2015 - handbook for Level III MB Workshop

NSF Math Bee LeveL III

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North South Foundation (NSF)

2015 - handbook for Level III MB Workshop

Prepared by:NSF Math Core team


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table of contents

About NSF ........................................................................................................................... 3

Goals of Math Bee ............................................................................................................ 4

Grade Groupings for JNV Math Bee .......................................................................... 4

JNV Math Bee - Contest Rules ...................................................................................... 4

Syllabus for Level 3 .......................................................................................................... 5

Why is Math Important? ................................................................................................ 6

What is Math? Where is it used? .................................................................................. 6

Role of Students ............................................................................................................... 7

Role of Schools ................................................................................................................ 7

Math as a Language ........................................................................................................ 8

Problem-Solving .............................................................................................................. 9

Common Strategies for Problem-Solving ................................................................ 9

Some Special Numbers and Patterns ........................................................................ 10

Workshop Problems ........................................................................................................ 13

Appendix A: Problem‐Solving Strategies ................................................................ 31

Appendix B: Facts, Formulas and Tricks ................................................................... 51

Appendix C: Practice Problems and Solutions ....................................................... 61

Sample Questions Paper ................................................................................................ 92


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North South FoundationNorth South Foundation (NSF) is a volunteer driven non-profit organization established in Illinois, USA in 1989.

NSF Mission is to• Encourage academic excellence among the poor by providing college scholarships in India regardless of religion, gender, caste and geographic

origin• Encourage academic excellence among the NRI children in USA in the areas of English spelling, vocabulary, math, science, geography, essay

writing, public speaking and brain bee by focusing on contests, workshops, coaching and educational games• Encourage academic excellence in India in English, math and science primarily in high schools and improve communication capabilities in

professional colleges by focusing on workshops, creative learning and contests.

The Scholarship Program in India is designed to encourage academic excellence among the poor particularly who want to pursue professional courses like Diploma Engineering, Engineering and Medicine etc. It is targeted at qualified, needy students entering college. Over 1600 scholarships have been awarded in 2017-18. Each scholarship is in the range Rs.10,000/- to 25,000/- per student per year. Awards are made following strict selection criteria based on merit and need. Students are selected from 33 centers all over India: Aurangabad, Ahmedabad, Baroda, Bengaluru, Bhopal, Bhubaneswar, Chandigarh, Kochi, Chennai, Guwahati, Hyderabad, Jammu & Kashmir, Jodhpur, Kanpur, Kochi, Kolkata, Kurnool, Madurai, Moradabad, Mumbai, Nagercoil, Noida/Delhi, Panchkula, Patiala, Patna, Pune, Rewa, Satna, Tanuku, Udaipur, Vijayawada, Visakhapatnam, Tanjavur. These NSF centers are run by dedicated local volunteers.

Educational Contests in USA is designed to encourage academic excellence among Indian American children. The spelling, vocabulary, math, science, geography, essay writing, and public speaking bees are conducted annually in two steps. Children initially participate in one of 85 regional centres. Winners of these local contests compete at the national finals. National top three rankers are awarded scholarships ranging from $1,000 to $250, redeemable in the winners’ freshman year of college. More than 120,000 contestants benefited from NSF contests so far. NSF introduced Human Values in 2014 to round out character building of a child to be a good citizen and a mantle of harmony.

Accomplishments of NSF Children: The winning streak in Scripps National Spelling Bee championship extended to 8th year in a row whereas National Geographic championship swung NSF way for the 4th year in a row in 2015. For the first time, NSF student won Math Counts championship in 2014. Similarly, NSF children won the first ever team championship in Middle School Science Bowl conducted by the U.S. government in Washington. These are like Super Bowl victories, unimaginable to repeat but faith and hard work favour the deserving.

Many of NSF students have been accepted into top ranking schools in USA like Harvard, MIT and Stanford. More importantly, these contests help children improve their communication skills, self-confidence and empower them to become better citizens of tomorrow.

Role Model Award: The Foundation awarded its inaugural Role Model Award, ‘Vishwa Jyothi’ to Rajiv Vinnakota in 2003, Nipun Mehta in 2004 and Nimo Patel in 2015 for their exceptional and inspiring work. It helps to showcase human values and academic excellence to the community. NSF will continue its efforts to identify and bring such individuals into limelight in both USA and India.

Educational Contests in India: NSF did a successful pilot programme in Hyderabad in late 2010 in Math, Science and Spelling. The positive experience and the demand for these contests has convinced NSF India, headquartered in Hyderabad, to deepen the commitment by bringing workbooks, training and contests in a streamlined process and introduce more modules in future as seen fit. It started with 125 registrants in Hyderabad and by 2015 the registrations jumped to over 6,000. The contests expanded to Bengaluru city and Rajiv Gandhi University of Knowledge Technologies (RGUKT) at Nuzvid campus and few others. These contests can be extended to other cities and states as NSF builds a bigger volunteer base to satisfy the need.

Educational contests in India are considered as another side of scholarship coin. Creative learning, focused interventions, and objective testing are part and parcel of NSF scholarship program. NSF is providing its services to both private and public sector institutions. Money is removed from being a hurdle for learning, much like cash scholarship program for higher education.

Expanding Reach: NSF teamed up with Navodaya Vidyalaya Samiti (NVS) in 2015 to improve scholastic achievements by conducting educational contests, in steps, in its eight regions of Jawahar Navodaya Vidyalaya (JNV) schools across the country.

JNV promotes creative learning as witnessed by its admission test which allocates 50% weight to “Mental ability Test”. Striving for student empowerment is central to JNV efforts, which is very much the philosophy of NSF, making it a natural partnership. NSF scholarship criteria for extending its services, is fully satisfied by the student population of JNV.

NSF looks forward to a long association with JNV by providing its focused interventions to sharpen the skills of JNV students.

Please visit www.northsouth.org for more information.


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Goals of Math Bee• To introduce important mathematical concepts

• To make Mathematics an easier subject to learn and excel at early age

• To teach major problem-solving strategies

• To strengthen mathematical intuition

• To develop mathematical flexibility in solving problems

• To stimulate enthusiasm and a love for Mathematics

• To better prepare students for excelling in college entrance exams like the SAT and ACT

• To foster Mathematical creativity and ingenuity

• To prepare students for national level mathematics competitions such as MATHCOUNTS and AMC

• To provide the satisfaction, joy, and thrill of meeting challenges

• To meet other students of Indian origin and build friendship

Grade groupings for JNv Math Bee• Participants in the NSF Math Bee are grouped according to their grade in school.

• The groupings are:

� Level 2 Grades 7 and 8

� Level 3 Grades 9, 10 and 11

• Each group will be tested separately using questions appropriate to their level.

• The questions will be based on the syllabus for each level.

JNv Math Bee – Contest Rules• Math Bee is held as a written tests.

• Calculators, Cell Phones are not allowed at any level.


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Logic

• Boolean algebra, venn diagram, logic

algebra

• Factorials

• Permutations and combinations

• Pascal’s Triangle

• Combinations and Probability

• LCM and GCF

• Applications of GCF and LCM

• Theory of exponents

• Introduction to logarithms

• Absolute Value

• Equalities and Inequalities Using Absolute

Value

• Solving Equations by Graphing

• Graphing an Inequality on a Coordinate Plane

• Graphing Pairs of Inequalities

• Solving Equations

• Definition of Linear, Increasing, and Decreasing

Functions

• Applications Solving Pairs of Equations by

Graphing

• Solving Systems of Equations by Substitution

• Solving Systems of Equations

• Quadratic equations and inequalities

Geometry

• Circles and Regular Polygons

• Advanced topics in Pythagorean Theorem

• Similar figures

• Applications of similarity

• Trigonometric Ratios

• Applications of Trigonometric Ratios

• Volume of a Circular Cylinder

• Volume of a Circular Cone

• Volume of a Sphere

• Volume of Prisms and Pyramids

• Applications of Volume

• Surface Area

• Surface Area of a Cylinder

• Surface Area of a Cone

• Area of a Sector

• Surface Area of a Sphere

• Applications of Surface Area

• Plans and Elevations

• Oblique Drawings

• Elements of coordinate geometry

arithmetic

• Advanced topics in fractions

• Rounding Off Decimals

• Squares and Square Roots

• Ordering the Rational Numbers

• Ratio and proportions

• Variations

• Irrational Numbers

• Approximating Square Roots

• Averaging Method

• Comparing Ratios

• Mean, Median, and Mode

• Taking Samples

• Reasonable Estimates

• Estimating Products

• Estimating Quotients

• Conversion between Systems of Measurement

• Kilograms and Liters

• Estimating Averages

Level 3 – Syllabus


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Why is Math Important?• Today’s students must master advanced skills in mathematics, science, and technology to stay on track

for college and for promising careers.

• Students who take algebra and geometry go on to college at much higher rates than those who do not

(83% vs. 36%)

• Most four-year colleges require three to four years each of high school math and science for admission

• Almost 90% of all new jobs require math skills beyond the high school level

• Entry-level automobile workers must use advanced mathematics formulas to wire a car's electrical circuits

• Strong math skills are needed for understanding graphs, charts, and opinion polls in a newspaper, for

calculating house and car payments, and for choosing a long-distance telephone service.

• Mastering challenging mathematics is not just a classroom skill -- it's a life skill!

Reference: Indiana State Teacher’s Association (http://www.ista-in.org/sam.cfm?xnode=2282)

What is Math? Where is it used?• The study of Mathematics is the study of patterns. We use patterns all the time to solve problems.

• Math teaches us different ways to analyze problems and solve them.

• Math is an universal language.

• Math is used everywhere, e.g.

� Calculate weekly shopping bill

� Budgeting household expenses

� Plan recipes

� Compute taxes each year

� Arts such as sculpture, drawing, and music

� Technological fields such as computers, rockets, and communications

� Plan vacations

� Reading a clock to tell time

� Biology, chemistry, physics

� Soft sciences, such as economics, psychology, and sociology

� Engineering fields such as civil, mechanical, and industrial engineering

� Anything that uses a computer uses mathematics

� Almost anything you can imagine requires some mathematics!


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Role of Students• Reviewthesyllabus

• Understandtheproblemsolvingstrategies

• Reviewthesetofpracticeproblems

• Setagoalofsolvingatleast10problemsfrom the practice set.

• Usethereferencematerialforadditionalproblems.

Role of School• encourage students in their endeavour.

• Provideexplanationsandoccasionallylendahelping hand.

• Makesurethehelpdoesnotbecomeacrutch.

• Objectiveistodevelopthestudent’sconfidence level and not merely him/her to pass the test

• Emphasizethespeedofproblem solving.


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Math as a Language• EveryproblemstatedintheEnglishlanguagemustbetranslatedintoamathematicalstatementbefore

we can solve the problem.

• Readcarefullyandthendecidewhatmathoperationtodo.

• Sometimesthesamestatementwillresultindifferentoperationsdependingontherestoftheproblem

statement.

Operation WordPhrase algebraic

addition A number plus 6; X + 6

Sum of a number and 6;

6 more than a number;

A number increased by 6

Subtraction A number minus 6; Y-6

The difference between a number and 6;

6 less than a number;

A number decreased by 6

Multiplication 6 times a number; 6xn

A number multiplied by 6; 6.n

The product of a number and 6

6n

Division A number divided by 6; a/6

The quotient of a number and 6

a 6


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Problem-SolvingProcessThisisaprocessthatyoubecomebetteratwithexperience.So practice often...

1 READ and UNDERSTAND

Read the problem carefully. - Underline the key words and units.Whatdoyouknow?- What are the key facts and details stated in the prob-

lem?- What do you know that is not obviously stated in the

problem?What are you trying to solve?- What is being asked to solve?

2 PLAN Chooseastrategybasedonyourexperience- Have you seen a similar problem before? If so, how is

this problem similar? How is it different?- How did you solve similar problems in the past?- What strategies are you familiar with?

3 SOLVE Usethestrategyyouselectedandsolvetheproblem.- If stuck, retrace steps in #1 and #2 and try a different

strategy perhaps.4 LOOK BACK

and CHECKCheckyouranswer.- Make sure that you answered what is being asked.- Is your answer in the correct units?Doesyouranswerseemreasonable?- Use estimation and reasoning to make sure that the

answer makes sense

CommonStrategiesforProblem-Solving1. Draw a Picture or Diagram

2. Make an Organized List /Table

3. Solve a simpler related problem

4. Find a pattern

5. Guess, Check and Revise

6. Experiment

7. Compute or Simplify

8. Use a Formula

9. Make a model

10. Work Backwards

11. Write an Equation

12. Eliminate


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SomeSpecialNumbersandPatterns

37 is a special number

3 × 37 = 111 and 1 + 1 + 1 = 3

6 × 37 = 222 and 2 + 2 + 2 = 6

9 × 37 = 333 and 3 + 3 + 3 = 9

Etc…..

Hereisanamazingpattern:

12 = 1

112 = 1 2 1

1112 = 1 2 3 2 1

11112 = 1 2 3 4 3 2 1

111112 = 1 2 3 4 5 4 3 2 1

What is 1111111112 =

Mountain Numbers

1

2 2

3 3

4 4

5

1

Continuethesepatternsandfindoutwhatmakesthemspecial.

1

1 + 3

1 + 3 + 5

1 + 3 + 5 + 7

1 + 3 + 5 + 7 + 9

1 + 3 + 5 + 7 + 9 + 11

1 + 3 + 5 + 7 + 9 + 11 + 13

1

3 + 5

7 + 9 + 11

13 + 15 + 17 + 19

21 + 23 + 25 + 27 + 29

31 + 33 + 35 + 37 + 39 + 41

43 + 45 + 47 + 49 + 51 + 53 + 55


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1089:AVerySpecialNumber

Followtheseinstructions:

1. Take any three digit number in which the first and last digits differ by 2 or more;

e.g., 335 would be okay, but not 333 or 332.

2. Reverse the number you chose in step 1. (Example: 533)

3. You now have two numbers. Subtract the smaller number from the larger one.

(Example: 533 – 335 = 198)

4. Add the answer in step 3 to the reverse of the same number. (Example: 198 + 891 = 1089)

Theanswerisalways1089.


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WORkSHOP

PROblEMS


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1. Howmanydifferentwords(includingnonsensewords)canbeformedusingthefourlettersoftheword“MATH”?

4 . 3 . 2 . 1 = 24

2. AClubhas10members.Inhowmanywayscantheychoose a slate of four officers, consisting of a president, vice president, secretary and treasurer?

P(10,4)=10.9.8.7=5040

Topic: Algebra: Permutations and Combinations



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3. the board of directors of a corporation has 10 members. Inhowmanywayscantheychooseacommitteeof3boardmembers to negotiate a merger?

10.9.8 C (10, 3) = -------------------- = 120 3.2.1

4. Supposethatanexperimentconsistsoftossingacoin10timesandobservingthesequenceofheadsandtails.Howmanydifferentoutcomesarepossible?

2 . 2 . 2 ……. 2 = 210 (= 1024)

1

h t

2 3 4 5 6 7 8 9 10


Topic: Algebra: Combinations and Probability


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5. Supposethatanexperimentconsistsoftossingacoin10times and observing the sequence of heads and tails.

Howmanydifferentoutcomeshaveexactlyfourheads?

10.9.8.7 C (10, 4) = -------------------- = 210 4.3.2.1

6. Supposethatanexperimentconsistsoftossingacoin10timesandobservingthesequenceofheadsandtails.Howmanydifferentoutcomeshaveatmosttwoheads?

–Exactly0heads: Thereis1outcome 1

–Exactly1head: C(10,1)=10ways 10

–Exactly2heads: C(10,2)=45ways 45

– total 56




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7. Supposethatanexperimentconsistsoftossingacoin10

timesandobservingthesequenceofheadsandtails.How

manydifferentoutcomeshaveatleastthreeheads?

C(10, 3) + C(10, 4) + ….. + C(10, 10)

– at least 3 outcomes = total number of outcomes – at most 2 outcomes = 1024 ‐ 56 = 968

8. 1 1 Ifx+––=2thenwhatisx3 + –– ? x x3

11 1

1 2 11 3 3 1

1 4 6 4 11 5 10 10 5 1

1 6 15 20 15 6 11 7 21 35 35 21 7 1

1 8 28 56 70 56 28 8 11 9 36 84 126 126 84 36 9 1

1 10 45 120 210 252 210 120 45 10 11 11 55 165 330 462 462 330 165 55 11 1

1 12 66 220 495 792 924 792 495 220 66 12 11 13 78 286 715 1287 1716 1716 1287 715 286 78 13 1

1 14 91 364 1001 2002 3003 3432 3003 2002 1001 364 91 14 1

Blaise Pascal1623-1662


Topic: Algebra: Pascal’s Triangle


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9. A pizza parlor offers a plain cheese pizza to which anynumberofthreepossibletoppings(SausageS,PepperoniP,MushroomsM)canbeadded.Howmanydifferentpizzascan be ordered?

– Number of Pizzas with no topping •BasicCheesePizza – Number of Pizzas with one topping •S P M

– Number of Pizzas with two toppings •SP PM SM

– Number of Pizzas with three toppings •SPM

23 = 8

(3)0

(3)1

(3)2

(3)3

10. Definex#ybytheequationx#y=xy–x.Thenevaluatethevalueof(2#3)#4.

2 # 3 = 6 – 2 = 4

(2 # 3) # 4 = 4 # 4 = 16 – 4 = 12

Topic: Algebra: Combinations

Topic: Algebra: Solving Systems of Equations by Substitution


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11. When the integer n is divided by 2, the quotient is u and the remainder is 1. When the integer n is divided by 5, thequotient is v and the remainder is 3. What is the value of 2u – 5v?

“The remainder is r when p is divided by z”

meansp = q z + r

Sum of the n numbers = (Average of numbers) . n

Sum of the 5 numbers = (–10) . 5 = –50Sum of the 3 numbers = 16

n = 2u + 1 n = 5v + 3

Answer:2

Answer:–33

12. If the average of five numbers is –10, and the sum of three ofthenumbersis16,thenwhatistheaverageoftheothertwonumbers?

