 into f ~ n c t i o nf The value of 5 ~ ~is }s

Question 24 . The correct answer is K. Th ere are 125 junior s who could be chosen. Fore c h of those 125 juniors, there are 100 sen iors who cou ld be chosen. That makes 125 . 100 =12,500 diff eren t pairs of stud ents who cou ld be chosen.

If yo u c h os ~ j . yo u probably added 12 5 and 100. If yo u chose G, perhaps yo u were figurin g thatonce 100 pairs were form ed, there would be no se ni ors left to ptit into a nother pair. Only 1 pa irwill be c ho se n , but there are many more than 100 pairs t hat a re possible ch o ices.

Question 25 . The correct answer is C . T he figure below show s the ramp.

2

= = ~ ; = ~ ~ I~ = = = = ~ 1 5F 100 iThe slope is g iven as rising 5 feet for every 100 feet of ho r izontal run. The ramp s rise IS 2 feet,

and th e horizont a l run is unknown. Let the ho r izontal run be rep resented as d Then there is a p ro-. , 5 I . d 2 100 40 fportion j = 100' ts so unon IS = - 5- = eet.

If YO li cho se A, you might ha ve si mplifi ed 5 to 0.5. Then th e proportion ~ = 0. 5 has the so lution 4 feet.

The most popular incorrect cho ic e was B, which hap p ens w hen the run is 5 times the rise. This gives

a reasonab le-look in g ramp. Ho w ever, the slope of suc h a ramp is t The required slope is 210

Question 26 . The correct answer is F You co uld ce rtainl y test all of th e answer choi ces tosolve thi s problem. Th e first cho ice, - 10, leaves the left s ide of the relation as 1- 10 - 241, whichsimp lifie s to 1-341, which is 34. And -34 :5 30 is false, so this is t he co rr ect a nswer.

Another way to so lve thi s prob lem is to th ink about t he int erpretation of abso lut e va lue as a di stance: It - 241 :5 30 m ea ns that the di st an ce on th e numb er lin e between t a nd 24 is at mo st 30unit s. This distance would be greater than 30 only when t was mo r e than 30 above 24 or morethan 30 below 24. Th at is when t is more than 5 4 or t is less than - 6. Answer choices G-K are a llcloser to 24 than this.

The mo st common wrong an swer is 54 (K), which is right at th e limit of how far away it can getfrom 24.

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MATHEMATICS I EXPLANATORY ANSWERS

Ques t on 27. The correct an sw er is D . Th e ph rase 5 tim es a num ber n ca n be r e p r e s e n t ed as 5n. If this is subtra cte d from 15, the ex press ion 15 - 5 n represent s the resu lt. Say ing t hat thisresul t is negati ve can be represented as 15 - 5n < O. Subtracting 15 fro m both sides gives - 5 n 3 (remembe r ,to reverse the directio n of th einequalit y) .

Most people w ho mi ssed this c ho se E. Thi s co uld come from not reve rsi ng the inequ alit y in t helast ste p of the solutio n, or it co uld come f rom subtra cting 15 from 5n rat her than subtrac ting 5nfrom 15. .

Answe r choice B is the res ult of so lving 15 - 5n = O T his woul d give the boundar y where t heex pressio n changes from positive to n egati ve, so it s closely related to the prob lem yo u were askedto solve. If yo u cho se this, you wou ld still n eed to find the values of t w here the express ion is neg-at ive, knowing that I t = 3 is the onl y place the ex press ion is zero.

Question 2 8 . The correct answer is J. A qu ick scan of th e an swer c ho ices s hould give y oua ~ ue that co mbinin g like terms is in o r e r

lx ' - 4x + 3 - 13x' - 4x - 3= lx 4x + 3 +1-3x ' + 4x + 3= lx + 1-3x )) + 1-4x + 4x + 13 + 3= -2 x 2 + Ox + 6

su btra c tin g is adding th e oppo s itereo r dering term scomb inin g li ke t erm s

If yo u cho se K, you probably subtracted t he 3x but added the -4x and th e -3. T he solution aboveto ok ca re of this explicitly on the seco nd line.

The other incor rec t answe rs res ult from variolls co mbination s of err or s wit h minu s s igns.

Question 29. The correct answ e r is B. Imagine th e orig inal 9 data item s in o rd er fromsma llest to largest, You mi ght ske tch a pictur e some thin g like tha t s hown below . The med ian is the5t h item in thi s list, because it is the middle va lue in the set .

When the addition a l 4 dat a items are put in to the list , there are 9 + 4 = 13 items on th e list. Beca use2 of th ese items are greater than the original med ian, and 2 of these items are less than the origina l median, the original median i s the midd le va lue in the new set. Th at mak es th e o riginal m edi '

an the new median.

