2ne1-fire

http://www.youtube.com/watch?v=uD-pmQohugI

bsed 2-s (mathematics)

_bharkada angels_

Urbana at Felisa

URBANA AT FELIZA*
(Excerpts)

ni Modesto de Castro

Ang Pagdalao

Manila….
FELIZA: Isa sa mga gaung tapat na ipinag-uutos nang paquiquipag capuoa tauo, ay ang pagdalao sa camaganac, caibigan o caquilala cun capanahonan cun dinadalao nang masamang capalaran o namamatayan caya ay carampatang dalauin at maquiramay sa catouaan, o aliuin caya sa harap. Ituro mo Feliza cay Honesto, cun papaano at cailan gagauin ang pagdalao.
Ang una una, i, gaun sa capanahonan, at pag di natama sa oras ay naiinip ang dinadalao, at magdadalang hiya ang dumadalao. Sa oras na may guinagaua, lalo’t cun nagagahol sa panahon cun cumacain, humahapon o nagdarasal ay di na uucol gauin, maliban na lamang cun magcadatihan ang nag-dadalauan. Caya dapat ipagtanong ang caugahan (nang bayan at sa lahat nang caquilala) at nang houang madalao sa panahong di uucol .
Cun nacasara ang pintoan nang daan, ay tugtuguing banayad at houag dalas-dalas.
Cun pamapanhic sa hagdanan, ay patatao, at cun may casamang tauo na ucol igalang ay ilalagay sa maguinhauang panhican.
Cun may casamang matanda, at di macacaya, ay tulungang pumanhic, at alalahanin ang panahong haharapin.
Cun pumapanhic na sa hagdanan, ay magdarahan nang patungtong sa baitang at huoag maingay at cun may masalubong na matanda o guinoo ay tumigil at paraanin sa canan o sa mabuting daanan.
Pagcapanhic nang hagdanan, ay huoang caracara, I tutuloy magpasabi sa alila cun mayroon, at cun uala ay tumugtog nang marahan sa pinto at nang mamalayang may tauo.
Cun datnang bucas ang pintoan nang salas o cabahayan, silid o iba cayang pitac nang bahay ay houang sisilip-silip, at sala sa cabaitan.
Cun macapagbigay na nang galang sa may bahay, icao ay patoloyin sa cabahayan at paupoin ca ay lumagay nang mahusay, houag magpaquita nang cagaslauan, na para baga nang manguyacoy, magpatong nang paa at magpaguiling-guiling.
Cun may dalang sombrero at tungcod, houag ilagay sa lamesa, canafe, cun di sa inaacala na mamatapatin nang may bahay lalong ibinabaual, na ilagay sa sa hihigan.
Cun ang dinadalao ay maquita na may gagauin, cacain o aalis caya, ay houag aabalahin, at ang catmpata, i, magpaalam.
Cun ang dumadalao ay mahal na tauo, ay samahan may hanggang sa daan, at sacali, t gabi ay tanglauan nang candilang may ningas cun mayroon.
Cun cayo ay dumadalao, at sa pag-alis ay sasamahan cayo nang may baha’y.
Cun sa inyong pagpupulong ay may dumating na ibang tauo na di caratihan, sa caratihan man sacali at inaacala ninyo na may ipahahayag na lihim, ay houag abalahin, magpaalam sa may bahay at sa lahat.
Cun lalaqui ang dumadalao sa isang babaye, ay di catungculan ihatid pa sa hagdanan o sa daan, maliban na lamang cun tooong mahal na tauo, nguni, ang lalaqui ay dapat maghatid sa babaye at catamtamang ilagay sa canan.
Cun cayo, i, dalauin ay dapat gumati, at cun cayo, i, anyayahan sa isang piguing nang iquinasal o ano mang pagcacatoua, ay nauucol na sa loob nang ualong arao ay gantihan namang dalauin na parang pagpapasalamat sa caniyang paquitang loob.
Cun cayo, i, anyayahan sa bahay nang binyag o libing at di nacapag bigay loob, cun macaraan na, ay carampatang dalauin at ipahayag ang cadahilanan.
Cun cayo naman ang pa sa libing o sa bahay caya nang namatayan, sila naman ang may catungculan na gumating dumalao.