Topic: Algebra: Solving Equations

Topic: Algebra: Solving Systems of Equations; Arithmetic: Averaging Method


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13. Jay drove his car for 2 hours at a rate of 70 kilometers per hour and for 5 hours at a rate of 60 kilometers per hour. Whatwashisaveragespeedforthe7‐hourperiod?

total distance Average Speed = ———————— total time

Total Distance = (2 hr) (70 kms/hr) + (5 hr) (60 kms/hr) = 440 kms

440 kms 6 Average Speed = ————— = 62 ------ kms / hr 7 hr 7

14. Find the length of the shortest altitude in triangle aBC.

• AreaofTriangleABC – ½ (3) (4) = 6

• AreaofTriangleABC – ½ (AC) (BO) – ½ (5) (BO)

• BO=2.4

A

B C

O3

4

Topic: Arithmetic: Mean, Median, Mode

Topic: Geometry: Area


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15. What is the sum of the squares of the roots?x2 - 3x +4 = 0

• Implicationsofquadraticformula

• SideResults:

- Sum of the roots p + q = - ab & p + q= 3

-Productsoftherootspq= ac & pq= 4

(p+q)2 = p2+2pq+q2

& p2+q2 = (p+q)2 - 2pq = 32 -2(4) = 1

16. Find thequotient of two integerswhosedifference is 10andwhoseproductisassmallaspossible.

• Lettheintegersbex,ysuchthaty–x=10. • Product P=xy = x (10 + x) = 10 x + x2 • P(x) is a quadratic function with a=1, b=10.

Hence minimum occurs at the vertex:

b x = - —— & x = - 5 2a

y = 10 + x = 5

Topic: Algebra: Quadratic Equations

Topic: Algebra: Quadratic Equations


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17. AcirclewithcenterOisinscribedinaregularhexagonofside 1 unit. What is the circumference of the circle?

30 - 60 - 90 triangle

Radius = 23

Circumference =

18. HowmanydiagonalsdoesthehexagonAbCDEFhave?

• Thenumberofdiagonalsisthreeless than n, number of vertices or sides, or (n – 3) = 6 – 3 = 3. • Sincetherearenvertices,thenumberof diagonals will be n (n – 3) = 18 • Sinceeachdiagonalhastwoends,wehave to divide this by two which yields the general formula for the number of diagonals of a regular polygon with n sides = ½ n (n – 3) = 9

a

F C

De

B

Topic: Geometry: Circles and Regular Polygons

Topic: Geometry: Regular Polygons


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19. Howmanyouncesofasolutionthatis30percentsaltmustbe added to a 50‐ounce solution that is 10 percent salt so that the resulting solution is 20 percent salt?

(x)(30%) + (50)(10%) = (x + 50)(20%)

x = 50 ounces

20. The sum of the digits of a two‐digit number is 11. If weinterchangethedigitsthenthenewnumberformedis45less than the original. Find the original number.

x = one’s digity = ten’s digit - x + y = 11

Original number = 10y + xInterchanged number = 10x + y10x + y = 10y + x – 45 y – x = 5

Answer:83

Topic: Arithmetic: Averages, Proportions

Topic: Algebra: Solving Systems of Equations


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21. Prakashcancleantheyardin40minutesandJasbircancleantheyardin60minutes.Howlongwillittakeforthemtomowthelawntogether?

22. IfRavicanmowthelawnin30minutesandwiththehelpofhisbrother,Shashi,theycanmowthelawnin20minutes,howlong(inhrs)wouldShashitakeworkingalonetomowthelawn?

• SincePrakashcancleantheyardin40min,hecancomplete1/40th of the yard in 1 minute. This is Prakash’s work rate.

• SinceJasbircanmowalawnin60min,hecancomplete1/60th of the lawn in 1 minute. This is Jasbir’s work rate.

• Workdone=workratextime

• Iftheyworkfor“t”minutestogether,thenwehavePrakash’swork done = t/40 and Jasbir’s work done = t/60. Since together they complete the “one” full job we have:

• Since Ravi canmow a lawn in 30min, he can complete1/30th of the lawn in 1 minute. This is Ravi’s work rate.

• SupposeShashicanmowalawnin“x”min,hecancomplete1/xth of the lawn in 1 minute. This is Shashi’s work rate.

• Workdone=workratextime

• Iftheyworkfor“20”minutestogether,thenwehaveRavi’swork done = 2/3 and Shashi's work done = 20/x. Since together they complete the "ONE" full job we have:

t t—— + —— = 1 40 40

2 20— + — = 1 3 x

Answer:t=24minutes

Answer:60minutes=1hour

Topic: Algebra: Solving Equation



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23. JayeshistwiceasoldashisfriendPran.Pranis5yearsolderthanAnil. In5years, Jayeshwillbe three timesasoldasAnil.HowoldisPrannow?

• LetPran’sagenow=x.ThenJayesh’sagenow=2xand Anil’s age now = x – 5.• In5years,Pran=x+5,Jayesh=2x+5,Anil=xandwe have, 2x + 5 = 3x which yields x = 5.

24. an equilateral triangle of side length 15 is completely filled in by non‐overlapping equilateral triangles of side length1.Howmanysmalltrianglesarerequired?

1 + 3 + 5 + 7 + …… + (2(15) –1) = 152 = 225

Answer:5years


Topic: Geometry


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25. a rectangle having area of 60 Square units, has sides of lengthx+3unitsandx+10meters.Whatisthelengthofthediagonal of the rectangle?

(x + 3)(x + 10) = 60 x2 + 13x + 30 = 60 x + 13x – 30 = 0 (x + 15)(x – 2) = 0 x=2

Hence, sides of the rectangle are 5, 12. Then the diagonal length = 13.

26. AclasscollectsRs60tobuyflowersforaclassmatewhoisin the hospital. Roses cost Rs 3 each, and Marigold cost Rs 2 each.Nootherflowersaretobeused.HowmanydifferentbouquetscouldbepurchasedforexactlyRs60?

• Letr=numberofroses,m=numberofmarigolds. • Giventhat3r+2m=60.Since“2m”isevenwethenhave "3r" is an even number which means "r" has to even.

The possible values of even values for r = 0, 2, …, 20.

Answer:11

Topic: Geometry and Algebra (Solving Equations)

Topic: Algebra: Solving Equations


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27. an advertising agency finds that of its 170 clients, 115 use television(T),100useradio(R),130usemagazines(M),75use televisionandradio,95use radioandmagazines,85use television andmagazines, and 70 use all three.Howmany clients use none of the media?

28. Supposethatwetossadiceandobservethesidethatfacesupward.What istheprobabilitythattheresults isoddorgreater than 4 ?

• SampleSpaceS={1,2,3,4,5,6}• EventE1={1,3,5}• EventE2={5,6}

• Pr(E1U E2) = Pr(E1) + Pr(E2) ‐ Pr(E1 E2) = 3/6 + 2/6 ‐ 1/6 = 4/6 = 2/3

Answer:10

T25

10

7015 25

5 R0

M 20

Topic: Logic: Venn Diagram

Topic: Algebra: Probability


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30. An urn contains eight white balls and two green balls. a sample of three balls is selected at random. What is the probability that the sample contains at least one green ball?

F: Experiment consisting of selecting at least one green ball.

Number of outcomes in F =

Pr (F) = 64 / 120 = 8 / 15

29. Anurn contains eightwhiteballs and twogreenballs.Asample of three balls is selected at random. What is the probabilityofselectingonlywhiteballs?

E: Experiment consisting of selecting 3 balls from 10

Pr(E) = Number of outcomes in E = N

8 3 7— = —10 15 3

( )( )

2 (1) 2

(2) 8 (2) 8

(1)+ =2(28) + 1(8) = 64




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31. Sita’sclockloses15minuteseveryhour.Theclockissetatthe correct timeat9A.M.What is the correct timewhenSita’sclockfirstshows11A.M.?

Consider a correct clock and a wrong clock both starting at9 A.M. As the correct clock moves 60 minutes, the wrongclock only moves 45 minutes. Since there are 120 minutesin two hours, when the wrong clock completes 120 minutes, the correct clock would complete 160 minutes = 2 hours and 40 minutes.

32. In an arithmetic sequence the 113th term is 786 and 125th term is 870. Find the 150th term

The nth term of an arithmetic sequence is given by: an = a1 +(n - 1)d •Hencewehave: a113 = a1 +112d = 786 a125 = a1 +124d = 870

•Solvingforthefirstterm“a1”andcommondifference “d” and then evaluating “a150” yields the answer.

Answer:11:40am

Answer:1045

Topic: Algebra; Logic

Topic: Arithmetic: Series & Algebra: Solving Equations


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APPENDIxA

Problem‐SolvingStrategies


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Problem-‐SolvingStrategies

1 FINDOUT Look at the problem.

Have you seen a similar problem before?

If so, how is this problem similar? How is it different? What facts do

you have?

What do you know that is not stated in the problem?

2 CHOOSEASTRATEGY How did you solve similar problems in the past? What strategies do

you know?

Try a strategy that seems as if it will work.

If it doesn't, it may lead you to one that will.

3 SOlVEIT Use the strategy you selected and solve the problem.

4 lOOkbACk Reread the question.

Did you answer the question asked? Is your answer in the correct

units? Does your answer seem reasonable?

StrategiesforProblem-‐Solving

basicstrategiesforsolvinganyproblem:

1. Drawing a Picture or Diagram

2. Making an Organized List

3. Making a Table

4. Solving a simpler related problem

5. Finding a pattern

6. Guess, Check and Revise

7. Experimenting

8. Compute or Simplify

9. Use a Formula

10. Make a model

11. Working Backwards

12. Writing an Equation

13. Eliminate


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DrawingaPictureorDiagramIf a problem is not illustrated, sometimes it is helpful to draw your own picture or diagram. A visual representation

of the situation may reveal conditions that may not be obvious when just read the problem. If the problem Is

not easily pictured, a simple diagram using symbols to represent the situation may help clarify the problem for

you.

ProblemThe eight teams of the city League will determine this season's champion with a single-elimination tournament.

That is, a team will be out of the tournament after one loss. How many tournament games will the championship

team have to play?

DrawingaPictureorDiagramIf a problem is not illustrated, sometimes it is helpful to draw your own picture or diagram. A visual representation

of the situation may reveal conditions that may not be obvious when just read the problem. If the problem Is

not easily pictured, a simple diagram using symbols to represent the situation may help clarify the problem for

you.


.. 34 ..

MakinganOrganizedlistA useful problem solving strategy is organizing information into some type of list, a technique that may serve

a variety of purposes. When a problem requires you to generate a large amount of data, a list may help you

account for all permutations and combinations and avoid repetition.

ProblemThree darts are thrown at the target shown below.

Assume that each of the darts lands within one of the rings

or within the bull's eye. How many different

point totals are possible.

MakinganOrganizedlist• Look at the problem carefully.

• What is being asked here?

• We are only interested in the totals. Hence the order in which the darts hit the target is not relevant.

• There are many ways to organize a list of possible totals. For our list, let us focus on the number of darts

that might hit the Bull's eye

MakinganOrganizedlist3 Darts hit the Bull's

eye

2 Darts hit the Bull's

eye


eye


eye

9+9+9 = 27 9+9+4 = 22

9+9+3 = 21

9+4+4 = 17

9+4+3 = 16

9+3+3 = 15

4+4+4 = 12

4+4+3 = 11

4+3+3 = 10

3+3+3 = 9

Answer: 10 Different point totals are possible

3

4

9


.. 35 ..

Making a table When a problem involves data that has more than one characteristic, an effective problem solving strategy is

to organize into a table. A table displays data so that it is easily located and understood. It is obvious to see the

missing data also. A table also serves as an aid in detecting invaluable patters.

ProblemSuppose that you roll two number cubes, each of which has faces numbered from 1 through 6. What is the

probability of rolling a sum of 8 in the uppermost faces.

Making a table • To solve this problem, you not only need to determine how many different numerical sums are possible,

but also in how many ways it is possible for the two number cubes to form the sum.

• A Table helps to produce and display data in an organized manner

Making a table

First Cube

S

e

c

o

n

d

C

u

b

e

Answer: 5/36


.. 36 ..

Solving a simpler related problem When a problem involves data that has more than one characteristic, an effective problem solving strategy is



ProblemThe houses on Main street are numbered consecutively from 1 to 150. How many houses numbers contain at

least one digit 7?

Solving a simpler related problem • You could solve this by examining every house number from 1 to 150. This involves more work than

necessary.

• Let us see how we can solve this by separating this problem in to two simpler problems

• How many house numbers contain the digit 7 in their unit place?

• How many house numbers contain the digit 7 in their tens place?

Howmanyhousenumberscontainthedigit7intheirunitplace?7 occurs once every 10 numbers in the unit place. For houses numbered 1 to 150, this happens 15 times.

How many house numbers contain the digit 7 in their tens place?

There are 10 such numbers, 70 to 79. Out of this 77 is already covered under the previous case.

Hence there are 9 such numbers.

Answer: 15+9 = 24 house numbers contain at least one 7 digit.


.. 32 ..





least one digit 7?


necessary.




how many house numbers contain the digit 7 in their unit place?7 occurs once every 10 numbers in the unit place. For houses numbered 1 to 150, this happens 15 times.





FinalMB3.indd 32 9/18/2012 1:52:38 PM


.. 37 ..

Finding a pattern At first, this looks like a tedious exercise in addition. Rather than proceeding with this lengthy calculation, let

us consider the sums of a few smaller series of numbers and try to identify a pattern

We see a pattern here.

The sum of 1 odd number is 1

The sum of first 2 odd numbers is 4 (2²)

The sum of first 3 odd numbers is 9 (3³)

The sum of first 50 odd numbers (1+3+5..+97+99) = 50² = 2500

Finding a pattern One of the most frequently used problem solving strategy is to identify a pattern. Often, this strategy is used

in conjunction with other strategies. Identifying patterns in a problem and simplifying the problem is a very

effective problem solving technique.

Problem

Whatisthesumofthefollowingseriesofnumbers?

1+3+5+7+….+97+99


.. 32 ..





least one digit 7?


necessary.




how many house numbers contain the digit 7 in their unit place?7 occurs once every 10 numbers in the unit place. For houses numbered 1 to 150, this happens 15 times.





FinalMB3.indd 32 9/18/2012 1:52:38 PM


.. 38 ..

Guess, check and ReviseAn effective way to solve certain problems is to make a reasonable guess of the answer, then check the guess

against the conditions of the problem. Sometimes, the first guess might lead you to the right answer and other

times, you might have to make a series of guesses before you succeed.

ProblemArrange the counting numbers from 1

to 6 in the circles at the right so that the sum of the numbers

along each side of the triangle is 10

Afewguesseswillleadyoutotheconclusionthatthereareonly3waystoobtainasumof10

1) 1+3+6 2) 2+3+5 3) 1+4+5

2) Out of these numbers, 1,3 and 5 appear in more than one combination. So they must be at the corners

of the triangle. Numbers 2,4,6 can be placed in of the remaining circles of the triangle.

18

1

4

5

6

3 2

3

6

1

2

5

3

2

5

6

1 4

1

6

3

4

5 2

5

2

3

4

1 6

5

4

1

2

3 6


.. 39 ..

Thisprobleminvolvesnotonlyrearrangementofthegivenobjects,butalsosomecarefulobservation

of geometric relationships. For a number of people, the simpler approach is to get 12 tooth picks and

experiment.

Step#1 Step#2 Step#3

ExperimentingProblems involving geometric configurations or spatial relationships can be solved by experimenting with a

physical model in which concrete objects can be manipulated.

ProblemThe figure at the right shows twelve tooth picks arranged to form three squares. How can you form 5 squares

by moving only 3 tooth picks?


.. 40 ..

Compute or Simplify

Manyproblemsarestraightforwardandrequirenothingmorethantheapplicationofarithmeticrules.

When solving problems, simply apply the rules and remember the order of operations.

Problem:Given(63)(54) = (N)(900), find N.

FINDOUT What are we asked? The value of N that satisfies an equation.

CHOOSEASTRATEGY Will any particular strategy help here? Yes, factor each term in the equation into

primes. Then, solve the equation noting common factors on both sides of the

equation.

SOlVEIT Break down the equation into each term's prime factors.

63 = 6 x 6 x 6 = 2 x 2 x 2 x 3 x 3 x 3

54 = 5 x 5 x 5 x 5

900 = 2 x 2 x 3 x 3 x 5 x 5

Two 2's and two 3's from the factorization of 63 and two 5's from the

factorization of 54 cancel the factors of 900. The equation reduces to

2 x 3 x 5 x 5 = N, so N = 150.

lOOkbACk Did you answer the question? Yes.

Does our answer make sense? Yes-since 900 = 302 = (2 x 3 x 5)2, we could have

eliminated two powers of 2, 3 and 5 to obtain the same answer.


.. 41 ..

UseaFormulaFormulas are one of the most powerful mathematical tools at our disposal. Often, the solution to a problem

involves substituting values into a formula or selecting the proper formula to use. When students encounter

problems for which they don't know an appropriate formula, they should be encouraged to discover the

formula for themselves.

Problem: The formula F = 1.8C + 32 can be used to convert temperatures between degrees Fahrenheit (F) and

degrees Celsius (C). How many degrees are in the Celsius equivalent of -22oF?

FINDOUT What are we trying to find? We want to know a temperature in degrees

Celsius instead of degrees Fahrenheit.

CHOOSEASTRATEGY Since we have a formula which relates Celsius and Fahrenheit temperatures, let's

replace F in the formula with the value given for degrees Fahrenheit.

SOlVEIT The formula we're given is F = 1.8C + 32. Substituting -22 for F in the equation

leads to the following solution:

-22 = 1.8C + 32

-22 32 = 1.8C

-30 = C

The answer is -30oC.

lOOkbACk Is our answer reasonable? Yes.


.. 42 ..

Make a Model Mathematics is a way of modeling the real world. A mathematical model has traditionally been a form of an

equation. The use of physical models is often useful in solving problems. There may be several models ap-

propriate for a given problem. The choice of a particular model is often related to the mathlete's previous

knowledge and problem-solving experience. Objects and drawings can help to visualize problem situations.

Acting out is also a way to visualize the problem. Writing an equation is an abstract way of modeling a problem

situation. The use of modeling provides a method for organizing information that could lead to the selection

of another problem-solving strategy.

UsePhysicalModels

ProblemFour holes are drilled in a straight line in a rectangular steel plate. The distance between hole 1 and hole 4 is

35mm. The distance between hole 2 and hole 3 is twice the distance between hole 1 and hole 2. The distance

between hole 3 and hole 4 is the same as the distance between hole 2 and hole 3. What is the distance, in mil-

limeters, between the center of hole 1 and the center of hole 3?

FINDOUT We want to know the distance between hole 1 and hole 3.

