If \(a^2 - b^2 = 30\) and \(a - b = 5\), then \(a =\)
Solution
Rewrite \(a^2 - b^2 = 30\) as follows:
\[ a^2 - b^2 = 30 \] \[ (a - b)(a + b) = 30 \]Use the fact that \(a - b = 5\) to find \(a + b\):
\[ a + b = \frac{30}{a - b} = \frac{30}{5} = 6 \]Add the right and left hand terms of the equations \(a - b = 5\) and \(a + b = 6\) to find:
\[ (a - b) + (a + b) = 5 + 6 \] \[ 2a = 11 \] \[ a = \frac{11}{2} \]Points \(A(8,0)\) and \(B(0,6)\) are points in the standard coordinate plane. Let \(a\) be the area, in squared units, of a circle that passes through A and B. Which of the following is the set of all possible values of \(a\)?
A) \(a < 25\pi\)
B) \(a > 48\)
C) \(a > 10\)
D) \(a \ge 100\pi\)
E) \(a \ge 25\pi\)
Solution
Since A and B are on the circle, the distance between them can be less than or equal to the diameter \(D = 2R\) (where \(R\) is the radius) of the circle. Hence:
\[ 2R \ge \sqrt{(8 - 0)^2 + (0 - 6)^2} \]Simplify:
\[ R \ge 5 \]Square both sides of the inequality and multiply them by \(\pi\):
\[ R^2 \ge 25 \] \[ \pi R^2 \ge 25\pi \]Since \(\pi R^2\) is the formula for the area \(a\) of the circle, then:
\[ a \ge 25\pi \]If \(4x + y = 5\) and \(2x + 3y = 12\), then \(3x + 2y =\)
Solution
Add the right hand sides and left hand sides of the two equations and simplify:
\[ (4x + y) + (2x + 3y) = 5 + 12 \] \[ 6x + 4y = 17 \]Divide all terms of the above equation by 2:
\[ 3x + 2y = \frac{17}{2} \]The price \(x\) of a t-shirt was first reduced by 10%. Then the new price was reduced by a further 20%. The price of the T-shirt after the two reductions is given by:
A) \(x - 0.3x\)
B) \(x - 30\)
C) \(x - 0.3x + 0.02x\)
D) \(x + 0.02x\)
E) \(x + 0.3x\)
Solution
Price after first reduction:
\[ x - 10\% x = x - 0.1x \]Price after the second reduction:
\[ (x - 0.1x) - 20\%(x - 0.1x) = x - 0.1x - 0.2x + 0.02x \] \[ = x - 0.3x + 0.02x \]If \(x + y = \sqrt{22}\) and \((x - y)^2 = 10\), then \(xy =\)
Solution
Square both sides of the equation \(x + y = \sqrt{22}\):
\[ (x + y)^2 = (\sqrt{22})^2 \]Expand \((x + y)^2\) and use \((x - y)^2 = 10\):
\[ x^2 + y^2 + 2xy = 22 \] \[ x^2 + y^2 - 2xy = 10 \]Subtract the left and right hand sides of the two equations above:
\[ (x^2 + y^2 + 2xy) - (x^2 + y^2 - 2xy) = 22 - 10 \] \[ 4xy = 12 \] \[ xy = 3 \]ABCD is a parallelogram and MN is parallel to DC. The length of BN is 1/3 of the length of BC. What is the ratio of the area of triangle BNM to the area of the parallelogram ABCD?
