QUANTITATIVE REASONING

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1 Quntittive Resoning QUNTITTIVE RESONING The quntittive resoning section tests your bility to use numbers nd mthemticl concepts to solve quntittive problems, nd your bility to nlyze dt presented in different wys, such s tble or grph form. This section requires only bsic knowledge of mthemtics (the mteril studied up to 9th-0th grdes in most Isreli high schools). Severl types of questions mke up the quntittive resoning section: Questions nd Problems, Grph or Tble omprehension, nd Quntittive omprisons (exmples of ech type pper lter on in the Guide). Questions nd Problems: These re multiple-choice questions ( question followed by four possible responses). They cover vriety of subjects, such s distnce problems, work problems, combintoril nlysis, probbility, equtions, geometry nd so on. Some re non-verbl questions in which the problem is presented numericlly; others re verbl questions, which require tht the problem be trnslted into mthemticl terms; other questions del with chrcteristics of geometricl figures, such s re, ngles nd so on. Grph or Tble omprehension: These re multiple-choice questions which relte to informtion ppering in grph or tble. tble presents numericl dt rrnged in columns nd rows. grph presents dt in grphic form, such s curve or br chrt. There re two min types of questions: - Questions involving the reding of dt, in which you re sked to find informtion ppering in the grph or tble. - Questions in which you re sked to mke vrious inferences bsed on the dt ppering in the grph or tble. Quntittive omprisons: These questions cover vriety of topics. They consist of pirs of quntities; in some cses dditionl informtion is provided. In ech question, you re sked to decide, on the bsis of the quntities nd the dditionl informtion (if provided), whether one of the quntities is lrger thn the other, whether the two quntities re equl, or whether there is not enough informtion to determine the reltionship between the two quntities.

2 Quntittive Resoning In generl, ll questions of given type re rrnged in scending order of difficulty. In other words, the esier questions, requiring less time to solve, pper first, with the questions becoming progressively more difficult nd requiring more time to solve. The figures ccompnying some of the questions re not necessrily drwn to scle. Do not rely solely on the figure's ppernce to deduce line length, ngle mesure nd so forth, unless these re specified in the figure (or in the question itself). ut if line in figure ppers to be stright, you my ssume tht it is, in fct, stright line. pge of Symbols nd Formuls ppers t the beginning of ech quntittive resoning section. This pge contins instructions, generl comments nd formuls, which you my refer to during the test. The pge of Symbols nd Formuls lso ppers on p. 5 of the Guide nd in the quntittive resoning sections of the prctice test. You should fmilirize yourself with its contents prior to tking the test. Pges 6-58 contin review of bsic mthemticl concepts, covering much of the mteril upon which the questions in the quntittive resoning sections re bsed. The ctul test my, however, contin some questions bsed on mthemticl concepts nd theorems tht do not pper on these pges. Pges contin exmples of different types of questions, ech followed by detiled explntion. 4

3 Quntittive Resoning This section contins 5 questions. The time llotted is 5 minutes. This section consists of questions nd problems involving quntittive resoning. Ech question is followed by four possible responses. hoose the correct nswer nd mrk its number in the pproprite plce on the nswer sheet. Note: The words ppering ginst gry bckground re trnslted into severl lnguges t the bottom of the pge. Generl omments bout the Quntittive Resoning Section * The figures ccompnying some of the questions re provided to help in nswering the questions, but re not necessrily drwn to scle. Therefore, do not rely on the figures lone to deduce line length, ngle mesure, nd so forth. * If line in figure ppers to be stright, you my ssume tht it is in fct stright line. * When geometric term (side, rdius, re, volume, etc.) ppers in question, it refers to term whose vlue is greter thn 0, unless stted otherwise. * When ( > 0) ppers in question, it refers to the positive root of. Symbols nd Formuls. The symbol represents 90 (right) ngle. The symbol «represents the ngle formed by line segments nd. b mens is prllel to b. b mens is perpendiculr to b.. Zero is neither positive nor negtive number. Zero is n even number. One is not prime number.. Percentges: % of x is equl to x 00 $ 4. Exponents: For every tht does not equl 0, nd for ny two integers n nd m -. n = n b. m + n = m n c. m n m n = ^ h (0 <, 0 < m) d. n m = ( n ) m 5. ontrcted Multipliction Formuls: ( ± b) = ± b + b ( + b)( b) = b distnce 6. Distnce Problems: time = speed (rte) mount of work 7. Work Problems: = output (rte) time 8. Proportions: If D E F then = DE EF nd DE = DF 9. Tringles:. The re of tringle with bse of length nd ltitude to the bse of length h is $ h b. Pythgoren Theorem: In ny right tringle, s in the figure, the following lwys holds true: = + c. In ny right tringle whose ngles mesure 0, 60 nd 90, the length of the leg opposite the 0 ngle is equl to hlf the length of the hypotenuse. leg hypotenuse leg 0. The re of rectngle of length nd width b is b. The re of trpezoid with one bse, the other bse b, nd ltitude h ] is + bg$ h. The sum of the internl ngles of polygon with n sides is (80n 60) degrees. In regulr polygon with n sides, ech internl ngle mesures b 60 80n l = b l degrees. n n. ircle:. The re of circle with rdius r is r ( =.4...) b. The circumference of circle with rdius r is r c. The re of sector of circle with centrl x ngle of x is π r $ ox (Rectngulr Solid), ube:. The volume of box of length, width b nd height c is b c b. The surfce re of the box is b + bc + c c. In cube, = b = c 5. ylinder:. The lterl surfce re of cylinder with bse rdius r nd height h is r h b. The surfce re of the cylinder is r + r h = r(r + h) c. The volume of the cylinder is r h 6. The volume of cone with bse rdius r π r $ h nd height h is x 5