What is the distance from hole 1 to hole 4? 35mm.

What is the distance from hole 1 to hole 2? Half the distance from hole 2 to hole 3.

What is the distance from hole 3 to hole 4? The same as from hole 2 to hole 3.

CHOOSEASTRATEGY Make a model of the problem to determine the distances involved

SOlVEIT Mark off a distance of 35mm.

Place a marker labeled #1 at the zero point and one labeled #4 at the 35-mm

point.

Place markers #2 and #3 between #1 and #4.

1) Move #2 and #3 until the distances between #2 & #3 and #3 & #4 are equal.

2) Is the distance between #1 & #2 equal to half the distance between #2 & #3?

Adjust the markers until both of these conditions are met.

Measure the distances to double check. The distance between #1 and #3 is

21mm.

lOOkbACk Does our answer seem reasonable? Yes, the answer must be less than 35.


.. 43 ..

MakeaModel-ActOuttheProblemThere may be times when you experience difficulty in visualizing a problem or the procedure necessary for its

solution. In such cases you may find it helpful to physically act out the problem situation. You might use people

or objects exactly as described in the problem, or you might use items that represent the people or objects.

Acting out the problem may itself lead you to the answer, or it may lead you to find another strategy that will

help you find the answer. Acting out the problem is a strategy that is very effective for young children.

ProblemThere are five people in a room and each person shakes every other person's hand exactly one time. How many

handshakes will there be?

Get five friends to help with this problem.

Make a list with each person's name at the top of a column.

Have the first person shake everyone's hand. How many handshakes were there? Four. Repeat this four more

times with the rest of the friends. Write down who each person shook hands with. Our table should look some-

thing like this:

FINDOUT We are asked to determine the total number of handshakes.

How many people are there? Five.

How many times does each person shake another's hand? Only once.

CHOOSEASTRATEGY Would it be possible to model this situation in some way? Yes, pick five friends

and ask them to act out the problem. Should we do anything else? Keep track of

the handshakes with a list.

SOlVEIT Rhonda Jagraj Rosario Kiran Margot

Jagraj Rosario Kiran Margot Rhonda

Rosario Kiran Margot Rhonda Jagraj

Kiran Margot Rhonda Jagraj Rosario

Margot Rhonda Jagraj Rosario Kiran

lOOkbACk Did we answer the question? Yes.

Does our answer seem reasonable? Yes.


.. 44 ..

MakeaModel-UseDrawingsorSketches

ProblemIf an eight-inch square cake serves four people, how many twelve-inch square cakes are needed to provide

equivalent servings to eighteen people?

Draw an 8 x 8 cake and cut it into 4 equal pieces. Since each piece is a square with side length of

4, the area of each piece is 4 x 4 = 16 square inches.

FINDOUT We are to find how many 12 x 12 cakes are needed.

How big is the original cake? 8 x 8. How many people did it feed? 4.

How big are the other cakes? 12 x 12.

How many people must they feed? 18

CHOOSEASTRATEGY How should we approach this problem? Diagram the cakes to understand the

size of the portions.

SOlVEIT So each person gets 16 square inches of cake.

18 people times 16 square inches per person equals 288 total square inches of cake

needed.

We know that a 12 x 12 cake contains 144 square inches of cake.

288 divided by 144 equals 2, so two 12 x 12 cakes are required to feed 18 people.

lOOkbACk Did we answer the correct question, and does our answer seem reasonable? Yes.

4

4


.. 45 ..

MakeaModel-UseEquations

ProblemLindsey has a total of $82.00, consisting of an equal number of pennies, nickels, dimes and quarters. How

many coins does she have in all?

FINDOUT We want to know how many coins Lindsey has.

How much money does she have total? $82.00.

How many of each coin does she have? We don't know exactly, but we know that

she has an equal number of each coin.

CHOOSEASTRATEGY We know how much each coin is worth, and we know how much all of her coins

are worth total, so we can write an equation that models the situation.

SOlVEIT Let p be the number of pennies, n the number of nickels, d the number of dimes,

and q the number of quarters.

We then have the equation p + 5n + 10d + 25q = 8200.

We know that she has an equal number of each coin, so p = n = d = q. Substituting

p for the other variables gives an equation in just one variable. The equation above

becomes p + 5p + 10p + 25p = 41p = 8200, so p = 200.

Lindsey has 200 pennies. Since she has an equal number of each coin,

she also has 200 nickels, 200 dimes and 200 quarters. Therefore, she has 800 coins.

lOOkbACk Did we answer the question asked? Yes.

Does our answer seem reasonable? Yes, we know the answer must be

less than 8200 (the number of coins if they were all pennies) and greater than 328

(the number of coins if they were all quarters).


.. 46 ..

Make a table, Chart, or List

Make a Chart

ProblemHow many hours will a car traveling at 45 miles per hour take to catch up with a car traveling at 30 miles per

hour if the slower car starts one hour before the faster car?

FINDOUT What is the question we have to answer? How long does it take for the faster car

to catch the slower car.

What is the speed of the slower car? 30 miles per hour. What is the speed of the

faster car? 45 miles per hour.

CHOOSEASTRATEGY What strategy will help here? We could model this on paper, but accuracy would

suffer. We could also use equations. But let's make a table with the time and

distance traveled since that will explicitly show what's happening

here.

SOlVEIT Make a table with two rows and four columns.

The rows will identify the cars, and the columns will mark the hours.

Where the rows and columns intersect will indicate distance traveled, since dis-

tance equals the speed times the amount of time traveled.

Car\Hour 1 2 3 4

Slow Car 30 60 90 120

Fast Car 0 45 90 135

At the end of the first hour, the faster car was just starting. At the end of the sec-

ond hour, the faster car had gone 45 miles. At the end of the third hour, the faster

car had gone 90 miles. This equals the distance traveled by the slower car in three

hours. So, the faster car only traveled for two hours.


Does our answer seem reasonable? Yes.


.. 47 ..

Workbackwards

A student needs at least a 95% average to receive a grade of A. On the first three tests the student averaged

92%. What is the minimum a student must average on the last two tests to receive a grade of A?

FINDOUT We are asked to find what a student must average on her last two tests to get an A.

What average is required for an A? 95%.

How many tests will be figured into the average? Five. How many test has she

taken so far? Three.

What is her average on the first three tests? 92%.

CHOOSEASTRATEGY What strategy would work well in this situation? Work backwards from the

minimum required average needed for an A to find the scores needed on the last

two tests.

SOlVEIT Work backwards from the required average on all five tests.

The average of the tests must be 95%. There are five tests so the

total number of points scored on the five tests must be, at least, 5 x

95 = 475.

So far the student averages 92% on three tests. While we don't know all of the

individual scores, the total number of points scored on the three tests must be 3

x 92 = 276.

475 points required minus 276 scored so far equals 199 required on the next two

tests.

199 divided by 2 equals 99.5.

The student must average 99.5% on her next two tests if she is to get an A.

lOOkbACk Did we answer the question asked? Yes. Does our answer seem reasonable? Yes,

we knew we were looking for a number

between 95 and 100.


.. 48 ..

Writing an equationAlgebra involves shorthand mathematical notation to represent different quantities and the relationship

between them. Usually letters of the alphabet or some other symbols are used to represent the unknown

quantity known as variables. Solving the equation or the inequality usually leads to the solution

ProblemThe triple of what number is 16 greater than that number?

Triple of a number

3n = n+16

is 16 greater than the

number

n = 16/2 = 8


.. 49 ..

eliminate The strategy of elimination is commonly used by people in everyday life. In a problem-solving context,

mathletes must list and then eliminate possible solutions based upon information presented in the problem.

The act of selecting a problem- solving strategy is an example of the elimination process. Logical reasoning

is a problem-solving strategy that is used in all problem-solving situations. It can result in the elimination

of incorrect answers, particularly in "if-then" situations and in problems with a listable number of possible

solutions.

ProblemWhat is the largest two-digit number that is divisible by 3 whose digits differ by 2?

FINDOUT What are we asked to find? A certain number.

What do we know about the number? The number is less than 100. It is divisible by

3. The digits of the number differ by 2.

CHOOSEASTRATEGY What strategy will help here? Working backwards from 99, list numbers and

eliminate those that do not satisfy the conditions given.

(Notice that we have already eliminated numbers greater than 99.)

SOlVEIT 99, 98, 97, 96, 95, 94, 93, 92, 91, 90,

89, 88, 87, 86, 85, 84, 83, 82, 81, 80,

79, 78, 77, 76, 75, 74, 73, 72, 71, 70, . . .

Eliminate those numbers that are not divisible by 3:

99, 98, 97, 96, 95, 94, 93, 92, 91, 90,

89, 88, 87, 86, 85, 84, 83, 82, 81, 80,

79, 78, 77, 76, 75, 74, 73, 72, 71, 70, . . .

From these, eliminate all numbers whose digits do not differ by 2:

99, 96, 93, 90, 87, 84, 81, 78, 75, 72, . . .

75 is the largest number that remains.


Do we have a two-digit number divisible by 3 whose digits differ by 2? Yes.


.. 50 ..

Reference (online)1. http://mathcounts.org2. http://www.unl.edu/amc/3. http://mathworld.wolfram.com4. http://edhelper.com5. http://mathforum.org6. http://www.artofproblemsolving.com/Alcumus/Introduction.php7. http://www.artofproblemsolving.com/Edutainment/FTW/documentation.php8. http://www.northsouth.org/st/kc/ProblemSolving.asp9. http://agmath.com10. http://MATHCOUNTS.saab.org/mc.cgi11. http://texasmath.org/

Reference (Books)1. MathCounts handbooks (Volume I and II) 2. The All Time Greatest Problems (MATHCOUNTS) 3. The Art of Problem Solving Volume 1:The Basics By Sandor Lehoczky and Richard Rusczyk 4. Practice Competitions for MATHCOUNTS By Josh Frost 5. Competitions Mathematics for Middle School By Jason Batterson 6. Introduction to Algebra by Richard Rusczyk 7. Introduction to Counting & Probability by David Patrick 8. Introduction to Geometry by Richard Rusczyk 9. Introduction to Number Theory by Mathew Crawford 10. Middle school Math Scott Foresman (Addison Wesley) 11. Mathematical Olympiads: Problems and Solutions from Around the World, 1999- ‐2000

(MAA Problem Book Series) by Titu Andreescu12. Geometry Revisited (New Mathematical Library) by 13. H. S. M. Coxeter; Samuel L.Greitzer (Published by MAA)


.. 51 ..

APPENDIxb

MATHCOUNTSTOOlbOx-Facts,FormulasandTricks


.. 52 ..

I. PRIME NUMBERS from 1 through 100 (1 is not prime!)2 3 5 711 13 17 19 23 29 31 37 41 43 47 53 5961 6771 73 7983 89 97

II FRACTIONS DECIMALS PERCENTS1/2 .5 50 %1/3 .3 33 .3 %2/3 .6 66 .6 %1/4 .25 25 %3/4 .75 75 %1/5 .2 20 %2/5 .4 40 %3/5 .6 60 %4/5 .8 80 %1/6 .16 16 .6 %5/6 .83 83 .3 %1/8 .125 12.5 %3/8 .375 37.5 %5/8 .625 62.5 %7/8 .875 87.5 %1/9 . 1 11. 1 %1/10 .1 10 %1/11 . 09 9. 09 %1/12 . 083 8 .3 %1/16 .0625 6.25 %1/20 .05 5 %1/25 .04 4 %1/50 .02 2 %

III. PERFECT SQUARES AND PERFECT CUBES

12 = 162 = 36112 = 121162 = 256212 = 441

22 = 472 = 49122 = 144172 = 289222 = 484

32 = 982 = 64132 = 169182 = 324232 = 529

42 = 1692 = 81142 = 196192 = 361242 = 576

52 = 25102 = 100152 = 225202 = 400252 = 625

13 = 1 63 = 216

23 = 873 = 343

33 = 2783 = 512

43 = 6493 = 729

53 = 125103=1000


.. 53 ..MATHCOUNTS Coaching Kit 42

IV. SQUARE ROOTS1 = 1 2 ≈ 1.414 3 ≈ 1.732 4 = 2 5 ≈ 2.2366 ≈ 2.449 7 ≈ 2.646 8 ≈ 2.828 9 = 3 10 ≈ 3.162

V. FORMULAS

Perimeter: Volume:Triangle p = a + b + c Cube V = s3

Square p = 4s Rectangular Prism V = lwh.

Rectangle p = 2l + 2w Cylinder V = πr2hCircle (circumference) c = 2πr Cone V = (1/3)πr2h

c = πd Sphere V = (4/3)πr3

Pyramid V = (1/3)(area of base)h.

Area:Rhombus A = (½)d1d2 Circle A = πr2

Square A = s2 Triangle A = (½)bh.a

Rectangle A = lw = bh Right Triangle A = (½)l1l2Parallelogram A = bhTrapezoid A = (½)(b1 + b2)h.

Total Surface Area: Lateral Surface Area:Cube T = 6s2 Rectangular Prism L = (2l + 2w)hRectangular Prism T = 2lw + 2lh + 2wh Cylinder L = 2πrhCylinder T = 2πr2 + 2πrhSphere T = 4πr2

Distance = Rate × Time

Slope of a Line with Endpoints (x1, y1) and (x2, y2): slope = m =y y

x x2 1

2 1

−−

� ��

Distance Formula: distance between two points or length of segment with endpoints (x1, y1) and (x2, y2)

D = x x y y2 12

2 12− + −� � � �

Midpoint Formula: midpoint of a line segment given two endpoints (x1, y1) and (x2, y2)

x x y y1 2 1 2

2 2+ +�

��,

Circles:

Length of an arc = xr

3602�

�� π� � , where x is the measure of the central angle of the arc

Area of a sector = xr

3602�

�� π� � , where x is the measure of the central angle of the sector

Equilateral Triangle A = (¼) s2 3


.. 54 ..

MATHCOUNTS Coaching Kit 43

Combinations (number of groupings when the order of the items in the groups does not matter):

Number of combinations = N

R N R

!!( )!−

, where N = # of total items and R = # of items being chosen

Permutations (number of groupings when the order of the items in the groups matters):

Number of permutations = N

N R

!( )!−

, where N = # of total items and R = # of items being chosen

Length of a Diagonal of a Square = s 2

Length of a Diagonal of a Cube = s 3

Length of a Diagonal of a Rectangular Solid = x y z2 2 2+ + , with dimensions x, y and z

Number of Diagonals for a Convex Polygon with N Sides = N N( )− 32

Sum of the Measures of the Interior Angles of a Regular Polygon with N Sides = (N − 2)180

Heron’s Formula:For any triangle with side lengths a, b and c,� Area s s a s b s c= − − −( )( )( ) , where s = ½(a + b + c)

Pythagorean Theorem: (Can be used with all right triangles)a2 + b2 = c2 , where a and b are the lengths of the legs and c is the length of the hypotenuse

Pythagorean Triples: Integer-length sides for right triangles form Pythagorean Triples – the largestnumber must be on the hypotenuse. Memorizing the bold triples will also lead to other triples that aremultiples of the original.

3 4 5 5 12 13 7 24 25

6 8 10 10 24 26 8 15 17

9 12 15 15 36 39 9 40 41

Special Right Triangles:45o – 45o – 90o 30o – 60o – 90o

hypotenuse = 2 (leg) = a 2 hypotenuse = 2(shorter leg) = 2b

leg = hypotenuse2

= c

2longer leg = 3 (shorter leg) = b 3

shorter leg = longer leg3

= hypotenuse2

a

45°

a

c

45° a

b

c

60°

30°


.. 55 ..


Geometric Mean: a

x

x

b= therefore, x2 = ab and x ab=

Regular Polygon: Measure of a central angle = 360n

, where n = number of sides of the polygon

Measure of vertex angle = 180 360−n

, where n = number of sides of the polygon

Ratio of Two Similar Figures: If the ratio of the measures of corresponding side lengths is A:B,then the ratio of the perimeters is A:B, the ratio of the areas is A 2 : B 2 and the ratio of thevolumes is A 3 : B 3 .

Difference of Two Squares: a b a b a b2 2− = − +� �� Example: 12 9 12 9 12 9 3 21 632 2− = − + = ⋅ =� ��

144 81 63− =Determining the Greatest Common Factor (GCF): 5 Methods

1. Prime Factorization (Factor Tree) – Collect all common factors2. Listing all Factors3. Multiply the two numbers and divide by the Least Common Multiple (LCM) Example: to find the GCF of 15 and 20, multiply 15 × 20 = 300,

then divide by the LCM, 60. The GCF is 5.4. Divide the smaller number into the larger number. If there is a remainder, divide the remainder into the divisor until there is no remainder left. The last divisor used is the GCF.

Example: 180 385 25 180 5 25 5 is the GCF of 180 and 385

5. Single Method for finding both the GCF and LCM Put both numbers in a lattice. On the left, put ANY divisor of the two numbers and put the quotients below the original numbers. Repeat until the quotients have no common factors except 1 (relatively prime). Draw a “boot” around the left-most column and the bottom row. Multiply the vertical divisors to get the GCF. Multiply the “boot” numbers (vertical divisors and last-row quotients) to get the LCM.

40 140 40 140 40 140 The GCF is 2×10 = 20 2 20 70 2 20 70 2 20 70 The LCM is 10 10 2 7 2×10×2×7 = 280

VI. DEFINITIONS

Real Numbers: all rational and irrational numbers

Rational Numbers: numbers that can be written as a ratio of two integers

Irrational Numbers: non-repeating, non-terminating decimals; can’t be written as a ratio of two integers (i.e. 7 , π )

360 175 2525 5 0


.. 56 ..MATHCOUNTS Coaching Kit 45

Integers: {…, -3, -2, -1, 0, 1, 2, 3, …}

Whole Numbers: {0, 1, 2, 3, …}

Natural Numbers: {1, 2, 3, 4, …}

Common Fraction: a fraction in lowest terms (Refer to “Forms of Answers” in the MATHCOUNTSSchool Handbook for a complete definition.)

Equation of a Line:

Standard form: Ax + By = C with slope = − A

BSlope-intercept form: y = mx + b with slope = m and y-intercept = b

Regular Polygon: a convex polygon with all equal sides and all equal angles

Negative Exponents: xx

andx

xnn n

n−−= =1 1

Systems of Equations: x y+ = 10 8 + y = 10 (8, 2) is the solutionx y− = 6 y = 2 of the system

2 16x =x = 8

Mean = Arithmetic Mean = Average

Mode = the number(s) occurring the most often; there may be more than one

Median = the middle number when written from least to greatest If there is an even number of terms, the median is the average of the two middle terms.

