From 1 to 2001 :

The number of multiples of 3 is : (2001 - 3) : 3 + 1 = 667

The number of multiples of 4 is : (2000 - 4) : 4 + 1 = 500

The number of common multiples of 3 and 5 is : 

(1995 - 15) : 15 + 1 = 133

The number of common multiples of 4 and 5 is :

(2000 - 20) : 20 + 1 = 100

The number of common multiples of 3 and 4 is :

(1992 - 12) : 12 + 1 = 166

The number of common multiples of 3, 4 and 5 is :

(1980 - 60) : 60 + 1 = 33

The answer is :

(667 - 133) + (500 - 166 - 100 + 33) = 801

\(0,\overline{009}=\dfrac{1}{111}\). So, the answer is 111

The area of the patio is : 20 x 14 = 280 (ft²)

There are 2 cases (because there's no figure) :

+ The garden is tangent to 2 20-ft sides and 1 14-ft side

The radius of the garden is : \(14:2=7\left(ft\right)\)

The area of the garden is : \(7^2\pi=49\pi\left(ft^2\right)\) 

The answer is : \(\dfrac{100\left(280-49\pi\right)}{280}\%\approx45\%\)

+ The garden is tangent to 2 14-ft sides and 1 20-ft side

The radius of the garden is : \(20:2=10\left(ft\right)\)

The area of the garden is : \(10^2\pi=100\pi\left(ft^2\right)\) (asburd because \(100\pi>280\))

So, the answer is 45%

\(\left(3b^2+2b+1\right)-\left(b^2+2b+3\right)=b^2+5b+4\)

\(\Leftrightarrow b^2-5b-6=0\Leftrightarrow b^2+b-6b-6=0\)

\(\Leftrightarrow b\left(b+1\right)-6\left(b+1\right)=0\Leftrightarrow\left(b-6\right)\left(b+1\right)=0\Rightarrow b=6\)

Let a, b be the number of bicycles and tricycles in the shop

\(\left\{{}\begin{matrix}a+b=67\\2a+3b=157\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2a+2b=134\\2a+3b=157\end{matrix}\right.\)\(\Rightarrow b=23\)

So, the answer is 23

Draw \(CK\perp AB\)

\(\widehat{CBK}=180^0-\widehat{ABC}=180^0-\dfrac{180^0\left(8-2\right)}{8}=45^0\)

\(\Rightarrow CK=\dfrac{BC}{\sqrt{2}}=\sqrt{2}\)

\(S_{ABC}=\dfrac{AB.CK}{2}=\dfrac{2\sqrt{2}}{2}=\sqrt{2}\left(cm^2\right)\)

1 h = 60 m

It takes : \(60.\dfrac{300}{1200}=15\) (m) to burn 300 cal when running at 12 m/h

It takes : \(60.\dfrac{300}{450}=40\) (m) to burn 300 cal when running at 5 m/h

The answer is : 40 - 15 = 25 (m)

We notice that the sum of 2 consecutive triangular numbers is always a square number : \(\dfrac{\left(k-1\right)k}{2}+\dfrac{k\left(k+1\right)}{2}=\dfrac{k.2k}{2}=k^2\)

So, we need to find the largest triangular number less than 100

\(\dfrac{n\left(n+1\right)}{2}< 100\Leftrightarrow n^2+n< 200\Leftrightarrow n^2+2n.\dfrac{1}{2}+\dfrac{1}{4}< 200\dfrac{1}{4}\)

\(\Leftrightarrow\left(n+\dfrac{1}{2}\right)^2< 200\dfrac{1}{4}\Rightarrow n< 14\)

So, n = 13 and the answer is : \(\dfrac{13.14}{2}=91\)

(One of them is 91 and the other is \(\dfrac{12.13}{2}=78\))

The number of miles per gallon of gasoline that her old car gets is :

\(24:\left(100\%+50\%\right)=16\)

The number of gallons of gasoline that her new car spend to travel 480 miles is : 480 : 24 = 20 

The number of gallons of gasoline that her old car spend to travel 480 miles is : 480 : 16 = 30 

The answer is : 30 - 20 = 10 (gallons of gasoline)

The answer is : \(\dfrac{30}{50}+\dfrac{30}{20}=\dfrac{21}{10}=2.1\) (hours)

The statement D is true because 5 - 1 = 4

Draw \(HI,GK\perp EF\)

