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Park Forest Math Team
Meet #3
Self-study Packet
Problem Categories for this Meet (in addition to topics of earlier meets):
1.
2.
3.
4.
5.
Mystery: Problem solving
Geometry: Properties of Polygons, Pythagorean Theorem
Number Theory: Bases, Scientific Notation
Arithmetic: Integral Powers (positive, negative, and zero), roots up to the sixth
Algebra: Absolute Value, Inequalities in one variable including interpreting line graphs
Important information you need to know regarding ALGEBRA
Absolute value; inequalities in one variable including interpreting line graphs
Absolute Value is the distance a number is from zero. Absolute value is never
negative.
The symbol for absolute value is
and
Inequalities
• To solve an inequality, solve as if it were a regular equation. Remember to switch
the direction of the inequality sign only if you multiply or divide by a negative!
Category 5
Algebra
Meet #3, January 2001
1. Find the positive difference between the two values of x that make the
following equation true.
2 x − 7 = 19
2. Find the set of all the integer values of x, such that
10
>3
x
3. The solution set of the equation 9 x + 3( 6 − x ) − C < 10 is given by the
graph shown below. What value of C will make this true?
x
-6
Answers
1. _______________
2. x =
{
}
3. _______________
0
6
Solutions to Category 5
Algebra
Meet #3, January 2001
Answers
1. To solve an equation with absolute value we
must pursue two possibilities. In this case,
2 x − 7 could equal 19 or − 19 .
{
2 x − 7 = 19
2 x = 26
x = 13
1. 19
2. x =
3. 44
−
−
−
}
3, 2, 11
, ,2 ,3
2 x − 7= − 19
2 x = − 12
x=− 6
The positive difference between these two
solutions is 19.
2. The absolute value of x must remain small so
that the quotient remains greater than 3. The six
possibilities are as follows:
10
1
10
x = − 3, − = 3 > 3
x = − 2, − = 5 > 3
3
3
2
10
10
x = − 1, − = 10 > 3
x = 1,
= 10 > 3
1
1
10
10
1
x = 2,
=5>3
x = 3,
=3 >3
2
3
3
The set, then, is x =
{
−
}
3,− 2,− 11
, ,2 ,3 .
3. First we distribute and combine like terms.
9 x + 3( 6 − x ) − C < 10
9 x + 18 − 3x − C < 10
6x < C − 8
C−8
x<
6
We see from the line graph that the solution is
C−8
x < 6 . That means that
= 6 . Now we
6
solve for C. C − 8 = 36 , so C = 44 .
Category 5
Algebra
Meet #3, January, 2003
1. Given the following set of numbers, find the absolute value of the difference
between the sum of the absolute values of the numbers and the absolute value of
the sum of the numbers.
{-10, 14, - 5, 7, - 3, - 9, 1}
28
2. For how many integer values of n is n − 3 a positive integer?
3. For what value of a is the solution set of the inequality below given by the
graph below?
3(2 x − 7) + 9 − 11x < 3a
-2 -1 0
Answers
1. _______________
2. _______________
3. _______________
1
2
3
4
5
6
7
8
9 10 11
Solutions to Category 5
Algebra
Meet #3, January, 2003
Answers
1. 44
2. 12
3. -14
1. First let’s find the sum of the absolute values of the
numbers in the set: -10 + 14 + -5 + 7 + −3 + −9 + 1
= 10 + 14 + 5 + 7 + 3 + 9 +1 = 49 . Now let’s find the
absolute value of the sum of the numbers in the set:
-10 +14 + -5 + 7 + -3 + -9 + 1 = −5 = 5 . Finally, the
absolute value of the difference is 49 − 5 = 44.
28
2. For the expression n − 3 to have an integer value, the denominator, n − 3 ,
must be a factor of 28. There are six positive factors of 28 (namely 1, 2, 4, 7, 14,
and 28) and the corresponding six negative factors of 28. Thus there are twelve
(12) different values of n for which the expression is an integer. It is not
necessary to determine those values of n, but they are -25, -11, -4, -1, 1, 2, 4, 5, 7,
10, 17, and 31.
3. The graph shows x > 6. Distributing and combining like terms, we get:
3(2 x − 7) + 9 − 11x < 3a
6 x − 21 + 9 −11x < 3a
−5 x − 12 < 3a
Adding 12 to both sides we get:
−5 x < 3a + 12
When we divide both sides of the equation by -5, the inequality switches direction,
so we have:
3a +12
x>
−5
Now we must find the value of a so that the expression on the right equals 6.
