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
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