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Practi ce ACT Tests

EXPL N TORY NSWERS Q u e s t i o n 30. The correct answer is K. T he shaded ar ea is the area of the larger circle minusthe area of the sma ller circle. Th e area of the latg er citcl e is n:r = n:(10)2 = OOn: squa re cent ime-te rs. T he area of the small er circle is n:(5)2 25n: squar e ce ntim eters. The d ifference is 75n: squarecentimeters.

I f you c ho se j , you likely found the difference in the perimet ers of the two ci rcles. You may haveused a perime ter formul a w hen you wanted a n area for mula . (Another possib iHty is that you ca lculated 10 2 as 2 1 0 and 52 as 2 . 5 . But, 10 2 is 10 . 10 and 52 is 5 5 .

Q u e s t i o n 31. The correct answer is E. Th e side lengt hs of a 30_60_90 triangle are in theratio 1: .J3: 2. I f yo u didn t remember t his, you co uld view a 30 _60_90 tr iang le as ha lf of anequilateral triang le, as shown below.

For the shad ed triang le, you know that t he ba se is ha lf as long as the hypo tenu se, because the baseof the equi lateral tri angle is the sa me length as t he oth er sides of the eq uilateral triang le. I f thehypote nuse is s uni ts long and the base is ts then t he ~ y t h g o r e ntheorem gives the height asJs 1 f ; JSI_; :=J f _s ~ J i ? = R = J 3 J f = J 3 1 = 1 s. .

Thi s s hQws that . th e ratios are t: -I{:1, whic h are equivalent to those g iven above , 1: .J3: 2.(O bviously , it s q uicker to know t he ratios than t o try to der ive the m eac h time you need them, but

do n t give up if you can t remembe r something - tr y to find it a different way.)

You could a lso use tri go nometry to de r ive th e s ide lengt h r at ios in a 300-60 -90 triangle.

If you chose B, yo u might have been thinkin g of a 45 _45_90 tri ang le, which has t his rat io ofside lengt hs.I f yo u chose D, you may have reaso ned tha t becau se the ang le meas ures are in th e rat io 1:2:3,

maybe th e side length s are i n the racio ..JI.:.Ji : .J3. This is a right tria ngle (it satisfies thePythagor ea n theorem), b ~ the ang les are close r t o 35-5SO-90.

from the Mak ers of the CT 38.


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Chapter 4

390

PRACTICE TEST 2

Th e triangles in A and C a re oot right triaogl es. You cou ld have elimi nated them becau se

t hey do not satisfy the Pythago rean theore m. For A, J1i li = J2 :t: 1. For C,)1 ( . )2 ) ~ ~ . 2 .

Question 32. The correct answer is G. If the x c o o r d i n t eis 20, th en the y ~ c o o r di n a t eca nbe fou nd by subs titutin g 20 for x, 0.005 (20)' - 2(20) 200 =0.005(400) - 40 200 =0.5 4)160 = 2 160.

In th eo ry, you co uld r ead the value off the grap h, but you would not be able to read it acc urat elyenou gh.

Man y incorrec t answers are caused by mistake s with th e first ter m , O 5x If you c hose F, youmay hav e mad e a decima l error calcu la ting 0.005{20) 2and gotten 0.2 rather th an 2.0. Oth er err or s

in ca lcul at ion l ea d t o H , J, andK

Question 33. The correct answer is D. The dist ance formula (or the P ythagor ean theorem) 'gives this distance directly. t is

J (200 - 0) ' + (0 - 200)' = .J200 + 2 00 ' = .J2 100 ' = 200. )2 ~ 200(1.4 14 ) =282.8 .Another way to s o l v ~this is to notic e that the length of F O is 200 unit s, and FG is longer. Also,the path from F to a to G is 400 unit s lon g, and the direct path alon g FG is s horter th an thispath . So, D is the only reasonab le answer amon g t hose given.

Question 34. The correct answer is F Th e shaded region is ire ly co ntained i n th e g iventriangle because the cu rve is below th e hypoten use of the triangl e, FG. Th e area of the tria ngle ismade up of the sh aded area plu s th e unshaded area above th e c urve an d inside the tr ian gle. So, th eshaded a rea is less than the area of the t rian gle. '

Question 35. The correct answer is B. With a = 4.2 , b = 5.0 , and a 50 measure for LCthen th e law of cosi nes gives the leng th of the third side of the triangle (the distanc e betwee n th ecargo sh ip and the fishin g boat ) as:

J(4.2) + (5.0)' - 2 4 .2 5 .0co s 5

If you cho se C. YOll probab ly missed the minu s sign in the middl e of the expression. Abo ut halfthe stu dents who do nor get th e co rrect an swer choose C.