Range = the difference between the greatest and least values

Measurements:1 mile = 5280 feet1 square foot = 144 square inches1 square yard = 9 square feet1 cubic yard = 27 cubic feet

VII. PATTERNS

Divisibility Rules: Number is divisible by 2: last digit is 0,2,4,6 or 8

3: sum of digits is divisible by 34: two-digit number formed by the last two digits is divisible by 45: last digit is 0 or 56: number is divisible by both 2 and 38: three-digit number formed by the last 3 digits is divisible by 89: sum of digits is divisible by 9

10: last digit is 0


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Sum of the First N Odd Natural Numbers = N 2

Sum of the First N Even Natural Numbers = N 2 + N = N(N + 1)

Sum of an Arithmetic Sequence of Integers: N

2× (first term + last term), where N = amount of

numbers/terms in the sequence

Find the digit in the units place of a particular power of a particular integer Find the pattern of units digits: 71 ends in 7

72 ends in 9(pattern repeats 73 ends in 3every 4 exponents) 74 ends in 1

75 ends in 7Divide 4 into the given exponent and compare the remainder with the first four exponents.(a remainder of 0 matches with the exponent of 4)Example: What is the units digit of 722? 22 ÷ 4 = 5 r. 2, so the units digit of 722 is the same as the units digit of 72, which is 9.

VIII. FACTORIALS (“n!” is read “n factorial”)n! = (n)×(n −1)×(n − 2)×…×(2)×(1) Example: 5! = 5 4 3 2 1× × × × = 1200! = 11! = 12! = 2 Notice 6! 6 × 5 × 4 × 3 × 2 ×13! = 6 4! 4 × 3 × 2 ×14! = 245! = 1206! = 7207! = 5040

IX. PASCAL’S TRIANGLEPascal’s Triangle Used for Probability:Remember that the first row is row zero (0). Row 4 is 1 4 6 4 1. This can be used todetermine the different outcomes when flipping four coins. 1 4 6 4 1 way to get ways to get ways to get ways to get way to get

4 heads 0 tails 3 heads 1 tail 2 heads 2 tails 1 head 3 tails 0 heads 4 tails

For the Expansion of (a + b)n , use numbers in Pascal’s Triangle as coefficients. 1 (a + b)0 = 1

1 1 (a + b)1 = a + b

1 2 1 (a + b)2 = a2 + 2ab + b2

1 3 3 1 (a + b)3 = a3 + 3a2b + 3ab2 + b3

1 4 6 4 1 (a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4

1 5 10 10 5 1 (a + b)5 = a5 + 5a4b + 10a3b2 + 10a2b3 + 5ab4 + b5

= = 30


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For 2n, add all the numbers in the nth row. (Remember the triangle starts with row 0.) 1 20 = 1

1 1 21 = 1 + 1 = 2

1 2 1 22 = 1 + 2 + 1 = 4

1 3 3 1 23 = 1 + 3 + 3 + 1 = 8

1 4 6 4 1 24 = 1 + 4 + 6 + 4 + 1 = 16

1 5 10 10 5 1 25 = 1 + 5 + 10 + 10 + 5 + 1 = 32

X. SQUARING A NUMBER WITH A UNITS DIGIT OF 5

(n5)2 = n × (n + 1) 2 5 , where n represents the block of digits before the units digit of 5Examples:

(35)2 = 3×(3+1) 2 5 (125)2 = 12×(12+1) 2 5= 3×(4) 2 5 = 12×(13) 2 5

= 1 2 2 5 = 1 5 6 2 5= 1,225 = 15,625

XI. BASES��

Base 2 = binary – only uses digits 0 – 1��

Base 16 = hexadecimal – only uses digits 0 – 9, A – F (where A=10, B=11, …, F=15)

Changing from Base 10 to Another Base:What is the base 2 representation of 125 (or “125 base 10” or “12510”)?We know 125 = 1(102) + 2(101) + 5(100) = 100 + 20 + 5, but what is it equal to in base 2?12510 = ?(2n) + ?(2n-1) + … + ?(20)The largest power of 2 in 125 is 64 = 26, so we now know our base 2 number will be:?(26) + ?(25) + ?(24) + ?(23) + ?(22) + ?(21)+ ?(20) and it will have 7 digits of 1’s and/or 0’s.

Since there is one 64, we have: 1(26) + ?(25) + ?(24) + ?(23) + ?(22) + ?(21)+ ?(20)We now have 125 – 64 = 61 left over, which is one 32 = 25 and 29 left over, so we have:1(26) + 1(25) + ?(24) + ?(23) + ?(22) + ?(21)+ ?(20)In the left-over 29, there is one 16 = 24, with 13 left over, so we have:1(26) + 1(25) + 1 (24) + ?(23) + ?(22) + ?(21)+ ?(20)In the left-over 13, there is one 8 = 23, with 5 left over, so we have:1(26) + 1(25) + 1(24) + 1(23) + ?(22) + ?(21)+ ?(20)In the left-over 5, there is one 4 = 22, with 1 left over, so we have:1(26) + 1(25) + 1(24) + 1(23) + 1(22) + ?(21)+ ?(20)In the left-over 1, there is no 2 = 21, so we still have 1 left over, and our expression is:1(26) + 1(25) + 1(24) + 1(23) + 1(22) + 0(21)+ ?(20)The left-over 1 is one 20, so we finally have:1(26) + 1(25) + 1(24) + 1(23) + 1(22) + 0(21)+ 1(20) = 11111012


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Now try What is the base 3 representation of 105?The largest power of 3 in 105 is 81 = 34, so we now know our base 3 number will be:?(34) + ?(33) + ?(32) + ?(31)+ ?(30) and will have 5 digits of 2’s, 1’s, and/or 0’s.Since there is one 81, we have: 1(34) + ?(33) + ?(32) + ?(31)+ ?(30)In the left-over 105 – 81 = 24, there is no 27 = 33, so we still have 24 and the expression:1(34) + 0(33) + ?(32) + ?(31)+ ?(30)In the left-over 24, there are two 9’s (or 32’s), with 6 left over, so we have:1(34) + 0(33) + 2(32) + ?(31)+ ?(30)In the left-over 6, there are two 3’s (or 31’s), with 0 left over, so we have:1(34) + 0(33) + 2(32) + 2(31)+ ?(30)Since there is nothing left over, we have no 1’s (or 30’s), so our final expression is:1(34) + 0(33) + 2(32) + 2(31)+ 0(30) = 102203

The following is another fun algorithm for converting base 10 numbers to other bases:

12510 = ?2 10510 = ?3 12510 = ?16

1 r.1 1 r.0 7 r.13(D)2 3 r.1 3 3 r.2 16 125

2 7 r.1 3 11 r.2

2 15 r.1 3 35 r.0 12510 = 7D16

2 31 r.0 3 105

2 62 r.1

2 125 10510 = 102203

12510 = 11111012 xyzn = (x × n2) + (y × n1) + (z × n0)

Notice: Everything in bold shows the first division operation. The first remainder will be the last digitin the base n representation, and the quotient is then divided again by the desired base. The process isrepeated until a quotient is reached that is less than the desired base. At that time, the final quotientand remainders are read downward.

XII. FACTORSDetermining the Number of Factors of a Number: First find the prime factorization (include the 1 if afactor is to the first power). Increase each exponent by 1 and multiply these new numbers together.

Example: How many factors does 300 have?The prime factorization of 300 is 22 × 31 × 52 . Increase each of the exponents by 1 and multiplythese new values: (2+1) × (1+1) × (2+1) = 3 × 2 × 3 = 18. So 300 has 18 factors.

Finding the Sum of the Factors of a Number:Example: What is the sum of the factors of 10,500?(From the prime factorization 22 × 31 × 53 × 71, we know 10,500 has 3 × 2 × 4 × 2 = 48 factors.)The sum of these 48 factors can be calculated from the prime factorization, too:(20 + 21 + 22)(30 + 31)(50 + 51 + 52 + 53)(70 + 71) = 7 × 4 × 156 × 8 = 34,944.

Starthere


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APPENDIxC

PracticeProblemsandSolutions


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APPENDIxCNSFMathbeeWorkshoplevel3HomeworkProblems

1. The surface area of a cube is 384 square centimeters. What is the ratio of square centimeters in the

area to the number of cubic centimeters in the volume of the cube? Express your answer as a common

fraction in lowest terms. (Topic: Geometry – area, volume, Arithmetic - fractions)

2. What is the greatest prime factor of 34101 + 34102 + 34103 ? (Topic: Arithmetic)

3. When the numerator of a common fraction in simplest form is decreased by 35, and the denominator is

decreased by 60, the resulting fraction is equivalent to the original fraction. Express the original number

as a common fraction. (Arithmetic – Fractions)

4. How many combinations of two positive two-digit integers have 540 as the product?

(Topic: Algebra – Combinations)

5. A large cube is dipped into red paint and then divided into 729 smaller congruent cubes. One of the

smaller cubes is then selected randomly. What is the probability that cube selected will have at least

25% of its area painted red? Express your answer as a common fraction in lowest terms.

(Topic: Algebra – Probability, Geometry)

6. A grocer stacks apples in the shape of a square pyramid. The bottom layer is 10 x 10 square, the top layer

is one apple and the nth layer is an n x n square. How many apples does she have in the pyramid?

(Topic: Arithmetic, Series)

7. Sita Jogs 7 kilometers per hour and Rani jogs 5 kilometers per hour. They start together at their campsite

and jog to an outpost 42 kilometers away. When Sita gets to the outpost, she immediately turns around

and heads back towards Rani. How many kilometers will they be from the outpost when they meet?

(Topic: Algebra – Equations)

8. How many different positive integer divisors does the number 200 have, knowing that

200=(52) * (23)? (Topic: Algebra, Combinations)

9. The sum of three integers is 418. The two smaller are consecutive integers and the two larger are

consecutive odd integers. Find the product of the smallest and the largest. (Topic: Algebra - Equations)

10. Rama leased a car for 5 years. The terms of the lease were Rs 8700 down and Rs 199 per month for the

length of the lease. In addition, she had to pay Rs 0.14 per kilometer for all miles driven in excess of

80000 kilometers. What was the total number of rupees in the cost of the lease given that she drove the

car 115635 kilometers? Round your answer to the nearest rupee. (Topic: Algebra – Solving Equations)


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11. Consider the following pattern:

sqrt ( 1+ 1*2*3*4) = 5

sqrt ( 1+ 2*3*4*5) = 11

sqrt ( 1+ 3*4*5*6) = 19

sqrt ( 1+ 4*5*6*7) = 29

sqrt ( 1+ 5*6*7*8) = 41

Determine sqrt ( 1+ 87*88*89*90).

(Topic: Arithmetic – Square roots)

12. Rama’s digital clock read 5:13 a.m. when she left for school. When she returned home 5 hours and 13

minutes later, the clock read 3:30 a.m. because the power had gone off during the day. If her clock

automatically reset to 12:00 a.m. when the power was restored, at what time that morning did the

power return? (Topic: Algebra)

13. Brian has 45 meters of fencing. He will use the fencing to enclose a play area for his puppy. What is the

maximum number of square meters he can enclose? Use 3.14 as an approximation of Pi. Express your

answer to the nearest whole number. (Topic: Geometry)

14. Out of 800 fish in an aquarium 90 % are guppies. How many guppies must be removed so that the

percent of guppies in the aquarium is 20%? (Topic: Algebra – Solving Equations)

15. The whole numbers 23 through 58 are arranged 6 x 6 table, with each number occurring exactly once,

and such that the sum of the entries of any row, any column, or any main diagonal is the same. What is

the common sum? (Topic: Algebra – Magic Squares)

16. A bag contains 5 red balls, 4 green balls, and 3 blue balls. If 3 balls are removed at random and

each ball is returned to the bag after removal, what is the probability that all 3 balls will be red?

(Topic: Algebra - Probability)

17. Consider the square having as vertices the following points (-1,1), (1,1), (1,-1), and (-1,-1) .

Find the probability that a randomly selected point inside the square will have the sum of its coordinates

greater than 1/2. (Topic: Algebra - Probability)

18. A point S is chosen inside the square MNPQ. What is the probability that the angle MSN is obtuse?

Express your answer as a decimal to the nearest hundredths. (Topic: Algebra - Probability)

19. A jar contains 6 red balls, 8 green balls and 8 yellow balls. Raja draws a ball at random from the jar and

returns it. Ranga draws a ball at random from the jar. What is the probability that they drew a ball of

the same color? Give your answer as a common fraction in lowest terms.

(Topic: Algebra – Probability, Fractions)


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20. A square is divided into three congruent rectangles. Draw one of the diagonals of the

original square, the square is now divided into four trapezoids and two triangles. If

the area of one of the smallest blue shaded trapezoid is 150 square centimeters, how

many centimeters are in the perimeter of the original square? (Topic: Geometry - Area)

21. Let ABC be right triangle in B. Suppose that AC= 212 cm and AB= 112 cm . Let M a

point in the plane of the triangle ABC such that the three points A, B, M are collinear and B is between

A and M. Suppose that BM= 56 cm. On AM, we draw the perpendicular at M. Let N be the point of

intersection of AC with this perpendicular line. What is the number of square centimeters in the area of

the trapezoid BCNM? The picture below is not drawn to scale. (Topic: Geometry - Area)

22. The obtuse angle of an isosceles triangle is bisected and each resulting angle

is 52 degrees larger than a base angle. How many degrees are in the measure

of the obtuse angle? (Topic: Geometry - Angles)

23. The measures of the three angles of a triangle form an arithmetic sequence. If the smallest angle

measures 42 degrees, what is the number of degrees in measure of the largest angle? (Topic: Geometry

- Angles, Arithmetic - Sequence)

24. 27 students took a final exam. The mean score of those who passed was 64, the mean score of those

who failed was 55, and the mean score of all the students was 61. How many students did not pass the

test? (Topic: Arithmetic – Mean, Median, Mode)

25. A car that originally sold for Rs 2,60,000 depreciates at a rate of 19% per year. What is the value of the car

at the end of 5 years? Round your answer to the nearest rupee. (Topic: Arithmetic)

26. A group of students at Rajeev Memorial Middle School consists of 5 students from the 6th, 7th and 8th

grades. This means that their ages are between 10 and 14. If the product of their ages is 247,104, what is

the sum of their ages? (Topic: Arithmetic)

27. What is the greatest integer that is less than the reciprocal of 0.105? (Topic: Arithmetic)

28. Raja's strategy for his first marathon (32.3 kms) was to run 2 kms, walk 1 km, run 2 kms, walk 1 km, and

continue this pattern until he completed the race. Raja’s average running pace is 6 minutes per km

and his average walking pace is 12 minutes per km. How many kms will it take Raja to complete the

marathon? Express your answer as a decimal to the nearest tenth. (Topic: Arithmetic)

29. A person weighing 113 kgs burns about 6.9 cal/min jogging and 10.9 cal/min running. Genevieve, who

weighs 113 kgs, wishes to exercise by jogging and running. She has 40 minutes to exercise and wishes

to burn exactly 360 calories. How many minutes should she spend running? (Topic: Algebra – Solving

equations)

A

NM

B C


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30. A school organization consists of 5 teachers, 9 parents and 7 students. A subcommittee of this group

must be formed by choosing 2 teachers, 5 parents and 5 students. How many different subcommittees

can be formed? (Topic: Algebra – Permutations and Combinations)

31. At a basketball tournament involving 7 teams, each team played 4 games with each of the other teams.

How many games were played? (Topic: Algebra – Permutations and Combinations)

32. Of a group of boys and girls at Netaji Public School's school party, 15 girls left early to play in a volleyball

game. The ratio of boys to girls then remaining was 18 to 5. Later, 45 boys left for a football game. The

ratio of girls to boys was then 5 to 9. How many students attended the party? (Topic: Algebra – Solving

Equations)

33. In a party attended only by 13 politicians and 6 lawyers, each politician shook hands exactly once with

everyone and each lawyer shook hands exactly once with each politician. How many handshakes took

place? (Topic: Algebra – Permutation and Combinations)

34. Two cubes of volumes 8 cm2 and 27 cm2 are glued together at their faces to form a solid with the smallest

possible surface area. What is the number of square centimeters in the surface area of the resulting

solid? (Topic: Geometry – Surface Area)

35. An equilateral triangle of side 14 centimeters is rotated about an altitude to form a cone. What is the

number of cubic centimeters in the volume of the cone? Express your answer to the nearest whole

number (Topic: Geometry - Volume)

36. What is the value of 993 + 3(992) + 3(99) + 1? (Topic: ???)

37. The mean of ten numbers is 7.2. When an 11th number is added to the set, the mean becomes 5. What is

the 11th number? (Topic: Arithmetic – Mean, Median, Mode)

38. Rani normally spends half an hour driving to work. When her average speed is 10 kmph slower than

usual, the trip takes ten minutes longer. How many kms does she drive to work? (Topic: Algebra – Solving

Equations)

39. What is the smallest three-digit number divisible by the first three prime numbers and the first three

composite numbers? (Topic: Arithmetic)

40. The five-digit number 246n8 is divisible by 9. What is the value of the digit n?

(Topic: Algebra – Solving Equations)

41. Find the value of x for which x degrees Fahrenheit is exactly the same as x degrees Celsius (remember

the formula for converting Celsius to Fahrenheit is to multiply the temperature by 9/5 and add 32).

(Topic: Algebra – Solving Equations)


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42. Find the smallest positive integer that is one less than a multiple of each of 3, 5, 7, 9, and 11.

(Topic: Arithmetic)

43. Compute: (2+12+22+32) + (8+18+28+38) (Topic: Arithmetic)

44. If 220 – 219 = 2x, what is the value of x? (Topic: Algebra – Solving Equations)

45. If 25*83*162 = 4m, what is the value of m? (Topic: Algebra – Solving Equations)

46. The perimeter of an isosceles triangle is 150 units and the altitude to the base is 45 units. What is the

number of square units in its area? (Topic: Geometry - Area)

47. What is the greatest number of right triangular sections, each of with a base of 3 inches and a height of

5 that can be cut from a rectangular piece of paper measuring 55 inches by 21 inches? (Topic: Geometry)

48. A bag contains 4 red balls, 2 green balls, and 4 blue balls. If 2 balls are removed at random and each

ball is returned to the bag after removal, what is the probability that both the balls will be red?