3a +12
=6
−5
3a +12 = −30
3a = −42
a = −14
Category 5
Algebra
Meet #3, January 2005
1. Find the positive difference between the two solutions of the equation below.
6 x − 4 = 36
2. How many integer values of n make the following inequality a true statement?
5<
28
n
3. For what value of B is the solution set of the equation below given by the graph
below?
x
3(4 x + 5) − 9x + B ≤ 48
-5
Answers
1. _______________
2. _______________
3. _______________
www.Imlem.org
0
5
Solutions to Category 5
Algebra
Meet #3, January 2005
Answers
1. 12
2. 10
3. 18
Average team got 15.88 points, or 1.3 questions correct
1. To solve the equation 6x − 4 = 36 , we need to solve
for both the positive and the negative results.
6x − 4 = 36
6x − 4 = −36
6x = 40
6x = −32
40
−32
x=
6
6
The positive difference between these two solutions is
x=
40 −32 40 32 72
−
+
=
= 12
=
6 6 6
6
6
28
is true for the following ten
n
(10) integer values of n: -5, -4, -3, -2, -1, 1, 2, 3, 4, 5.
2. The inequality 5 <
3. The solution set given by the line graph is x ≤ 5 . If
we solve the inequality for x, we get:
12x + 15 − 9x + B ≤ 48
3x + 15 + B ≤ 48
3x ≤ 33 − B
33 − B
x≤
3
B
x ≤ 11 −
3
We now want to know when 11 – B/3 is equal to 5. We
can solve the related equation 11 – 5 = B/3, for which B
must be 18.
www.Imlem.org
Category 5
Algebra
Meet #3, January 2007
1. How many integer values of n make the following statement true?
6
≥1
n
2. Solve the following inequality. Write your solution with the x on the left, the
appropriate inequality sign in the middle, and a mixed number in lowest terms on
the right.
−3(6x + 7) ≤ 54
3. Let A be the sum of the absolute values of the numbers in the set below. Let B
be the absolute value of the sum of the numbers in the same set. List all possible
positive differences between A and B if x is an integer.
{-5, 7, x}
Answers
1. _______________
2. _______________
3. _______________
www.imlem.org
Solutions to Category 5
Algebra
Meet #3, January 2007
Answers
1. 12
2. x ≥ −4
1
6
3. 10, 12, 14
1. If n has a value greater than 6 or less than –6, then the
6
absolute value of the fraction will be less than one and the
n
statement will be false. For any value of n between –6 and 6
(including these end points), the statement will be true. There
are 6 + 1 + 6 = 13 integers in that range, but we cannot include
6
zero since is undefined. Therefore, there are just 12 integer
0
values of n that make the statement true.
2. Let’s divide both sides of the inequality by –3 first,
remembering that we must switch the direction of the
inequality sign when we divide by a negative.
−3(6x + 7) ≤ 54
6x + 7 ≥ −18
Next, we subtract 7 from both sides of the inequality, divide by
6, and make a mixed number as follows.
6x ≥ −25
−25
6
1
x ≥ −4
6
x≥
3. If x = 0, then A = −5 + 7 + 0 = 12 and B = −5 + 7 + 0 = 2 . The positive
difference between A and B is 10. If x = 1, then A = −5 + 7 + 1 = 13 and
B = −5 + 7 + 1 = 3, and the difference is still 10. The difference remains 10 for all
positive values of x. If x = -1, then A = −5 + 7 + −1 = 13 and B = −5 + 7 + −1 = 1,
and the difference is 12. If x = -2, then A = −5 + 7 + −2 = 14 and
B = −5 + 7 + −2 = 0 , and the difference is 14. If x = -3, then A = −5 + 7 + −3 = 15
and B = −5 + 7 + −3 = 1, and the difference is still 14. For all values of x less than
or equal to –2, the difference is 14. The possible differences are 10, 12, 14.
www.imlem.org
Category 5
Algebra
Meet #3, January 2009
1.
What is the least possible solution for in the inequality below?
2.
For what value of does the solution to the inequality below match the
graph below?
0
3.
6
Mike chose 3 distinct numbers from the set below and took the absolute
value of their sum. Sean chose 3 distinct numbers (not necessarily different
from Mike’s) from the set below and took the sum of their absolute values.
What is the greatest possible absolute value of the difference between
Mike’s and Sean’s result?
Answers
1. _______________
2. _______________
3. _______________
Solutions to Category 5
Algebra
Meet #3, January 2009
1.
So – 4 is the least integer solution for .
Answers
1.
2.
3.
30
2.
