Assuming that each small division represents one unit, which of the labeled points in the figure below is at a distance of 10 units from the origin of the rectangular coordinate system?
Solution
We first write the coordinates of all 5 points:
\( M(-7 , 7) \), \( P(2 , 4) \), \( N(8 , 7) \), \( Q(-6 , -8) \), \( R(7 , -7) \).
The distances from the origin to each point are:
\[ d(O,M) = \sqrt{(-7 - 0)^2 + (7 - 0)^2} = 7\sqrt{2} \]
\[ d(O,P) = \sqrt{(2 - 0)^2 + (4 - 0)^2} = 2\sqrt{5} \]
\[ d(O,N) = \sqrt{(8 - 0)^2 + (7 - 0)^2} = \sqrt{113} \]
\[ d(O,Q) = \sqrt{(-6 - 0)^2 + (-8 - 0)^2} = 10 \]
The distance from point Q to the origin is 10 units.
If \( y = kx + 2k \), where \( k \) is a constant, and if \( y = -14 \) when \( x = 5 \), what is the value of \( x \) when \( y = -24 \)?
Solution
First find \( k \) using \( y = -14 \) when \( x = 5 \):
\[ -14 = 5k + 2k \]
\[ k = -2 \]
Substitute \( k = -2 \) into the equation:
\[ y = -2x - 4 \]
Now substitute \( y = -24 \) and solve for \( x \):
\[ -24 = -2x - 4 \]
\[ -2x = -20 \]
\[ x = 10 \]
\( m \) and \( n \) are integers greater than zero such that \( 3^m = 9^{1/n} \). What is the value of \( m \cdot n \)?
Solution
Rewrite both sides with the same base:
\[ 3^m = (3^2)^{1/n} \]
\[ 3^m = 3^{2/n} \]
Therefore:
\[ m = \frac{2}{n} \]
\[ m \cdot n = 2 \]
If \( f \) is a function such that \( f(x+1) = 2x - 1 \), then \( f(2x) = \)
Solution
Let \( t = x + 1 \), so \( x = t - 1 \). Substitute into the given equation:
\[ f(t) = 2(t - 1) - 1 = 2t - 3 \]
Now substitute \( t = 2x \):
\[ f(2x) = 2(2x) - 3 = 4x - 3 \]
A large rectangle is made up of 4 congruent smaller rectangles. What is the length of a small rectangle if the area of the larger rectangle is 432 square units?
Solution
The area of one small rectangle is:
\[ \frac{432}{4} = 108 \]
Let \( L \) and \( W \) be the length and width of one small rectangle. From the diagram, \( L = 3W \). The area is:
\[ L \cdot W = 3W \cdot W = 108 \]
\[ 3W^2 = 108 \]
\[ W^2 = 36 \]
\[ W = 6 \]
Thus:
\[ L = 3W = 18 \]
The result of \( 2 + 25\% + 1.6 \) is equivalent to:
A) 3.625
B) 3.25
C) \( 3\frac{1}{4} \)
D) 385%
E) 28.6%
Solution
Convert 25% to decimal and add:
\[ 2 + 0.25 + 1.6 = 3.85 \]
\( 3.85 = 385\% \). Answer: D.
Which of these points is inside the closed curve defined by \( (x - 2)^2 + (y + 3)^2 = 4 \)?
A) \( (0, -3) \)
B) \( (3, -2) \)
C) \( (4, 1) \)
D) \( (1, 4) \)
E) \( (0, 0) \)
Solution
The equation represents a circle with center \( O(2, -3) \) and radius \( R = 2 \). A point is inside if its distance to \( O \) is less than 2.
Distance from \( O \) to \( A(0, -3) \):
\[ d = \sqrt{(2-0)^2 + (-3+3)^2} = 2 \] (Not less than 2)
Distance from \( O \) to \( B(3, -2) \):
\[ d = \sqrt{(2-3)^2 + (-3+2)^2} = \sqrt{2} \approx 1.414 \] (Less than 2)
Thus point B is inside. Answer: B.
The perimeter of a rectangular field is 8 times its width. The area of the field is 48 m². What is the perimeter, in meters, of the field?
Solution
Let \( L \) and \( W \) be length and width. Perimeter \( P = 2L + 2W = 8W \).