4 Quntittive Resoning REVIEW OF SI MTHEMTIL ONEPTS SYMOLS elow is list of commonly used symbols tht my pper on the test. Symbol b b Mening stright lines nd b re prllel stright line is perpendiculr to stright line b 90º ngle (right ngle) «x = y x y x < y x y the ngle formed by sides nd x equls y x does not equl y x is smller thn y x is smller thn or equl to y 0 < x, y both x nd y re greter thn 0 x = ± x x : y x my be equl to or to (-) the bsolute vlue of x: if 0 < x, then x = x if x< 0, then -x = x 0 = 0 the rtio of x to y TYPES OF NUMERS Integer: n integer is number composed of whole units. n integer my be positive or negtive; 0 is lso n integer. For exmple:..., -4, -, -, -, 0,,,, 4,... Non-integer: number tht cnnot be expressed in whole units. For exmple:.7, 6 onsecutive numbers: Even number: Odd number: Integers tht follow in sequence in differences of. For exmple, 4 nd 5 re consecutive numbers; (-) nd (-) re lso consecutive numbers. In generl, if n is n integer, then n nd (n + ) re consecutive numbers. n integer which, when divided by, produces n integer (in other words, it is evenly divisible by ). Note tht bsed on this definition, 0 is n even number. In generl, if n is n integer, then n is n even number. n integer which, when divided by, produces non-integer (in other words, it is not evenly divisible by ). In generl, if n is n integer, then n+ is n odd number.

5 Quntittive Resoning Prime number: Reciprocl numbers Opposite numbers n integer tht is evenly divisible by only two numbers itself nd the number. For exmple, is prime number becuse it is evenly divisible only by nd. Note tht is not prime number. pir of numbers which, when multiplied, equl. Exmples: For 0, b 0 nd re reciprocl numbers; $ = b nd b b re reciprocl numbers; $ = b pir of numbers whose sum equls zero. For exmple, nd (-) re opposite numbers. In other words, (-) is the opposite number of ( + (-) = 0). RITHMETIL OPERTIONS WITH EVEN ND ODD NUMERS even + even = even odd + odd = even odd + even = odd even even = even odd odd = even odd even = odd even odd = odd even even = even odd odd = odd odd even = even There re no similr rules for division. For exmple, the quotient of two even numbers my be odd b 6 = l, even b 4 = l, or non-integer b 6 4 = l. DIVISORS ND MULTIPLES Fctor (Divisor): The fctor of positive integer is ny positive integer tht divides it evenly. For exmple, the numbers,,, 4, 6, 8, nd 4 re fctors of 4. 7

6 Quntittive Resoning ommon Fctor (ommon Divisor): common fctor of x nd y is number tht is fctor of x nd lso fctor of y. For exmple, is common fctor of both 4 nd 0. Prime Fctor (Prime Divisor): prime fctor is number tht is fctor (divisor) of some other number nd is itself prime number. For exmple, nd re the prime fctors of 4. ny positive integer (greter thn ) cn be written s the product of prime fctors. For exmple, 4 = = Multiple: multiple of n integer x is ny integer tht is evenly divisible by x. For exmple, 6, nd 88 re multiples of 8. MTHEMTIL OPERTIONS WITH FRTIONS Reduction: When the numertor nd the denomintor of frction hve common fctor, ech cn be divided by the sme number, nd the resulting frction is equivlent to the originl frction with smller 6 denomintor. For exmple, if we divide the numertor nd the denomintor of by 4, the result 4 is b 6 4 = l. Multipliction: To multiply two frctions, multiply the numertors by ech other nd the denomintors by ech other. For exmple: Or in generl: 5 $ 5 0 $ 7 = $ 7 = b $ c d $ c = b$ d Division: To divide number (integer or frction) by frction, multiply the number by the reciprocl of the divisor. (The reciprocl of b is b ) 8 $ 8 6 For exmple: : 5 8 = $ 5 = = 5$ 5 Or in generl: b c d $ d = : = $ = c b d b c b$ c d To multiply or divide n integer by frction, the integer cn be regrded s frction whose denomintor is, for exmple: = 8

7 Quntittive Resoning ddition nd Subtrction: To dd or subtrct frctions, they must be converted into frctions tht hve common denomintor. common denomintor is number tht is evenly divisible by the denomintor of ech of the frctions. fter finding suitble common denomintor, ech of the frctions must be converted into frction tht hs this common denomintor. To do so, multiply the numertor nd denomintor of ech of the frctions by the sme integer, so tht the number obtined in the denomintor will be the number tht ws chosen to be the common denomintor. Since the numertor nd denomintor re multiplied by the sme number, the frction hs ctully been multiplied by, nd its vlue hs not chnged. fter converting the frctions so tht they hve common denomintor, dd or subtrct the new numertors tht were obtined, nd reduce to lowest terms where possible. 5 For exmple, to solve the problem: is possible common denomintor, since it is evenly divisible by the denomintors of ech of the frctions: 4 : 4 = 6, 4 : 6 = 4, 4 : 8 = We will now convert ech of the frctions into frctions with this common denomintor: To convert 4 into frction whose denomintor is 4, multiply the numertor nd the denomintor $ 6 8 by 6: 4$ 6 = 4 To convert 6 into frction whose denomintor is 4, multiply the numertor nd the denomintor $ 4 4 by 4: 6$ 4 = 4 5 To convert 8 into frction whose denomintor is 4, multiply the numertor nd the denomintor 5$ 5 by : 8$ = Next, dd up only the numertors: = 4 = 4 Percentges Percentges re specific cse of frctions: % of x is 00 $ x. In these questions, convert the percentges to hundredths, nd solve s in norml frction problems. Exmple : Wht is 60 percent of 80? (or: Wht is 60% of 80?) Insted of 60 percent, substitute 60 hundredths, express the question in mthemticl terms, nd solve it s you would norml multipliction of frctions: $ $ 80 = 00 = 6$ 8 = 48 Thus, 60% of 80 is 48. Exmple : Joe hd to py 5 shekels tx on the 50 shekels tht he erned. Wht percent is the tx? The question is ctully "Wht percent of 50 is 5?" onvert the question into mthemticl expression: x 00 $ 50 = 5 nd solve the eqution for x: x = 5 Thus, x = 0. In other words, 5 is 0% of 50, which is the percentge of the tx. 9