Th e 85 0 angle in answer choices D and E w ould be the angle meas ure by the fishin g boat , but o nlyif the a ngle by the cargo s hip was a right n e It turns alit t hat it isn't a rig ht angle.

from th e Makers of the CT, . /


14/27

Practice ACT Tests

MATHEMATICS PRACTICE TEST 2

Question 36. The correct answer is F Because aJ = b, then b1 = a J 2. S.\.Ibstituting this intothe equation c = b 2 gives c = a 3 2. ~ c a u s ea J 2 = a 6, the result is c = a 6

The most common incorrect answer is G. If you chose that answer, you probably wrote (a l )2 as a s.H you write a 3 2 out as aJ) a l and then a . a . a) a . a . al you can see that it is a 6

If you c hose H perhaps you wro te a 3 2 as 2(a l ).

Question 37. h ~ correct answer is E. Thi s sequence decreases by 17 - 12 = 5 eac h term.Here are some more terms:

Term 1 2 3 4 5 6 7 8 9Value 17 12 7 2 - 3 - 8 -13 - 18 . -23

Each of A- D can be ve r ified from the cha rt . T he commo n differ ence for D is defined so that it ispo sit ive if t he seque nce is increasing an d nega t ive if the sequence is decreasing. In symbo ls, if thesequence is rep resented by ter ms a p a 2 , a l , ' , then t he difference between two terms is a;+ I - a;for all i. If this difference is constant for all i, then it is ca lled the common difference and thesequ ence is ca lled rithmeti .

at I fThat leaves only E tha t could be fa lse, T he ratio of consec utiv e terms is defin ed by a o r all i.,

For the fir st twO terms, th e ratio is : ~ which i s a bit more than 0.70. For the seco nd and third

te rm s, the ratio is ;i, w hich is a bit less than 0.59. That means the ra t ios are not equal for allterms, and so there 'is no commo n ra tio. That means E is false.

The most co mmon incorrect answer is B. If you chose this, per haps you reasoned t ha t, because thesum of the first 4 terms is 38, the Slim of the firsr 5 terms cannot be anything less. Writing outmore te rm s makes it clear that this can be so. Or, you may have made an ar ith me tic m istake finding the fifrh term and so a rrived at the wrong sum.

If you c ho se C, and had t he correct sum for the first 5 terms, th en pe rhap s you ma de an arithmeti cmistake sub trac tin g 5 for each term, or you miscounted terms. If you have a different sum for Band also ha ve a different eigh th term than C, yo u know that there is some thin g amiss in your workand you should go back and t ry to find your mistake if there i s time.

If you cho se D, perhaps yOll di d not understand rhe concepts of common difference and comm o n ratio.

. .S

RThe ONLY Official Prep Guide AGa from the Makers of the ACT 391

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Chapter 4

392

Question 38. The correct answ er is G. The area of a parallelogram is given by bh wh ereis t he lengt h of the base an d h is clle height (here , th e distance betw een rhe bottom and th e top).

1 sometimes help s to pictur e this formul a geometr icall y

~

/ ,,o f ~ ~ ~ xa b

If you c ut off t he right tr ian gle from the right s ide, it fits exac tly onto th e left s ide to fo rm a r

tangle. The recta ng le has the same area as die parall

elogram.Th

e rec rang le's ar ea isbh

w hereb

is t he length of the base and h is the height. Thi s is even th e sa me formu la as for the para llelo gra m.Th e diff erence is that h is the lengt h of a side of the rectang le, but it is not the lengt h of a side ofthe parallelo gra m.

And so, for eith er the rectangle or t he parallelogra 'm, =10 coordinate units and h =6 coordinateunits, maki ng the area bh =10 . 6 =60 square coor d ina te un its.Th e distance fro m (0,0) to (3,6) is 2 + 62 '= I2 + 2 2 = 3..J5 coo rdinat e unit s. I f you cho se j ,yo u probably mul tipli ed thi s side lengt h b y the length of the ba se 10 coord inate unit s. Thi s is th estereotypical mistake when figur ing th e area of a para llelogram. Th e picture above shows w hy th eheigh t is the right thing to use, not the length of t he s ide.

If you c ho se H yo u may ha ve ca lculated t he leng th from (0 ,0) to (3,6) as2 _3 2 = ~ = 3.J3 a nd multiplied by the leng th of the ba se.

Question 39. The correct answer is D. The normal a mo unt of lead is 1.5 x 10-5 miJIigramsper liter. T his can be w ritten as 0.000015 mill igrams per lite r.

To day's level, 100 t imes t he normal a moun t, is 100 (1.5 x 10 -5 . = 1.5 x 10-5 X 102 =1.5 x 10-5 2 =1.5 X 10-3 mi lligra ms per lit er. Thi s can be written as 0.001 5 mill igrams per liter. This is larger th a nthe norma l amo un t .

I f you chose A, you lik ely added - 100 to the expo nent. Thi s a nswe r is 0.00000000 .. 000 000001 5milli g rams per liter, where there are 104 zer os between the decimal point and the 15. Thi s issma ller than th e normal a mount.

