Question
What is the area, in square feet, of the triangle whose sides have lengths equal to 10, 6 and 8 feet?
Solution
If given the three sides of a triangle, we can use Heron's formula. However, this is a special triangle. Note:
\[ \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \]This means the triangle is a right triangle with hypotenuse 10 and legs 6 and 8. The area \(A\) is:
\[ A = \frac{1}{2} \times 6 \times 8 = 24 \text{ square feet} \]Question
Simplify the expression:
\[ 3^{102} + 9 \times 3^{100} + \frac{3^{103}}{3} \]Solution
Rewrite each term:
\[ 9 \times 3^{100} = 3^2 \times 3^{100} = 3^{102} \] \[ \frac{3^{103}}{3} = 3^{103} \times 3^{-1} = 3^{102} \]Substitute back:
\[ 3^{102} + 3^{102} + 3^{102} = 3 \times 3^{102} = 3^{103} \]Question
Of the 80 students in class, 25 are studying German, 15 French and 13 Spanish. 3 are studying German and French; 4 are studying French and Spanish; 2 are studying German and Spanish; and none is studying all 3 languages. How many students are not studying any of the three languages?
Solution
Using a Venn diagram:
- German only: \(25 - (3 + 2) = 20\)
- French only: \(15 - (3 + 4) = 8\)
- Spanish only: \(13 - (4 + 2) = 7\)
Number studying at least one language: \(20 + 8 + 7 + 3 + 4 + 2 = 44\)
Number not studying any: \(80 - 44 = 36\)
Question
In the figure below, AB is a diameter of the large circle. The centers C1 and C2 of the smaller circles are on AB. The two small circles are congruent and tangent to each other and to the larger circle. The circumference of circle C1 is \(8\pi\). What is the area of the large circle?
Solution
Circumference of C1 is \(8\pi\), so its diameter is:
\[ \frac{8\pi}{\pi} = 8 \]The radius \(R\) of the large circle equals the diameter of C1, so \(R = 8\). Area:
\[ A = \pi R^2 = \pi \times 8^2 = 64\pi \]Question
Round \(202^2\) to the nearest hundred.
Solution
\[ 202^2 = 40804 \]Rounded to nearest hundred: \(40800\)
Question
If \(w\) workers, working at equal rates, can produce \(x\) toys in \(n\) days, how many days does it take \(c\) workers, working at same rates, to produce \(y\) toys?
Solution
Rate per worker per day:
\[ r = \frac{x}{n \cdot w} \]Let \(N\) be days needed for \(c\) workers to produce \(y\) toys:
\[ r = \frac{y}{N \cdot c} = \frac{x}{n \cdot w} \]Solve for \(N\):
\[ N = \frac{y \cdot n \cdot w}{c \cdot x} \]Question
A number of the form \(213ab\), where \(a\) and \(b\) are digits, has a remainder less than 10 when divided by 100. The sum of all digits is 13. Find \(b\).
Solution
Number can be written as:
\[ 213ab = 21300 + 10a + b \]Remainder when divided by 100 is \(10a + b\). For remainder < 10, we need \(a = 0\). Sum of digits:
\[ 2 + 1 + 3 + 0 + b = 13 \]Solve: \(b = 7\)
Question
Find a negative value of \(x\) that satisfies:
\[ \sqrt{(x+1)^2 - (2x + 1)} + 2|x| - 6 = 0 \]Solution
Simplify inside square root:
\[ (x+1)^2 - (2x+1) = x^2 + 2x + 1 - 2x - 1 = x^2 \]Equation becomes:
\[ \sqrt{x^2} + 2|x| - 6 = 0 \] \[ |x| + 2|x| - 6 = 0 \] \[ 3|x| = 6 \] \[ |x| = 2 \]Solutions: \(x = 2\) or \(x = -2\). Negative solution: \(x = -2\)
Question
The equation \(\frac{1}{a} + \frac{1}{|a|} = 0\) has:
A) infinite solutions B) no solutions C) 1 solution D) 2 solutions E) 3 solutions
Solution
Case 1: \(a = 0\) → undefined.
Case 2: \(a > 0\) → \(|a| = a\), equation: \(\frac{1}{a} + \frac{1}{a} = \frac{2}{a} = 0\) → no solution.
Case 3: \(a < 0\) → \(|a| = -a\), equation: \(\frac{1}{a} + \frac{1}{-a} = 0\) → identity true for all negative \(a\).
