Download n m + n p -m

Comparing and Ordering Fractions – Black Problems
1. Benson has Golden Delicious apples, each of which weighs .6 pounds, and Jonathan
apples, each of which weighs .8 pounds. He wants to make applesauce such that 1/3
of the weight is from Golden Delicious apples and 2/3 of the weight is from
Jonathan Apples. He wants to use all 12 of his Golden Delicious apples. How many
Jonathan apples should he use?
2. All Digital Divisors. What is the least positive whole number that is divisible by all
of the whole numbers from 1 through 9?
3. Mary will pick a positive integer less than 80 that is a multiple of 7. Susan will pick a
positive integer less than 80 that is a multiple of 9. What is the probability that
they both will pick the same number? Express your answer as a common fraction.
4. Using the following three clues, can you figure out which integer I am?
(1) If I am not a multiple of 4, then I am between 60 and 69.
(2) If I am a multiple of 3, I am between 50 and 59.
(3) If I am not a multiple of 6, I am between 70 and 79. What integer am I?
5. The greatest common factor of two numbers is 14, and the least common multiple is
168. If the two numbers are not 14 and 168, what is their sum?
6. What is the sum of the positive integers k such that k is greater than 2 and less
3
27
8
than ?
9
7. How many multiples of 7 are between 30 and 790?
8. What is the sum of all of the multiples of 3 between 100 and 200?
p m
9. Given: 5 = n = m + n = - . What is the value of p?
104
13 39
156
10. Use All Digits. What is the largest multiple of 12 that can be written using each
digit 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9 exactly once?
11. Shake a Three. What is the probability of rolling a multiple of 3 with a pair of fair
dice?
12. What is the 11th term in the geometric sequence 3, -3, 3 , -3, ...? Express your
2 4 8
answer as a common fraction.
Comparing and Ordering Fractions – Black Solutions
1. The least common multiple of the weights of the two kinds of apples is 2.4 pounds, which means that 4
Golden Delicious apples weigh the same as 3 Jonathan apples. A ratio of 4 Golden Delicious to 6
Jonathans would give the 1/3 to 2/3 weight ratio that Benson wants for his apples sauce. Since Benson
plans to use all 12 of his Golden Delicious, he will need to use 18 Jonathan apples.
2. All Digital Divisors. 2520. The number would have to be the lowest common multiple of the digits 1
through 9 or the product of 5 x 7 x 8 x 9 = 2520
3. There are 11 multiples of 7 less than 80 and eight multiples of 9 less than 80. The only multiple of
both 7 and 9 that is less than 80 is 63. There is a 1/11 chance that Mari will pick 63 and a 1/8 chance
that Susan will pick 63. The probability that they both pick 63 is (1/11) x (1/8) = 1/88.
4. Let’s assume our mystery number is a multiple of 3, so it must be either 51, 54 or 57 according to the
second statement. However, it can not be 51, 54 or 57 because then it would not be a multiple of 4,
and if it is not a multiple of 4, according to the first statement it must be between 60 and 69. Now we
know the mystery number is not a multiple of 3. Because it is not a multiple of 3, it certainly can’t be a
multiple of 6 either, so it must be either 71, 73, 74, 75, 76, 77 or 79, according to the third
statement. This means the mystery number is a multiple of 4, since if it is not a multiple of 4, it would
have to be a number between 60 and 69. Of the numbers 71, 73, 74, 75, 76, 77 and 79, only 76 is a
multiple of 4. Our mystery number is 76.
5. The greatest common factor of the two numbers is 14, and the least common multiple is 168. The
greatest common factor can be prime factored as 14 = 2 • 7; the least common multiple prime factors
as 168 = 2 • 2 • 2 •3 • 7. A Venn Diagram will help to organize the information.
Draw two circles, one for each of the unknown numbers, as shown in diagram A. Enter the two factors
of the greatest common factor in the section that belongs to both circles. This place the two factors
for the least common multiple, and three factors remain.