I f you chose B,.possib ly yo u mu ltiplied the expo nent 2 fro m 102 by th e exponent - 5 from the normal amount. This is 0.00000000015 mil ligrams pe r liter, wh ich i s sma ller than the normal a mount.

I f yo u cho se C, yo u co uld have subtr ac ted the expone nt 2 from the exp on ent -5 in the nor mal'amount. Th is g ives 0.0000 001 5 m illigra ms per lite r which is sma ller tha n the no rm a l amount .

\ ~ . . /from the Makers of the ACT


16/27

Pract ice ACT Tests

MATHEMATICS PRACTICETES 2 EXPLANATOR ANSWERS

Quest ion 40 . The correct answer is G. If y ll don t see a way to approa ch this problemright off, a usefu l ge nera l technique is to look for so me conc rete numbers that illu strate th7 con-dition s. One thought is to start c hec king numbers from 1,000 o lip to find a per fect squar e. Using

a calcu lator wo u ld be a goo d idea. JI,OOO 31.62, ,OOt 31.64, JI,002 3 1. 65, clear-ly. t hi s will take a while. A perfect squa re between 1,000 and 9,999 wo uld have to have its sq uareroot between ... 1,000 and J9 999 because the sq uar e root function is an increasing function. Thi s

will.help yo u find examples more quickly: 25 2 = 625 is roo low, 35 2 =1,225 is in the interval. Thesma llest perfec t square in this interval is 32 2 = 1,024. Because 100 2 = 10,000, the largest perfectsquare in this interval ha s to be 99 2 =91801. The sq uare root s of all of these perfec t squares areth e integers f rom 32 up to 99. All these sq uar e roots hav e 2 digits.

A qui cker way is to Hnd . .jl OOO 31.62 and J9 999 99.99, which is a range that co ntains a llpossi ble sq uare roots of the perfect sq ua res. Any integer in this ran ge (from 32 to 99) has 2 digits.

The most com mon incorrect answer is K. If you chose th is answer, pe rh aps you made a calcu la-tion erro r that led you to believe yo u 'd found a perfec t squa re between 1,000 and 9,999 whosesquar e root had 3 dig its. And thep you correctly found one w hose square root had 2 digits. Yourlog ic was correct. Some students probably choose K because they do no t know how to solve theproblem. Looking fo r concrete numbe rs would be a goo d st rat egy if you have time.

Question 41. The correct answer is B. Thi s prob lem is in the ge nera l form a + b) 2 wh ichis eq uiv ale nt to a2 + 2ab + b2 by the following derivation .

a + b)2 = a + b) a + b) = a . a + a . b + b . a + b . b a 1 + 2ab + b2

If a = t x and b = -y, then a + b)2 = t x - yj2. Th is is equiv a lent to a2 + 2ab + b2 =

tX)2 T 2{ tx) -y) + -y)2 = tx2 - xy + yl.ff you chose A, you probably squared the fi rst term and squ ar ed the second term. This is theste reotyp ica l erro r, an d it's someth ing that co llege math teachers (an d h ig h sc hoo l math teac herswant you to know not to do. Some thin g that mi ght hel p you reme m be r is to have a concre te exam-ple: 1 + 3)2 = 4 2 = 16, but 1 2 + 3 2 =, 1 + 9 = 10.

If yo u chose C, you probably did everything correctly except remembering to square the t to get t.

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17/27

Chapter 4

394 ,

M THEM TICS EXPL N TORY NSWERS

Question 42 The correct answer is F To calculate a matrix product, you go across eachrow in th e first matr ix and down eac h column in the second matrix. You multipl y the terms {ra in

a row by the correspond ing terms from th e co lumn, and you a d d a ll of those terms toget her forth e row-column comb inat ion and put the sum in that row-co lu mn of the result matrix. You do t h isfor eac h row-column co mb inat ion. .

That ex planation is a ll corre ct, but pre tty abst ract. Your mat h text might have so me exa mpl es thatyou can look ov er to mak e things more co ncrete.

In thi s case, the re is only 1 element in eac h row of the first matrix, and 1 element in each columnin th e secon d m a tr ix. The matrix product is:

[

a] [ a . 1 aO a (-1)] [ 0 0 0 ]2a [I 0-1]= 2a 1 2a0 2a(-I) = 2a 0-2a3a 3a 1 3a0 3 ( -1 3a 0-3a

The co mput at ions in G are correct, but the terms were put in the wrong places.

If you c ho se K, you may h ave reasoned t hat whatever role the 1 played in the second mati :ix, the- 1 would cancel it out, and t he 0 would not change that. If all of the terms were comb ined, thatwo uld be the case. But a ma t ri x can keep the terms sepa ra te.

Question 43 The correct answer is D. Because the ba se is a str aigllt line, th ese two anglemeasu res a dd up to 180 . In the langu age of algeb ra, th is ca n b e represe nt ed as 4x 6 2x) =180. Solving this for x gives 6x 6 = 180, then x 1 =30, then x = 29. This mak es one anglemeasu re 4(29) 6 = 122 degrees and the other angle meas ur e 2(29) = 58 degrees. (C hec k: Is 12 2

58 equa l to 180 ? Yes.