Ang anyaya, i, magagaua sa bibig o sa sulat at ang inaanyayahan, cun ibig pahinuhod ay houag magpairiri at sa ibig sa ayao ay pasasalamatan ang gayong paquitangloob. Cun sacali’t aayao ay tumangi nang mahusay, at ang dahilanin ay ang manga gauain at iba pang bagay na paniualaan, ngunit ciilag sa cabulaanan.
Pag napaoo, ay houag sumala sa oras na taning, at cun di macatupad nang pangaco, ay sabihin ang pinacadahilanan, at di carampatang sumira nang pangungusap, maliban na lamng cun may tunay ana pinagcabalahan. Alamin ang oras nang houag dumating na maaga, at houag namang mahuli.
Feliza, si Honesto palibhasa, i, bata, hindi malayo na sa paquiquipagcapouatauo, ay magpaquita nang capusucan nang loob, cun macarinig nang capusucan nang loob, cun magcbihira, i, di macapag paparaan, caya pangungusapan mo na iilagan ang paquiquipagtalo.
Cun macarinig nang di catouiran at di mapangaralan ang ngaungusap, ay pa-raanin, cun ang macarinig ay linis sa cato-tohanan, ay houag sagasain, nang di pagtalunan.
Si Honesto, i, na sa panahon pa nang pagaaral nang ucol sa Dios, ganag sarili at sa paquiquipagcapuoa tauo, caya turoan mong tumupad, nitong tatlong catungculan, na catampatang pag-aralan nang isang cristiano.
Cun matutong cumilala, sumamba, t, mamintuho sa Dios, ay di lalaquing bulag ang isip, matututong umilag sa casalanan, mamimihasang gumaua nang cabanalan, matiticman ang caguinhauhan at magandang capalaran quinacamtan dito sa ibabao nang lupa nang isang catoto nang Dios, na sampong sa hirap at casaquitan ay nacaquiquita nang toua’t caligayahan, inaaring cruz na magaang pasanin, minamatamis sa loob, tinatanggap nang boong pagibig, at palibhasa, i, natatanto na di macasususnod nang langit, cin di magpasan nang Cruz.
Natatanto rin naman na mag cacadalaua ang hirap nang di marunong umayon sa calooban nang Dios, hirap na ang dumarating ay naghihirap pang lalo sa di pagcatutong magtiis, sapagca’t iquinagagality, at cun minsa, i, ipinangangalit ay lalong nalulubog sa hirap. Nguni ang marunong magtiis, ay magtiis, nacararanas nang toua sa hirap, sapagca, t, inaaring cruz na bigay nang Dios, at minamatamis sa loob.
Cun si Honesto, i, matuto nang catungculang ucol sa caniyang sarili, cahit ualang tauong sucat sacsing cahihiyan, cahit hualang taoung sucat macquita na sa caniya, i, sumisi, ay di pahihinuhod ang loob sa isang licong isip, di mapapanibulos sa isang gaua cayang lihis sa matouid, at palibhasa, i, natatanto ang boong cahulugan nang masama at magaling.
Cahit paghandogan nang boong cayamanan at carangalan sa mundo, cahit pagpisanan nang lahat nang hirap, cahit icatapos nang buhay, ay di magpapahamac gumaua nang icacasira nang sariling puri, palibhasa, i, natatanto yaong matouid na hatol nang Dios Espirito Santo na isinulat Ni Solomon; magpilit cang mag-ingat nang magandang pangalan, ang cahuluga, i, magmahal ca sa asal.
Cun Matutong maquipagcapoua tauo, i, magpapacailag sa quilos, asal at pangungusap na macasusucal sa mata nang iba, at di man cusain, ay calulugdan at iibiguin nang lahat.
Ipagcaloob naua nang Dios na matandaan at itanim sa dibdib ni Honesto itong maicling hatol na isinulat co sa iyo, at nang malagui sa pag-ibig sa Dios, at matutong magpacamahl sa asal. Adios, Feliza, hanggang sa isang sulat. URBANA.