Because we’ve divided by if is a negative number, we would
have to reverse the to . According to the graph, the solution is which
means the inequality was reversed and must be negative. Therefore we
actually have
and since we know the equation below finds .
3. To get the largest absolute value of a sum Mike wants to pick numbers that are
all positive or all negative so that he is adding their values. The largest he could
get would be . The smallest value Mike
could get would be or .
For Sean to get the largest sum of the absolute values he just needs the three
numbers with the largest absolute values. In this case he wants
for which the sum of their absolute values is
The smallest value he could get would be .
The greatest possible difference between Mike’s result and Sean’s result would be
by taking Sean’s largest (30) and Mike’s smallest (0).
Meet #3 January 2011 Category 5 ± Algebra
1. How many integers satisfy the inequality below? ܰ
ฬ ฬ െ ͳʹ ൏ Ͷ ʹ
2. The solution of the inequality ȁ ݔെ ܯȁ ʹ is given by: ͵ ݔ . What is the value of the parameter ?ܯ 3. The solution to the inequality ȁ ݔ ͳȁ Ͷ is given by the line graph below. What is the value of ܰ? N N+8 Answers
1. _______________
2. _______________
3. _______________
www.imlem.org Meet #3 January 2011 Solutions to Category 5 - Algebra Answers ே
ே
ଶ
ଶ
1. We can rewrite the inequality ቚ ቚ െ ͳʹ ൏ Ͷ as ቚ ቚ ൏ ͳ or ȁܰȁ ൏ ͵ʹ. This is true for the integers in the range ሼെ͵ͳǡ െ͵Ͳǡ ǥ Ǥ ǡͲǡ ǥ ǡ͵Ͳǡ ͵ͳሽ - a total of ͵ integers. 1. ͵ 2. ͷ 3. െͷ 2. When solving ȁ ݔെ ܯȁ ʹ, if the argument inside the absolute value is positive we get ݔ ܯ ʹ, and if it is negative we get ݔ ܯെ ʹ. Comparing this to the given solution ͵ ݔ we see that ܯൌ ͷ. 3. When solving ȁ ݔ ͳȁ Ͷ, if the argument is positive we get ݔ ͳ Ͷ ݎ ݔ ͵DQGLILW¶VQHJDWLYHZHJHW ݔ ͳ െͶ ݔݎ െͷ. Comparing these to the line graph we conclude that ܰ ൌ െͷ. www.imlem.org Category 5 Algebra Meet #3, January 2013 1. Solve the following inequality. Write your solution with the x on the left, the appropriate inequality sign in the middle, and a mixed number in lowest terms on the right. 7 5x + 19 −2 > 40 €
2. For his birthday, Jarod received $54 from his Aunt Edna and three gift cards from his Uncle Joe. Although the gift cards are for different stores, each gift card has the same dollar value. If the absolute value of the difference between the money from Aunt Edna and the value of the three gift cards from Uncle Joe is $18, what is the absolute value of the difference between the two possible amounts that each gift card is worth? 3. How many integer values of n make the following inequality true? 20
2<
n
€
Answers (
1. _____________________ 2. $ ___________________ 3. ____________ integers )
Solutions to Category 5 Algebra Meet #3, January 2013 1. €
(
)
(
Answers 3
x > −2
5 1. )
7 5x + 19 −2 > 40
7 5x + 19 −2 > 40
7 5x + 19 > 42
35x + 133 −2 > 40
(
)
5x + 19 > 6
5x > −13 or −13
x>
5
3
x > −2
5
2. $12 €
3. 18 integers 35x + 131 > 40
35x > −91 5x > −13
x > −2
3
5
2. We can capture −3x = 18 , € this idea in the absolute value equation 54
where x is the unknown value of each gift card. To solve this, we consider two separate equations as shown above. The absolute value of the difference between $24 and $12 is $24 – $12 = $12. €
54 −3x =18
54 −3x = −18
54 −18 = 3x
54 +18 = 3x
36 = 3x
72 = 3x
x =12
x = 24
3. Since there is an absolute value sign in this inequality, we need to solve t€
wo separate inequalities € as shown below. 20
20
2>−
n
n
and 2n > −20 2n < 20 n < 10
n > −10
We need to find values of n that satisfy both inequalities, so it helps to write the solution as the compound inequality −10
< n < 10 . The 18 integers in t€
his range are n = -‐9, €-‐8, -‐7, -‐6, -‐5, -‐4, -‐3, -‐2, -‐1, 1, 2, 3, 4, 5, 6, 7, 8, and 9. We have to skip n = 0, since we cannot divide by zero.
2<
€