Th e problem asks for th e meas ure of the smallest of these two angles. T hat is 58.

If you cho se C, you pr o bably just stopped as soon as yo u found the value of x. The problem as ksfor so met hin g di fferent. The ot her popu lar incor rect a nswer is E.

Question 44 The correct answer is G. One way to solve this problem is to list out all thenumb ers in thi s range that you think might be prim e, then check to see if any of them facto r. Youprobably know th at you do n t hav e .to check the even numbe rs. T hat leaves the followin g list:

31 33 35 37 39 41 43 45 47 49

If one of these does factor, it will have a prime factor of at mo st .J49, whic h is 7. You have a lreadyeliminated all m\l ltip les of 2.If you elimin ate all multiple s of 3, 5, a nd 7, a nythin g left on the listis a prime number. The m ultiples of 3 on the list are 33, 39, a nd 45. You can eli minate 35 becau seit is a multiple of 5. You ca n eliminate 49 because it is a multiple of 7. Th e n all the numb ers

m i n i nname ly 3 1, 37, 41 43, and 47, are prime number s.

If you chose H you must have counte d a number .as prime th at really isn t prime . You might want tofigu re out whic h o ne tha t was. People who miss this problem t end to cou nt extra numbers as prim es.

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Practice ACT Tests

I .

,PRACTICE TEST 2 EXPLANATORY ANSWERS , 14i ~uestion 45 T h e correct answer is E The cotangen t of L in this right triangle is .rhe length of

the leg adjacent to the angle d ivided by the lengt h of the leg opposite the angle . That ratio is J-

Xl

. .f you chose D, yo u cho se the tangent of the angle. Answer cho ice C represent s the sine of theangle. Answer choice B is the cosecant of the angle. f you want to get problems like this correct ,you need to have a way to keep the trig functions straig ht,

Question 46. The correct answer is J.

Because no booths can be empty, imag ine 1 person sitti ng in each booth. That leaves 10 people standing around wait ing for you to tell them where to sit. How many booths can you fill up with another3 peop le each? We ll, you can get 3 gro ups of 3 people from the 10 who are st ill sta ndin g, with1 person left over . That means tha t you can fiU at most 3 boot hs with 4 people. The re w ill be 1 boothwith 2 people, and the other 6 boot hs w ill have 1 person, (Check: Are there 20 people? 3 4) 0 3)

1{2)6(

1)=

12 2 6=

20.Yes. Are

there 10 booths? 3 0 1 6=

10.Yes.)

1 you cho se K, you fill ed up 5 booths w it h 4 people each . But, you left the ~ h e r5 booths w ith110 ne s itting there. The problem specif ied that NO boot hs are empty. (T his is the most commonwrong answer,)

f you chose H, yo u may hav e just miscounted. Make sure you check your work, You might wa ntto draw a diagram for problems l ike this so that you can chec k your work easily.

from the Makers of the ACT, / 395


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Chapter 4

396

PR CTICE TEST 2

uestion 47. Th e correct answer is B. The area of the trapezoid is t b , + b h =

t 8 + 4)3. = 6 . 3 = 18 sq u?re inches. The area of the unshaded recta ngle is 4 . 3 = 12 sq uareinches. The t riangles and the e ~ n g l etoge the r for m the trapezoid , so the a rea of the trapezoid m inus

the area of the rectangle is the area of the triangles. And th is area is 18 - 12 =6 sq uare inche s.Another way to so lve this problem is to slide thl ; two shaded triangles together a nd calcu late t hearea of the new triangle. From the o r iginal figure, notice t h at the comb ined base of the trianglesis the amount left over from (he t rapezoid s botto m ba se when t he wi dth of .the rectangle isremoved. That is 8 - 4 = 4 inches. The figure below shows the co m bined tr iang les. Thei r comb inedarea is h = t 4)( 3) =6 square inch es .

4

The most commo n incorrect answer is D , which coul d come up in a variety ~ f ways. Many st udents have t ro uble finding the area of trapezo ids and resort to ca lculating bh as if t he t rapezo idwere a rectang le or pa ra llelogram. The figure below shows the t rapezo id inside a recta ngle wit hbase 8 inches and height 3 inches. T he rectangle has a rea bh w hich is clea rly larger t ha n the areaof the t rapezoid.

If you chose E, pe rhaps you calculated the area of the trapezoi d and stopped. Per haps you t houghtthe prob lem was askin g for t he area of the triangles com bined with the area of the rectangle , bu tit asks for the comb ined area of the triangles only.

The t r ia ngle on t he lef t looks like it is half of a square. If so, it has bas e 3 inches and heig ht 3 inch

es and t hen area t 3) 3) = i sq uare inches. If the other . tr iangle has the same area, the combi nedarea of the tr iangles is i + = 9 square inc hes. Perhaps you reasoned th is way if you c hose C.

