1 Logic and Set Theory2.Pdf

February 1, 2018 Any calculator permitted on N.Y.S. Regents examinations may be used. The word “compute” calls for an exact answer in simplest form. 1 - Logic and Set Theory

1. Let P be the set of single-digit prime numbers and let O be the set of single-digit odd numbers. Compute the number of elements in O ∪ P .

2. In a survey of 120 third-graders, 80 of them liked the color blue and 60 of them liked the color red. If 50 of them liked blue and red, how many of the third-graders didn’t like red or blue?

February 2018 1 - Logic and Set Theory

1. Let P be the set of single-digit prime numbers and let O be the set of single-digit odd numbers. Compute the number of elements in O ∪ P . SOLUTION: 6 The set O ∪ P is {1, 2, 3, 5, 7, 9}, which has 6 elements.

2. In a survey of 120 third-graders, 80 of them liked the color blue and 60 of them liked the color red. If 50 of them liked blue and red, how many of the third-graders didn’t like red or blue? SOLUTION: 30 Together, there are 140 third-graders who liked blue and red, but 50 of those were double-counted, so there are only 140 − 50 = 90 of them who liked either red or blue (or both). Subtracting from 120, there are 120 − 90 = 30 third-graders who don’t like red or blue.

3. In Metropolis, there are Truthies (who always tell the truth) and Fibbies (who always lie). Three citizens of Metropolis make the following statements. Jenny says, “Kenny or Lenny is a Truthie (but not both).” Kenny says, “Jenny or Lenny is a Truthie (but not both).” Lenny says, “Both Jenny and Kenny are Fibbies.” Which of the three people are Fibbies? Note: List any member who is a Fibbie or write NONE if all three are Truthies. SOLUTION: Lenny (only) Suppose Lenny is a Truthie. Then, both Jenny and Kenny are Fibbies. Then, only Lenny is a Truthie, which makes Jenny a Truthie, and that’s a contradiction. So, Lenny is a Fibbie. Therefore, at least one of Jenny and Kenny is telling the truth. If Jenny is a Truthie, then exactly one of Jenny or Lenny is a Truthie, so Kenny is a Truthie also. Similarly, if Kenny is a Truthie, so is Jenny. Therefore, there is only one Fibbie, and that’s Lenny. February 2, 2017 Any calculator permitted on N.Y.S. Regents examinations may be used. The word “compute” calls for an exact answer in simplest form. 1 - Logic and Set Theory

February 2017 1 - Logic and Set Theory

1. Suppose the following four statements are true: If you study and you get lucky, then you will succeed. If you succeed, then you will be famous. Jimmy is not famous. Jimmy studied. Draw a valid conclusion that uses all of this information. SOLUTION: Jimmy did not get lucky. From the ﬁrst two sentences, one can conclude “If you study and you get lucky, then you will be famous.” Incorporating the third given sentence, one can conclude that “You study and you get lucky” is false. Because Jimmy did study, the only way that “You study and you get lucky” is false is if Jimmy did not get lucky.

2. In Mcmland, there are four kinds of people: Amoricans, Braths, Crews, and Dwurfs. All Amoricans are Braths, all Crews are Dwurfs, all Dwurfs are Amoricans, and all Crews are Braths. If none of the people groups are equal in number, which group is the smallest? SOLUTION: Crews All Amoricans are Braths, so the Braths are not the smallest. All Dwurfs are Amoricans, so the Amoricans are not the smallest. All Crews are Dwurfs, so the Dwurfs are not the smallest. The Crews are the smallest.

3. Students in the ninth grade at Springﬁeld High were polled to see if they were fans of the Mets and/or the Yankees. There are 70 students who do not cheer for the Mets. There are 95 students who do not cheer for the Yankees. There are 127 students who cheer for the Mets or for the Yankees (or both). There are 90 students who cheer for the Mets. Compute the number of students who cheer for the Mets but not the Yankees. SOLUTION: 62 Because there are 90 students who cheer for the Mets and 70 who do not, there are 160 students in the ninth grade at Springﬁeld High. Subtracting the 127 who cheer for either team, there are 33 students who do not cheer for either team. There are 95 students who do not cheer for the Yankees, so there are 95 − 33 = 62 who cheer for the Mets. February 4, 2016 Any calculator permitted on N.Y.S. Regents examinations may be used. The word “compute” calls for an exact answer in simplest form. 1 - Logic and Set Theory

1. Let P be the set of prime numbers less than 10. Let Q be the set of single-digit odd numbers. Compute the number of numbers in P ∪ Q.