(Topic: Algebra - Probability)

49. Let ABCD be a square such that AB=16 cm. Let M be the midpoint of AD and N the midpoint of AB. What

is area of the quadrilateral MNCD? Give your answer to the nearest integer. (Topic: Geometry - Area)

50. 1/5 of the cars manufactured by a company are Vans and the rest are ambassador cars. If 1/13 of the cars

produced are blue and 1/3 of the vans are blue, then what fraction of the ambassador cars are blue?

Give your answer is a common fraction in lowest term. (Topic: Arithmetic - Fractions)

51. Let ABCD be a square. We draw a circle of center C and of radius CB. A point P is on the arc BD that

is situated inside the square and is such that the distance from P to the side AB is 3 centimeters

and the distance from P to the side AD is 6 centimeters. Find the side of the square in centimeters.

(Topic: Geometry)

52. The surface area of a cube is 486 square centimeters. What is the ratio of square centimeters in the

area to the number of cubic centimeters in the volume of the cube? Express your answer as a common

fraction in lowest terms. (Topic: Geometry – Area, Volume)

53. The sequence 1, 2, 4, 7, . . . is generated by adding 1 to the first term to get the second, adding 2 to the

second term to get the third, adding 3 to the third term to get the fourth, and so on. What is the value

of the 107th term in the sequence? (Topic: Algebra – Introduction to logarithms)

54. The 9 sides of a regular 9-gon are extended to form 9 congruent isosceles triangles.

How many degrees are in the measure of the angle CAB in one of these triangles

such that its vertex is not a vertex of the polygon? (Topic: Geometry)

CA

B


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55. Let a, b and c be the lengths of the three sides of a right triangle. Suppose that

a+b+c= 26 and a2 + b2 + c2= 288. What is the number of square units in the area of this triangle ?

(Topic: Geometry – Area)

56. Three circles of radii 8, 10 and 72 are tangent to each other such that the center of each circle is outside

the two other circles. Find the number of square units in the area of the triangle whose vertices are the

centers of the three circles? (Topic: Geometry – Area)

57. The original price of an item was 300 rupees. The store deducted 20% , and then deducted an additional

20% off the reduced price. How many rupees would a consumer save if the store had simply reduced the

original price by 40%? (Topic: Arithmetic)

58. Paul made a round trip between two points A and B in 6 hours. His speed going up is 3 kmph. His

speed going down is 6 kmph . His speed going on a level surface 4 kmph . What is the distance in miles

between A and B ? (Topic: Arithmetic – Averaging Method)

59. If 87(1) +87(7) +87(21) +87(35) +87(35) +87(21) +87(7) +87(1) = 2x . Find x. (Topic: Algebra – Solving System of

Equations)

60. We select 3 numbers at random, with replacement, from the set of integers from 1 to 600 inclusive. What

is the probability that the product of the 3 numbers is even ? Express your answer as a common fraction

in lowest terms. (Topic: Algebra – Probability)

61. What is the greatest product obtainable from two integers whose sum is 2000? (Topic: Algebra –

Applications of GCF and LCM)

62. Every camper at EKO is required to take exactly two of the three crafts classes offered. One summer,

41 campers took basket weaving, 46 took cabinet making, and 49 took pottery. How many campers

attended camp EKO this summer? (Topic: Algebra – Permutations and Combinations)

63. A caterer offers 5 different appetizers, 6 different drinks and 7 different sandwiches. How many

combinations of 2 appetizers, 2 drinks and 2 sandwiches can someone choose for his party? (Topic:

Algebra – Permutations and Combinations)

64. Compute the following sum:

1+ 13 + 13!/( 2! (13-2)!)+ ... +13!/( k! (13-k)!) + ...+ 13 +1 (Topic: Algebra – Permutations and Combinations)

65. A Hollow piece of cylindrical pipe has an outside radius of 1.9 centimeters and an inside radius of 1.8

centimeters The pipe is 4 meters long. How many square centimeters are in the total surface area of the

pipe? Express you answer as a decimal to the nearest tenth. Use Pi=3.14 (Topic: Geometry – Area)


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66. A group of 42 people has a mean income of exactly Rs 30,100 . Given that one new person is added

to the group with an income of Rs 55,900, by how many rupees does the mean income of the group

increase? (Topic: Arithmetic – Averaging Method; Mean, median and Mode)

67. When each side of a square is increased by 10 meters, the area is increased by 1320 square meters. By

how many feet does each side of the original square have to be decreased in order to decrease the area

of the original square by 1320 square meters? (Topic: Geometry –Area)

68. Find |8 - 15 | + |15 - 8 | (Topic: Algebra – Absolute value)

69. If 11 < x < 16, what is the value of

| |x - 6| + |x - 29| | ? (Topic: Algebra – Absolute value)

70. If 85 % of a number is 595. What is 45 % of that number? (Topic: Arithmetic – ratio and Proportions)

71. A convex polygon has n sides and 10 n diagonals. Find the value of n. (Topic: Geometry – Circles and

Regular Polygons)

72. How many degrees are in the measure of the smaller angle that is formed by the hands of a clock when

it is 2 o'clock? (Topic: Geometry)

73. Each interior angle of a regular polygon measures 165 degrees. How many sides does the polygon have?

(Topic: Geometry – Circles and Regular Polygons)

74. Given that 243m = 3 and mn = 125 . Evaluate mn . Express your answer as a common fraction in lowest

terms. (Topic: Algebra – Theory of exponents)

75. If x=99 , compute the value of x4 + 4 x3 + 6 x2 + 4 x + 1. (Topic: Algebra – Solving Equations)

76. Shira uses 6 small skeins of yarn for each scarf she knits. In order to be ready for the next holiday, she

needs to make 15 scarves. How many small skeins of yarn does she need? (Topic: Algebra – Solving

Equations)

77. Two arithmetic sequences A and B both begin with 30 and have common differences of absolute value

10, with sequence A increasing and sequence B decreasing. What is the absolute value of the difference

between the 51st term of sequence A and the 51st term of sequence B? (Topic: Algebra – Solving Equations;

Absolute Value)

78. A rectangular prism measures 10 inches by 20 inches by 10 inches. What is the length, in inches, of the

internal diagonal? Express your answer in simplest radical form. (Topic: Geometry – Volume of Prisms and

Pyramids)


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79. Cindy wishes to arrange her coins into X piles, each consisting of the same number of coins, Y. Each

pile will have more than one coin and no pile will have all the coins. If there are 13 possible values for

Y given all of the restrictions, what is the smallest number of coins she could have? (Topic: Algebra –

Permutations and Combinations)

80. Each of the integers 1, 2, 3, ... , 16 is written on a separate slip of paper and these slips are placed in a pile.

Jillian will randomly draw slips from the pile without replacement and will continue drawing until two of

the numbers she has drawn from the pile have a product that is a perfect square. What is the maximum

number of slips that Jillian can draw without obtaining a product that is a perfect square? (Topic: Algebra

– Permutations and Combinations)

81. Eight kilograms of feathers and two grams of gold together cost Rs. 932. Fourteen kilograms of feathers

and three grams of gold together cost Rs. 1402. What is the cost of five kilograms of feathers and five

grams of gold? (Topic: Algebra – Solving Equations)

82. When rolling two standard six-sided dice, what is the probability of getting a sum larger than 10? Express

your answer as a common fraction. (Topic: Algebra – Combinations and Probability)

83. Ten pictures, each measuring 4 cm by 6 cm, are mounted on a piece of green poster board so that there

is no overlap. The poster board measures 20 cms by 17 cms. All the shapes are rectangular. How many

square inches of the green background will show after the pictures are mounted? (Topic: Geometry –

Surface Area)

84. Let n equal the number of sides in a regular polygon. For 3 ≤ n < 10, how many values of n result in

a regular polygon where the common degree measure of the interior angles is non- integral? (Topic:

Geometry – Circles and Regular Polygons)

85. Five players on the basketball team all scored a different number of points in last night’s game. Each

player scored more than 9 points. Rani scored the fewest points and Tara scored the second fewest

points. If Tara scored 16 points and 16 points was the average for the five players, how many points did

Rani score? (Topic: Arithmetic – Averaging Method)

86. The ratio of domestic stamps to foreign stamps in Rama’s stamp collection is 3:1. If Rama sold thirty of

his domestic stamps, the ratio of domestic stamps to foreign stamps would be 1:2. How many foreign

stamps does Rama have in his collection? (Topic: Arithmetic – Ratio and proportions)

87. What is the 200th term of the increasing sequence of positive integers formed by omitting only the

perfect squares? (Topic: Arithmetic – Squares and Square Roots)

88. Sita has 45 coins in her piggy bank. She has one fewer 5 paise coins than she has 10 paisa coins, and one

fewer 25-paisa coin than three times the number of 5-paisa coin. If Sita has only 5 paisa, 10 paisa and 25

paisa coins, how many 5 paisa coins does she have? (Topic: Logic – logic)


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89. The mean of three numbers is 5/9. The difference between the largest and smallest number is 1/2, which

happens to be one of the numbers. What is the smallest number? Express your answer as a common

fraction. (Topic: Arithmetic – Mean, Median and Mode)

90. A bag contains red marbles, white marbles, green marbles and blue marbles. There are an equal number

of red marbles and white marbles, and five times as many green marbles as blue marbles. There is a 35%

chance of selecting a red marble first. What is the fewest possible number of green marbles in the bag?

(Topic: Algebra – Combinations and Probability)

91. A rectangular box has interior dimensions 6 cms by 5 cms by 10 cms. The box is filled with as many solid

3 cms cubes as possible, with all of the cubes entirely inside the rectangular box. What percent of the

volume of the box do the cubes take up? (Topic: Geometry – Volume of Prisms and Pyramids)

92. Planets X, Y and Z take 360, 450 and 540 days, respectively, to rotate around the same sun. If the three

planets are lined up in a ray having the sun as its endpoint, as shown, what is the minimum positive

number of days before they are all in the exact same locations again?

(Topic: Geometry – Circles and Polygons)

93. The speed of a stream is 3 km per hour. A boat travels upstream 12 km and then returns to its original

position downstream along the same route. If the speed of the boat in still water is 9 km per hour, what

is the average speed of the boat for the entire round trip? (Topic: Arithmetic – Averaging method)

94. Ramu has a box containing only red marbles, blue marbles and green marbles. He needs to select at

least 17 marbles without replacement to be sure at least one of them is green. He needs to select at

least 18 marbles without replacement to be sure at least 1 of them is red. He needs to select at least 20

marbles without replacement to be sure all three colors appear among the marbles selected. How many

marbles are in the box? (Topic: Algebra – Permutations and Combinations)

95. Six students are being grouped into three pairs to work on a science lab. How many different

combinations of three pairs are possible? (Topic: Algebra - Permutations and Combinations)

96. An integer X has the following properties:

1.) X is a multiple of 17

2.) X is less than 1000

3.) X is one less than a multiple of 8.

What is the largest possible value of X? (Topic: Algebra - Permutations and Combinations)

97. Srinu, Raja and Ramu are the three participants in a race. In how many different ways can the three

finish if it is possible for two or more participants to finish in a tie? (Topic: Algebra – Permutations and

Combinations)

98. If three standard, six-faced dice are rolled, what is the probability that the sum of the three numbers

rolled is 9? Express your answer as a common fraction. (Topic: Algebra – Combinations and Probability)


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99. A rectangular garden has a length that is twice its width. The dimensions are increased so that the

perimeter is doubled and the new shape is a square with an area of 3600 square meters. What was the

area of the original garden, in square meters? (Topic: Geometry - Circles and Regular Polygons)

100. Given that the diagonals of a rhombus are always perpendicular bisectors of each other, what is the area

of a rhombus with side length units and diagonals that differ by 6 units? (Topic: Geometry - Circles and

Regular Polygons)


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1. Ans: ¾ Let a be the side of the cube Surface Area = 6 * a2 = 384, so a=8. Volume= a3 = 512 (Surface Area)/Volume= 384/512

2. Ans: 397 Factor the expression as 34101 ( 1 + 34 + 342) and find the greatest prime factor of

(1 + 34 + 342)=1191. It is easy to see that the answer is 397.

3. Ans: 7/12 Let a/b be the original fraction then a/b = (a-35)/(b-60) this implies that a b - 60 a = a b - 35 b

hence 60 a = 35 b and a/b= 35/60=7/12.

4. Ans: 5 {10,54},{12,45},{15,36},{18,30},{20,27}

5. Ans: 92 / 729 We have to count all the small cubes that were part of

the 12 edges. We have 8 on the corners plus 12*(9-2) that are on the edges but not on the corners for a total of 8+ 12*(9-2)=12*9 -

16=92. Hence the answer is 92/(93)=92/729.

6. Ans: 385 The answer is 12 + 22 + 32 + ... +102=385. Here one can use the formula: 12 + 22 + 32 + ... +n2= 1/6 * n * (n +1) (2*n +1) with n=8.

7. Ans: 7 Sita will take 42/7=6 hrs. to reach the outpost. At that

time Rani would have jogged 6*5 kms. Hence they are now separated by 42-6*5=12 kms. Let x be the distance from the outpost when they meet. Then

(12-x)/5=x/7, which gives x=7

8. Ans: 12 A divisor of (52) * ( 23) is of the form (5s) * ( 2t) where 0<= s<= 2 and 0<= t<= 3. Hence the number of divisors is (2+1)*(3+1)=12.

9. Ans: 19458 Let 2 n+1 be the middle number so: 2 n + 2 n + 1 + 2 n + 3 = 418 6 n + 4= 418 6 n = 418 - 4= 414 n = 414/6 = 69. Hence the three numbers are : {138,139,141}andtheansweris138*141=19458

10. Ans: 25629 8700 + 12*5* 199 + 0.14 *0.14*(115635 - 80000) =25629

to the nearest rupees

11. Ans: 7831 sqrt( 1+ 1*2*3*4) = 5 = 2*3 -1 sqrt( 1+ 2*3*4*5) = 11 = 3*4

-1 sqrt( 1+ 3*4*5*6) = 19 = 4*5 -1 sqrt( 1+ 4*5*6*7) = 29 = 5*6 -1 sqrt( 1+ 5*6*7*8) = 41 = 6*7 -1

So, sqrt( 1+ 87*88*89*90)= 88*89-1=7831.

12. Ans: 6:56 (5:13) +(5:13)-(3:30)=6:56

13. Ans: 161 The maximum area is attained when we have a circle. Hence 2*Pi*R = 45. Hence R= 45/(2*Pi) and Hence the area

is: Pi*R2= Pi * (45/2*Pi)2 = 452/(4*Pi) = 161 if Pi=3.14 to the nearest whole number.

14. Ans: 700 Let x be the number of guppies to be removed. Then one

has to solve the following equation: 800 * 90/100 - x = (800 -x) * 20/100 and find x = 700.

15. Ans: 243 This is a generalization of the magic square problem. If

the numbers were from 1 to 6*2=36 than the common sum would have been

36 * (36+1)/6=111. The numbers are now from 23 to 58, so we have to add 22

to each number in the table, then the common sum would be 111+ 6*22=243.

16. Ans: 125/1728 (5/12)3=125/1728

17. Ans: 9/32 One has to find the area above the line x + y = 1/2 and

inside the square. This is an isosceles right triangle. Each of its legs measures

1+ (2-1)/2. Its area is 1/2 * (1+ (2-1)/2)2. Hence the answer is: (1/2 * (1+ (2-1)/2)^2)/4= 9/32.

18. Ans: 0.39 The point S must be inside the semi-circle of diameter MN

inside the square. Let us suppose that R is the radius of this semi-circle. The area of this semi-circle is = Pi R2/2. The area of the

square is ( 2 R )2 = 4 R2. Hence the answer is: (Pi R2/2)/( 4 R2) = Pi/8= 0.39 to then nearest hundredths.

19. Ans: 41/121 (6/22)2 + (8/22)2 + (8/22)2=41/121.

Answers


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20. Ans: 120 Let x be the side of the square. Then the two bases of the

shaded trapezoid are x/3 and 2x/3 and the height of it is x/3. Hence its area is: (1/2)*(x/3)*( x/3 + 2x/3) = (x2)/6 = 150.

This implies that x2 = 6*150 = 900 and x = Sqrt[900] = 30. The answer is 4*30=120.

21. Ans: 12600 BC=Sqrt[2122-1122]=180. AB/AM=BC/MN= 112/168.

Hence MN= (168/112)*180=270. Area of BCNM= (1/2) * 56 * (180+ 270) = 12600

22. Ans: 142 Let x be the measure of the obtuse angle. The measure of the base angle will be x/2- 52. One has to solve x + (x/2- 52)2 =180 x + x -2 * 52 = 180 2 x = 180 + 2 * 52 x =90 + 52 = 142.

23. Ans: 78 The angles will be 42, 42 + x and 42 +2 x degrees. So 42+ 42 + x + 42 +2 x= 180 Solving for x gives x= 60 - 42=18. The largest angle will be

42 + 2* 18= 78 degrees.

24. Ans: 9 Let x be the number of students who failed. Then 27-x is the number of the students who passed. Then solve the following equation: 64(27-x) + 55 x = 61*27=1647 to find x=9.

25. Ans: RS 90660 The answer is 260000 * ((100-19)/100)5 = 260000 * 0.815 = 90660 to the nearest rupees.

26. Ans: 60 the prime factorization of 247,104 is: 247,104= (26) * (33) * (111) * (131). So there is 1 student of age 13,1 student(s) of age 11, 3 students of age 12. Hence the sum of their ages is: 13 * 1 + 11 * 1 + 12 * 3 = 60.

27. Ans: 9 0.105=105/1000 so 1/0.105 = 1000/105= 9 plus a remainder of 55 so the

answer is 9.

28. Ans: 255.6 Notice that 32.3=3*10+ 2 +0.3. So the time taking will be

(2*6 + 12)*10 + 2*6 + 0.3*12=255.6

29. Ans: 21 Let x be the number of minutes she should spend

running, then one has to solve the equation:

10.9 x + 6.9 (40-x) = 360 and find x=21.

30. Ans: 26460 (5 choose 2 ) * (7 choose 5 ) * (9 choose 5 )= (5!)/( 2! (5-2)!) * (7!)/( 5! (7- 5)!) * (9!)/( 5! (9- 5)!) 10 * 21 * 126 = 26460.