-------------------------------------------------------------------------------------

* P. Modesto de castro, Pag susulatan nang dalawang binibini na si Urbana at si Feliza na magtuturo ng mabuting caugalian, (Reprint; Manila Libreria J. Martinez, 1902). p. 74-77.

pictures...hekhek

answer part 2

ANSWERS:
1. Let x = the smaller number
y = the larger number
Equation:
Four times the smaller is five less than the larger
4x = 55 - x – 5
Solving:
4x = 55 - x - 5
4x + x = 55 - 5
5x = 50
x = 10 - the smaller number
55 - x = 45 - the larger number

2. Let u = units digit of original number
t = tens digit of original number
10t + u = original number
10u + t = new number
Equation:
(a) The sum o the digits is 10
t + u = 10
(b) The new number is less than the original number
10u + t = 10t + u - 18 or 9u - 9t = -18
Solving:
9u + 9t = 90 (multiply eq. 1 by 9)
9u - 9t = 18
18u = 72
u = 4
t + u = 10
t = 10 - 4
t = 6
Therefore, 10t + u = 10(6) + 4
= 60 + 4
= 64 -- original number

3. Let x = Gustav’s present age
y = Father’s present age

PRESENT PAST (A year ago)
GUSTAV x x - 1
FATHER y y - 1
Equations:
(a) Father is seven times as old as Gustav
y = 7x
(b) Father was nine times as old as Gustav
y - 1 = 9(x - 1) or y - 9x = -8

Solving:
Substitute 7x in place of y in eq. 2.
7x - 9x = -8
-2x = -8
x = 4
Therefore, Gustav’s present age is 4 years.

4. SECRET/NOTE:
Subtract 444 from the final answer, and then group the digits of the resulting number in pairs from right to left.

______ ______ ______ _______ __________
(month) (day) (day)
Suppose:
a. Multiply the number of the month of your birth by 100. (9 x 100 = 900)
b. To this product, add the day of the month of your birth. (900 + 24 = 924)
c. Multiply this sum by 2. (924 x 2 = 1,848)
d. Add 8 to this product. (1,848 + 8 = 1,856)
e. Multiply the sum by 5. (1,856 x 5 = 9,280)
f. Add 4 to the product. (9,280 + 4 = 9,284)
g. Multiply the sum by 10. (9,284 x 10 = 92,840)
h. Add 4 to this product. (92,840 + 4 = 92,844)
i. Add your age to this sum. (92,844 + 18 = 92,862)
j. Tell me your final answer. (92,862)
k. You are ____ years old and your birthday is on (month) and (day) (18, 09, 24)

Therefore, you are 18 years old and your birthday is on 09(month) and 24(day).

5. 19,419 – 13,882 = 5537
.’. Then, turn your calculator upside down, you will read there LESS

6. 2,101 x 18 = 37818
.’.Turn your calculator upside down, the word displayed is BIBLE

7. 256 x 209 = 53504
.’.Turn your calculator upside down, the word displayed is HOSES

8. The sum of the product is always
(B – 1) • 10,000 + A
that is, 1 less than the volunteer’s second number placed next to his or her first number.






Suppose:
If the volunteer chooses 4562 and 8234, the sum will be 82,334,562 since;
(B – 1) • 10,000 + A
= (8234 – 1) • 10,000 + 4562
= (8233) • 10,000 + 4562
= 82, 330, 000 + 4562
= 82, 334, 562

9. SECRET/NOTE:
Hint: Find the difference between the first and second values of the expression and the difference of the second and third values of the expression. Then find the difference of the 2 resulting number.

A = third difference/2
B = first difference – A
C= first value given by volunteer

10. Since 6 = VI and 11 = XI, therefore;
Show 6 and 6 make 11
VI and VI make XI
XI makes XI

11. We can analyze in this way. Two tiles are added to make each new figure – one tile to the top and another to the right side. To make the tenth figure, two tiles must be added six times to the seven tiles of the fourth figure. So there will be 7 + (2 x 6) = 7 + 12, or 19 tiles in the tenth figure.


12. 9th figure…









1 2 3 4 5 6 7 8 9

NUMBER OF
DOTS
1
4
9
16
25
36
49
64
81


13. a. I + I = II
b. VI – I = V
c. X – IV = VI
d. IX – IV = V
e. VI + V
f. V + VI = IX

14. * There are 8 squares in the given figure.