8 Quntittive Resoning For questions tht involve chnge expressed s percentge, convert the question into one of the two generl formts presented in exmples nd (wht is x percent of y, or wht percent of y is x), nd solve s frction problem. Exmple : Exmple 4: The price of n item tht cost 80 shekels ws rised by 5%. Wht is the new price? Questions deling with chnge expressed s percentge generlly involve percent of the originl price unless otherwise specified. Since 5% ws dded to the old price, the new price is 5% of the old price (00% + 5%). Therefore, you must clculte wht 5% of 80 is (s in exmple ). 5 Substitute hundredths for percent nd solve 00 $ The new price is 00 shekels. The chnge in the price of certin item is given, nd you re sked to clculte the chnge s percentge. For exmple, the price of n item dropped from 5 shekels to shekels. y wht percentge did the price drop? The difference in the price is shekels out of 5 shekels. You hve to clculte wht percent of 5 is (similr to exmple ). onvert the question into mthemticl expression: x 00 $ 5, nd solve the $ 00 eqution for x: x 0 5 Thus, the price dropped by 0%. Rtio The rtio of x to y is written s x : y. For exmple, the rtio between the number of pirs of socks nd the number of shirts tht Eli hs is :. In other words, for every pirs of socks, Eli hs shirts. Stting it differently, the number of socks tht Eli hs is greter thn the number of shirts tht he hs. rithmetic Men The rithmetic men (verge) of set of vlues is the sum of the vlues divided by the number of vlues. For exmple, the verge of the set of vlues,, 5, 0, nd is 8 becuse If the verge of set of vlues is given, their sum cn be clculted by multiplying the verge by the number of vlues. Exmple: Dnny bought 5 items t n verge price of 0 shekels. How much did Dnny py for ll of the items? If we multiply the verge by the number of items, we will obtin 0 5 = 50. Thus, Dnny pid totl of 50 shekels for ll of the items tht he bought. 40

9 Quntittive Resoning In generl, the term "verge" will be used in the questions rther thn "rithmetic men." weighted verge is n verge tht tkes into ccount the reltive weight of ech of the vlues in set. Exmple: Robert's score on the mid-term exm ws 75, nd his score on the finl exm ws 90. If the weight of the finl exm is twice tht of the mid-term exm, wht is Robert's finl grde in the course? The set of vlues in this cse is 75 nd 90, but ech hs different weight in Robert's finl grde for the course. The score of 75 hs weight of nd the score of 90 hs weight of. To clculte the weighted verge, multiply ech score by the weight ssigned to it, nd divide by the sum of the weights: Thus, Robert's grde in the course is 85. $ 75+ $ 90 + = 85. This clcultion is identicl to the clcultion of simple verge of the three numbers 75, 90 nd 90. POWERS ND ROOTS Rising number to the nth power (n is positive integer) mens multiplying it by itself n times. For exmple: = Or in generl: 444 $... $ $ 444 = n n times The expression n is clled power; n is the exponent; nd is the bse. positive number rised to the 0th power equls. Thus, for ny 0, 0 =. When number is rised to negtive power, the result is the sme s tht obtined by rising the - reciprocl of the bse to the opposite power. For exmple: = b l = $ $ = 8 n - n Or in generl: = b l = n n The nth root of positive number, expressed s, is b, which if rised to the nth power, will give s follows: For exmple: 6 = 4, becuse 4 = = 5, becuse 5 = 5 8 =, becuse 4 = 8 It should be stressed tht when (0 < ) ppers in question, it refers to the positive root of. When the root is not specified, squre (nd-order) root is intended, for exmple, 8 = 8 = 9. root cn lso be expressed s power in which the exponent is frction. This frction is the reciprocl of the order of the root, = n ] 0 < g. n 4

10 Quntittive Resoning sic rules for opertions involving powers (for ny n nd m): Multipliction: To multiply powers with the sme bse, dd the exponents: m n = m+n Division: To divide power by nother power with the sme bse, subtrct the exponent in the m ] m ng denomintor from the exponent in the numertor: Note: When the powers do not hve the sme bse, the exponents cnnot be dded or subtrcted. n Rising to power: n To rise power to power, multiply the exponents: _ m i ] Rising product or quotient to power: m$ ng ( b) m = m b m ; m m b l b m b Since roots cn lso be expressed s powers, the lws for solving problems involving powers cn lso be pplied to roots. m For exmple, to clculte the product $ n ] 0 g, express the roots s powers: m n m n $ $ The next step is the sme s when multiplying powers; in other words, dd the exponents: ` $ m n m n j elow re number of bsic rules tht pply to inequlities involving powers: If 0 < b < nd 0 < n then b n < n If 0 < b < nd n < 0 then n < b n If < nd m < n then m < n If 0 < < nd m < n then n < m ONTRTED MULTIPLITION FORMULS To multiply two expressions enclosed in prentheses, ech of which is the sum of two terms, multiply ech of the terms in the first expression by ech of the terms in the second expression, then dd the products. For exmple: ( + b) (c + d) = c + d + bc + bd This generl formul cn be used for finding the product of ny two expressions, but to sve time, you might wnt to memorize some common formuls: 4

11 Quntittive Resoning ( + b) = ( + b) ( + b) = + b + b ( b) = ( b) ( b) = b + b ( b) ( + b) = b OMINTORIL NLYSIS The Number of Results in Multi-Stge Experiment The number of possible results of n experiment consisting of severl independent stges (tht is, they do not ffect ech other) is the product of the number of possible results in ech stge. For exmple, if we toss die nd then toss coin, wht is the number of possible results of this experiment? The number of possible results of tossing die is 6, nd the number of possible results of tossing coin is. Thus, the number of possible results of this experiment is 6 =. One of the possible results is the number on the die nd tils on the coin. It mkes no difference whether we first toss the die nd then toss the coin, or toss both t the sme time. In either cse, there re possible results. Ordered Smples n ordered smple is one in which the order of the results obtined in multi-stge experiment is importnt. Exmple: bsket contins 9 slips of pper, numbered through 9. If slips of pper re drwn t rndom from the bsket one fter nother, nd their numbers written in row, three-digit number will be obtined. How mny different numbers cn be obtined in this wy? To nswer the question, we hve to know the smpling method tht is being used. In ny cse, the order in which the results re obtined is importnt; for exmple, the number is different from the number.. Smpling with replcement: Ech slip of pper is replced in the bsket fter it is drwn, mking it possible for it to be drwn gin. The number of possible numbers tht cn be obtined ech time slip of pper is drwn from the bsket is 9. Therefore, the number of three-digit numbers tht cn be formed is = 79. b. Smpling without replcement: The slips of pper tht were drwn from the bsket re not replced. The number of possible numbers tht cn be obtined when the first slip is drwn is 9; when the second slip is drwn, only 8 (since one slip hs lredy been withdrwn from the bsket); nd when the third slip is drwn, 7. Thus the number of possible numbers is = 504. In generl, the number of possibilities for creting n ordered row of r items out of set of n items ( out of 9 in the bove exmple) is:. n r, if ech item cn be drwn more thn once (smpling with replcement). b. n (n )... (n r + ), if ech item cn be drwn no more thn once (smpling without replcement). Number of Possible rrngements (Permuttions) of n Ordered Smple The number of different possible rrngements of the 9 slips of pper, i.e., the number of possibilities for creting n ordered row of ll 9 slips of pper, with ech slip ppering only once (n = r), equls = 6,880. 4