2. At a party, there are 40 people. Of the 40 people at the party, 22 people are girls, 5 people are left-handed, and 2 people are left-handed girls. How many right-handed boys attended the party?

3. In the trunk of Juan’s car are four bags: a bag of apples, a bag of bananas, a bag of carrots, and a bag of dates. Together, the bag of apples and the bag of carrots weigh the same as the bag of bananas and the bag of dates do together. Together, the bag of apples and the bag of dates weigh more than the bag of bananas and the bag of carrots do together. The bag of bananas weighs as much as the bags of carrots and dates do together. Using A, B, C, and D to abbreviate the bags of apples, bananas, carrots, and dates, arrange their weights in order from greatest to least. Note: If you think the bag of apples weighs more than the bag of bananas, which weighs more than the bag of carrots, which weighs more than the bag of dates, write ABCD as your answer. Monroe County Math League Contest #5. SOLUTIONS

February 2016 1 - Logic and Set Theory

1. Let P be the set of prime numbers less than 10. Let Q be the set of single-digit odd numbers. Compute the number of numbers in P ∪ Q.

SOLUTION: 6 The set P = {2, 3, 5, 7}. The set Q = {1, 3, 5, 7, 9}. Therefore, P ∪ Q = {1, 2, 3, 5, 7, 9}, which has cardinality 6.

2. At a party, there are 40 people. Of the 40 people at the party, 22 people are girls, 5 people are left-handed, and 2 people are left-handed girls. How many right-handed boys attended the party?

SOLUTION: 15 At the party, G is the set of girls and L is the set of left-handed people. We know that the cardinality of G ∪ L is 22 + 5 − 2 = 25. The right-handed boys are the people not in G ∪ L, so there are 40 − 25 = 15 right-handed boys at the party.

SOLUTION: ABDC Swapping out dates for carrots makes the apple side heavier, so D ¿ C. The bag of bananas weighs more than the carrots and dates do together, so B ¿ D. Now, adding the sentences that come from the ﬁrst two problem statements – A + C = B + D and A + D ¿ B + C – we obtain 2A + C + D > 2B + C + D, which implies A ¿ B. Therefore, A ¿ B ¿ D ¿ C, so write down ABDC. February 5, 2015 Any calculator permitted on N.Y.S. Regents examinations may be used. The word “compute” calls for an exact answer in simplest form. 1 - Logic and Set Theory

1. Let A = {1, 3, 5, 7, 9} and B = {1, 4, 9}. Compute the number of elements of A ∪ B.

2. In a club, there are 40 people who are lefthanded or who are blondes (or both). There are 3 lefthanded blondes in the club. There are L left-handed people in the club. There are B blondes in the club. Compute L + B.

3. Aaron, Beth, and Carol were each given a positive integer. No one knew any other person’s number, but the three people were told that the sum of their three numbers was 16. Aaron could deduce that all three numbers were diﬀerent. Beth could only deduce that Carol’s number was diﬀerent than Aaron’s. When Aaron and Beth told Carol about their deductions, Carol could determine all three numbers. Compute the product of their three numbers. Monroe County Math League Contest #5. SOLUTIONS

February 2015 1 - Logic and Set Theory

1. Let A = {1, 3, 5, 7, 9} and B = {1, 4, 9}. Compute the number of elements of A ∪ B.

SOLUTION: 6 We have A ∪ B = {1, 3, 4, 5, 7, 9}, so the number of elements is 6.

SOLUTION: 43 We know that L ∪ B = L + B − (L ∩ B), so L + B = (L ∪ B) + (L ∩ B), and our answer is 40 + 3 = 43.

SOLUTION: 54 Aaron knows the three numbers are diﬀerent, so his number A is odd and greater than or equal to 9. Beth must also have an odd number because the other two may not be equal, but her number B must be less than or equal to 7. Therefore, possible triples (A, B, C) are (9, 1, 6), (9, 3, 4), (9, 5, 2), (11, 1, 4), (11, 3, 2), or (13, 1, 2). Since Carol knows which triple it is uniquely, her number can’t be 2 (which occurs in three possibilities) or 4 (which occurs in 2 possibilities). The numbers are 9, 1, and 6, whose product is 54. February 6, 2014 Any calculator permitted on N.Y.S. Regents examinations may be used. The word “compute” calls for an exact answer in simplest form. 1 - Logic and Set Theory

1. Suppose that Mrs. Vipper told Alan, “If you wash the windows, then I will give you $10.” Suppose also that Alan does not receive $10 from Mrs. Vipper. Compute the probability that Alan washed the windows.