Solution
Since MN is parallel to DC, triangles BMN and BDC are similar and therefore:
\[ \frac{BM}{BD} = \frac{BN}{BC} = \frac{1}{3} \]The area \(A_1\) of triangle MBN is given by:
\[ A_1 = \frac{1}{2} \sin(S) \cdot BM \cdot BN \]The area \(A_2\) of the parallelogram ABCD is twice the area of triangle DBC and is given by:
\[ A_2 = 2 \cdot \frac{1}{2} \sin(S) \cdot BD \cdot BC = \sin(S) \cdot BD \cdot BC \]The ratio \(A_1 / A_2\) is given by:
\[ \frac{A_1}{A_2} = \frac{\frac{1}{2} \sin(S) \cdot BM \cdot BN}{\sin(S) \cdot BD \cdot BC} \] \[ = \frac{1}{2} \cdot \frac{BM}{BD} \cdot \frac{BN}{BC} = \frac{1}{2} \cdot \frac{1}{3} \cdot \frac{1}{3} = \frac{1}{18} \]If \(T = \frac{rR}{r + R}\), then \(R =\)
Solution
Cross multiply:
\[ T(r + R) = rR \]Solve for \(R\):
\[ Tr + TR = rR \] \[ TR - rR = -Tr \] \[ R(T - r) = -Tr \] \[ R = \frac{-Tr}{T - r} = \frac{Tr}{r - T} \]If \(-2 < x < 2\), then \(\frac{|4 - x^2|}{x - 2} =\)
Solution
If \(-2 < x < 2\) then:
\[ -2 + 2 < x + 2 < 2 + 2 \quad \Rightarrow \quad 0 < x + 2 < 4 \] \[ -2 - 2 < x - 2 < 2 - 2 \quad \Rightarrow \quad -4 < x - 2 < 0 \]\(|4 - x^2|\) may be written as:
\[ |4 - x^2| = |2 - x||2 + x| = |x - 2||x + 2| \]Because of the inequalities \(0 < x + 2 < 4\) and \(-4 < x - 2 < 0\), we can write:
\[ |x - 2||x + 2| = -(x - 2)(x + 2) \]Hence:
\[ \frac{|4 - x^2|}{x - 2} = \frac{-(x - 2)(x + 2)}{x - 2} = -(x + 2) \]For all real numbers \(x\), \(|-x^2 - 1| =\)
Solution
Rewrite as follows:
\[ |-x^2 - 1| = |-(x^2 + 1)| \]Simplify:
\[ = x^2 + 1 \]If \(\frac{3}{7}\) of the students in a school are girls, what is the ratio of boys to girls in this school?
Solution
Let \(x\) be the total number of boys and girls in the school. The number of boys is given by:
\[ x - \frac{3}{7}x = \frac{4}{7}x \]The ratio of boys to girls is given by:
\[ \frac{\frac{4}{7}x}{\frac{3}{7}x} = \frac{4}{3} \]If \(\frac{7}{5}\) of a number is 6 more than 20% of the number, what is \(\frac{11}{10}\) of the number?
Solution
Let \(x\) be the number. The statement "\(\frac{7}{5}\) of a number is 6 more than 20% of the number" is mathematically translated as:
\[ \frac{7}{5}x = 20\% x + 6 \]The above equation may be written as:
\[ \frac{7}{5}x = \frac{1}{5}x + 6 \]Solve for \(x\):
\[ x = 5 \]Calculate \(\frac{11}{10}x\):
\[ \frac{11}{10} \cdot 5 = \frac{55}{10} = 5.5 \]What is the degree measure of the smaller angle formed by the hour and the minute hands of a clock at 2:50?
Solution
Each "5 minutes" division corresponds to a central angle of:
\[ \frac{360}{12} = 30^\circ \]
The minute hand at 50 minutes corresponds to:
\[ \text{Minute hand: } 50 \text{ minutes} \rightarrow \frac{30 \times 50}{60} = 25^\circ \]The size of the smaller angle is:
\[ 4 \times 30^\circ + 25^\circ = 145^\circ \]AC is a diameter of the circle shown below and B is a point on the circle such that triangle ABC is isosceles. If the circle has a circumference of \(8\pi\), what is the area of the shaded region?