12 Quntittive Resoning In generl, if n is the number of items in set, then the number of possible rrngements is... (n ) n. This number is written s n!, nd is clled "n fctoril." Non-Ordered Smples If the order of the results obtined in multi-stge experiment is not importnt, the smple is nonordered smple. The number of non-ordered smples equls the number of ordered smples divided by the number of possible rrngements. Exmple: bsket contins 9 pens, ech of different color. Three pens re drwn t rndom from the bsket nd not replced. How mny smples (sets) of different colored pens cn be obtined? The number of ordered smples is = 504. The number of possible rrngements (in ech smple) is = 6. The number of non-ordered smples is = 84. PROILITY Probbility theory is mthemticl model for phenomen (experiments) the occurrence of which is not certin. Such situtions cn hve number of possible scenrios or outcomes. Ech possible outcome is clled "simple event," nd the collection of outcomes n "event." (For the ske of brevity, we will use the term event to men simple event.) Ech event is ssigned number from 0 to, which reflects the probbility (likelihood) tht the event will occur. The higher the probbility, the greter the chnce the event will occur. n event tht is certin to occur hs probbility of, nd n event tht hs no possibility of occurring hs probbility of 0. Sometimes, ech of the possible outcomes of prticulr experiment hs n equl probbility (in other words, ech of the simple events hs n equl probbility). Exmples of experiments of this type The tossing of coin: The probbility of "heds" coming up is equl to the probbility of "tils" coming up. This probbility is. The tossing of die: The probbility of obtining ech of the numbers ppering on the fces of the die is 6. These re cses of tossing fir die/fir coin. The rndom removl of bll from bg contining 5 blls of equl size: The probbility of rndomly removing ech of the blls is 5. When ll possible outcomes hve n equl probbility, the probbility of n outcome occurring is clculted s follows: The number of possible outcomes of prticulr event, divided by the totl number of possible outcomes of the experiment (phenomenon). For exmple, the probbility tht in tossing single die we will obtin the event "the outcome is less thn or equl to " is 6 or, becuse this event hs possible outcomes (outcomes, nd ), nd the experiment of tossing die hs totl of 6 possible outcomes. 44

13 Quntittive Resoning The probbility tht two events will occur When two events occur t the sme time or one fter the other, two situtions re possible:. The events re independent, i.e., the probbility of one event occurring is not ffected by the probbility of the other event occurring. The probbility of both events occurring is equl to the product of the probbilities of ech individul event occurring. For exmple, in tossing two fir dice, the probbility tht number tht is less thn or equl to will turn up twice is equl to the product of the probbilities of number tht is less thn or equl to turning up in ech of the tosses, since the outcome of tossing one die does not ffect the outcome of tossing the other die. This probbility is equl to $ 4.. The events re dependent, tht is, the probbility of prticulr event occurring is ffected by the occurrence of different event. In other words, the probbility of prticulr event occurring fter (or given tht) we know tht nother event hs occurred is different from the probbility of tht prticulr event occurring without such knowledge. The probbility of the event "the outcome is less thn or equl to " (we will cll this event ), given tht we know tht in tossing the die the event "outcome is even" hs occurred (we will cll this event ), is clculted s follows: The probbility of occurring is the number of outcomes in which both nd occurred (in the exmple, is the only outcome tht is both even nd less thn or equl to ), divided by the number of outcomes in which occurred (outcomes, 4 nd 6 re even). Therefore, the probbility is. This probbility is different from the probbility of event without knowledge of condition (which equls, s clculted previsouly). Distnce, Speed (Rte), Time The speed (rte) t which n object trvels is the distnce tht the object covers in unit of time. The formul for the reltionship between the speed, the distnce the object covers nd the mount of time it requires to cover tht distnce is: v s t where v = the speed (rte) s = distnce t = time ll possible reltionships between distnce, speed nd time cn be derived from this formul: s t v nd s = v t Using these reltionships, ny unknown vrible out of the three cn be clculted if the other two vribles re known. For exmple, trin trveled 40 kilometers t speed of 80 kilometers per hour. How long did the journey tke? You re given v (80 kph) nd s (40 km), nd you hve to determine t. Substituting the given informtion s 40 into the formul t v, we get t 80. Thus, the journey took hours. 45