2. Each of the four Teenage Mutant Ninja Turtles made a statement about the four of them, as follows. Michaelangelo: “Exactly one of us is lying.” Raphael: “Exactly two of us are lying.” Leonardo: “An odd number of us are lying.” Donatello: “An even number of us are lying.” How many of them were lying (with the rest telling the truth)? Note: there are TWO correct answers... give them BOTH...

3. Let T = {3, 6, 9, ··· , 3n, ··· 99}, F = {5, 10, 15, ··· , 5n, ··· 100}, and S = {7, 14, 21, ··· , 7n, ··· 98}. Compute the number of elements in T ∪ F ∪ S. Monroe County Math League Contest #5. SOLUTIONS

February 2014 1 - Logic and Set Theory

SOLUTION: 0 By using the contrapositive, we see that if Mrs. Vibber didn’t give Alan $10, then he didn’t wash the windows. Therefore, Alan absolutely did not wash the windows. The probability of an impossibility is 0.

SOLUTION: 2 or 3 (need both) Leonardo and Donatello cannot both be telling the truth, so at least one person is lying, and it is impossible for all four to be lying. Suppose that Leonardo is lying (that an even number of them are lying). That even number would have to be two. If that number were 2, then the two would be Michaelangelo and Leonardo. Suppose instead that Donatello is lying (that an odd number of them are lying). Then Raphael must also be lying, so if an odd number of them are lying, that number is 3. It is possible that 3 of them are lying if Michaelangelo is also lying. Our answers are 2 and 3.

3. Let T = {3, 6, 9, ··· , 3n, ··· 99}, F = {5, 10, 15, ··· , 5n, ··· 100}, and S = {7, 14, 21, ··· , 7n, ··· 98}. Compute the number of elements in T ∪ F ∪ S.

SOLUTION: 55 We have that |T | = 33, |F | = 20, and |S| = 14. We also have some double-counted elements, namely the 6 elements in T ∩ F , the 4 elements in T ∩ S, and the 2 elements in F ∩ S. There are no elements in T ∩ F ∩ S, since 3, 5, and 7 are relatively prime with product 3 · 5 · 7 = 105 > 100. Therefore, our answer is 33 + 20 + 14 − 6 − 4 − 2 = 55. February 7, 2013 Any calculator permitted on N.Y.S. Regents examinations may be used. The word “compute” calls for an exact answer in simplest form. 1 - Logic and Set Theory

2. Each person below made one true and one false statement about the order in which the five people spoke. Write the correct order in which they spoke as a sequence of five letters, from first to last. For example, you would write ABCDE if you think Adam spoke before Beth, who spoke before Carol, and so on. Adam: I was first. Beth was second. Beth: I was third. Adam was last. Carol: I was second. Diane was fourth. Diane: I was fourth. Ethan was first. Ethan: I was third. Carol was second.

3. Let {2} represent the set of all positive integer multiples of 2 that are less than or equal to 100. Let {3} represent the set of all positive integer multiples of 3 that are less than or equal to 100. Let {5} represent the set of all positive integer multiples of 5 that are less than or equal to 100. Compute the number of elements in {2} ∪ {3} ∪ {5}. Monroe County Math League Contest #5. SOLUTIONS February 2013 1 - Logic and Set Theory