15.
. . . . . . . . . . . .
. . . . . . . . . . . . .
. . . . . . . . . . . . . .
. . . . . . . . . . . . . .
. . . . . . . . . . . . . .
. . . . . . . . . . . . . .
. . . . . . . . . . . . . .





16. One strategy for problem solving is to make a list of possible solutions. Forming a list of four digit numbers using digits 9, 6, 8, and 1 can be done directly on a class graph. Place different number on each line. List all possibilities keeping the thousands place constant.

9681 6981 8961 1968
9618 6918 8916 1986
9861 6891 8691 1698
9816 6819 8619 1689
9186 6198 8196 1896
9168 6189 8169 1869

There are 24 different four-digit numbers possible using digits 9, 6, 8 and 1.






17. The last year that the numbers actually reversed was 1881 (which was also a “mirror image” year). The next year with palindromic number will be 1991, then 2002 (neither of which are dates with “mirror images”).


1881


.
1500 1600 1700 1800 1900 2000

18. One method of justification is to mark the two guesses on a number line and surround each point with two possible answers by adding and subtracting the “amounts off” and find out which ones correspond.

30 30 120 120

. .

470 500 530 650 770

Therefore, there were 530 jawbreakers in the jar.

19. The class graph should take on the appearance of a bell curve. Emphasize the pattern by cutting out the strip of Xs from each student’s graph. The bell curve will begin to take shape as move results are added to graph. Note that there are more ways to make a sum of 5(0 + 5, 1 + 4, 2 + 3, 3 + 2, 4 + 1, 5 + 0) than the sum of 0(0 + 0). Sums 19, 20, 21 are impossible. Should 9 or 10 have the greater number of responses? Ask the class to chart the different combinations for the sums 0-18 by placing combinations to the left of the sums. They should then graph the sums to the right. A similar shape should appear on both sides.
















*The graph shows the frequency of the sum, with 9 having the greatest response.
20. Students can demonstrate the answer using a long paper or line with a length of 98 or more units. If they accordian fold the length, with the first fold on unit 23, and each successive fold point extending one unit over the linear length of the previous fold, the final fold will occur on 98. This is also a good problem for visualizing a number, increasing that number 1, 2, 3 then obtaining a sum of 98:

beginning page number
beginning page number + 1
beginning page number + 2
beginning page number + 3
4 x (beginning page number) + 6 = 98
4 x (beginning page number) = 92
beginning page number = 23
*The drawings began on page 23.

21. Students can justify their answers in many ways, including a branching diagram, that are commonly used for tournament or elimination schedules or by marking out alternate Xs (to indicate the losing team in each game) and counting the marks.












Answers may be justified mathematically by considering the number of games in each round. Remember that each game involve two teams.

Beginning Round 16 teams = 8 games
Second Round 8 teams = 4 games
Third Round 4 teams = 2 games
Final Round 2 teams = 1 game
15 games total
*Therefore, fifteen (15) games would be played in a single-elimination tournament with 16 teams.


22. The weight of the nails can be separated from the weight of the can by linear representation. The nails ca then be graphically divided in half.



CAN 15 NAILS 15 NAILS




0 23 143 265
(can weight) (can w/ 15 nails) (can w/ 30 nails)

Some students may begin by working with the difference between the two weights; others may notice that half of the nails were lost and concentrate on the half that remains inside the can.

If half of the nails are gone, students can determine the difference in weight;
263 (original weight)
- 143 (remaining weight)
120 ounces (weight of 15 missing nails)

This means that the 30 original nails weighed 240 ounces (15 nails weigh 120 ounces). Hence, the weight of the can is determined from either the first or the second weighing of the can by subtracting the weight of the nails:
263 143
- 240 or - 120
23 23

*Therefore, the empty can (with no nails) weighs 23 ounces.

23. Let the students list several sports they know and ask them to determine which sport has the most “class value” using the table that shows the letter and the corresponding value of each letter. And write it on the table provided for.

*Answers will vary with the sports depending on what kind of sport the students choose.

24. Some students will want to justify immediately with numbers while others will want to act out the problem. The beginning question that arises when acting the rounds is how many marbles each individual has to start. Pick an arbitrary amount, such as 100 marbles each.