Each angle of the octagon is : \(\dfrac{180\left(8-2\right)}{8}=135^0\)

\(\Rightarrow\widehat{HAF}=\widehat{GFA}=180^0-135^0=45^0\)

Let a be the side length of the octagon, then \(HI=IA=GK=KF=\dfrac{a}{\sqrt{2}};IK=HG=a\)

\(\Rightarrow AF=a\left(1+\sqrt{2}\right)\Rightarrow S_{AHF}=\dfrac{a\left(1+\sqrt{2}\right).\dfrac{a}{\sqrt{2}}}{2}=\dfrac{a^2\left(2+\sqrt{2}\right)}{4}\)

\(S_{ADF}=S_{ABCDEFGH}-S_{AFGH}-S_{ABCD}-S_{FED}\)

\(=S_{AFGH}+S_{ABEF}+S_{BCDE}-S_{AFGH}-S_{ABCD}-S_{FED}\)

\(=S_{ABEF}-S_{FED}=a.a\left(1+\sqrt{2}\right)-a.\dfrac{a}{\sqrt{2}}:2\)

\(=\dfrac{a^2\left(4+3\sqrt{2}\right)}{4}\)

The answer is : \(\dfrac{4+3\sqrt{2}}{4}:\dfrac{2+\sqrt{2}}{4}=1+\sqrt{2}\approx2.41\)

Let a, b be the number of 3-ounce packages and 10-ounce packages B-Rite sold 

\(\left\{{}\begin{matrix}3a+10b< 16.16\\a+b=76\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}3a+3b+7b< 256\\3a+3b=228\end{matrix}\right.\)

\(\Rightarrow7b< 28\Rightarrow b< 4\)

So, the answer is 3

\(x-2y=8\Leftrightarrow y=\dfrac{1}{2}x-4\)

Let the equation of the line through point P be \(y=\dfrac{1}{2}x+a\left(a\ne-4\right)\)

\(3=\dfrac{1}{2}.4+a\Leftrightarrow a=1\)

So, the answer is \(y=\dfrac{1}{2}x+1\)

Let a, b be the side length of each poygon respectively \(\left(a,b\in Z^+\right)\)

\(\left\{{}\begin{matrix}6a=8b\\6a+8b< 200\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{a}{b}=\dfrac{4}{3}\\12a< 200\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}a⋮4\left(b⋮3\right)\\a< 17\end{matrix}\right.\)

The answer is : (16 - 4) : 4 + 1 = 4

3 diagonals of the hexagon which are the diameters of the circle divide the hexagon into 6 congruent equilateral triangles. The area of each equilateral triangle is : \(\dfrac{\left(4\sqrt{3}\right)^2.\sqrt{3}}{4}=12\sqrt{3}\) (cm²)

The answer is : \(12\sqrt{3}.6=72\sqrt{3}\) (cm²)

Let \(\overline{ab}\) be the original integer \(\left(a,b\in N;a>0;a,b\le9\right)\)

\(\left\{{}\begin{matrix}a+b=8\\\overline{ba}=4\overline{ab}+3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a+b=8\\a+10b=40a+4b+3\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}b+a=8\\6b-39a=3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2b+2a=16\\2b-13a=1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}15a=15\\b=8-a\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=1\\b=7\end{matrix}\right.\)

So, the answer is 71

Let a be the smallest angle in degree, then the largest angle is 4a, the third angle is 2a + 10, the fourth angle is 3a - 60. We have :

\(a+4a+2a+10+3a-60=360\Leftrightarrow10a=410\Leftrightarrow a=41\Leftrightarrow4a=164\)

So, the answer is 164⁰

The number of students who participate in at least 1 activity is :

73 + 65 + (114 - 32 - 12) = 208 

The answer is : 208 + 558 = 766 (students)

Let n be the number of sides of the polygon\(\left(n\in N;n\ge3\right)\)

\(\dfrac{n\left(n-3\right)}{2}=20\Leftrightarrow n^2-3n=40\Leftrightarrow n^2+5n-8n-40=0\)

\(\Leftrightarrow n\left(n+5\right)-8\left(n+5\right)=0\Leftrightarrow\left(n-8\right)\left(n+5\right)=0\)

\(n\ge3\Rightarrow n+5\ge8\). So, n = 8 (satisfied)

Hence, the polygon has 8 sides