14 Quntittive Resoning Meters cn be converted to kilometers nd seconds to hours, nd vice vers. There re,000 meters in every kilometer ( meter =, 000 kilometer). There re,600 seconds, which equl 60 minutes, in every hour ( second = hour)., 600 5, 000 speed of kilometer per hour is equl to speed of 8 meters per second (or, 600 meters per second). speed of meter per second is equl to speed of.6 kilometers per hour. Work (Output) Output is the mount of work per unit of time. The formul for the reltionship between output, mount of work nd the time needed to do the work w is p = t, where p = output (rte ) w = mount of work t = time ll possible reltionships between output, mount of work nd time cn be derived from this formul: w t = p nd w = p t This formul cn be used to clculte ny unknown of the three vribles if the other two re known. For exmple, builder cn finish building one wll in hours. How mny hours would be needed for two builders working t the sme rte to finish building 5 wlls? We re given the mount of work of one builder ( wll), nd the mount of time spent working ( hours). Therefore his output is of wll in n hour. Since the question involves two builders, the output of both is $ = We re lso given the mount of work which both builders re required to do 5 wlls. We cn 5 therefore clculte the mount of time they will need: t = 5 : = 5 $ = = 7. Thus, they will need 7 hours. PRLLEL (STRIGHT) LINES Prllel lines tht intersect ny two stright lines divide the stright lines into segments tht re proportionl in length. b c c Thus in the figure, = c, = d b nd d + b = c+ d. Other reltionships between the segments cn be deduced bsed on the b c d given reltionships. ngles n ngle is right ngle if it mesures 90. n ngle is n cute ngle if it mesures less thn 90. n ngle is n obtuse ngle if it mesures more thn

15 Quntittive Resoning djcent ngles The two ngles tht re formed between stright line nd ry extending from point on the stright line re clled djcent ngles. Together they form stright ngle nd their sum therefore equls 80. For exmple, in the figure, x nd y re djcent ngles; therefore, x + y = 80. y x Verticl ngles When two stright lines intersect, they form four ngles. Ech pir of non-djcent ngles re clled verticl ngles nd they hve the sme mesure. In the figure, x nd z re verticl ngles nd therefore hve the sme mesure, s do y nd w; in other words, x = z nd y = w. When stright line intersects two prllel lines (trnsversl), eight ngles re formed, s in the figure:, b, c, d, e, f, g, h. x y w z orresponding ngles orresponding ngles re ngles locted on the sme side of trnsversl nd on the sme side of the prllel lines. orresponding ngles hve the sme mesure (see figure). Thus, in the figure, = e, c = g, b = f, d = h b c d e f g h lternte ngles lternte ngles re locted on opposite sides of trnsversl nd on opposite sides of the prllel lines. lternte ngles hve the sme mesure. Thus, in the figure, c = f, d = e, = h, b = g. Other reltionships between the different ngles cn be deduced bsed on the given reltionships. For exmple: since c nd d re djcent ngles (c + d = 80 ), nd since c nd f re lternte ngles (c = f), then obviously, d + f = 80. Similrly, we cn prove tht c + e = 80, nd so on. TRINGLES ngles of Tringle The sum of the interior ngles of ny tringle is 80. In the figure, + + = 80. n ngle djcent to one of the tringle's ngles is clled n exterior ngle, nd it is equl to the sum of the other two ngles of the tringle. For exmple, in the figure, is the ngle djcent to, nd therefore = +. 47

16 Quntittive Resoning In ll tringles, the longer side lies opposite the lrger ngle. For exmple, in the figure on the previous pge, if < <, it follows tht side (which is opposite ngle ) is longer thn side (which is opposite ngle ), nd side is longer thn side (which is opposite ngle ). Medin of Tringle is line tht joins vertex of tringle to the midpoint of the opposite side. For exmple, in the tringle in the figure, D is the medin to side (therefore D = D). D ltitude of Tringle The ltitude to side of tringle is perpendiculr line drwn from vertex of the tringle to the opposite side. For exmple, in ech of the tringles in the figures, h is the ltitude to side. h h re of Tringle The re of tringle equls hlf the product of the length of one of the sides multiplied by the ltitude to tht side. For exmple, the re of ech tringle in the bove figures is $ h. Inequlity in Tringle In every tringle, the sum of the lengths of ny two sides is greter thn the length of the third side. For exmple, in the tringles in the bove figures ( + ) >. ongruent Tringles Two geometric figures re congruent if one of them cn be plced on the other in such wy tht they both coincide. ongruent tringles re specific cse of congruence. If two tringles re congruent, their respective sides nd ngles re equl. For exmple, tringles nd DEF in the figure re congruent. Therefore, = DE, = EF, = DF, nd =, =, =. There re 4 theorems tht enble us to deduce tht two tringles re congruent: () Two tringles re congruent if two sides of one tringle equl the two corresponding sides of nother tringle, nd the ngle between these sides in one tringle is equl to the corresponding ngle in the other tringle. E D F 48

17 Quntittive Resoning For exmple, the tringles in the figure re congruent if = DE, = DF nd =. (b) Two tringles re congruent if two ngles of one tringle re equl to the two corresponding ngles of nother tringle, nd the length of the side between these ngles in one tringle equls the length of the corresponding side in the other tringle. For exmple, the two tringles in the figure re congruent if =, = nd = DE. (c) Two tringles re congruent if the three sides of one tringle equl the three sides of the other tringle. (d) Two tringles re congruent if two sides of one tringle equl the corresponding two sides of the other tringle, nd the ngle opposite the longer of the two sides of one tringle is equl to the corresponding ngle in the other tringle. For exmple, the tringles in the figure re congruent if = DE, = DF, nd = (when > nd DE > DF). Similr Tringles Two tringles re similr if the three ngles of one tringle re equl to the three ngles of the other tringle. In similr tringles, the rtio between ny two sides of one tringle is the sme s the rtio between the corresponding two sides of the other tringle. For exmple, tringles nd DEF in the figure re similr. Therefore, DE = nd so on. DF D F It follows tht DE = =. DF EF E 40 TYPES OF TRINGLES n equilterl tringle is tringle whose sides re ll of equl length. For exmple, in the figure, = =. In tringle of this type, ll of the ngles re lso equl (60 ). If the length of the side of such tringle is, then its ltitude is nd its re is n isosceles tringle is tringle with two sides of equl length. For exmple, in the figure, =. The two ngles opposite the equl sides re lso equl. For exmple, in the figure, =. n cute tringle is tringle in which ll ngles re cute. 49