1. Given the following two true statements: If I was on time for school, then my homework was handed in and my teacher was happy. My homework was not handed in. Multiple choice: Which of the following statements must be true? Answer A, B, C, or D. (A) I was on time for school. (B) I was not on time for school. (C) My teacher was happy. (D) My teacher was not happy. SOLUTION: B The contrapositive of the first given statement is “If my homework was not handed in or my teacher was not happy, then I was not on time for school.” Since the first part of the contrapositive is true (based on the second statement), we conclude that I was not on time for school. 2. Each person below made one true and one false statement about the order in which the five people spoke. Write the correct order in which they spoke as a sequence of five letters, from first to last. For example, you would write ABCDE if you think Adam spoke before Beth, who spoke before Carol, and so on. Adam: I was first. Beth was second. Beth: I was third. Adam was last. Carol: I was second. Diane was fourth. Diane: I was fourth. Ethan was first. Ethan: I was third. Carol was second. SOLUTION: CBEDA Suppose Adam’s first statement was true; then Adam is first. This means Beth’s first statement is true, and she is third. This means that Ethan’s second statement is true, and Carol is second. This means that Diane’s not fourth, so she is fifth. This means that Ethan is fourth, and herein lies our problem: if the order is ACBED, then both of Diane’s statements are false. If Adam’s second statement is the truth, then Beth is second, and Adam is not first. This means that Adam is last (Beth’s second statement). Carol’s second statement is true, so Diane is fourth. This means that Ethan is not first, so he must be third, and the order is CBEDA. 3. Let {2} represent the set of all positive integer multiples of 2 that are less than or equal to 100. Let {3} represent the set of all positive integer multiples of 3 that are less than or equal to 100. Let {5} represent the set of all positive integer multiples of 5 that are less than or equal to 100. Compute the number of elements in {2} ∪ {3} ∪ {5}. SOLUTION: 74 We have |{2}| = 50, |{3}| = 33, and |{5}| = 20. We also have |{6}| = 16, |{10}| = 10 and |{15}| = 6. Lastly, we have |{30}| = 3. Thus, our answer is 50 + 33 + 20 − 16 − 10 − 6 + 3 = 74. October 13, 2011 Any calculator permitted on N.Y.S. Regents examinations may be used. The word “compute” calls for an exact answer in simplest form. 5 - Logic and Set Theory

1. Let A be the set of positive multiples of 3 that are less than 20. Let B be the set of positive multiples of 4 that are less than 20. Let C be the set of positive multiples of 5 that are less than 20. Compute the number of elements in the set A ∪ B ∪ C.

2. On the Island of Perpetual Thinking, there are only Truthies (who always tell the truth) and Lieies (who always lie). Suppose that Colombo (just a detective, not a Truthie or a Lieie) runs into Bert and Ernie on the Island, and Bert says, “I am a Lieie or Ernie is a Truthie.” Between Bert and Ernie, how many of them are Truthies? SOLUTION: 2 If Bert was a Lieie, then his statement would be false (thus Bert would have to be a Truthie and Ernie would have to be a Lieie). However, that would cause a conﬂict which implies that Bert must be a Truthie. In that case at least one part of his statement is true and as it cant be the ﬁrst one, Ernie must be a Truthie, too. The number of Truthies is 2.

3. Suppose that last year, in the Logic and Set Theory category for the Monroe County Math League Meet, 300 students competed. 20 students got perfect scores, 15 students scored 5 points, 30 students got 4 points, 35 students got only question #3 wrong, 105 students answered question #1 correctly, 90 students answered question #2 correctly, and 85 students answered question #3 correctly. Compute the number of students who scored 0 points in Logic and Set Theory last year. SOLUTION: 140 We consider the union of Ones, Twos, and Threes (the sets of students who answered the three individual questions correctly). We know that 1 ∪ 2 ∪ 3 = 1 + 2 + 3 − (1 ∩ 2) − (1 ∩ 3) − (2 ∩ 3) + (1 ∩ 2 ∩ 3), so we compute 105 + 90 + 85 − (35 + 20) − (30 + 20) − (15 + 20) + 20 = 160, so our desired number is 300 − 160 = 140.

OCTOBER 14, 2010

CALCULATORS MAY BE USED ON THIS TOPIC

#5 LOGIC & SET THEORY ANSWERS:

1. Given: If I have Chemistry lab and I have a study period, then today is not Wednesday. If I complete my English essay, then I have a study period. I have Chemistry lab if today is Wednesday. Today is Wednesday. Draw a final conclusion.

2. Television News sent a letter to selected subscribers asking which of the following three shows they watched on a regular basis – Grey’s Anatomy, 24, or Mad Men. The result of the 420 questionnaires that were returned showed that

223 selected Grey’s Anatomy 192 selected 24 151 selected Mad Men 86 selected Grey’s Anatomy and 24 54 selected Grey’s Anatomy and Mad Men 68 selected 24 and Mad Men 24 selected all three

How many enjoyed watching exactly one of these shows?

3. Four children, Emad, Ken, Sue, and Tim were playing when suddenly I heard a loud noise followed by much shouting. I rushed in to find a valuable lamp had been knocked off the table and broken beyond repair. Emad and Ken spoke almost at the same time:

Emad saying, “It wasn’t me!” Ken saying, “It was Tim!” Sue yelled, “No, it was Ken!” Tim said, with a straight face, “Ken’s a liar.”

Only one of them was telling the truth, so who knocked over the lamp?