A
2
7
B
2
2
2
7
C
3
2
2
7
2
3
There are two different ways to place these three factors in the empty parts of the circles. Both
factors of 2 could go in one circle with the 3 in the other (diagram B) , or one 2 and the 3 could go in
one circle with the second 2 in the other (diagram C). Diagram C doesn’t work, though, because there
is a common factor of 2 in the non-overlapping parts of the circles. Under that scenario, the GCF
would be 28, not 14. Diagram B is the correct one, and the numbers are 2 • 2 • 2 • 7 = 56 and 2 • 7 • 3 =
42. The sum of 56 and 42 is 98.
6. 105
7. Knowing our multiples of 7 may quickly lead us to the fact that our first multiple of seven within the
desired range is 7x5 = 35. We could continue identifying and counting multiples of 7 until we reach
790, but that would take a very long time! Dividing 790 by 7 gives us a quotient of 112 with a
remainder of 6, so we know that 7x112 < 790 and 7x113 >790. We’ve discovered that multiplying 7 by
any integer from 5 through 112 will give us a multiple of 7 within the desired range. How many integers
are there from 5 through 112? Be careful! Don’t make the mistake of just using the difference
between 112 and 5… you’ll be one integer short! Think of it this way: we want to include all of the first
112 positive integers except for the first four positive integers. This means there are 112 – 4 = 108
integers from 5 through 112, and therefore, there are 108 multiples of 7 between 30 and 790.
8. Adding the first and last multiples of 3 between 100 and 200, we get 102 + 198 = 300. The second and
the second-to-last multiples of 3 also have sum of 300. Of the 33 multiples of 3 between 100 and 200,
we can make 16 pairs that have a sum of 300 and then add the number 150 that is alone in the middle.
The sum is 16 x 300 + 150 = 4950.
9. Multiplying through by the LCM of 13, 39, 156 and 104 (which is 312), we obtain: 120 = 8n = 2(m + n) =
3(p – m). Solving 120 = 8n for n, we get n = 15. Substituting n = 15 into 120 = 2(m + n), we get 120 = 2(m
+ 15) & 60 = m + 15 & m = 45. Using m = 45 in the equation 120 = 3(p – m), we get 120 = 3(p – 45) &
40 = p – 45 & p = 85.
10. Use All Digits. 9, 876 543 120. Any number that uses all digits exactly once will be divisible by 3. We
also need it to be divisible by 4. If the number named by the last two digits is divisible by 4, then so is
the number. So we arrange the digits in order but reverse the positions of the 1 and 2 so that the last
two digits name a number divisible by 4.
11. Shake a Three. 1/3. There are 36 different equally likely pairs that can be rolled. Twelve pairs add to
a multiple of three: (1, 2), (2, 1), (1, 5), (5, 1), (2, 4), (4, 2), (3, 3), (4, 5), (5, 4), (3, 6), (6, 3), and (6, 6).
The probability is 12/36 = 1/3.
12. For geometric sequence, a common ratio is multiplied with each term to determine the next term. If
this common ratio is positive, then the terms will stay positive or stay negative. If the common ration
is negative, the terms will alternate between positive and negative
0
values. If we are multiplying each term by a common ratio with an
absolute value greater than one, the absolute value of the terms will
-3
get larger. However if we are multiplying each term by a common
2
ratio with an absolute value less than one, the absolute value of the
3
terms will get smaller. For the geometric sequence in problem #3, we
go from 3 to - 3 . What ratio was multiplied with 3 to get - 3 ? We
2
2
can solve 3x = - 3 . Dividing both sides by 3 yields x = - 1 . From here
2
we can see that each term has been multiplied by
-3
4
8
2
-1
2
to get the next
3
16
term. Because this common ratio is negative, our terms will alternate
between positive and negative values. Because this common ratio has
an absolute value less than 1, the absolute values of our terms will get smaller. Here is the visual
representation of the terms of the sequence. The vertical line represents 0, and the terms of the
sequence are shown from top to bottom. Notice how the segments representing the terms are each
half the size of the previous segment, and the values are alternating from one side of zero to the
other. If we continue this pattern, we can determine the 11th term of the sequence. Notice, also, the
numerator is always 3 while the denominators are successive power of 2. The 11th term will be on the
right side of 0 (positive) and will have a numerator of 3 and a denominator of 210. This is
3 .
1024
Bibliography Information
Teachers attempted to cite the sources for the problems included in this problem set. In some cases,
sources were not known.
Problems
Bibliography Information
2 , 10 , 11
Collier, C. Patrick. Menu Collection Problems
Adapted from Mathematics Teaching in the
Middle School. New York: National Council of
Teachers of Mathematics, 2000. Print.
1, 3 – 9 , 12
Math Counts (http://mathcounts.org)