18 Quntittive Resoning n obtuse tringle is tringle with one obtuse ngle. right tringle is tringle with one ngle tht is right ngle (90 ). The side opposite the right ngle (side in the figure) is clled the hypotenuse, nd the other two sides re the legs (sides nd in the figure). ccording to the Pythgoren theorem, in right tringle the squre of the hypotenuse is equl to the sum of the squres of the legs. For exmple, in the figure, = +. This formul cn be used to find the length of ny side if the lengths of the other two sides re given. leg leg hypotenuse In right tringle whose ngles mesure 0, 60 nd 90, the length of the leg opposite the 0 ngle equls hlf the length of the hypotenuse. For exmple, the length of the hypotenuse in the figure is. Therefore, the length of the leg opposite the 0 ngle is. It follows from the Pythgoren theorem tht the length of the leg opposite the 60 ngle is In n isosceles right tringle, the ngles mesure 45, 45 nd 90, the two legs re of equl length, nd the length of the hypotenuse is times greter thn the length of the legs QUDRILTERLS qudrilterl is ny four-sided polygon. For exmple: TYPES OF QUDRILTERLS Rectngles nd Squres rectngle is qudrilterl whose ngles re ll right ngles. The opposite pirs of sides in rectngle re equl in length. The perimeter of the rectngle in the figure is + b or ( + b). The length of the digonl of rectngle is b (bsed on the Pythgoren theorem). b +b D The re of the rectngle (S) is the product of the lengths of two djcent sides. For exmple, in the figure, S = b. b squre is rectngle whose sides re ll of equl length. The perimeter of the squre in the figure is 4. The length of the digonl of the squre in the figure is D The re of squre is equl to the squre of the length of the side. For exmple, in the figure, S =. 50

19 Quntittive Resoning TRPEZOID trpezoid is qudrilterl with only one pir of prllel sides. The prllel sides re clled bses, nd the other two sides re clled legs. The bses of trpezoid re not equl, nd re therefore referred to s the long bse nd short bse. The ltitude of trpezoid is segment joining the bses of the trpezoid nd perpendiculr to them. The re of trpezoid is equl to the sum of the length of the bses multiplied by hlf the ltitude. For exmple, in the figure: The length of the long bse () is. The length of the short bse (D) is b. The length of the ltitude is h. The re of the trpezoid is S h $ ] b g. b D h n isosceles trpezoid is trpezoid whose legs re of equl length. For exmple, in the figure, = D. The bse ngles of n isosceles trpezoid re equl. For exmple, in the figure, D = D = = D =. In this type of trpezoid, if two ltitudes re drwn from the ends of the short bse to the long bse, rectngle nd two congruent right tringles re obtined. D right trpezoid is trpezoid in which one of the bse ngles is right ngle (see figure). D PRLLELOGRMS ND RHOMUSES prllelogrm is qudrilterl in which ech pir of opposite sides is prllel nd of equl length. For exmple, in the prllelogrm in the figure: D, D = D, D = b h b D The digonls of prllelogrm bisect ech other. s stted, ech pir of opposite sides in prllelogrm is of equl length. Therefore, the perimeter of the prllelogrm in the figure is + b. The re of prllelogrm equls the product of side multiplied by the ltitude to tht side. For exmple, the re of the prllelogrm in the figure is h. rhombus is qudrilterl whose four sides re ll equl. Ech pir of opposite sides in rhombus is prllel, nd it cn therefore be regrded s prllelogrm with equl sides. 5

20 Quntittive Resoning Digonls of Rhombus Since rhombus is type of prllelogrm, its digonls bisect ech other. In rhombus, the digonls re lso perpendiculr to ech other. Since ll of the sides of rhombus re of equl length, the perimeter of the rhombus in the figure is 4. re of Rhombus Since rhombus is type of prllelogrm, its re too cn be clculted s the product of side multiplied by the ltitude. For exmple, the re of the rhombus in the figure is h. In ddition, the re of rhombus cn be clculted s hlf the product of its digonls. $ D For exmple, the re of the rhombus in the figure is. KITE (DELTOID) kite is qudrilterl formed by two isosceles tringles joined t their bses. For exmple, kite D in the figure is composed of tringles D nd D ( = D, = D). D The digonl joining the vertices of the two isosceles tringles bisects the digonl tht is the bse of the two isosceles tringles nd is verticl to it. (For exmple, in the figure, bisects D nd D) b b The perimeter of the kite in the figure is + b. The re of kite equls hlf the product of the lengths of the digonls. For exmple, the re of the kite in the figure is $ D. REGULR POLYGON regulr polygon is polygon whose sides re ll of equl length nd whose interior ngles hve the sme mesure. Exmples: regulr pentgon is five-sided regulr polygon. regulr hexgon is six-sided regulr polygon. regulr octgon is n eight-sided regulr polygon. The size of the interior ngle of regulr polygon with n sides cn be 60c clculted using the formul b80c l n For exmple, the figure shows regulr hexgon. The size of ech of its 60c interior ngles is 0, becuse 80c 6 0c 5

21 Quntittive Resoning IRLE rdius is line segment tht joins the center of circle to point on its circumference. chord of circle is line segment tht psses though the circle nd joins two points on its circumference. dimeter of circle is chord tht psses through its center. The length of circle's dimeter is twice the length of its rdius. If the rdius of circle is r, the dimeter of the circle is r. n rc is the prt of the circle between two points on its circumference. The circumference of circle with rdius r is r. (The vlue of is pproximtely.4.) The re of circle with rdius r is r. Inscribed ngle n inscribed ngle is n ngle whose vertex lies on the circumference of circle nd whose sides re chords of the circle. Inscribed ngles intercepting the sme rc hve the sme mesure. For exmple, in the figure, ngles nd re inscribed ngles, both of which intercept rc ; therefore, =. n inscribed ngle tht lies on the dimeter of circle (tht is, on n rc whose length equls hlf the circle's circumference) is right ngle. entrl ngle centrl ngle is n ngle whose vertex is the center of the circle nd whose sides re rdii of the circle (in the figure, is centrl ngle). centrl ngle is twice the size of ny inscribed ngle tht intercepts the sme rc. For exmple, in the figure, is centrl ngle nd is n inscribed ngle, nd both intercept the sme rc. Therefore, =. rc Two points on the circumference of circle define two rcs. For exmple, in the figure, points nd define two rcs one corresponding to centrl ngle nd one corresponding to centrl ngle. The smller rc corresponds to, the smller of the two ngles. α The length of this rc is πr $ 60 (r is the rdius of the circle). Sector sector is the prt of circle bounded by two rdii nd n rc. The ngle between the two rdii is centrl ngle. For exmple, the shded region in the figure is the sector of circle with x centrl ngle. The re of the sector of the circle is π r $ 60 r xº r 5

