Choice C
$$Length = \frac{100}{5}=20 \ feet$$
Choice J
$$p = \frac{5}{2+9+5}=\frac{5}{16}$$
Choice E
$$Number \ of \ arrangement = 4\times3\times2=24$$
Choice F
30 pens still remain, enough for 6 packages;
20 notebooks still remain, enough for 10 packages;
24 envelopes still remain, enough for 8 packages;
84 cookies still remain, enough for 7 packages;
45 candy bars still remain, enough for 9 packages;
Apparently, pens are the bottleneck.
Choice C
$$V=\frac{1}{3}\cdot \frac{22}{7}\cdot 6^2\times 28=1056$$
Choice G
Choice B
$$(0.1x^2+3x+80) +(0.5x^2-2x+60)=(0.1x^2+0.5x^2)+(3x-2x)+(80+60)=0.6x^2+x+140$$
Choice H
Only d = 3t + 15 fulfils all the data in the table.
Choice C
$$$40\times(1-d)=$30 \\ $$
$$\Rightarrow d=25\%$$
Choice G
$$|-6|-|7-41|=6-34=-28$$
Choice C
$$Value=$6,880\times \frac{70}{70+50+40}=$3,010$$
Choice G
Only the number line in G fulfils:
$$18\leq age <30$$
Choice D
$$\frac{AC}{AB}=\frac{AE}{AD} \\$$
$$ \Rightarrow \frac{AC}{6+5+4}=\frac{12}{6} \\$$
$$ \Rightarrow AC=30$$
Choice G
$$\frac{\sqrt{50}}{2}\approx 3.54 \approx 4$$
Choice D
$$9k+2k=44 \\$$
$$ \Rightarrow k=4 \\$$
$$ 9k=36$$
Choice H
$$1-\frac{1}{9}-\frac{1}{6}=\frac{13}{18}$$
Choice A
$$35\times(10+x)=42\times10 \\$$
$$ \Rightarrow 10+x=12 \\ $$
$$\Rightarrow x=2$$
Choice J
$$Area \ of \ 2 \ shaded \ regions = 8\times 10 - \frac{8\times10}{2}=40$$
Choice C
Only C fulfils all three inequations.
Choice H
The range of y = sin (x-π) is :
$$-1\leq y \leq 1$$
The range of y = -5sin (x-π) is :
$$-5\leq y \leq 5$$
The range of y = 3-5sin (x-π) is :
$$-2\leq y \leq 8$$
Choice D
$$g(-3)=(-3)^2-4=5 \\$$
$$ \Rightarrow f(g(-3))=f(5)=2\times5+1=11$$
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Choice G
$$3f+4c=$25 \\$$
$$f+2c=$11 \\$$
$$\Rightarrow …
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Choice C
$$71,\ 77,\ 78,\ 80,\ \underline{86, …
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Choice H
Let the length of the …
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Choice E
$$AB\times AD=1 \\$$
$$ \Rightarrow …
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Choice G
$$\frac{16+x}{16+x+7+19}=\frac{3}{5} \\$$
$$ \Rightarrow x= …
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Choice E
$$a^{-2}=(a^{2})^{-1}=\frac{1}{a^2}$$
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Choice F
If Jamie's claim is true, …
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Choice D
$$Area = 3 \times 5 …
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Choice H
$$BD=\sqrt{3^2+4^2}=5$$
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Choice A
The slope of a horizontal …
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Choice J
When A is reflected over …
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Choice A
$$8^2\cdot 4^{0.5}=(2^3)^2\cdot (2^2)^{0.5}=2^6\cdot 2=2^7$$
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Choice K
$$Number \ of \ applicants …
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Choice B
$$\log_{2}{\sqrt{8}}=\log_{2}{\sqrt{2^3}}=\log_{2}{2^{\frac{3}{2}}}=\frac{3}{2}$$
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Choice F
$$\frac{\frac{90-15}{3}}{90}\times 360°=100°$$
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Choice E
$$\frac{4}{5}+\frac{7}{x}= \frac{4x}{5x}+\frac{7\times5}{5x}=\frac{4x+35}{5x}$$
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Choice K
$$\frac{353}{85+353+47}=\frac{353}{485}$$
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Choice D
$$\frac{14+10+85+353}{560}\approx 0.83$$
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Choice H
Of the 15 incorrectly classified …
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Choice A
To simplify matters, let’s make …
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Choice G
sin(-x°) = -sin x°
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Choice D
Volume of box = 8 …
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Choice G
Since ΔBCD is a right …
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Choice D
200 days ago is equivalent …
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Choice G
∠1 = ∠2
→3x - …
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Choice E
The best way to solve …
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Choice H
The Scalene Triangle has no …
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Choice D
A sportswriter estimates the team …
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Choice J
The denominator of f(x) = …
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Choice E
555_ _ _ _
Each …
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Choice F
The circle represents all dots …
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Choice B
The question asks what x …
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Choice F
The slope represents the copying …
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Choice E
Let r be revenue, n …
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Choice K
The quickest way would be …
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Choice A
To make one banner, ¼ …
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Choice F
What you need to know: …
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Choice D
According to the graph, the …
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Choice K
Requirements for Matrix Multiplication:
In …
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