Answer: A) infinite solutions
Question
The inequality \(x^2 - 2x + 1 \le 0\) has:
A) no solutions B) a set of solutions C) 1 solution D) 2 solutions E) 3 solutions
Solution
\[ x^2 - 2x + 1 = (x-1)^2 \le 0 \]Square is always ≥ 0, so equality only when \(x = 1\).
Answer: C) 1 solution
Question
In the figure below, AC is parallel to DE. AE, FG and CD intersect at B. FG is perpendicular to AC and DE. DE = 5 inches, BG = 8 inches, AC = 6 inches. Find area of triangle ABC.
Solution
Triangles ABC and DBE are similar. Ratio:
\[ \frac{DE}{AC} = \frac{BG}{FB} \] \[ \frac{5}{6} = \frac{8}{FB} \Rightarrow FB = \frac{48}{5} \]Area of triangle ABC:
\[ \frac{1}{2} \times AC \times FB = \frac{1}{2} \times 6 \times \frac{48}{5} = 28.8 \text{ in}^2 \]Question
Points A(1,1), B(4,b), C(6,1). What positive value of \(b\) makes triangle ABC a right triangle with AC as hypotenuse?
Solution
Using Pythagorean theorem:
\[ AB^2 + BC^2 = AC^2 \] \[ (4-1)^2 + (b-1)^2 + (6-4)^2 + (1-b)^2 = (6-1)^2 + (1-1)^2 \] \[ 9 + (b-1)^2 + 4 + (b-1)^2 = 25 \] \[ 2(b-1)^2 = 12 \Rightarrow (b-1)^2 = 6 \] \[ b = 1 \pm \sqrt{6} \]Positive value: \(b = 1 + \sqrt{6}\)
Question
Line L passes through (-2,0) and (0,a). Line LL passes through (4,0) and (6,2). What value of \(a\) makes them parallel?
Solution
Slope of L: \(m_1 = \frac{a-0}{0-(-2)} = \frac{a}{2}\)
Slope of LL: \(m_2 = \frac{2-0}{6-4} = 1\)
Parallel if slopes equal: \(\frac{a}{2} = 1 \Rightarrow a = 2\)
Question
\(a, b, c, d\) are distinct numbers such that:
\[ a + b = d \] \[ a \cdot b \cdot c = 0 \]Which number MUST be 0?
Solution
Since \(a\) and \(b\) are distinct and \(a+b=d\), neither \(a\) nor \(b\) is 0. Then from second equation, \(c\) must be 0.
Question
Simplify:
\[ \frac{10^4(5^4 - 2^4)}{21} \]Solution
Factor difference of squares:
\[ 5^4 - 2^4 = (5^2 - 2^2)(5^2 + 2^2) = (25-4)(25+4) = 21 \times 29 \]Substitute:
\[ \frac{10^4 \times 21 \times 29}{21} = 10^4 \times 29 = 290,\!000 \]Question
For what value of \(k\) do the equations \(2x + 4 = 4(x-2)\) and \(-x + k = 2x - 1\) have the same solution?
Solution
Solve first equation:
\[ 2x + 4 = 4x - 8 \Rightarrow 12 = 2x \Rightarrow x = 6 \]Substitute into second:
\[ -6 + k = 12 - 1 \Rightarrow k = 17 \]Question
An object travels at 15 feet per minute. How many feet does it travel in 24 minutes and 40 seconds?
Solution
Convert to seconds: 24 min 40 sec = \(24 \times 60 + 40 = 1480\) seconds.
Speed in ft/sec: \(\frac{15}{60} = 0.25\) ft/sec.
Distance: \(1480 \times 0.25 = 370\) feet.
Question
Simplify:
\[ \frac{4}{\sqrt{20} - \sqrt{12}} \]Solution
Multiply numerator and denominator by conjugate:
\[ \frac{4(\sqrt{20} + \sqrt{12})}{(\sqrt{20})^2 - (\sqrt{12})^2} = \frac{4(\sqrt{20} + \sqrt{12})}{20 - 12} \] \[ = \frac{4(2\sqrt{5} + 2\sqrt{3})}{8} = \sqrt{5} + \sqrt{3} \]Question
In the figure, DE is parallel to CB and \(\frac{AE}{EB} = 4\). Area of triangle AED is 20 in². Find area of triangle ABC.