31. Ans: 84 The first team plays with the other 6 teams 24=6 * 4

games The second team plays with the other 6 teams 24=6 *

4 games and so on. In total we will have 168= 6 * 4 * 7 games. This way we have counted each team twice. Hence the answer is 84=168/2

32. Ans: 130 Let b be the number of boys and g be the number of girls.

b/(g-15) = 18/5 (g-15)/(b-45)=5/9 Cross multplying, simplifying and solving gives: b=90 and g=40. Hence the answer is b+g=130

33. Ans: 156 Letthe13politiciansbe{p1,p2,...,p13} p1 shook 19-1 hands; p2 shook 19-2 hands; (not counting the hand shaking he

did with p1) p3 shook 19-3 hands; (not counting the hand shaking he

did with p1 and p2) p13 shook 19-13 hands; (not counting the hand shaking

he did with p1,p2, ... and p5) Now every lawyer's only handshake with the other

politicians has been counted. So we have to add (19-1)+ (19-2)+ (19-3)+ ... + (19-13)= 13*19 - 13*(13+1)/2=156

34. Ans: 70 sq. cm The side lengths of the two cubes are 2 and 3. They would

be glued together so that the 2 side length cube covers up 4 square units of the 3 side length cube. 4*5 is the surface area of the 2 side length cube. 9*6-4 is the surface area of the three side length cube. 50+20= 70 sq.cm.

35. Ans: 1866 cm3

The base of the cone is a circle with radius 7. So the area of the base is 49 π. The height is 7√3. So the volume is 49 π * 7 √3 / 3. Estimated to the nearest whole number that is 622 cm3.

36. Ans: 10 laks 99 = x x3 + 3x2 + 3x + 1 = (x + 1)3 (99 + 1)3 = 1003 = 10,00,000

37. Ans: -17 The ten numbers have to add up to 7.2*10= 72 the 11 numbers add up to 11*5 = 55 so the next number

is 55-72= -17


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38. Ans: 20 x is the normal speed. The equation is 30x=(x-10)40 solving x=40. to answer the question multiply 40 by .5 rather than 30

because it is 30 minutes and that is .5 hours as in kmph. 40*.5=20

39. Ans: 120 The three digit number has to be divisible by: 2 3 4 = 2*2 5 6 = 2*3 8 = 2*2*2 the LCM of those numbers is 2*2*2*3*5, or 120.

40. Ans: 7 246n8 is divisible by 9 when 2+4+6+n+8 is divisible by 9. 20+n is divisible by 9. n=7

41. Ans: -40 (9x/5) + 32 = x (4x/5) = -32 4x = -160 x = -40

42. Ans: 3464 First, find the LCM of 3, 5, 7, 9, and 11. Any number that is

divisible by nine is automatically divisible by 3. The LCM is : 5x7x9x11 = 3465 Answer 3465 - 1 = 3464

43. Ans: 160 Grouping 8's and 2's we have (2+8) + (12+18) + (22+28) + (32+38) = 10+30+50+70=160

44. Ans: 19 Taking 219 common we have, 219 (2-1) =219*1 therefore x=19

45. Ans: 11 25*83*162 =22*23*83*162 =22*163*162 =4*46*44 =411 therefore m=11

46. Ans: 1080 Let x be the length base for this triangle. Then (150-x)/2 will be the length of the two equal sides.

Then one can write the following equation: ((150-x)/2)2 = (x/2)2 + 452 and get x=48. Hence the area

is: 48 * 45/2 = 1080.

47. Ans: 154 The number is clearly twice the number of rectangles of

sides 3 and 5. To get the greatest number of rectangles, one has to

subdivide 55 into 11 parts of 5 inches each and 21 in 7 parts of 3 inches each. Hence the answer is:

2 * 7 * 11 = 154.

48. The answer is: (4/10)2=4/25.

49. Area of Square - Area of AMN - Area BNC= 162 - (16/2) * (16/2) * (1/2) - (16/2) * 16 *(1/2)= (5/8)* 162=

160 exactly.

50. Let x be the numbers of all cars. We have x/5 vans and (1-1/5)*x sedans, x/( 3 * 5) blue

vans. Hence we have x/13 - x/(5 * 3)= x*( 1/13 - 1/(5 * 3)) blue

ambassador cars. Hence the fraction of blue ambassador cars is: (x * ( 1/13 -

1/(5 * 3)) /( (1 -1/5) * x) = ( 1/13 - 1/(5 * 3)) /(1 -1/5) = 1/78.

51. Let x be the side of the square. From the point P, draw a perpendicular PH to BC. It is easy to see that PH= x - 6 and HC=x - 3.

In the right triangle PHC one can write: x2 = (x-6)2 + (x-3)2

Solving for x and choosing the appropriate root, we find that x=15.

52. Let ‘a’ be the side of the cube. Then Surface Area = 6 * a2 = 486. This implies that a=9.

Volume= a3 = 729. Hence (Surface Area)/Volume= 486/729= 2/3 after

simplification if needed.

53. Let us call a[n] this sequence. Notice that: a[2] = 1 + 1 a[3] = 1 + 1 + 2 = 1 + 3 *(3-1)/2 a[4] = 1 + 1 + 2 + 3= 1 + 4 *( 4-1)/2 ... a[n]= 1 + n * (n-1)/2 Hence a[107] = 1 + 107 * (107-1)/2 = 5672.

54. Let n be the number if sides of this polygon. Each interior angle measures: (1/n) * 180*(n-2). Hence the measure of the angle C in the triangle CAB is:

180 - (1/n) * 180*(n-2) = (1/n )*360. Therefore, the angle CAB measures: 180 - 2 * ( 1/n) * 360= 180 -720/n= 100 if n=9.

55. Suppose that the hypotenuse is a, then a2 = b2 + c2=288/2=144.

This gives that a = Sqrt[144]=12, which in turns gives that


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b + c = 26-12=14. Square both sides to get: b2 + c2 + 2 b c = 142 =196

144 + 2 b c = 196 b c = (196-144)/2=26 Area of triangle = (b c)/2 = 26/2 =13.

56. Notice that each side of the triangle is obtained by summing two of the radii of the three circles. Hence the sides of the triangle are 82, 80 and 18.

Notice that 822= 802 + 182. Hence the triangle is right triangle and the area is: (80 * 18)/2 = 720

57. After 20% reduction, the price would be .8 * 300= 240 rupees. Another 20% of 240 is 48 rupees. The consumer would pay 240-48=192 rupees under this first method of reduction. The second method of reduction makes the price .6 * 300=180 rupees. The saving is 192-180=12 rupees.

58. Let x be the number of kms going up between A and B. Let z be the number of kms going down between A and B. Let y be the number of kms of level surface between A and B.

For the round trip, Paul will have gone up x+z kms and will have gone down x+z miles. Of course, he would have travelled 2 y kms on a level surface. Hence the equation:

(x+z)/3 + (x+z)/6 + 2 y/4 = 6 Simplify the above equation to get Ń ( x + y + z) = 2 * 6 * 3 ( x + y + z) = 36/3=12

59. Notice that 87(1) +87(7) +87(21) +87(35) +87(35) +87(21) +87(7) +87(1) = 87 (1+1)7 =221 27= 221+7=2x. Hence x=21+7=28.

60. For the product to be odd, all 3 numbers must be odd. The probability of this event is (1/2)3=

1/8. For the product to be odd, one of the 3 numbers must be even. The probability of this event

is 1- 1/8 = 7/8.

61. If the sum of two numbers is constant, then their product is maximum if the two numbers are equal. Hence 1000+1000=2000 and so the answer is 1000*1000=1000000.

62. Let x be the number of campers who took basket weaving and pottery. Let y be the number of campers who took cabinet making and pottery. Let z be the number of campers who took basket weaving and cabinet making.

Since every camper is required to take exactly two of the three crafts classes offered, then one has:

x + y = 49 x + z = 41 y + z = 46 Add the three equation to get: 2*(x + y + z) = 136. Hence x + y + z= 68 and this number

of campers.

63. The answer is: (5!)/(2!*(5-2)!) * (6!)/(2!*(6-2)!)* (7!)/(2!*(7-2)!)= 5 * 6 * 7=3150

64. Think about the binomial expansion and realize that this sum is (1+1)13=213=8192.

65. The area is the sum of 4 parts: The inner part is=2 * 3.14 * 1.8 * 100 * 4=4521.6 square

centimeters. The outer part is=2 * 3.14 * 1.9 * 100 * 4=4772.8 square

centimeters. The top and bottom (washers) are= 2*(3.14 * 1.92-3.14 *

1.82)=2.3236 square centimeters. Hence the total surface area of the pipe is= 4521.6+ 4772.8 +2.3236=9296.7236 square inches.

This is equal 1117.7 square inches to the nearest tenth.

66. It increases by (42*30100 + 55900)/43 - 30100=600

67. Let x be the original side of the square. Then (x+10)2 -x2=1320

Solving for x, gives x=61. Let y be the number of feet that each side of the original

square has to be decreased in order to decrease the area of the original square by 1320 square feet. Then 612 - (61-y)2 =1320

y2 - 122 y + 1320=0 . This equation is quadratic in y. Solving for y, gives y=12 or y=110. The second value

y=110 is not acceptable since it is bigger than the length of the side of the square which is 61 feet. So the answer is 12.

68. Notice that |8-15|=|-7|= 7 and |15-8|=|7|= 7 then |8 - 15 | + |15 - 8 |= 7 + 7= 14

69. | |x - 6| + |x - 29| |=| x - 6 -x + 29 |=29-6=23

70. Let x be the number. Then x= 100/85 * 595=700. Hence the answer is: 45/100 * 700=315.

71. The number of diagonals in a convex n-sided polygon is n*(n-3)/2. So one has to solve

n*(n-3)/2= 10*n n-3 = 2*10 n = 2*10 + 3 = 23.

72. Each hours amounts to 360/12=30 degrees, then 2 hours will give rise to an angle of

30*2=60.

73. Let n be the number of sides, then (n-2)* 180= 165 * n. Solving n we find that n=24

74. Notice that 35=243 and hence m = 1/5. Now (1/5)n= 125 implies that n=- 3. Hence mn= -3/5.


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75. Notice that x4 + 4 x3 + 6 x2 + 4 x + 1= (x+1)4 =(99+1)4 = 1004=100000000

76. It takes 6 small skeins of yarn to knit a scarf. We are asked to determine how many skeins it takes to make 15 scarves. Therefore 6 × 15 = 90. (Answer)

77. The 51st term of sequence A is 30 + (10 × 50) = 30 + 500 = 530. The 51st term of sequence B

is 30 – (10 × 50) = 30 – 500 = –470. Therefore 530 – (–470) = 530 + 470 = 1000 (Answer)

78. The internal diagonal is given by (Answer)

79. What this question boils down to is that we’re looking for a number which has a certain number of divisors. But how many? Remember that we can’t have a single pile, (i.e., no 1 × X ) which means we’re looking for 15 divisors including 1 and X. But when we normally create count up the divisors it’s an even number (say divisors of 6 are 1 × 6 and 2 × 3 for 4 divisors). What this means is that X is a perfect square (so we have the square root of X times the square

root of X). So what perfect square has 15 divisors? Let’s look at the first 15 or so perfect squares:

1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225. Any perfect squares that are created from prime numbers (like 3 × 3) would never have 15 divisors. This leaves 16, 36, 64, 81, 100, 144, 196 and 225. Any that are powers of a prime also won’t get us there. That leaves 36, 100, 144, 196, and 225. The divisors of 36 are 1, 2, 3, 4, 6, 9, 12, 24 and 36. No. The divisors of 100

are 1, 2, 4, 5, 10, 20, 25, 50, 100 No. The divisors of 144 are 1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48, 72, 144. That’s 15 divisors. X = 144. Note that one can do the prime factorization and use the simple formula to find the number of divisors also (See Appendix B for the formulas)

80. We need to start by figuring out which sets of 2 integers have a product that form a perfect square: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16. The perfect squares must be less than 16 × 15 = 240 List them: 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225. We can’t make 1 because we don’t have two 1’s. 4 = 1 × 4; 9 = 1 × 9; 16 = 1 × 16, 2 × 8; 25 has none; 36 = 3 × 12, 4 × 9; 49 has none; 64 = 4 × 16; 81 has none; 100 has none; 121 has none; 144 = 9 × 16; 169 has none; 196 has none; 225 has none. So let’s list them again: 1 × 4, 1 × 9, 1 × 16, 2 × 8, 3 × 12, 4 × 9, 4 × 16, 9 × 16. This means that the “unsafe” numbers are 1, 2, 3, 4, 8, 9, 12 and 16. The safe numbers are 5, 6, 7, 10, 11, 13, 14, and 15. That’s 8 numbers. We must choose those first. Then what? Since 2, 8, 3, and 12 appear only once (against one of the values in the list), we can choose any two of these as long as we don’t choose any of the other 2. So we’re up to 10 numbers. That leaves 1, 4, 9, and 16. If we choose 1, we can’t choose 4, 9 or 16. If we choose 4,

we can’t choose 1, 9 or 16. If we choose 9, we can’t choose 1, 4 or 16. If we choose 16, we can’t choose 1, or 9, but we can choose 4. Let’s choose 16. We have 11 numbers! (Answer)

81. Let f = the cost of a kg of feathers and let g = the cost of a gm of gold. We then have,

8f + 2g = 932 and 14f + 3g = 1402 which can be solved to yield f=2 and g=458. We then have 5f

= 5(2) = 10 and 5g = 5(458) = 2290. Therefore 2290 + 10 = 2300 (Answer).

82. We must find the probability of getting either 11 or 12. For 11 we could roll either 6 and 5 or

5 and 6. For 12 we could roll only 6 and 6. The probability asked = 3/36 = 1/12 (Answer).

83. The area of the poster board is 20 × 17 = 340. The area covered by the 10 pictures is 10 × 4 ×

6 = 240. Therefore: 340 – 240 = 100. (Answer).

84. The sum of the interior angles of a polygon is 180(n – 2). We can dispense with n = 3 or n = 4. We know those are 60 and 90. For n=5,6,..10, only 1 n-value will give non-integer angle

measure which is for n=7. Hence the answer is 1. (Answer)

85. The players scored a total of 16 × 5 = 80 points. Subtract Kara’s score and you have 80 – 16 = 64. Since everyone scored a different number of points the three players who scored higher than Kara must have scored something like 17, 18 and 19. Will this work?

17 + 18 + 19 = 54 54 + 16 = 70 80 – 70 = 10 This works but could one of the higher scoring players

scored even more? What if the player who scored 19 actually scored 20?

17 + 18 + 20 = 55 55 + 16 = 71 80 – 71 = 9 but we were told that each player scored more than 9

points. Hence the answer is 10. (Answer)

86. Let d = the number of domestic stamps. Let f = the number of foreign stamps given that d=3f. If John makes the sale then: (d – 30)/f = 1/2 which gives 2d-60=f. Since d=3f, solving the system of equations we get d=36 which yields f=12. (Answer)

87. Let’s enumerate the number of perfect squares ≤ 200. 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196 That’s 14 perfect squares. So if we removed them from the

first 200 numbers we’d be left with 186 numbers. What’s the next perfect square? It’s 225. That’s too far. Hence, 200 + 14 = 214 (Answer).

88. Let n = the number of 5 paiset, d = the number of 10


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paise and q = the number of 25 paise . Given that: n + d + q = 45; d = n + 1 and q = 3n – 1. Solving the system of three equations yields n=9. (Answer).

89. The sum of the 3 numbers is 3(5/9)=15/9. Since 1/2 is one of the numbers the other two must sum to 15/9 - 1/2 = 7/6. Let x=the first number and y=the second number. We then have the equations x + y = 7/6 and x – y = 1/2. Solving we get x=5/6 and y=1/3 (Answer).

90. Let r = the number of red marbles, w = the number of white marbles, g = the number of green marbles and, b = the number of blue marbles. Given that, r = w and g = 5b. Since r = w, there is also a 35% chance of selecting a white marble first. That means there is only a 100 – (35 + 35) = 30% chance of selecting a green or blue. But we know that there are 5 times as many green marbles as blue so there is a 25% chance of selecting a green marble and 5% chance of selecting a blue marble. 35% of any number less than 20 will not result in an integer so let’s look at whether 20 total marbles works. Yes! We can have r = 7; w = 7; g = 5; b = 1. Answer=5.

91. Looking at the first level of boxes in the 6 x 5 dimension, we can only have two boxes. Those will take up 3 × 2 = 6 across. We can’t have a second set of two boxes because in the length, there is only 5 – 3 = 2 inches left and we need 3. Well, how many levels of 2 boxes can we have? Only 3 levels because 3 × 3 = 9 and 10 – 9 = 1. Therefore, we have 6 3x3x3 cubes. The volume of the cubes is 6 × 3 × 3 × 3 = 162. The volume of the 6 x 5 x 10 box is 6 × 5 × 10 = 300 Hence we have 162/300 (100) = 54%. (Answer).

92. This is just the LCM of 360, 450 and 540. 360 = 2 × 2 × 2 × 3 × 3 × 5 450 = 2 × 3 × 3 × 5 × 5 540 = 2 × 2 × 3 × 3 × 3 × 5 The LCM is: 2 × 2 × 2 × 3 × 3 × 3 × 5 × 5 = 8 × 27 × 25 =

5400 (Answer).

93. One way (whichever doesn’t matter), the boat is going with the stream. It takes advantage of the additional speed of the water so it actually goes 12 km/hr. It travels 12 km which takes 1 hour. Going the other way, the boat has to fight the stream so it only goes 9 – 3 = 6 km per hour. To go 12 km takes 2 hours. So, the boat travels 24 km in a total of 3 hr. Hence average speed =

24/3 = 8 (Answer)

94. Let r = the number of red marbles, b = the number of blue marbles, g = the number of green marbles. Given that r + b = 16 (the 17th would be green); g + b = 17 (the 18th would be red) and; r + g = 19. Solving the system of three equations then yields, g=10, r=9 and b=7. Hence we have

r + b + g = 9 + 7 + 10 = 26 (Answer).

95. Let’s look at one example which always has 1 and

2 grouped together: (1, 2), (3, 4), (5, 6) The other combinations are (1, 2), (3, 5), (4, 6) and (1, 2), (3, 6), (4, 5). So when grouping (1,2) we have 3 combinations. The same will hold true for (1, 3), (1, 4), (1,5) and (1,6).

So that is 3 × 5 = 15 combinations. Are we done? Yes, because enumerating each of the others

will cover all combinations. Hence the answer is 15. (Answer).