. .

from the Makers of the ACT


20/27

Practice ACT Tests

MATHEMATICS

Question 48. The correct answer is G. Each of the corner triangles a re right triangles

because they share an a ng le with the sq uar e. Both legs of these right triangles are 6 inches longbecause they are half the length of the ,side of the s quar e. So, the hypotenuse of each of th ese tri

angles is J + 62 = ~ = 6.J2 inches. The perimeter of EFGH is made up of 4 of thesehypotenuses, so the perimeter of EFGH is 4 . 6.J2 = 24.J2 inches.

YO\l could have used the ratio of sides in an isosceles right triangle, rather than the Pyt ha go reantheo rem, to get the hypotenuse of the corne r triangles. The basic flow of the solution is the sam e.

I f you c hose F perhaps you noticed that the area of EFGH is hal f the area of ABCD That s a goo dobservation. That does not mean, though, that the pe r imeter of EFG H is ha lf the perimete r ofABCD If that were true, then each side of EFGH would be 6 inc hes long, and the co rn er u ian

g les would have 3 sides of length 6 inc hes. The triangle must then be equilateral a nd have a rightangle. That can t happen. (T he perimete r of a figure that is geometrica lly similar to th e origina land has area in the ratio 2:1 has perimeter in the ra t io .J2 :1.)

Question 49. The correct answer is A. To see this, start wi th the statement. J = I Iw h ich is tru e fo r all real values of y. If yo u wonder why there is an absolute value in this eq ua-

t ion, test y -5. Substitute x for y in the eq uati o n. The result i s J _X)l = I x I which is exp res-sion I

On the other hand, U is always positive or zero, a nd [ is always negative or zero, so the only t imethey are equal is when x is zero. For examp le, when x =1 1-1 1=1 but -I l l = 1. So II a nd IIIare not eq uivalent.The most common incorrect ansWer is C. Even though expressio ns II a nd III look a lot alike - theyha ve the exact same symbols in almost t he same order he y are not equivalent. Express io n II isnever negative and III is neve r po si tive.

I f you answered E, you likely saw that II and III are not equivalent. Expression I looks so muchdiff ere nt than eithe r of th e others that it is tempting t o just say it ca n t be the same. Oneapproach would be to test a few numbers in the exp ress ions. Be sure YOll test a pos itive numberand a negative number. I f yo u sub stitute these into I and into II correctly, you will f ind theymatch. Then, yo u should suspect that I and cou ld be equivalent and look for a way to convince you rself of th is.

f rom the Makers of the ACT 397


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Chapter 4

398

PR CTICE TEST 2 EXPL N TORY NSWERS

Question 50. he correct answer is K Statement 11 , that AB and EF are parallel, is true

because both are perpendicular to th e same other line, A DBecause L DEB is marked with a right an gle, DE is perpendicular to BE, and so G is tr ue.

H is true because the two a ngles, ~ C Band L FCE are ve rtical angles .

. Stateme nt J is true because th e angles in BAC are co ngruent to the a ngles in f-.EFC. First, LA iscongruent to LE FC because they are botn rig ht angles. Next, L ACB and L FCE are verti cal ang les.And third, the remaining angles have to have the sa me measure because the sum of the interiorangle measures in a ny trian g le is 180 0

Because a ll of the other stateme nt s are true, you shou ld expec t that statement K could be false.The following d iagram shows s uch a case. On ly if LE CF is c o n g r ~ e n tto LEDF can CE becongruent to DE.

E

D

8

Question 51. he correct answer is D. Thi s geome tric figure has 6 sides, but you are onlyg iven the length of 4 of thos e sides. One of the slickest ways to find the perimeter is to move two ofthe sides to form a rectangle with the same perimeter. The .drawing ~ o w shows t he new recta ngle

3

2

4 4

h e bottom of thi s rectang le is 4 inches long, and the right side is 3 inches l o ~ gSo, th e perimeteris 4 3 4 3 = 14 inches.

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Chapter 4

400

M THEM TICS

Question 53. T h e correct answer is D. Marshall made 24 ca lls on the first day. H e makes5 more ca lls eac h day than he had th e day befor e. That mea ns he m ade 29 ca lls on the second day.The tab le below show s the numb er of ca lls he mad e on each of th e 20 da ys.

day 1 2 3 4 5 6 7 8 9 10calls 24 29 34 '39 44 49 54 59 64 69

day 11 12 13 14 15 16 17 18 19 20ca lls 74 79 84 89 94 99 104 109 114 119

You cou ld try to add a ll of these up. You mi ght make a mist ak e, even with a calc ulacor. But, thatm et hod is str aig ht forwa rd and would work.