Solution
Triangles AED and ABC are similar. Ratio of sides:
\[ \frac{AB}{AE} = \frac{AE + EB}{AE} = \frac{4EB + EB}{4EB} = \frac{5}{4} \]Ratio of areas = square of side ratio: \(\left(\frac{5}{4}\right)^2 = \frac{25}{16}\)
Area of ABC = \(\frac{25}{16} \times 20 = 31.25\) in²
Question
If \(f(n) = n + \sqrt{n}\), where \(n\) is a positive integer, which could be a value of \(f(n)\)?
I) 5 II) 10 III) 12
Solution
Let \(x = \sqrt{n} \ge 0\), integer. Then \(n = x^2\), and \(f(n) = x^2 + x\).
Check: \(x^2 + x = 5\) → \(x^2 + x - 5 = 0\), no integer solution.
\(x^2 + x = 10\) → \(x^2 + x - 10 = 0\), no integer solution.
\(x^2 + x = 12\) → \(x^2 + x - 12 = 0\) → \((x+4)(x-3)=0\) → \(x=3\) (positive).
Only III is possible: \(f(9) = 9 + 3 = 12\).
Question
If \(a\) and \(b\) are even, which could be an odd integer?
A) \((a+b)^2\) B) \(a^2+b^2\) C) \((a+1)^2+(b+1)^2\) D) \((a+1)(b+1)-1\) E) \(\frac{a+1}{b+1}\)
Solution
Let \(a=2m\), \(b=2n\).
A) \((2m+2n)^2 = 4(m+n)^2\) (even)
B) \(4m^2+4n^2 = 4(m^2+n^2)\) (even)
C) \((2m+1)^2+(2n+1)^2 = 4m^2+4m+1+4n^2+4n+1 = 2(2m^2+2m+2n^2+2n+1)\) (even)
D) \((2m+1)(2n+1)-1 = 4mn+2m+2n+1-1 = 2(2mn+m+n)\) (even)
E) Example: \(a=2, b=4\) → \(\frac{3}{5}\) not integer; \(a=8, b=2\) → \(\frac{9}{3}=3\) (odd integer possible)
Answer: E
Question
If \(x < 0\), which must be true?
I) \(x^5 < |x|\) II) \(x < \sqrt{-x}\) III) \(x - \frac{1}{|x|} < 0\)
Solution
I: \(x^5\) negative, \(|x|\) positive → true.
II: \(x\) negative, \(\sqrt{-x}\) positive → true.
III: \(x\) negative, \(\frac{1}{|x|}\) positive → sum negative → true.
All are true.
Question
If \(n\) is a positive integer and \(\frac{n!}{(n-2)!} = 342\), find \(n\).
Solution
\[ \frac{n!}{(n-2)!} = n(n-1) = 342 \] \[ n^2 - n - 342 = 0 \] \[ (n-19)(n+18) = 0 \]Positive solution: \(n = 19\)
Question
Find sum of reciprocals of solutions to:
\[ x^2 - \frac{3}{5}x = -\frac{11}{3} \]Solution
Rewrite: \(x^2 - \frac{3}{5}x + \frac{11}{3} = 0\)
Let roots be \(x_1, x_2\). Sum \(S = x_1 + x_2 = \frac{3}{5}\), product \(P = x_1 x_2 = \frac{11}{3}\).
Sum of reciprocals:
\[ \frac{1}{x_1} + \frac{1}{x_2} = \frac{x_1 + x_2}{x_1 x_2} = \frac{3/5}{11/3} = \frac{9}{55} \]Question
A number is 987562153ab where a and b are digits, \(a+b=11\) and \(a < b\). Find values such that the number is divisible by 4.
Solution
Divisibility by 4: last two digits \(10a+b\) divisible by 4.
Given \(a+b=11\) ⇒ \(b=11-a\). So \(10a + (11-a) = 9a+11\) divisible by 4.
\(9a+11 ≡ 0 \pmod{4}\) ⇒ \(9a ≡ 1 \pmod{4}\) ⇒ \(a ≡ 1 \pmod{4}\)
Possible \(a\): 1,5,9 with \(a < b\).
Check:
- \(a=1\) → \(b=10\) invalid digit
- \(a=5\) → \(b=6\), \(56÷4=14\) valid
- \(a=9\) → \(b=2\), but \(a < b\) false
Answer: \(a=5, b=6\)