96. X is one less than a multiple of 8 so X is a multiple of 17 that is odd. The largest multiple of

17 less than 1000 is 986 which is even. 986 – 969 is odd. Is this 1 less than a multiple of 8? No.

Subtract 34 from 969. That gives 935. Is this 1 less than a multiple of 8? Yes. Hence the answer is 935.

97. Let’s start without a tie. Then the number of ways is 3! = 6. Now let’s do the ties. If all 3 tie, then there’s only 1 way. If only 2 tie, then we have 3 ways (Alfred and Brandon tie, Alfred and Charles tie and Brandon and Charles tie.) when 2 people tie for first and 3 ways when 2 people tie for last. 6 + 1 + 3 + 3 = 13 (Answer).

98. The combinations that one can consider include (1, 2, 6) (2, 2, 5) (3, 3, 3) (1, 3, 5) (2, 3, 4) (1, 4, 4) When all 3 numbers are different there are 6 ways to get

that combination. So (1, 2, 6) and (1, 3, 5) and (2, 3, 4) each have 6 combinations. That’s 18. When there are only 2 different numbers there are 3 combinations. So (1, 4, 4) and (2, 2, 5) each have 3 combinations for a total of 6 combinations. When all numbers are the same there is only 1 combination. (3, 3, 3) has 1 combination. So we have 18 + 6 + 1 = 25 combinations. The total number of combinations is 6 × 6 × 6 = 216. Hence we have the probability 25/216 (Answer).

99. If the new garden is a square with 3600 sq. feet, then each side of the square is 60 feet. The perimeter of the new garden is 240 feet. The perimeter of the old garden was 2(l + w) = 2(2w + w) = 6w. This was doubled to 12w or 240. This implies w = 20 and hence l = 2w = 40. The area of the old garden was 20 × 40 = 800 (Answer).

100. The diagonals of a rhombus are always perpendicular bisectors of each other. We have a rhombus with side length and diagonals that differ by 6 units. Let 2x be one diagonal and 2x + 6 the other. Then we can create a right triangle with sides x and x + 3 and hypotenuse 89.

Using Pythagoras theorem then gives: x2 + (x + 3)2 = 89. Solving for x gives x = –8; x = 5. Since

x cannot be negative so x = 5. Therefore, the diagonals 2x and 2x + 6 are 10 and 16, respectively. The area of the rhombus is ½ d1 d2 where d1 and d2 are the 2 diagonals of the rhombus. Hence we have ½ × 10 × 16 = 80. (Answer)

Thank you to Mousumi Bhattacharya, Vasanthi Nagareddi and Padmanabhan Seshaiyer for this problem set and solutions.


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SAMPlEPAPER-1(Sample questions assorted by Bharti Gupta)

Section 11. According to the official Superbowl Halftime Honor Roll, there have been 32 halftime shows

requiring a producer. Disney has been listed as the producer of six of these shows. What

is the ratio of the number of shows produced by Disney to the number of shows with a

different producer? Express your answer as a common fraction.

2. Kavita’s resolution is to get more exercise. She decided that starting with the first full week in

2005, she will do a 45- ‐minute workout four times each week (four days during every seven- ‐

day period). This means she will have to exercise four of the days during the week of Jan. 1

through Jan. 7, and then four of the days during the week of Jan. 8 through Jan. 14, and so

on. If Kavita sticks to her resolution, what is the first possible date on which she could reach

a total of 25 hours of exercise for the year?

3. A competition problem requires one hour to fully develop (write, proofread, edit and

typeset). This problem is then given to 30,000 students, each working an average of 24 seconds

to solve the problem. What is the ratio of a problem’s development time to the total time

spent by the students to solve the problem? Express your answer as a common fraction.

4. On Pankaj’s 12- ‐hour clock, the tip of the hour hand reaches three inches out from the center

of the clock and the tip of the minute hand reaches five inches out from the center of the

clock. In order to set the clock ahead, Pankaj must manually move the hands of the clock

through the hour of motion. What is the total distance, in inches, that the tip of the hour hand

and the tip of the minute hand travel when Pankaj sets the clock ahead one hour? Express

your answer to the nearest whole number.

5. Dussehra is reported to be the busiest day of the year for many restaurants and the peak day

of the year for long distance telephone calls. It is also estimated that about 96% of citizens

take part in some way in Dussehra celebration. According to this last statistic, what is the

probability that three randomly selected citizens will each take part in some way in Dussehra

celebration? Express your answer as a percent to the nearest whole number.

6. Themedianofthesetofintegers{5,8,12,7,x}isthesameasthemean.Whatisthesumof

all the distinct possible values of x?

7. After the cookout, Manju and her family are going to the local fireworks show. Manju has heard

that the show is a total of 45 minutes long with an 11- ‐minute finale incorporated. She also

knows that the average number of fireworks per minute during the finale is double the average

number of fireworks per minute during the entire show. What percent of the show’s fireworks

are shot off during the first 34 minutes of the show? Express your answer to the nearest

tenth.


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8. If twice a number is equal to 6 more than half the number, what is the number?

9. The Space Shuttle launched last week and met up with the International Space Station (ISS).

Among other things, the shuttle is bringing important equipment to the ISS and taking away

discarded gear and trash. If they are dropping off 15 tons of equipment onto the ISS and

taking away 13 tons of trash and gear from the ISS, what is the percent decrease from 15

tons to 13 tons? Express your answer to the nearest tenth.

10. At MATHCOUNTS Middle School, we don’t want to miss out on any time that we could be in

math class! Since we missed our 45- ‐minute math class on Labor Day Monday, we are going to

extend the math classes on Tuesday through Friday so that we have the same amount of

total math time this week as we would have during any regular five- ‐day school week. By how

many minutes must we extend the math class on each of the four days we are in school this

week? Express your answer as a decimal to the nearest hundredth.

11. Manish decided that he would try to make some extra money by raking the leaves in his

neighbors’ yards. Manish figured he would need approximately seven bags per yard in which

to collect the leaves. The bags are sold 10 to a box, and Josh is planning on raking 14 yards.

How many boxes of bags must he buy?

12. Rita has a jar of coins containing the same number of nickels, dimes and quarters. The total

value of the coins in the jar is $13.20. How many nickels does Jamie have?

13. Mitu noticed that the high temperature on October 21st was 66 degrees, the high temperature

on October 26th was 63 degrees and the high temperature on October 31st was 60 degrees.

It appears that the high temperatures on every fifth day are forming an arithmetic sequence.

If this sequence were to continue, on what date would the high temperature be 33 degrees?

14. In one study, 82% of merchants responding to a survey said sales during the Thanksgiving

weekend grew or stayed even from last year’s Thanksgiving weekend. The remaining merchants

reported lower sales. If there were 28 respondents for the survey, how many respondents

reported that sales decreased from last year? Express your answer to the nearest whole

number.

15. What is the median of the first 2006 positive integers? Express your answer as a decimal to the

nearest tenth.

16. In the arithmetic sequence –7, –3, 1, …, what is the first term greater than 2006?

17. For the reception, the bride changed from her Christian Dior dress to a Vera Wang design.

Perhaps this was because her wedding dress weighed 20 kgs, and the train was 5 meter long.

Assuming that the train accounted for 70% of the weight of the dress, what was the average

weight per linear meter of the train? Express your answer in kgs per linear meter and as a

decimal to the nearest tenth.


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18. Savi decided that she wants to give each of her friends a small pouch of candy hearts.

She’ll use fabric and ribbon to make and tie the pouch, and each one will contain four

candy hearts. The hearts come in six colors: white, orange, pink, green, purple and yellow. If

each pouch contains exactly four hearts, such that no two hearts are the same color, how

many possible combinations of hearts could be in a pouch?

19. Sunita painted a mural for Durga Puja and mixed her own green paint. She used a ratio of

3 parts yellow to 2 parts blue to mix the paint. Manisha wanted to also paint a mural with

the same green color, but she currently has 8 cups of green paint that is a mixture of 40%

yellow paint and 60% blue paint. In order to get the same shade of green as Sunita, how

many cups of yellow paint must she add to her mixture?

20. Vivek jogs 7 km per hour and Mamta jogs 5 km per hour. They start together at their

campsite and jog in opposite directions for 2 hours. How far away are they at the end of

the jog?


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Section2:Multiple-Choicequestions

1. What is the product of all the positive integer factors of 15?

a) 15! b) 225 c) 0 d) 210

2. When the sum of 2/3 and 5/ 24 is expressed as a common fraction, what is its denominator?

a) 3 b) 24 c) 8 d) 72

3. What is the value of 34 − 24?

a) 1 b) 4 c) 65 d) None of the above

4. How many pairs of distinct integers chosen from the set of odd integers between 6 and 16

have a sum greater than 23?

a) 2 pairs b) 8 pairs c) 1 pair d) 4 pairs

5. If 8 − x = −2x − 8 + 5x, what is the value of x?


6. Two standard dice are tossed together. What is the probability of obtaining exactly 6? Express

your answer as a common fraction.

a) 1/6 b) 1/18 c) 5/36 d) 1/12

7. Out of 800 fish in an aquarium 90 % are guppies. How many guppies must be removed so that

the percent of guppies in the aquarium is 20% ?

a) 560 b) 700 c) 112 d) 448

8. What is the value of 993 + 3(992) + 3(99) + 1 ?

a) 100,000 b) 10,000,000 c) 1,000,000 d) None of the above

9. If 25*83*162 = 4m, what is the value of m ?


10. An operation z is defined as a z b = 6a − b. What is the value of 4 z 22?

a) 88 b) 128 c) 2 d) 108

11. If 7 apples cost x rupees, how many rupees do 70 apples cost? Express your answer in terms

of x.

a) 10 b) 10x c) 490x d) None of the above

12. What percent of the positive integers less than or equal to 36 are factors of 36?

a) 100% b) 25% c) 22.22% d) None of the above


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13. Four numbers are written in a row. The mean of the first two numbers is 10, and the mean

of the last two numbers is 20. What is the mean of all four numbers?

a) 60 b) 15 c)10 d) 20

14. What is the 200th term of the increasing sequence of positive integers formed by omitting only

the perfect squares?


15. Bobby has 45 meters of fencing. He will use the fencing to enclose a play area for his

puppy. What is the maximum number of square meters he can enclose? Use an 3.14 as an

approximation of Pi. Express your answer to the nearest whole number.

a) 125 b) 161 c) 87.5 d) 50

Section1:1. Answer: 3/13 There were six shows produced by Disney and 32 – 6 =

26 that were not. Therefore, the ratio we are looking for is 6:26, which simplifies to 3/13 when written as a common fraction.

2. If kavita sticks to her resolution, she will do four 45-minute workouts each week, which is 4 x45 = 180 minutes each week, which is 180 /60 = 3 hours per week. By the end of her eighth week she will have exercised 3 x8 = 24 hours. The start dates of these 8 weeks are Jan 1, 8, 15, 22, 29, (remember January has 31 days) Feb. 5, 12, 19. Therefore, during the week starting on Feb. 26 she will need only one more hour of exercise. If she exercises on the 26th, she’ll be up to 24 hours and 45 minutes. If she exercises on the 27th, too, that will put her over the 25-hour mark. Therefore, the first possible date is February 27 if she sticks to her resolution.

3. If there are 30,000 students each spending 24 seconds on the problem, this is a total of 30,000 x 24 = 720,000 seconds. This is equivalent to 720,000 / 60 = 12,000 minutes or 12,000 / 60 = 200 hours. The ratio of the time for development to the total time students spend to solve it is 1 hour to 200 hours or 1/200.

4. The tip of the minute hand will be moving through one full circle with a radius of 5 inches. The circumference of a circle this size is the product of the diameter and pi, which is 10p inches. The hour hand will just be moving from one number to the next. This is one-twelfth of the entire way around the circle with a radius of three inches, or (1/12)(6p) = p/2. The sum of these two distances is 10.5p, which is approximately 33 inches.

5. If we were to just select one random citizen, the answer would be 96%, according to the data we were given. But we have three people, and each of them has a 96% chance of participating in Dussehra celebration. (Since we randomly chose them, we can assume that if the first person participates in Dussehra celebration that has no impact on whether the other two people participate.) The probability of three unrelated events occurring is equal to the product of the three individual probabilities, which is .96 x.96 x .96 = .88 = 88% for this scenario.

6. We have been given a set of integers, and we know that the mean and the median are playing a role in this problem. In order to determine the median, we are going to need to get the integers into numerical order, which is {5,7,8,12}…anduntilweknowthevalueofx,wedon’tknow where it would fit into the list. Notice, though, that whatever integer value x has, the median can only possibly be 7 or 8. If the median is 7, then the mean is 7, so the sum of the values is 5(7) = 35. Since 5 + 7 + 8 + 12 = 32, x would have to be 3. (And if x = 3, then the median is still 7, so this works.) Our other alternative is for the median to be 8, so the mean is 8, so the integers add to 5(8) = 40, and x = 8. (And if x = 8, then the median is still 8, so this works.) The sum of the distinct possible values of x is 3 + 8 = 11.

7. We can let the average number of fireworks per minute for the entire show be x. This means there will be 45x fireworks during the entire show. It also means that the average number of fireworks during the final 11 minutes is 2x, so there will be 11(2x) = 22x fireworks during those 11 minutes. That leaves 45x – 22x = 23x fireworks for the first 34 minutes of the show. This is 23x / 45x = 23 / 45 = 51.1% of the fireworks for the entire show.

SamplePaper1-Answerkey


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8. 4

9. Going from 15 tons to 13 tons is a decrease of 15 – 13 = 2 tons. Two tons out of 15 tons is 2 /15 = 13.3%, so this is a decrease in weight of 13.3%.

10. Since we missed class on Monday, we missed 45 minutes of math. We then need to gain a total of 45 minutes over the next four days. If we divide this amount of time into four equal pieces, we see that we must add 45 /4 = 11.25 minutes to our math class during each of the remaining four days of school this week.

11. If the average number of bags per yard is seven and he is raking 14 yards, he will need 7 x 14 = 98 bags.

The bags are sold in boxes of 10, so he will use the number of bags in 98 /10 = 9.8 boxes, so he must purchase 10 boxes of 10 bags to rake the 14 yards.

12. We know that Rita has the same amount of each coin in the jar. One nickel, one dime and one quarter is equivalent to 40 cents. Now we just need to figure out how many groups of 40 cents are needed to get a total of $13.20. By performing (13.20 / 0.40 ) = 33 we see that there are 33 of each type of coin, and therefore 33 nickels.

13. We could certainly list out the next temperatures (subtracting 3 degrees each time) with their respective dates (counting out five days each time), but let’s approach this from a different angle. We need the temperature to go down 60 – 33 = 27 more degrees after October 31. This is 27 ¸ 3 = 9 additional 3-degree decreases. This 9th decrease will take place on the 9 x5 = 45th day after October 31. November has 30 days, so it will be complete on the 15th day of December. On December 15, we can expect a high temperature of 33 degrees if the arithmetic sequence continues.

14. Since 82% of the people reported a raise or staying even, the remaining 18% would have been the folks who saw a decrease. Since there were 28 respondents, 18% of this number is 0.18 ´ 28 = 5 respondents, to the nearest whole number.

15. Since we are considering the first 2006 positive integers, we can divide this in half and see that there are the first 1003 and then the second 1003, with no integer right in the middle. The median is then the average of the two middle-most terms, which would be the 1003rd and 1004th terms, or 1003 and 1004. The average of these two values is 1003.5.

16. We can see that the terms are increasing by four each time. If we continue the pattern, we see the sequence is –7, –3, 1, 5, 9, 13, … and each term is one greater than a multiple of 4. (Notice that starting with 1, we can consider that we need to add 4x to get as close to 2006 as we can, and 4x +

1 will always be one greater than a multiple of 4.) Knowing our divisibility rules for 4, we know 2004 is divisible by 4 (since 04 is divisible by 4). This means 2005 would be a term in the sequence. We want the first term greater than 2006, so adding four to 2005 gets us to our answer of 2009.

17. If the train accounted for 70% of the weight of the entire dress, then the train weighed 0.7 x 20 = 14 kgs. Divided equally among the 5 m of the train, the train weighed approximately 14/5 = 2.7 kgs per meter.

18. This is a problem that is really easier if we answer a different question. Rather than counting the number of different groups of four hearts to put into the pouch, it’s easier to count the number of combinations of two hearts to leave out of each pouch. Each of the six colors could be matched with one of the five remaining colors, to make a total of 6 x5 = 30 ways to chose the two colors to leave out. However, this way counts yellow/pink as different from pink/yellow. So every pair of two colors is represented twice. Therefore, there are only 15 combinations of two colors since the order doesn't matter. Each of these combinations of hearts to leave out has a corresponding combination of hears to put in, so there are also 15 different four- heart groupings that could be used.

19. Sunita currently has (0.40)(8) = 3.2 cups of yellow paint and (0.60)(8) = 4.8 cups of blue paint in her mixture. We know that she needs more yellow paint, so she will eventually have 3.2 + x cups of yellow paint and 4.8 cups of blue paint, and we want this to be a ratio of 3:2. So we can set up the proportion (3.2 + x) / 4.8 = 3 / 2. If we find the cross-products, we get 6.4 + 2x = 14.4. Subtracting 6.4 from both sides leads to 2x = 8 and x = 4. Therefore, Manisha needs to add 4 cups of yellow paint to the current mixture to get Sunita’s shade of green.

20. 14+10=24 km

Section2:21. 225

22. 8

23. 65

24. 4 pairs

25. 4

26. 5/36


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27. 700. Solution: Let x be the number of guppies to be removed.

Then one has to solve the following equation: 800 * 90/100 - x = (800 -x) * 20/100 and find x = 700

28. One million Solution: 99 = x x3 + 3x2 + 3x + 1 = (x + 1)3

(99 + 1)3 = 1003 = 1,000,000

29. 11 Solution: 25*83*162

=22*23*83*162

=22*163*162

=4*46*44 = 411 therefore m=11

30. 2

31. 10 x

32. 25% Solution: Factors are: 1,2,3,4,6,9,12,18 and 36 – nine factors

33. 15 Solution: (10+10+20+20) /4

34. 214 Solution: Let’s enumerate the number of perfect squares ≤

200, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196 That’s 14 perfect squares. So if we removed them from the

first 200 numbers we’d be left with 186 numbers. What’s the next perfect square? It’s 225. That’s too far. Hence, 200 + 14 = 214

35. 161 Solution: The maximum area is attained when we have a

circle. Hence 2*Pi*R = 45. Hence R= 45/(2*Pi) and Hence the area is:

Pi*R2= Pi * (45/2*Pi)2 = 452/(4*Pi) = 161 if Pi=3.14 to the nearest whole number.