As rounds are acted out between two students, ask a third to write the number and result of each on the chalkboard.

Have the class vary the game winner and starting numbers of marbles until they are satisfied that the results of the 15 rounds do not depend upon the quantities.




Round Winner Evanna Antonia
1 Evanna 101 99
2 Evanna 102 98
3 Antonia 101 99
4 Antonia 100 100
5 Antonia 99 101
6 Antonia 98 102
7 Evanna 99 101
8 Evanna 98 102
9 Evanna 99 101
10 Evanna 100 100
11 Evanna 101 99
12 Evanna 102 98
13 Antonia 103 97
14 Evanna 104 96
15 Evanna 105 95

*Evanna and Antonia played 15 rounds of marbles.

25. Mathematical justification could be demonstrated by either of these methods.

100 (total questions) – 26 (difference between correct and incorrect answers) = 74; then
half of the 74 are incorrect = 37 incorrect answers, and
100 questions – 37 incorrect answers = 63 correct answers
OR
half of the 74 are correct = 37 correct answers, and
37 correct answers + 26 correct answers = 63 correct answers

*Therefore, with a score of 63, reed’s letter grade would have been D.

26. one approach might be to label the teams A, B, C, D, E and arrange combinations-such as AB indicating a game of team A against B. one order of listing might be;
AB BC CD DE
AB BC CD DE
AC BD CE
AC BD CE
AD BE
AD BE
AE
AE
Or, using another approach grouping approach; AB AC AD AE
BA BC BD BE
CA CB CD CE
DA DB DC DE
EA EB EC ED

*Therefore, there would be 20 games in the tournament: 8 + 6 + 4 + 2 = 20
(or 5 x 4 = 20)

27. In the following listing, the corresponding number of legs is shown in parenthesis.
Pigeons Cats Total Number
1 (2) 34 (136) 35 (138)
2 (4) 33 (132) 35 (136)
3 (6) 32 (128) 35 (134)
*Answering:
20 pigeons + 15 cats = 35 heads
(2 legs x 20 pigeons) + (4 legs x 15 cats) = 40 + 60 = 100 legs

28. Encourage students to develop a plan for symbolizing each item (unless, of course, they want to write out the complete name each time). One method might be to let letters represent specific food offerings:
First Course Second Course Third Course
O = Soup C = Chicken I = Ice Cream
A = Salad B = Beef K = Cake
F = Fish M = Mousse
P = Pork

In abbreviated form, then, the listing would appear:

1st course O A
2nd course C B F P
3rd course I K M






Listing carefully to alternative methods that student may have of justifying their answers.
A combination for a meal might now be represented by OCI: representing soup, chicken, and ice cream. An organized listing can then be formed of all such meals:

OCI OCK OCM ACI ACK ACM
OBI OBK OBM ABI ABK ABM
OFI OFK OFM AFI AFK AFM
OPI OPK OPM API APK APM

*Therefore, there are 24 different possible meals given the choices of the above three courses: 2 (first course choices) x 4 (second course choices) x 3 (third course choice) = 24 meals.

29. Ask students to explain the method they used to arrive at their solution. Perhaps some assumed that Concha was one year old and proceeded to build chart possible ages for Concha and Ralph.

Concha’s age 1 2 3 4 5 6 7 8 9…
Ralph’s age 3 6 9 12 15 18 21 24 27…

A record of their ages in six yars can be placed below this chart for comparison.

Concha’s age 1 2 3 4 5 6 7 8 9…
Ralph’s age 3 6 9 12 15 18 21 24 27…
Concha’s age in 6 yrs 7 8 9 10 11 12 13 14 15…
Ralph’s age in 6 yrs. 9 12 15 18 21 24 27 30 36…

From the information it is possible to see that when Concha is 12 Ralph will be twice her age. That makes 18 Ralph and Concha 6 at the present age.
*Ralph is 18 years old.

30. The problem might be graphically represented using a line of set length.











Since there are 4 tires turning at one time for 4000 miles, 16000 miles are covered by all the tires. Miriam, however, used 5 tires to cover the 16000 miles, so 16000 ÷ 5 = 3200 miles covered by each tire.