Solutions October 2010

LOGIC & SET THEORY

1.) All Givens are true. Given: Let c represent I have Chemistry lab. (cs∧→ )~ w Let s represent I have a study period. es→

Let e represent I complete my English essay. wc→ Let w represent today is Wednesday. w

(c∧→ s )~ w = true w= true e→= s true ()true∧→ s false = true w→= c true e→= false true ∴()true ∧= s false ∴=c true ∴=e false ∴=s false Conclusion: I did not complete my English essay.

2.) Grey’s Anatomy 24

107 62 62

24 30 44

Mad Men

107+62+53=222

3.) Let’s assume Ken is telling the truth. If Ken’s statement is true, then Emad’s statement would also be true. (Contradiction) Therefore Ken is lying, Tim is telling the truth and Emad and Sue are also lying. Emad knocked over the lamp.

OCTOBER 15, 2009

CALCULATORS MAY BE USED ON THIS TOPIC

#5 LOGIC & SET THEORY ANSWERS:

1. Sergeant

2. Yellow

3. {6, 8}

1. The last box of donuts is missing from the Police Station Coffee Room. The culprit was the patrolman, the detective, or the sergeant. During the investigation, each made the following statement: Sergeant: The patrolman ate the donuts. Patrolman: The sergeant’s statement is true. Detective: I did not eat the donuts. As it happened, at least one of them lied and at least one of them told the truth. Who ate the donuts?

2. A bag contains marbles that are red, green, blue, or yellow. They are identical in size, have different numbers of each, and a single marble is drawn. Three people are asked to guess the color of the marble that is drawn. Otto says, “It is red or green.” Oriole says, “Two marbles were drawn.” Oscar says, “It is not yellow or it is blue.” If all three statements are not true, then what is the color of the marble?

3. Let U = { x | x ∈ N and x < 10} A = { x | x ∈ N and x is odd and x < 10} B = { x | x ∈ N and x is even and x < 10} C = { x | x ∈ N and x < 6} If N is the set of natural numbers and U is the universal set, find (B ∩ C) ′ ∩ [(A ′ ∪ C) ∩ B].

OCTOBER 16, 2008

CALCULATORS MAY BE USED ON THIS TOPIC

#5 LOGIC & SET THEORY ANSWERS:

1. a

2. butler

3. zero

1. Which of the following conclusions make the argument valid? Supply a valid conclusion to the following argument using all the premises.

If you are a baby, then you are illogical. If you can manage a crocodile, then you are not despised. If you are illogical, then you are despised.

a) If you are a baby, then you cannot manage a crocodile. b) If you cannot manage a crocodile, then you are a baby. c) If you are not a baby, then you can manage a crocodile. d) If you are illogical, then you can manage a crocodile.

2. The gold was missing. The thief was either the butler, the maid, or the cook. During the investigation, the employees made the following statements in this order.

Butler: The maid stole the gold. Maid: That is true! Cook: I did not steal the gold.

As it happened, at least one of them lied and at least one of them told the truth. Who stole the gold?

3. An examination of 19 hospital patients turns up 9 cases of fever, 4 cases of jaundice, and 11 cases of fatigue. 7 people have both fever and fatigue, 2 have fever and jaundice, and 2 have jaundice and fatigue. 6 people have none of the three problems. How many people have just fever or just jaundice?

to sS,tte- !.[*=\,r-trs N4O)SrR@rEC@U)ilrY NMAT'F{][,EAGT.T]E ocToBER16,2008 CALCALATORSMAY BE USED OIYTHIS TOPIC

#s LOGIC& SETTHEORY ANSWERS:

1.a

2. butler

a zero

1. Whichof thefollowing conclusions make the argument valid? Supplya valid conclusionto the following argumentusing all the premises.

If you area baby,then you areillogical. B '2 tr ,r,Q If you canmanage a crocodile,then you arenot despised.C --? z--r.,D D -) If you areillogical, then you aredespised. 5- -" b' G-t= (et ff you area baby,then you cannotmanage a crocodile E-?A b) If you cannotmanage a crocodile,then you area baby. D-)\:s c) If you arenot a baby,then you canmanage a crocodile, d) If you areillogical, then you canmanage a crocodile.