Solve for \( L \):
\[ 2L + 2W = 8W \]
\[ 2L = 6W \]
\[ L = 3W \]
Area is 48:
\[ L \cdot W = 3W \cdot W = 48 \]
\[ 3W^2 = 48 \]
\[ W^2 = 16 \]
\[ W = 4 \]
Then \( L = 12 \) and \( P = 2(12) + 2(4) = 32 \) meters.
The area of a triangle with side lengths 2.4, 1.8 and 3.0 is equal to:
Solution
Use Heron's formula. First, semi-perimeter \( s \):
\[ s = \frac{2.4 + 1.8 + 3.0}{2} = 3.6 \]
Area:
\[ A = \sqrt{s(s-a)(s-b)(s-c)} \]
\[ A = \sqrt{3.6(3.6-2.4)(3.6-1.8)(3.6-3.0)} \]
\[ A = \sqrt{3.6 \times 1.2 \times 1.8 \times 0.6} = 2.16 \]
The number of books \( N \) to be sold at a book fair is given by:
\[ N(p) = \frac{25000}{3p + k} \]
where \( p \) is the price per book and \( k \) is a constant. If 1000 books are sold at $7 per book, how many books will be sold at $10 per book?
Solution
Given \( N = 1000 \) when \( p = 7 \):
\[ 1000 = \frac{25000}{3(7) + k} \]
\[ 1000(21 + k) = 25000 \]
\[ 21000 + 1000k = 25000 \]
\[ 1000k = 4000 \]
\[ k = 4 \]
Now for \( p = 10 \):
\[ N = \frac{25000}{3(10) + 4} = \frac{25000}{34} \approx 735.29 \]
The number of books must be an integer, so 735 books.
For what value(s) of the parameter \( m \) does the equation \( -2x^2 + mx = 2 \) have one solution only?
A) 0
B) -2, 2
C) -1, 1
D) 16
E) -4, 4
Solution
Rewrite: \( -2x^2 + mx - 2 = 0 \). Multiply by -1: \( 2x^2 - mx + 2 = 0 \).
For one solution (double root), discriminant \( \Delta = 0 \):
\[ \Delta = (-m)^2 - 4(2)(2) = m^2 - 16 = 0 \]
\[ m^2 = 16 \]
\[ m = \pm 4 \]
Answer: E.
The average of the first 4 data values of a data set is 21. The average of the last two data values is 27. What is the average of the 6 data values?
A) 24
B) 22
C) 23
D) 20
E) 21
Solution
Sum of first 4 values: \( 4 \times 21 = 84 \). Sum of last 2 values: \( 2 \times 27 = 54 \). Total sum: \( 84 + 54 = 138 \).
Average of all 6: \( \frac{138}{6} = 23 \). Answer: C.
If the values of the 4th and 7th terms of an arithmetic sequence are 6.5 and 11, then the value of the 20th term is:
A) 32
B) 31
C) 31.5
D) 30.5
E) 30
Solution
Let the first term be \( a_1 \) and common difference \( d \). Then:
\( a_4 = a_1 + 3d = 6.5 \)
\( a_7 = a_1 + 6d = 11 \)
Subtract first from second: \( 3d = 4.5 \), so \( d = 1.5 \). Then \( a_1 = 6.5 - 3(1.5) = 2 \).
20th term: \( a_{20} = a_1 + 19d = 2 + 19(1.5) = 2 + 28.5 = 30.5 \).
Answer: D.
A store offers a 15% reduction such that the new price of a TV is $680. What was the price before the reduction?
A) 591
B) 800
C) 665
D) 780
E) 900
Solution
Let original price be \( x \). After 15% reduction: \[ x - 15\% x = x - 0.15 x = 0.85x = 680 \].
\[ x = \frac{680}{0.85} = 800 \]
Answer: B.