31. One way for students to organize their thoughts on this problem might be to graph a chart showing which petal is left on daisies with varying numbers of petals. Students should soon recognize that petal 1 is left several times. Rewriting these facts and beginning a new group time “petal 1” is encountered leads to the following chart.


















# of petals to start petal # remaining # of petals to start petal # remaining # of petals to start petal # remaining
1 1 8 1 16 1
9 3 17 3
2 1 10 5 18 5
3 3 11 7 19 7
12 9 20 9
13 11 21 11
14 13 22 13
15 15 23 15
24 17
25 19
26 21
27 23
28 25
29 27
30 29
31 30

*Demonstration shows that the number 15 is the last remaining petal

32. Alfred likes numbers that are the square of integers, so he would like 25 (5 x 5), 64 (8 x 8), and 100 (10 x 10).

33. The average eggshell has a thickness of 1/100 inch.










Prepared by:
bhabygenuiz_24

answer(non-routines problems)

ANSWERS:
1. Let x = the smaller number
y = the larger number
Equation:
Four times the smaller is five less than the larger
4x = 55 - x – 5
Solving:
4x = 55 - x - 5
4x + x = 55 - 5
5x = 50
x = 10 - the smaller number
55 - x = 45 - the larger number

2. Let u = units digit of original number
t = tens digit of original number
10t + u = original number
10u + t = new number
Equation:
(a) The sum o the digits is 10
t + u = 10
(b) The new number is less than the original number
10u + t = 10t + u - 18 or 9u - 9t = -18
Solving:
9u + 9t = 90 (multiply eq. 1 by 9)
9u - 9t = 18
18u = 72
u = 4
t + u = 10
t = 10 - 4
t = 6
Therefore, 10t + u = 10(6) + 4
= 60 + 4
= 64 -- original number

3. Let x = Gustav’s present age
y = Father’s present age

PRESENT PAST (A year ago)
GUSTAV x x - 1
FATHER y y - 1
Equations:
(a) Father is seven times as old as Gustav
y = 7x
(b) Father was nine times as old as Gustav
y - 1 = 9(x - 1) or y - 9x = -8

Solving:
Substitute 7x in place of y in eq. 2.
7x - 9x = -8
-2x = -8
x = 4
Therefore, Gustav’s present age is 4 years.

4. SECRET/NOTE:
Subtract 444 from the final answer, and then group the digits of the resulting number in pairs from right to left.

______ ______ ______ _______ __________
(month) (day) (year)
Suppose:
a. Multiply the number of the month of your birth by 100. (9 x 100 = 900)
b. To this product, add the day of the month of your birth. (900 + 24 = 924)
c. Multiply this sum by 2. (924 x 2 = 1,848)
d. Add 8 to this product. (1,848 + 8 = 1,856)
e. Multiply the sum by 5. (1,856 x 5 = 9,280)
f. Add 4 to the product. (9,280 + 4 = 9,284)
g. Multiply the sum by 10. (9,284 x 10 = 92,840)
h. Add 4 to this product. (92,840 + 4 = 92,844)
i. Add your age to this sum. (92,844 + 18 = 92,862)
j. Tell me your final answer. (92,862)
k. You are ____ years old and your birthday is on (month) and (year) (18, 09, 24)

Therefore, you are 18 years old and your birthday is on 09(month) and 24(day).

5. 19,419 – 13,882 = 5537
.’. Then, turn your calculator upside down, you will read there LESS

6. 2,101 x 18 = 37818
.’.Turn your calculator upside down, the word displayed is BIBLE

7. 256 x 209 = 53504
.’.Turn your calculator upside down, the word displayed is HOSES

8. The sum of the product is always
(B – 1) • 10,000 + A
that is, 1 less than the volunteer’s second number placed next to his or her first number.






Suppose:
If the volunteer chooses 4562 and 8234, the sum will be 82,334,562 since;
(B – 1) • 10,000 + A
= (8234 – 1) • 10,000 + 4562
= (8233) • 10,000 + 4562
= 82, 330, 000 + 4562
= 82, 334, 562

9. SECRET/NOTE:
Hint: Find the difference between the first and second values of the expression and the difference of the second and third values of the expression. Then find the difference of the 2 resulting number.