22 Quntittive Resoning Tngent to ircle tngent is line tht touches the circumference of circle t only one point, the "point of tngency." The ngle formed by the rdius nd the tngent t tht point is right ngle. For exmple, in the figure, line segment is tngent to the circle with rdius r. Two Tngents to ircle Two tngents to circle tht intersect t (prticulr) point re lso clled two tngents tht originte t one point. r The length of ech tngent is the length of the segment tht joins the tngents' point of intersection nd the point of tngency. Tngents to circle tht originte t one point re equl in length. For exmple, in the figure, is the point of intersection, nd re the points of tngency, nd =. Polygon ircumscribing ircle polygon tht circumscribes circle is polygon whose sides re ll tngent to the circle. Polygon Inscribed in ircle polygon inscribed in circle is polygon whose vertices ll lie on the circumference of the circle. Inscribed Tringle Every tringle cn be inscribed in one nd only one circle (tht is, circle with the vertices of the tringle lying on its circumference). If the inscribed tringle is right tringle, the center of the circle tht circumscribes it is the midpoint of the tringle's hypotenuse. Qudrilterl Inscribed in ircle Not every qudrilterl cn be inscribed in circle. The sum of the opposite ngles of qudrilterl inscribed in circle lwys equls 80. For exmple, in the qudrilterl in the figure, + = 80 + = 80 54

23 Quntittive Resoning Qudrilterl ircumscribing ircle When qudrilterl circumscribes circle, the sum of the lengths of ech pir of opposite sides is equl. For exmple, in the qudrilterl in the figure, + c = b + d. When squre circumscribes circle, the length of the side of the squre equls the length of the dimeter of the circle (see figure). SOLIDS ox (Rectngulr Prism) nd ube box is three-dimensionl figure with six rectngulr fces. The box's three dimensions re its length, width nd height (, b nd c respectively, in the figure). c b The surfce re of box is the sum of the res of its fces. The surfce re of the box in the figure is b + c + bc + b + c + bc or b + c + bc. The volume (V) of box is the product of its length, width nd height. The volume of the box in the figure is V = b c. cube is box whose three dimensions re ll equl. ll of the fces of cube re equl in re. The re of ech fce of the cube in the figure is d. Therefore, the surfce re of the cube is 6d. The volume of the cube in the figure is V = d. d d d ylinder cylinder is three-dimensionl figure whose two bses re congruent circles on prllel plnes. In right cylinder, the line joining the centers of the circles is verticl to ech of the bses. h The lterl surfce re of cylinder with bse rdius of length r nd with height h is the product of the circumference of the bse multiplied by the height, tht is, r h. r The totl surfce re of cylinder is the sum of the res of the bses nd the lterl surfce. The re of ech bse is r nd the lterl surfce re is r h. Thus, the totl surfce re is r h + r = r (h + r). The volume of cylinder is the product of the re of one of the bses multiplied by the height, tht is, r h. 55

24 Quntittive Resoning one cone is figure formed by joining the points on the circumference of circle with point outside the plne of the circle. right cone is formed when the point outside the circle lies on line tht psses through the center of the circle nd is perpendiculr to the plne of the circle. πr h The volume of cone with bse rdius r nd height h is V $ =. h r Right Prism right prism is three-dimensionl figure whose two bses re congruent polygons on prllel plnes nd whose lterl fces re rectngles. The type of prism is defined by the number of sides of its bse. For exmple, tringulr prism hs three-sided bses, qudrngulr prism hs four-sided bses, nd so on (see figures). The height of prism is the length of the segment tht joins the bses nd is perpendiculr to them. It is the distnce between the bses of the prism. The lterl surfce re of the prism is the sum of the res of ll the lterl fces. The lterl surfce cn lso be clculted by multiplying the perimeter of the prism's bse by its height. The totl surfce re of prism is the sum of the lterl surfce re nd the res of the two bses. The volume of prism equls the re of one of the bses multiplied by the height. Pyrmid pyrmid is figure formed by joining the vertices of ny polygon to point outside the plne of the polygon clled the vertex or pex of the pyrmid. The polygon is clled the bse of the pyrmid. The lterl fces of the pyrmid re tringles. pyrmid is referred to by the number of sides of its bse. For exmple, tringulr pyrmid hs three-sided bse, qudrngulr pyrmid hs four-sided bse, nd so on (see figure). The height of pyrmid is the line segment extending perpendiculrly from the pyrmid's vertex to its bse. This is the distnce between the pyrmid's vertex nd bse (see figure). If S is the re of the pyrmid's bse nd h is the pyrmid's height, then the pyrmid's volume is V = S $ h h 56

25 Quntittive Resoning Edge The edge of three-dimensionl figure is the stright line formed where two fces meet. For exmple, box hs edges. The bold line in the pyrmid on the previous pge is one of its edges. NUMER LINE (XIS) number line is geometric representtion of the reltionships between numbers. * The numbers long the xis increse to the right. * The distnce between points long the xis is proportionl to the difference between the numericl vlues corresponding to the points. For exmple, the distnce between the points corresponding to vlues (-4) nd (-) is equl to the distnce between the points corresponding to vlues nd 5. rtesin oordinte System rtesin coordintes on coordinte plne hve two number lines (xes) tht re perpendiculr to ech other. The horizontl line is clled the x-xis nd the verticl line is clled the y-xis. The numbers long the x-xis increse to the right. The numbers long the y-xis increse upwrds. The xes divide the plne into four qudrnts, designted in the figure by Romn numerls I, II, III, IV. Ech point in the coordinte plne corresponds to pir of x nd y vlues. For exmple, the x-vlue of point in the figure is 4, nd its y-vlue is. The x-vlue of point in the figure is (-) nd its y-vlue is. It is customry to write the x- nd y-vlues of the points in prentheses, with the x-vlue to the left of the y-vlue, s follows: (x, y). For exmple, point is written s (4, ) nd point is written s (-, ). The x- nd y-vlues for point re sometimes clled the coordintes of tht point. The point in the plne corresponding to (0, 0) is the point of intersection of the two xes nd is clled the origin. ll points on line prllel to the x-xis hve the sme y-coordinte, nd ll points on line prllel to the y-xis hve the sme x-coordinte. For exmple, in the figure, line k is prllel to the y-xis. Thus, ll of the points on line k hve the sme x-coordinte. In the figure, x =.5 Line m is prllel to the x-xis. Thus, ll of the points on line m hve the sme y-coordinte. In the figure, y =.5 II 4 y III m y k 4 I IV x x 57