Another approach i s to not ice that the numb er o f ca lls for the first da y a nd the last da y add up to14 3, as do the number of calls for Da y 2 and Day 19 , as do the number fo r D ay 3 and Da y 18, asdo a ll th e other pai rs of days, working forward from the beginning and backwards from the end .Th ere are 10 pai rs of da ys, eac h with 143 ca lls. That is 1,430 ca lls for the 20 days .

A thir d approach relies on accurate ly re memb ering th e fo llow ing fo rm u la for the sum of a n arit hmetic series: S = a (1/ - l)dJ, whe re the first term in th e ser ies is Q, the common d ifferencebetwee n terms is d there are n terms in the series , and the sum is S . Subst ituting the ~ n t t i s

you know gives l [2 24 (20 - 1) ,5\ = 10[48 95 ] = 10[143] = 1,430 ca lls.I f yo u chose B, you may hav e calculated th e numb er of ca Us for Da y 10 a nd multipli ed this by 10.

I f yo u chose C, yo u may have broken the problem into 2 parts, the base 24 ca lls and the additionalcaUs. Thi s is a goo d st rategy. Th e add itional ca lls now are a geometric se ries (5 10 15 ... 95).Th is has 19 terms (remember that on Day 1 there are no additional ca lls). The sum of the ser ies is19 5;95 =950. G reat . Th en 950 24 = 974 ca lls. Unfortu nately, yo u 've only ad ded in t he ba se 24ca lls for 1 da y. Marshall made the base 24 calls on a ll 20 da ys. You d need co add 20 . 24 = 480instead of just 24. This gives 950 480 = 1,430 ca lls, which is cor rect.

Question 54. T h e correct answer is K. Thi s is ca lled a piecewise -defined unction becauseit is pieces of different functions put together to form a s ingl e function. When x S 1, the equatio nis {x) = - 2, so ({ 1) = P - 2 = - 1 and ({O) =02 - 2 = 2. (Actuall y, thi s information is enoughto eliminate all the g ra ph s except the co rrect c;>ne But y'ou might want ro read the rest of th is explanation anyhow.)

Th e graph of y =x 2 - 2 is a parabola, op en ing upward. Its g rapb is s ho w n on the next page. Theonly part of the graph that appl ies fo r th is piecewise defined funct ion is the parr where x :s; 1. Th erest of the g rap h is repr esent ed with dashes.

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Practic e CT Te s t s

-4 -2

y

64

2

-4

2 4 6 8 t O ~

W hen 1 < x < 5, t hen x) = x - 7, so 2) =2 - 7 = 5 and 4 ) =4 - 7 = 3. T his is .a str a igh tline. A gra ph is shown below, with the patts outsi de 1 < x < 5 show n wit h das hes an d ope ncircl es

y

42

8 lOx

When x 5, t he equ at ion is x) =4 - x so 5) =4 - 5 = 1 and 6) =4 - 6 = 2. T his is also astr a ight line. Its gra ph is s how n be low, w ith the pa rt outside x 5 s how n with d as hes

K puts a ll of these pieces together.

y

2

- 4 H _ r 2 t - + - f + 2 c - + 4 k ; : ~ t - 6 t + l 8 + + C [x-2-4

- 6

-8

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Chapter 4

402

Question 55 . The correct answer is D. The tJ b le below shows the vJlue of the cosine function at values of a that are the endpoints of th e intervals from eac h of the answer choices.

e 0 , , , 2 ,. J , T ncos a 1 II I 0 I - 1, -,approximation of cos a I 0.866 0.5 0 - 0.5 1

The value -0 .385 is betwee n 0 and -0.5, wh ich is say ing that th e val ue of cos a is betwee n

cos ~ and cos 23 Because cos a is con,tinuo us, that mean s there i s a value of a between I and23

that sa t isfies th e cond it ions of the problem.

If you chose C, you might have been looking for a place where cos a = +0.385 rathe r than --0.385.

If you chose A or E, YOll might have been looking for a place where si n a = 0.385.

Question 56. The correct answer is J. Th e equation x - 6a) x + 3b) = 0 has these two so lutions. You ca n .check this by substituting 6a in for X and substituting -3b in for x .) MultiplyingOll t this cquarion gives x 2 + -6a + 3b)x + -6a) 3b) = 0, which is the same asx + -6a + 3b)x - 18ab = O.

If you chose F, you ma y have gotten the initia l equatio n right , x - 6a) x + 3b) = 0, but then mu ltip lied incorrectly to get x) x) + -6a) 3a) = 0 .

Mo st of the other inco rrect answers could be du e to mist ak es with negat ive signs. G comes fromthe init ial equation x + 6a) x - Jb) =0, where the sig ns a re opposite what they shou ld be. If yousu bst itut e 6a in for x, you will not get zero on th e left side of the eq uation.) If you chose H , youmay have st a rted with the equat ion x - 6a) x - 3b) = O

Question 57. The correct answer is A. The midpo ints of the sides of t he square a re o n thecircle. These points hav e coord inates (0,3), .P,6), (6,3), and (3,0). The first poi nt, (0,3), satisfies A,but non e of r,he other eq uation s.