A = third difference/2
B = first difference – A
C= first value given by volunteer

10. Since 6 = VI and 11 = XI, therefore;
Show 6 and 6 make 11
VI and VI make XI
XI makes XI

11. We can analyze in this way. Two tiles are added to make each new figure – one tile to the top and another to the right side. To make the tenth figure, two tiles must be added six times to the seven tiles of the fourth figure. So there will be 7 + (2 x 6) = 7 + 12, or 19 tiles in the tenth figure.

9th figure…

Non-Routines Problems

1. The sum of the two numbers is 55. Four times the smaller is 5 less than the larger. Find the numbers.

2. The sum of the digits of a two digit number is 10. If the digits are reversed, the new number is 18 less than the original number. Find the original problem.

3. Gustav’s father is seven times as old as he is. One year ago, he was nine times as old as Gustav. Find Gustav’s present age.

4. Guessing Age and Date of Birth

PROCEDURE: Tell your volunteer to do the following:

a. Multiply the number of the month of your birth by 100.

b. To this product, add the day of the month of your birth.

c. Multiply this sum by 2.

d. Add 8 to this product.

e. Multiply the sum by 5.

f. Add 4 to the product.

g. Multiply the sum by 10.

h. Add 4 to the product.

i. Add your age to this sum.

j. Tell me your final answer.

k. You are ____ years old and your birthday is on (month) and (day).

5. Is 13882 greater than or less than 19419? To know the answer, perform the operation:

*19419-13882= __?_

6. What book must you read to have peace of mind? You can find the answer my multiplying 2101 by 18.

7. The fire in the LS building reached 256°. There were 209 firefighters who showed up to battle the blaze. They were not successful because they forgot their _____________. For you to get the answer perform the indicated operation: *Find the product of 256 and 209.

8. Lightning Multiplication and Addition

PROCEDURE:

a. Have a volunteer select two four-digit numbers, say A and B. Your choice for the two four-digit numbers must be (10,000-A) and (B-1).

b. Have the volunteer multiply his or her pairs of numbers and then add the two products.

c. You are able to do the calculator’s faster than the volunteers, even if the volunteer uses calculator.

9. Think-of-a-Quadratic Expression

PROCEDURE: Tell your volunteer to do the following:

a. Think of a quadratic expression of the form (ax²+bx+c).

b. Find the values of the expression when x=0,1 and 2 in that order and tell me your results in that order.

c. Your expression is _____________.

10. Using Roman Numerals, show that 6 and 6 make 11.

11. How many tiles will be in the tenth form?

. . . ?

1 2 3 4 10

12. Find the ninth (9^th) of the given figurative number using the dots as the pattern and complete the table and complete the figure below.

. . . . ?

1 4 9 16 . . . ?

1^st 2^nd 3^rd 4^th . . . 9^th

	1	2	3	4	5	6	7	8	9
NUMBER OF DOTS	1	4	9	16	?	?	?	?	?

13. Move one stick to make each of the equations correct. (NOTE: Equations are written in Roman Numerals.)

a. I – I = III

b. IV – I = V

c. IX – V = VI

d. X + IV = V

e. VI – IV = XI

f. VI – IV = IX

14. How many squares can you find?

15. Dissect an equilateral triangle into one rectangle and four congruent triangles.

16. How many different four-digit numbers could you make using each of the following digits?

9 6 8 1

17. Sanford has been thinking about palindromes. Numerical palindromes are numbers like 22 which, when the digits are reversed, still show the same number. He is trying to think of the last time that the numbers in a year made a palindrome. Can you help by finding that number and marking it on the number line?

15001600 1700 1800 1900 2000

18. The Sweethome Drug Store had a contest. They filled a jar with jawbreakers and whoever came closest to guessing the number of jawbreakers in the jar got a new radio. Candace tried to estimate the weight of the jar, and then made a guess of 650-jawbreakers, which were 120 off the actual count. Basil tried to count the jawbreakers around the bottom of the jar and estimate the number of layers. He finally guessed 500-which was only 30 off. How many jawbreakers were really in the jars?

19. If you add the last two digits of telephone numbers, will certain sums always appear more frequently than others in any phonebook? If so, what are they? What makes you think they will be more common? Can you test your answer without adding all the numbers in all the phonebooks? How?

