Key: Let $a$ is the difference between two consecutive terms.
Fourth term: $61$
Fifth term: $61+a$
Sixth term: $61+2a$, therefore, $61+2a=93$. Solve for $a$ to get $a=\dfrac{93-61}{2}=16$
Ninth term: $61+5a=61+5×{16}=141$
Key: $x^2 = 9y^2$. Take the square root of both sides yields $x=\pm{3y}$. Taking the positive solution gives $3y=5+3y$, which is a contradiction (no solution). Substituting the negative solution to $x=5+3y$ yields $-3y=5+3y$. Solve for $y$ gives $y=-\dfrac{5}{6}$. Finally, solve for $x$ gives $x=5+3\left(-\dfrac{5}{6}\right)=\dfrac{5}{2}$.
Key: The square has length of 2. Since the circle is inscribed in the square, its diameter is equal to the side of the square. The radius is 1. The area of the square is $\pi{r^2}=\pi$. The shaded area is $\dfrac{3}{4}$ of the square or $\dfrac{3\pi}{4}$.
(A) $AB = BC$ (B) $BC = AC$
(C) $x=y$ (D) $x=z$ (E) $y=z$
(Answer: (E))
Note: Figure not drawn to scale.
Key: Since the triangle is isosceles and $AB > AC$, therefore either $AB = BC$ ((A) is true) or $BC = AC$ ((B) is true). If $AB = BC$, then $x=z$ ((D) is true). If $BC = AC$, then $x=y$ ((C) is true). Since $AB > AC$, $z>y$ ((E) is FALSE).
The equation of the line above is $y=-2x+4$. Which of the following is the graph of $y=|-2x+4|$ ?
(Answer: )
Key:
I. $f(a+b)=2f(b)$
II. $f(a+b)=f(a)\cdot{f(b)}$
III. $f(a)+f(a)=f(2b)$
(Answer: I and III only)
Key:
If $a=b$, then $f(a+b)=f(b+b)=f(b)+f(b)=2f(b)$. Choice I is true. Also, $f(a)+f(a)=f(2a)=f(2b)$, choice III is also true.
If the following 5 cards are placed in a row so that the card number 3 is never at either end, how many different arrangements are possible ?
(Answer: 72)
Key:
The number of arrangements for 4 cards 1, 2, 4, 5 is $4!=4×3×2×1=24$. Since card number 3 can be placed only after the first, second, and third positions (1, 2, or, 4), the number of possible arrangements are $24×3=72$.
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