Solution
The circumference is given, we can find the radius \(R\):
\[ 2\pi R = 8\pi \quad \Rightarrow \quad R = 4 \]Since AC is the diameter, ABC is a right triangle (Thales theorem). Using Pythagorean theorem:
\[ AB^2 + BC^2 = 8^2 \]Since triangle ABC is isosceles, \(AB = BC\):
\[ 2AB^2 = 64 \quad \Rightarrow \quad AB^2 = 32 \quad \Rightarrow \quad AB = BC = 4\sqrt{2} \]Let \(A_s\) be the area of the shaded region. Subtract the area of triangle ABC from the area of the semicircle:
\[ 2A_s = \frac{1}{2}\pi \cdot 4^2 - \frac{1}{2} \cdot AB \cdot BC = 8\pi - 16 \] \[ A_s = 4\pi - 8 \]ABCD is a square of side \(x\) feet. MCD is a triangle such that \(MC = \frac{x}{3}\) feet. Find \(x\) so that the area of the trapezoid ABMD is equal to 30 square feet.
Solution
The area of the trapezoid is given by:
\[ \frac{1}{2} \cdot x \cdot \left(x + \frac{2}{3}x\right) = \frac{x}{2} \cdot \frac{5x}{3} = \frac{5}{6}x^2 \]The area is known to be 30:
\[ \frac{5}{6}x^2 = 30 \quad \Rightarrow \quad x^2 = 36 \quad \Rightarrow \quad x = 6 \]The average (arithmetic mean) of 3 numbers is 42. The average of another 5 numbers is 50. What is the average of all 8 numbers grouped together?
A) 46
B) 47
C) 48
D) 49
E) 50
What is 5% of 6%?
A) 0.3
B) 3
C) 0.03
D) 0.0003
E) 0.00003
At a high school, where each student takes one foreign language only, \(\frac{3}{8}\) of the students take French and \(\frac{1}{5}\) of the remaining students take German. If all the other students study Italian, what percent of all students take Italian?
A) 50
B) 80
C) 10
D) 40
E) 30
On a certain day, a student spent \(\frac{1}{4}\) of his homework time reviewing his lessons, \(\frac{7}{12}\) of the time solving the assigned homework and the rest of the time revising for his coming tests. What is the ratio of the time spent revising for his tests to the time spent solving his assigned homework?
A) 1:7
B) 1:3
C) 2:7
D) 7:12
E) 7:4
From 2003 to 2004 the profits made by a company increased by 10%. A year later in 2005, the profits of this company decreased to their 2003 value. By what percent did the profits decrease from 2004 to 2005?
A) 10
B) 5
C) 1
D) 20
E) 9
In a bookshop, the sales of scientific books increased by 40% while the sales of engineering books decreased by 50% from 2001 to 2002. If \(R\) is the ratio of the number of scientific books to the number of engineering books in 2001 and \(r\) the same ratio in 2002, what is \(k\) if it is given by \(k = r/R\)?
A) 2.8
B) 1.25
C) 0.2
D) 1
E) 0.8
If \(f(x - 2) = g(x + 1) + \frac{1}{x} + 2\), what does \(f(0)\) equal?
A) \(g(1) + 2\)
B) undefined
C) \(g(3) + \frac{5}{2}\)
D) \(g(0) + 2\)
E) \(g(1)\)
Which of the following is not equivalent to \(\frac{7}{10}\)?
A) 0.7
B) \(\frac{35}{50}\)
C) \(\frac{9}{10} \times \frac{15}{18}\)
D) 70%
E) \(\frac{3}{10} \div \frac{3}{7}\)
If \(3x - 2 = 41\), what is the value of \(\sqrt{3x + 6}\)?
A) \(\sqrt{\frac{43}{3}}\)
B) \(\sqrt{49}\)
C) \(\sqrt{41}\)
D) \(\sqrt{43}\)
E) \(\sqrt{\frac{41}{43}}\)
If \(n\) and \(m\) are integers such that \(n^2 = \frac{\sqrt{|m|}}{2}\), which of the following CANNOT be the value of \(m\)?
A) -4
B) 0
C) 64
D) -1024
E) 9
The system of equations \(2x + y = 2\) and \(-x + ky = 5\) does not have a solution if \(k =\)
A) 0
B) 1
C) \(\frac{1}{2}\)
D) \(-\frac{1}{2}\)
E) 5