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Section 1

1) (n‐1) ! / (n+1) !

2) The product of fifteen consecutive whole numbers is 0. What is the greatest possible sum of the whole

numbers?

3) Raj wrote a list of six positive integers on his paper. He chose the first and second integers randomly ,

but the third integer was the sum of first and second, and each of the remaining integers was the sum

of the two previous integers in the list . He then found the sum of all six integers . What is the ratio of the

fifth integer in his list to this sum ?

Express your answer as a common fraction.

4) When the points scored by six basketball players during a game were arranged in order, the difference

between any two consecutive totals was 9. The average of the six players' points was 36.5. How many

points were scored by the highest scoring player?

5) At Amar’s tree farm they sell Neem and Cashew trees. They plant saplings of these two kinds of trees

that are 8 cm and 5 cm tall, respectively. The Neem trees grow at a constant rate of 12 cm per year, and

the cashew trees grow at a constant rate of 14 cm per year. After how many years will these trees be the

same height? Express your answer as a common fraction.

6) The surface area of a cube is 450 sq cms. Find the length of an interior diagonal of the cube?

7) At a factory, the machines are run for 10 hours per day. With the current equipment, part A was pro-

duced at a rate of 50 parts per minute with an accuracy rate of 96% (96% of the parts produced met

quality control standards). The factory just purchased a new machine that is supposed to produce part

A at a rate of 48.5 per minute with 99% accuracy. At the end of a 10 hour day, how many more parts A

should meet the quality standards when using the new machine than when using the current machine?

8) Define m # n as Im ‐ nI ‐ m2 for all integers from 1 to 10 inclusive. If m and n are distinct integers, what

is the smallest possible value of m # n?

9) Dr. Math’s four‐digit house number ABCD contains no zeroes and can be split into two different two‐dig-

it primes “AB” and “CD” where the digits A, B, C and D are not necessarily distinct. If each of the two‐digit

primes is less than 40, how many such house numbers are possible?

10) Three‐and‐a‐half hours ago it was 10:15 a.m. How many minutes is it from now to the next occurrence

of noon?

SAMPlEPAPER-2(by Deepa Kanere )


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11) M and N are both perfect squares less than 100. If M – N = 27, what is the value of M + N ?

12) Anand has four test scores. The median is 80 points, and the range is 12 points. What is the maximum

possible score he could have received on a test?

13) When the diameter of a pizza increases by 2 cms, the area increases by 44%. What was the area, in

square inches, of the original pizza? Express your answer in terms of π.

14) 281 – 280 = 220y What is value of y ?

15) If x + y + z = 7 and x2 + y2 + z2 = 19, what is the arithmetic mean of the three products xy, yz and xz?

16) If 5/6 of the kids ahead of Shubham and 1/5 of the kids behind Shubham totals the same as the number

of kids on the merry go round , how many kids are on the merry go round ?( Hint: on a merry go round,

a person is ahead of you and behind you at the same time)

17) If (x+3) x‐3 =1 , find the 2 values of x that satisfy the solution.

18) What is the average of all integers from –n to n+1?

19) Ninety cookies were divided into three piles in the ratio 1: 1/3 : 1/6 . How many cookies are there in the

largest pile ?

20) Vedant takes a square piece of paper and folds it in half. He then folds that half into thirds and then that

small piece in thirds again . The area of the small piece of paper in front of him is how much fraction of

the area of the original piece?


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Section 2

1) The ratio of blue marbles to red marbles in the bag is 3 to 2. What percent of the marbles are blue?

A. 33% B. 60% C. 150% D. 55%

2) Given the illustration below, what is the length of chord AC of circle B?

A. r B. 2r C. r2 D. There's no way to tell the information given.

3) Find a3+b3 , if (a+b)=7 and ab=5

A. 47 B. 238 C. 105 D. 147

4) Solve: 4 √18 x 2√2

A. 26 B. 36 C. 54 D. 48

5) Which is a correct list of equivalent numbers?

A. 2 4/5 = 2.8 = 280% B. 7/10 = .07 = 7%

C. 1 1/2 = 112 = 11.2% D. 1/4 = .25 = 2.5%

6) This is a sketch of the sector of a circle

120 0

9 unit

Calculate the area of this sector ( correct to one decimal point )

A. 9.4 unit2 B. 18.8 unit2

C. 36.8 unit2 D. 84.8 unit2


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7) Sonia has written letters to four of her friends and sealed the letters in envelopes . Now she does not

know which envelope contains which letter . If she address the envelopes to her four friends at random,

what is the probability that each envelope contains the correct letter ?

A. 1 /256 B. 1 /24 C. 1/16 D. 1 / 4

8) Vijay worked for 35 hours at the normal hourly rate of pay and five hours at double time . He earned

561.60 in total for this work. What was the normal hourly rate of pay?

A. 7.02 B. 12.48 C. 14.04 D. 16.05

9) Which of the following correctly expresses n as the subject of v= 3 m n2 / r

A. n = +/‐ sqrt(rv) / 3m B. n=+/‐ r x sqrt( v/3m)

C. n= +/‐ r x sqrt (v) /3m D. n=+/‐sqrt( rv/ 3m)

10) Arrange the numbers 5.6 x10‐2 , 4.8 x10‐1 , 7.2 x 10‐2 from smallest to largest

A. 5.6 x10‐2 , 7.2 x 10‐2 , 4.8 x10‐2 B. 4.8 x10‐1, 5.6 x10‐2 , 7.2 x 10‐2

C. 7.2 x 10‐2 , 5.6 x10‐2 , 4.8 x10‐1 D. 4.8 x10‐1, , 7.2 x 10‐2, 5.6 x10‐2

11) Manav has a credit card with an interest rate of 0.05% per day and no interest – free period . Manav used

the credit card to pay for car repairs costing 470 . He paid the credit card account 16 days later. What is

the total amount ( including interest ) that he paid for the repairs?

A. 480.24 B. 483.84 C. 504.00 D. 864.00

12) Two dice are rolled . What is the probability that only one dice shows a six?

A. 5/36 B. 1/6 C. 5/18 D. 11/36

13) The number represented by a 1 followed by one hundred zeros is called a googol .Which of the follow-

ing is equal to a googol?

A. 10 m2 B. 10 10 C. 10 99 D. 10 100

14) The price of CD is 22.00 , which includes 10% Tax, What is the amount of tax included in this price?

A. 2.00 B. 2.20 C. 19.8 D. 20.00

15) Simplify 10(x+3) ‐2(4x +2)

A. 2x + 5 B. 2x+26 C. 6x+5 D. 6x+26


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SamplePaper2-Answerkey

Section 1Sr No Answer Solutions

1 1 / [ n * (n + 1) ] (n ‐ 1)! / (n + 1)! = [ 1 * 2 * 3...(n ‐ 1) ] / [ 1 * 2 * 3...(n ‐ 1) * n * (n + 1) ] = 1 / [ n * (n + 1) ]

2 105 To have a product 0, one of the 15 consecutive whole numbers has to be 0. And since the smallest whole number is 0, the 15 consecutive whole numbers are 0,1,2,3,...,14 i.e. Sum = 0+1+2+3+4+5+6+7+8+9+10+11+12+13+14 =105 Alternatively Formula : Sum from 1 to n = n(n+1)/2 = 14*(14+1)/2 = 14*15/2=105

3 1/ 4 Let x1 and x2 be the first two integers. x3 = x1 + x2 x4 = x2 + x3 = x2 + x1 + x2 = x1 + 2 x2 x5 = x3 + x4 = x1 + x2 + x1 + 2 x2 x5 = 2x1 + 3x2 x6 = x4 + x5 = x1 + 2x2 + 2x1 + 3x2 x6 = 3x1 + 5x2 We have the expression for the 5th integer. So let’s get the sum of all 6. x1 + x2 + x3 + x4 + x5 + x6 = x1 + x2 + x1 + x2 + x1 + 2x2 + 2x1 + 3x2 + 3x1 + 5x2 = 8x1 + 12x2

The ratio = (2x1 + 3 x2 ) / (8x1 + 12x2 ) = (2x1 + 3 x2 ) / 4((2x1 + 3 x2 ) = 1/ 4

4 61 points It's an arithmetic sequence with d=9 To get the average you need the Sum/6 or Sum=6*average S(6)=(6/2)(a+L), where a is the 1st term and L is the last. last term= a+5d = a+5*9 = a+45 Substitute that into the Sum equation to get: S(6)=(3)(a+a+45)=6*average 3(2a+45)=219 6a+1135=219 6a=84 a=16 The last=a+45=16+45=61 ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ The highest scoring player scored 61 points.

5 3/2 years = 1.5 years Let n = the number of years. When these trees will be of same size Then, 8 + 12n = 5 + 14n 3 = 2n n = 3/2 years


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6 15 cms If the surface area of a cube is 450,then the surface area of one face = 450/6 = 75 Length of one side = Sqrt(75) You can use pythagoras to find an external diagonal on one face: L^2 + L^2 = H^2 75+75=H^2

H=Sqrt(150) Then the internal diagonal now has it's height as the same length as a side and it's base the same length as an external diagonal. So the internal diagonal H can be found using pythagoras again: H^2 = 75 +150 ; H^2 = Sqrt (225) = 15

7 9 more part A 50 x 60 x 10 x 0.96 = 28,800 48.5x60x10x0.99=28,809 28,809‐28,800 = 9 more part A

8 ‐99 Since the first part of the expression involves an absolute value, we know the first part of the expression will have to be positive. If we want to make the final value as small as possible we need the result of Im ‐ nI to be as small as possible. Since m and n must be distinct integers we can't get a result of 0 but an pair of consecutive integers will produce a result of 1 in this situation. Next we need to think about what will happen when we subtract the squared value of m. Since we are looking for the smallest value we want to subtract the largest possible value from 1, thus m2should be 102 = 100. Thus, the smallest possible value is 1 ‐ 100 = ‐99.

9 56 If Dr. Math’s four‐digit house number contains no zeroes and can be split into two different two‐digit primes “AB” and “CD” (each less than 40) where the digits A, B, C and D are not necessarily distinct, possible options for AB and CD are 11, 13, 17, 19, 23, 29, 31 and 37. Since AB and CD should be different from each other, there are 8 options for the first two‐digit integer and 7 options for the other two‐digit integer. This means that there are 8(7) = 56 possible house numbers.

10 1335 mins Three and one half hours ago it was 10:15 am. That means it is now 1:45 pm. We must find how many more minutes it is from now until the next noon. Once we get past midnight we’ll have 12 hours. Then 1:45 pm to midnight is 10 hours and 15 minutes. That’s a total of 22 hours and 15 minutes. (22 × 60) + 15 = 1320 + 15 = 1335

11 45 M and N are both perfect squares less than 100. If M – N = 27, then what is the value of? Let’s list all the squares less than 100. 1, 4, 9, 16, 25, 36, 49, 64, 81, 100. If we add 27 to each square we get 28, 31, 36, 43, 52, 63, 76, 91, 108 and 127. 36 is the only square. Therefore, M = 9 and N = 36 Sqrt M + Sqrt N = Sqrt 9 + Sqrt 36 = 3+ 6 = 9 Therefore M +N =36+9 =45

12 92 There are four test scores which have a median of 80 and a range of 12. We are asked to find the maximum possible score that could have been received on a test. Since there is an even number of tests, either both middle tests were 80 or their sum was 2 × 80 = 160. The largest number we can choose for test #1 is 80 so that the range will the maximum score for test #4. Test #2 and #3 would then be 80 each. Since the range is 12, then 80 + 12 = 92 is the largest test score.

13 25 π When the diameter of a pizza increased by 2 cm, the area of the pizza increases by 44%. We are asked to find the area of the original pizza. Let x = the radius of the original pizza. Then the area of the original pizza is πx2. Increasing the diameter by 2 cm is the same increasing the radius by 1 in. π (x + 1)2 / π x2 = 1.44 (x + 1)2 / x2 = 1.44 (x + 1) / x = 1.2 (x + 1) = 1.2 x 0.2x=1 2x =10 X=5 The area of the original pizza is 25 π


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14 4 280 (2‐1) = 220y 280=220y

20y=80 y=4

15 5 x + y + z = 7 x2 + y2 + z2 = 19 We are asked to find the mean of the three products, xy, yz and xz. (x + y + z)2 = 72 = 49 (x + y + z)2 = x2 + xy + xz + yx + y2 + yz + zx + zy + z2 =49 (x2 + y2 + z2) + 2xy + 2xz + 2yz =49 19 + 2(xy + xz + yz) = 49 2(xy + xz + yz) = 49 – 19 = 30 xy + xz + yz = 15 (xy + xz + yz )/3 = 15 /3 = 5

16 31 Let x= the number of kids ahead of Shubham . x= the number of kids behind Shubham. Therefore, x+1= the no. of kids on the merry‐go‐round. ( 5/6 x )+ (1/5 x )= x +1 ; X=30 And x+1 = 31 kids on the merry‐go‐round

17 3 and ‐ 2 (x +3 )x‐3 =1 when the exponent =0 i.e.(x=3) Or When x+3= 1 i.e. (x= ‐2)

18 1/2 21) –n, .. ,‐1 ,0 ,1,…n , n+1 n n The sum of the integers is n+1 . There are 2n+2 integers in the list . (n+1) / (2n+2) = 1/2

19 60 cookies The ratio is the same as 6:2:1 . Divide the 90 cookies by 9 There are 10 in the smaller pile , 20 (10x2) in the midde pile and 60 (10X6 ) in the large pile.

20 1/18 ½ x 1/3 x 1/3 = 1/18

Section 2Sl. No Answer

1 B

2 A

3 B Hint: a3+b3=(a+b)3‐3ab(a+b)

4 D

5 A

6 D

7 B

8 B

9 A

10 B

11 B

12 D

13 D

14 A

15 B


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SAMPlEQUESTIONSPAPER

1. On a number line, A is at 11,B is at 38, C is at c and D is at d. Points C and D trisect segment AB. What is the value of c + d ?

2. The two arithmetic sequences, 1,5,9,13, … and 1,6,11,16, ..., have infinitely many terms in common. What is the sum of the first three common terms?

3. A dart is randomly thrown and lands within the boundaries of a 6 foot by 6 foot square. The unshaded regions are each a quarter of an inscribed circle. What is the probability that the dart lands in one of the shaded regions? Express your answer as a common fraction in terms of π

4. How many positive integers x satisfy the inequality 100 ≤ x2 ≤200 ?

5. An oil tanker containing 108,000 gallons of oil releases one third of its remaining volume every two hours. How many gallons have been released after the first six hours?

6. A bag contains five red socks and eight blue socks. Lucky reaches into the bag and randomly selects two socks without replacement. What is the probability that Lucky will get different-colored socks? Express your answer as a common fraction.

7. Eight women of different heights are at a party. Each woman decides to only participate in a handshake with women shorter than herself. How many handshakes take place?


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8. A unit circle is inscribed in a regular hexagon. What is the number of square units in the area of the hexagon? Express your answer in simplest radical form.

9. Three 3-digit numbers are formed using the digits 1 through 9 exactly once each. The hundreds digit of the first number is 1. The tens digit of the second number is 8. The units digit of the third number is 5. The ratio of the first number to the second number to the third number is 1:3:5, respectively. What is the sum of the three numbers?

10. According to the graph, 22.2% of the counties have a population of 0-9999 people. How many counties have a population of one-quarter million or more? Express your answer to the nearest whole number.

11. Positive integers B and C satisfy B(B - C) = 23. What is the value of C ?

12. The area of the shaded region is 276π square feet. The radius of the larger circle is 6 feet more than the radius of the smaller circle. What is the number of feet in the radius of the larger circle?

Distribution of Counties by Population(percent of counties: Total number of counties is 3141)

County Population Size

0-9999

10,000-24,999

25,000-49,000

50,000-99,000

100,000-249,999

250,000-449,999

500,000-999,999

1,000,000 and over


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13. Each of the letters A, B, C and D represents a different odd integer between two and ten.

What is the least possible value of A x B - CD

? Express your answer as a common fraction.

14. How many ordered triples of three prime numbers exist for which the sum of the members of the triple is 24?

15. A regular hexagon and an equilateral triangle have equal perimeters. What is the ratio of the area of the hexagon to the area of the triangle? Express your answer as a common fraction.

16. If the sum of two unequal positive integers is 2000, then the sum of their reciprocals is smallest when the two integers are

A) 996 & 1004 B) 997 & 1003 C) 998 & 1002 D) 999 & 1001

17. In 4 hours, through how many degrees does the hour hand of a circular clock move? A) 1440° B) 120° C) 90° D) 48°

18. What year will it be 3700 days from June 30, 1995?

A) 2000 B) 2005 C) 2006 D) 2045

19 What is 32 of 10% of 10?

A) 32 B)15 C)3.2 D)1.5


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20. When 2 lines intersect, the sum of the 4 angles thus formed is A) 90° B) 180° C) 270° D) 360°

21. What is the ratio of the number of seconds in 45 minutes to the number of seconds in one hour?

A) 1:2 B) 1:4 C) 3:4 D) 4:3

22. How many of the terms in the sequence 1 x 12, 2 x 12, 3 x 12, 4x 12, . . . , 15 x 12, 16 x 12 are multiples of 16?

A)1 B)2 C)4 D)8

23. A semicircle of radius 2 has the same area as a circle of radius 1

A) 14 B)1 C) √2 D) 2

24. How many of the first 1000 positive integers are multiples of all four of the number 2, 3, 4, and 5?

A)8 B)9 C)16 D)17

25. Marina types 5 pages in 4 minutes. Ruth types 1 page in 1 minute. In 7 hours, how many more pages will Marina type than Ruth?

A)105 B)84 C)15 D)140

1. 49 2. 63

3. 8 - π8

4. 5 5. 76, 000 (gallons) 6. 20

39

7. 0 8. 2√3 (square units) 9. 1161 10. 229 11. 22 12. 26 (feet) 13. 6

7

14. 15 (triples)

15. 32

16. D 17. B 18. B 19. D 20. D

21. C 22. C 23. C 24. C 25. A

Answer Key


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2015 - handbook for Level III MB Workshop