\( (xy)^{4n} - (xy)^{2n} \) is equivalent to:
A) \( (xy)^{6n} \)
B) \( (xy)^{2n} \)
C) \( [(xy)^{2n} - (xy)^n][(xy)^{2n} + (xy)^n] \)
D) \( [(xy)^{2n} - (xy)^n]^2 \)
E) \( [(xy)^{2n} - (xy)^n][(xy)^{2n} - (xy)^n] \)
Solution
Write as a difference of two squares:
\[ (xy)^{4n} - (xy)^{2n} = ((x y)^{2n})^2 - ((x y )^n)^2 \]
and write as a product of the sum and the difference: \[ = ((x y)^{2n} - (x y )^n)((x y)^{2n} (x y )^n) \]Answer: C.
The pie chart shows the expenses of the Taylor family. If they spent $600 on food, how much did they spend on clothing?
A) 350
B) 400
C) 500
D) 600
E) 700
Solution
From chart, food is 30% and mortgage is 35%, hence the percentage of food expenses is: \[ 100 \% - 30 \% - 35 \% = 35 \% \] Let total expenses be \( T \) and 30% are spent on food, hence \[ 0.3T = 600 \] so \[ T = 2000 \].
Expenses on clothing: \[ 0.35 \times 2000 = 700 \].
Answer: E.
Which of these is not correct?
A) \( \sqrt{x^2 + 2x + 1} = |x + 1| \)
B) \( |-x^2 - 1| = x^2 + 1 \)
C) \( \sqrt{x^4} = x^2 \)
D) \( \frac{-x^2 - 6x - 9}{-x - 3} = -x - 3 \)
E) \( |e - \frac{1}{2} - 3| = 3.5 - e \)
Solution
Check each:
A: Correct, since \( x^2+2x+1 = (x+1)^2 \).
B: Correct, since \( -x^2-1 \) is always negative, absolute value gives \( x^2+1 \).
C: Correct only if \( x^2 \) is non-negative, which it is. Actually \( \sqrt{x^4} = |x^2| = x^2 \) since \( x^2 \ge 0 \).
D: Simplify numerator: \( -x^2-6x-9 = -(x^2+6x+9) = -(x+3)^2 \). Denominator: \( -x-3 = -(x+3) \). Then fraction = \( \frac{-(x+3)^2}{-(x+3)} = x+3 \). But right side is \( -x-3 \). So D is false.
E: Left: \( e - 3.5 \). Since \( e \approx 2.718 \), \( e-3.5 \) is negative, so absolute value is \( 3.5 - e \). Correct.
Thus D is not correct. Answer: D.
Functions \( f \) and \( g \) are graphed below. What are the values of \( x \) between -1 and 5 (inclusive) for which \( f(x) > g(x) \)?
A) \( [-1, 5] \)
B) \( [-1, 0) \cup (0, 4) \cup (4, 5] \)
C) \( [-1, 0) \cup (4, 5] \)
D) \( [2, 5] \)
E) \( [0, 4] \)
Solution
From graph, \( f(x) > g(x) \) when the graph of \( f \) is above \( g \). This occurs approximately for \( x \) in \( [-1,0) \) and \( (4,5] \). At \( x=0 \) and \( x=4 \), they are equal. Answer: C.
A circle of center C is shown. A, B and D are points on the circle such that A, C and D are collinear. If AB = BC, then the size of angle BDC is:
A) 10°
B) 30°
C) 50°
D) 60°
E) 90°
Solution
Since AB = BC, triangle ABC is equilateral (CB and CA are radii). \[ \angle BCD = 180^{\circ} - 60^{\circ} = 120^{\circ} \] CB and CD are radii hence triangle BCD is isosceles , hence angle BDC is given by \[ \dfrac{180-120}{2} = 30^{\circ} \] Answer: B
Twice the difference of the squares of two consecutive positive integers is \( 4n \). What is the sum of the two integers?
A) \( 2n \)
B) 2
C) 4
D) \( 4n \)
E) 1
Solution
Let integers be \( x \) and \( x+1 \). Then:
\[ 2[(x+1)^2 - x^2] = 4n \]
\[ 2(x^2+2x+1 - x^2) = 4n \]
\[ 2(2x+1) = 4n \]
\[ 4x+2 = 4n \]
\[ 2x+1 = 2n \]
\[ 2x = 2n - 1 \]
\[ x = n - \frac{1}{2} \]
But \( x \) must be integer, so \( n \) must be half-integer. The sum of the two integers is \( x + (x+1) = 2x+1 = 2n \).
Answer: A.