2. The gold wasmissing. The thief waseither the butler,the maid,or the cook. Duringthe invpstigation,the employeesmade the following statementsin this order. L{ Butler:The maid stole the gold. n^d^"A S*'llC ure{"J-t t€ ^J{3 Maid: Thatis true! '5\Ar- ; *f'ffi-gr 'Cook: 'I did not stealthe gold. ""td It Lboh *4* "r1*;nt",,;*Ut;;"tt-[{a.U+ _i n*rf-C*uL* -rr As it happenbd,at leastone of themlied andat-lg4qlone of ttrerr0-Igld_lbglnfth. '- ' whottotathegffi -----r"! - -

3. An examinationof 19 hospitalpatients turns up 9 casesof fever,4 casesofjaundice, and 11 casesof fatigue. 7 peoplehave both feverand fatigue, 2have feverand jaundice, and2 havejaundice and fatigue. 6 peoplehave none of the threeproblems. fa,r*&t C{ How.manypeople have justiust feverfbverorjus(jaungice?or iust taundice'/ l? % '- \---l

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#5 LOGIC& SETTHEORY ANSWERS:

A nail is not lost

1. Supposea Kingdom is not lost. Form a valid conclusion,using all the statementsfor each argument. ,l If a nail is lost,then a shoeis lost. n.---) -5 If a shoeis lost,thpn a;horseis lost. 5 -=-r h If a horseis lost, tlren rider is lost. h-) r {I If a rider is lost, then a battleis lost. V-4 L- If a battle is lost, then a Kingdom is lost. t-+K trr6 l'1 but- -r-

2. I-et A(m, n) be the set of n consecutivepositive integers whose least element is ru. What is the greatestinteger in A(17,49)n A(49,17X tr'rrmLd+fl+t ft= tT,rg,11... K-11+"1*t+E &" 14::{aS } +" ta^rg'rF 'b t'\arf( *.' tc qq , , , t? ft rri":^6i b = ,5o,5t 5e rn-tft*ts J 6"*4

3. Of 75 studentssurveyed, 48 like basketball,45 like football,58 like baseball,28 like basketballand football, 37 like footballand baseball, 40 like basketballand baseball, and25like all three sports.How manystudents like no or only onesport? ?1 4l'O zg +8 - iq *-1# - >t L lL tS e gL t5 {p s ,9 l> lz 2' Z7 ?3 7l y3 ^42- a3 -?5 5L T' -?l4F { nor".l -;=@ &oilJt 5+s+6++

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2. After the math meet,20 teachersate at Mac's Place. Threeof the teachershad only a burger;three had a burgerand fries; onehad fries anda shake;five teachershad only a shake;four had a burgerand a shake;two teachershad q.burgera shakeand fries. How many teachersate only fries? fl g- {:^r.r; $*vqerl 3+ 3'{tto ?+ I { ! r rB,}, }0-!k: ? Q- S\oJct* 3.LetA: {everqnon-negative whole numbers}, B integersless than 12), C: {integersthat are not multiples of three},and the universal set [./: {all integers}. Find(C'.' A)nB. * !\rr-:rr::", $*; ii 'V/' Ct ft"rrr\-{.,p1e'; ] \- {]'n A i lg*s1"4..n,:l-'ll* f'cnnre"- r+\^.^ r'b 3 /J uhr"+ LJ t'r 1,, a-u..,,,../,,\i,.. l+. lrr. f .Ll}l?'r",'r \ ,_-*i1"r., tl *n 1^t,.or*e*{";.,*^* ,r,uq q{g + ]"'*t,' ANSWERS [c 5l -i i e- &.',1., ,:]..L

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(2 pts.) 2. L t :,'* (3 pts.) J.

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1.Suppose a snailis climbingup a slippery3O-inch wall. Eachminute the mail climbs 5 inches,then it slidesback 4 inches.How manyminutes will it takethe snailto n^r.t\ egct'''. i+ ur E* ; f/rd- 2. -pennies balance I cup. How manypennies balance the cup? tg+tC3laP ro rgarrf-tS a*A ICs*f+f.f flr'*nft4sutd qg-|L?:rc lL?-t$a tS{-Zf = tC 3. Assumethat for a certain ,o*o}, tulflrg;.?u I. Somestudents are not honest II. All fastfood workers are honest. Srr,a 4$ $asf {nqt wnk *ra,a* hru.uf Basedupon the above,whioh of tho following arenecessary cnnclusions? A. Somestudents are fast food workers. * S Orrr,l- 5.{aa-daarL aa-*" ,\Ift' B. Somefast food workers are not stude,nts. L.a5srr* @So-" studentsare not fast food workers. iJt D. No fast food worker is a student. E. No studentis a fast food worker.

ANSWERS

(1pt.) 1. 2U uxnutes

(2 pts.) 2. lo peruues

(3 pts.) 3. TahantqShrewsbury, Bumcoat