Alternately, because the cir cle centered at h,k) with radiu s r has th e equation x - h)2 + y - k)2 = ,you ca n find the equation by finding the center and radiu s. From the diagra m, you can see that thecenter of the circle is the same as the center of the sq uar e, which is (3,3). Also, the radiu s of thecircle is the distance from (0,3) to (3,3), which is 3 coord inate units. So, x - 3)2 + y - 3)2 = 2 is anequation of the circle.

One com mon mistake is to rem ember the equation of the c ircle incorrectly, wit h plu s s igns w hereth ere sho uld be minu s s igns (C) . Another commo n misrake is to not squ are the r ~ i u son the rights ide of the equa tio n (B). Or, some people do both of these things E). If you c ho se D, you may ha veused the d iameter on the rig ht side of the eq uation, or you may have u sed the larg est coordinatefrom the figur e a nd lI sed plus s igns instead of minus).

ohe ONL: OfficIal Prep GuIde from the Makers or the ACT


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Pract ic e ACT Tests


u s t on 58. The correct answer is J. Let Ll feet be the lengt h of Pendulum 1 and tl sec-on ds be the t ime fo r a comp lete swi ng of Pendulum 1. Let L2 and t 2 describe Pendulum 2 in thesame way. T he time fo r a comp lete sw ing of Pendulum 1 is tr iple t he time req ui red for a completeswi ng of Pen dulum 2. This m e ~ st ha t tl =3tr .The most common a p p r ~ c h e sto solvi ng problems like th is invo lve fin ding an equation that con-tains the two variables the question asks about, here L l and L 2 and solv ing for one variable orsolving for the ratio of the variables.

By the e q u ~ i on that re lates L to t the eq uation t l = 3t 2 becomes 2n Jfi = 3 . 2n.Jfi (which is an

equation t hat co nta ins bot h L l and L 2 D iv ~ d n gboth sides by 2n gives the equation Jfi= 3 Jfi

Squaring bot h sides gives

m= 9 M ultip lying both sides by 32 g ives L] = 9L

rSo, Pendulum l 's str ing is 9 times the length of Pen d ulum 2's string .

The most common inco rrect answer .is G. I f you chose that, you migh t have rea soned that if on eva riable triples, then any other variable m ust also tri p le. Thi s happens fo r so me function s, no tab lylinear func tions, but the fu nction in this prob lem is no t linear. Th e val ue of L m ust go up by a fac-tor of 9 so that the s q uare root wi ll go up by a factor of 3.

If you c hose H, you may hav.e rea soned that, in order for t he square roo t to go up by a factor of3, the qua ntity un der the square root must go up by a factor of 6. That wou ld mean that L wouldgo up by a factor of 6. This turns o ut not to be eno ugh, because if L goes up by a facror of 6, thesquare root will o nly go lip by a factor of J6, w h ich is less tha n 3.

uestion 59. The correct answer is A. By p ropert ies of2 log, xy).= 2( log, x log,yl = 2(, II .

logarit hms, log xyF =. .. If you c hose D, yo u may have done aU of the ste ps above correc tl y, except for t h inking that

loga xy) = log x' Jogny (us ing mul ti plica tion rather th an add ition on t he right side).

Answer C can come from t hinki Qg that loga . ~ y ) 2= loga X2y l) = logaXl . io's., y = 2 log x . 2 Jog y.T h ~ first and last ste ps are correct, but not t he middle one:

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Chapter 4

MATHEMATICS

Question 60 The correct answer is F Let d be Jennifer's distance in 1990. l er distance in1991 would be 1 .1 . d. And her distance in 1992 would be 1.2 (1.1 . d , which simplifi es to1.32 . d This represents an increase of 32 from 1990 to 199 2.Th e most common incorrect answer is G. I f you chose thi s, you p robably added the 10 and the20 to get 30 . This would work fine if the perce nt s were percents of the sa me ciling. But thefirst increase is a perce nt of the 19 90 distance while th e seco nd i s a percent of the 1991 distance.

One good strategy for investigating prob lems like this is t o choo se a specific numb er to representthe initi a l distance. Say you choo se 10 feet for her 1990 distan ce. The 1991 distance wou ld b e theoriginal 10 feet plus a 10 increase, which is 1 more foot, for a total of 11 feet. To find the 1992dista nce, tak e the 11 feet and add 20 of this amount, which is 2.2 feet. Then, the 1992 distanceis 11 2.2 = 13.2 feet. The percent increase from the original 10 feet is 1 3 ~o I O 1 =(0.32 )(100 ) = 32 . Thes e are the exact same operations as were done in the first method, but cilecalculations are done with concrete numbers rat her than abstract variables.

act 2 explanatory answers - math

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