26 Quntittive Resoning Only one line cn be drwn through ny two points on plne. The prt of the line between the two points is clled line segment. If the line segment is prllel to the x-xis, its length is the difference (in bsolute vlue) between the x-coordintes of the points. For exmple, in the figure, line segment D is prllel to the x-xis. The x-coordinte of point is 4 nd the x-coordinte of point D is (-). The difference between the x-coordintes of the points is 5. Therefore, the length of line segment D is 5. y 4 D x If the line segment is prllel to the y-xis, its length is the difference (in bsolute vlue) between the y-coordintes of the points. For exmple, in the figure, line segment is prllel to the y-xis. The y-coordinte of point is 4 nd the y-coordinte of point is (-). The difference between the y-coordintes of the points is 7. Therefore, the length of line segment is y x If the line segment is not prllel to one of the xes (for exmple, line segment EF in the figure), its length cn be clculted using the Pythgoren theorem: Drw right tringle such tht the segment is the hypotenuse nd the legs re prllel to the x-xis nd the y-xis. The length of the leg prllel to the x-xis equls the difference between the x-coordintes of points E nd F (4 = ), nd the length of the leg prllel to the y-xis equls the difference between the y-coordintes of points E nd F ( = ). Using the Pythgoren theorem, we cn clculte the length of the hypotenuse: ( EF = + = 8 ) y 4 E F x 58

27 Quntittive Resoning QUESTIONS ND PROLEMS These questions cover vriety of topics, such s distnce problems, work problems, combintoril nlysis nd probbility, equtions, geometry, nd so on. Some re verbl questions which hve to be converted into lgebric expressions nd the solution given in numericl form; some re non-verbl questions tht lredy hve the formt of lgebric expressions; nd some del with chrcteristics of geometric shpes, such s re, volume, ngles, nd so on. elow re some smple questions, together with solutions nd explntions. Note: The exmples in the Guide re rrnged by type, but this is not the cse in the ctul exm. VERL QUESTIONS. driver covered third of the distnce from Hif to Eilt t speed of 75 kph. He covered fifth of the remining distnce in one hour, nd the rest of the distnce t speed of 80 kph. The distnce between Hif nd Eilt is 450 kilometers. If the driver hd driven the entire distnce t constnt speed, t wht speed would he hve driven so tht the journey from Hif to Eilt would tke exctly the sme mount of time? (kph = kilometers per hour) () 70 kph () 75 kph () 80 kph (4) None of the bove This question is mthemticl problem presented in verbl form; therefore the first step is to convert it into lgebric expressions. Strt by clerly defining wht you re sked to find: the speed t which to drive in order to cover the distnce between Hif nd Eilt in the sme mount of time tht it took the driver in the question. Therefore, this is distnce problem, nd the formul s v = t, which connects distnce, speed nd time cn be pplied since the distnce (s) is given, the time (t) cn be clculted, nd the speed (v) is the unknown tht you hve to find. The question provides the informtion tht the distnce between Hif nd Eilt is 450 kilometers. The totl mount of time needed by the driver to cover the entire distnce between Hif nd Eilt cn be clculted s follows: The distnce is divided into three segments. The time it took the driver to cover ech segment is s follows:. third of the distnce is 50 km, becuse 450 $ kilometers equls 50 kilometers. It took the driver two hours to cover this segment, becuse it tkes two hours to trvel 50 kilometers t speed of 75 kph b 50 = l. 75 b. fifth of the remining distnce is 60 kilometers, since the remining distnce is = 00 kilometers, nd 00 $ 5 kilometers equls 60 kilometers. The question provides the informtion tht the driver covered this segment of the journey in one hour. 59

28 Quntittive Resoning c. The rest of the distnce is 40 kilometers, since = 40. The driver covered this distnce in three hours, becuse it tkes three hours to trvel 40 kilometers t speed of 80 kph. Thus, the journey from Hif to Eilt took totl of 6 hours (two hours, plus one hour, plus three hours). y substituting the dt into the formul, it is now possible to compute the constnt speed t which it is necessry to drive in order to cover 450 km in 6 hours: t = 6, s = 450, s 450 v = t = 6 = 75. Thus, the speed is 75 kph, nd the correct response is ().. t the ge of 0 dys, bby elephnt ets 5 cndies. From this ge onwrds, its ppetite grows, nd ech dy it ets twice the number of cndies it te the previous dy. How mny cndies will the bby elephnt et t the ge of 4 dys? () 40 () 80 () 00 (4) 0 On the tenth dy, the bby elephnt ets 5 cndies. Ech dy fter tht it ets twice the number of cndies tht it te the previous dy. Thus, on the th dy it ets 0 cndies (5 ), on the th dy it ets 0 cndies (5 ), nd so on. In generl, if n is positive integer, then on dy (0 + n) the bby elephnt will et 5 n cndies. Thus, on the 4th dy it will et 80 cndies (5 = 5 4 = 80), nd the correct response is ().. resturnt offers different first courses nd 4 different min courses. In ddition to the first course nd the min course, it lso offers choice of soup or dessert. How mny different combintions of three-course mels cn be put together t this resturnt? () () 4 () 8 (4) 4 There re three possible choices for the first course, nd four different min courses tht cn be dded to ech first course chosen. Thus, there re 4 different combintions of first courses nd min courses. To ech of these combintions, either soup or dessert cn be dded. In other words, there re totl of different combintions of the three courses, which equls 4 different possibilities. The correct response is therefore (4). 60

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