20. Werner noticed that the sum of the page numbers of four special pages of drawings in his favorite book is 98. If the pages follow each other, on what page do the drawings begin?

21. Felix plays baseball. His team is one of the sixteen teams that will play in the city tournament. When any team loses one game, it will be out of the tournament. Assuming there are no tie games, how many games will be played before the winner can be announced?

22. Penny bought a can nails at the hardware store. The can held 30 nails, and the can and nails together weighed 263 ounces. On her way home Penny dropped the can and spilled 15 nails onto the ground. After she got home, she weighed the can (with 15 fewer nails) again, and again it weighed 143 ounces. What would the empty can weigh?

23. Look at the name of your favorite sport and the letters in its name. If A=1€, B=2€, C=3€… Z=26€, what is the value of the name of your favorite sport?

A = 1€	B = 2€	C = 3€	D = 4€	E = 5€	F = 6€
G = 7€	H = 8€	I = 9€	J = 10€	K = 11€	L = 12€
M = 13€	N = 14€	O = 15€	P = 16€	Q = 17€	R = 18€
S = 19€	T = 20€	U = 21€	V = 22€	W = 23€	X = 24€
		Y = 25€	Z = 26€

EXAMPLE: HOCKEY = 67€ 8(H) + 15(O) + 3(C) + 11(K) + 5(E) + 25(Y)

VALUE
01 - 10€ value
11 - 20€ value
21 - 30€ value
31 - 40€ value
41 - 50€ value
51 - 60€ value
61 - 70€ value
71 - 80€ value
value > 80€

24. Antonia and Evanna like to shoot marbles. They draw a circle and put a rock at the center, and then each shoots a marble from the outside of the circle to see who can come closest to the rock without hitting it. At the end of each round, the loser gives the winner a marble. Today Antonia has won five games and Evanna has five marbles more than she had when they started to play. Assuming there was a winner for each round, how many rounds did they play?

25. Reed is taking an advanced science class this semester.

The last test that he took had 100 problems on it. Letter

A	90 – 100
B	80 – 89
C	70 – 79
D	60 -69
E	Below 60

grades were based on the ranges of correct answer shown to the right. Reed answered all the questions and had 26 more answers correct than incorrect. What was his letter grade?

26. Jennifer and Lamont didn’t make the baseball team this year, but they always go watch the teams their friends play for. Several of their friends’ teams are playing in a tournament this week. If there are 5 baseball teams in this tournament, and if each team plays each of the other teams twice, how many games will be played altogether before the tournament ends?

27. Jose saw some pigeons and cats in the parking lot this morning. He counted 35 heads and 100 legs on these animals. Did he count more pigeons or more cats in the parking lot?

28. Courtney attends a boarding school and eats dinner in the school’s cafeteria. If, for each dinner, she is allowed to

1st course	soup or salad
2nd course	chicken, beef, fish or pork
3rd course	ice cream, cake, or mousse

choose only one item from each of the following courses, how many different meals can Courtney make?

29. Today, Ralph is three times as old as Concha. Six years from now, Ralph will be twice as old as Concha. How old is Ralph right now?

30. Miriam decided to take an automobile trip of 4,000 miles across the United States. She rotated her tires (the four that were on the car and the one spare) so that at the end of the trip each tire had been used for the same number of miles. How many miles had been driven on each tire?

31. Perry found a large daisy with many petals- 15 petals to be exact. Starting with petal number 1 and going in a clockwise order, he plucked every other petal until only one petal remained. If the first petal plucked was petals number 2, which petal number was the last one plucked?

32. Alfred has some definite likes and dislikes when it comes to numbers.

49 1 16 47 83 51

4 9 81 3 44 64

He likes these… He does not like these…

Which of the following numbers do you think he would like? Put an X in the column under each of your choices.

86 25 72 40 64 100

33. Evan has been buying eggs from nearby chicken farm. When he was making breakfast yesterday, he found that one of the eggs was hard to break. The shell appeared to b thick to break easily! How thick do you think the average eggshell is?

NOTE: Key to correction is at the back. Thank you.

Prepared by:

bhabygenuiz_24