math-eval/TAL-SCQ5K huggingface dataset:
Problem: $$99+99+99+99+99+99+99+99+99+99=$$． 
Options:
aoVal: A
content: $$10\times 99$$ 
aoVal: B
content: $$10\div 99$$ 
aoVal: C
content: $$9\times 99$$ 
aoVal: D
content: $$99\times 99$$ 
Answer: A
Analysis: The sum of ten $$99$$\textquotesingle s is the same as $$10\times 99$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Addition and Subtraction
Problem: $$100-98+96-94+92-90+\ldots +8-6+4-2+0=$$． 
Options:
aoVal: A
content: $$26$$ 
aoVal: B
content: $$50$$ 
aoVal: C
content: $$52$$ 
aoVal: D
content: $$100$$ 
Answer: B
Analysis: $$\left( 100-98 \right)+\left( 96-94 \right)+\ldots +\left( 8-6 \right)+\left( 4-2 \right)+0=2\times 25+0=50$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Addition and Subtraction
Problem: What is $$4\frac{7}{20}$$ as a decimal? 
Options:
aoVal: A
content: $$4.035$$ 
aoVal: B
content: $$4.14$$ 
aoVal: C
content: $$4.35$$ 
aoVal: D
content: $$4.7$$ 
aoVal: E
content: $$4.72$$ 
Answer: C
Analysis: $$4\frac{7}{20}=4\frac{35}{100}=4.35$$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Fractions->Operations of Fractions
Problem: When three positive integers $a$, $b$, and $c$ are multiplied together, their product is 100. Suppose $a \textless{} b \textless{} c$. In how many ways can the numbers be chosen? 
Options:
aoVal: A
content: $$0$$ 
aoVal: B
content: $$1$$ 
aoVal: C
content: $$2$$ 
aoVal: D
content: $$3$$ 
aoVal: E
content: $$4$$ 
Answer: E
Analysis: NA 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: If $*abcd*=a\times d+b\times c$ ,then $*2543*=$ ． 
Options:
aoVal: A
content: $$14$$ 
aoVal: B
content: $$22$$ 
aoVal: C
content: $$26$$ 
aoVal: D
content: $$120$$ 
aoVal: E
content: $$23$$ 
Answer: C
Analysis: $*2543*=2\times3+5\times4=6+20=26$.  So the answer is $\rm C$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Operations with New Definition->Operating Directly
Problem: \textbf{According to a report for veterinarians in the United States, 36.5 percent of households in the United States own dogs and 30.4 percent of households in the United States own cats. If one household in the United States is selected at random, what is the probability that the selected household will own a dog or a cat?} 
Options:
aoVal: A
content: $$0.111$$ 
aoVal: B
content: $$0.331$$ 
aoVal: C
content: $$0.558$$ 
aoVal: D
content: $$0.669$$ 
aoVal: E
content: \textbf{Not enough information is given to determine the probability} 
Answer: E
Analysis: \textbf{The sample in the report is about veterinarians. The population cannot be a household in the United States.} 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: An amusement park has a collection of scale models, with ratio $1: 35$, of buildings and other sights from around the country. If the height of a signal tower is $2250$ feet. What is the height in feet of its replica to the nearest whole number? (adapted from 2018 AMC 8, Question 1) 
Options:
aoVal: A
content: $$62$$ 
aoVal: B
content: $$63$$ 
aoVal: C
content: $$64$$ 
aoVal: D
content: $$65$$ 
aoVal: E
content: $$66$$ 
Answer: C
Analysis: You can see that since the ratio of real building\textquotesingle s heights to the model building\textquotesingle s height is $1: 35$. If the height of the tower is $2250$ feet, to find the height of the model, we divide by $35$ . That gives us $64.28$ which rounds to $64$ . Therefore, to the nearest whole number, the duplicate is (C) $64$ feet. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Ratios and Proportions
Source: 其它
Problem: The total value of $$25$$ dimes is $$125$$ times to total value of． 
Options:
aoVal: A
content: $$1$$ penny 
aoVal: B
content: $$1$$ nickel 
aoVal: C
content: $$2$$ pennies 
aoVal: D
content: $$2$$ nickels 
Answer: C
Analysis: $$25$$ dimes $$=250$$￠$$=125\times 2$$￠$$=125\times 2$$ pennies. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Ratios and Proportions
Problem: If the 9-digit number $$2017122\square2$$ can be divisible by $$4$$, then the number in $$\square $$ can be ． 
Options:
aoVal: A
content: $$0$$ 
aoVal: B
content: $$2$$ 
aoVal: C
content: $$4$$ 
aoVal: D
content: $$7$$ 
aoVal: E
content: $$8$$ 
Answer: D
Analysis: We check if it is divisible by $$4$$ by looking at the last two digits.  $$72$$ is divisible by $$4$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Fractions
Problem: The last four digits of Linda\textquotesingle s phone number are $2022$. If the numbers $2,0,2,2$ are summed, what is the final result?~(adapted from $$2011$$ Math kangaroo Problems, Level $$1-2$$, Question \#$$7$$) 
Options:
aoVal: A
content: $$2022$$ 
aoVal: B
content: $$202$$ 
aoVal: C
content: $$22$$ 
aoVal: D
content: $$6$$ 
aoVal: E
content: $2026$ 
Answer: D
Analysis: $2+0+2+2=6$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Addition and Subtraction ->Addition of Whole Numbers->Addition in Horizontal Form
Problem: $$\dfrac{8}{9}\div \dfrac{4}{7}\div \dfrac{1}{3}=$$~\uline{~~~~~~~~~~}~，$$\dfrac{7}{9}\div \dfrac{1}{3}\div 1\dfrac{5}{9}=$$~\uline{~~~~~~~~~~}~． 
Options:
aoVal: A
content: $$\dfrac{13}{3}$$，$$2$$． 
aoVal: B
content: $$\dfrac{14}{3}$$，$$2$$． 
aoVal: C
content: $$\dfrac{14}{3}$$，$$\dfrac{3}{2}$$． 
aoVal: D
content: $$\dfrac{13}{3}$$，$$\dfrac{3}{2}$$． 
Answer: C
Analysis: $$\dfrac{8}{9}\div \dfrac{4}{7}\div \dfrac{1}{3}=\dfrac{4\times 2}{3\times 3}\times \dfrac{7}{4}\times \dfrac{3}{1}=\dfrac{2\times 7}{3}=\dfrac{14}{3}$$．  $$\dfrac{7}{9}\div \dfrac{1}{3}\div 1\dfrac{5}{9}=\dfrac{7}{9}\times \dfrac{3}{1}\times \dfrac{9}{14}==\dfrac{7}{9}\times \dfrac{3}{1}\times \dfrac{9}{2\times 7}=\dfrac{3}{2}$$． 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Fractions
Source: 其它
Problem: Amelia cuts a $75$ cm by $30$ cm rectangle into identical squares without any leftovers. If the length of the squares is a whole number of cm, what is the least number of squares can she get? 
Options:
aoVal: A
content: $$2$$ 
aoVal: B
content: $$5$$ 
aoVal: C
content: $$10$$ 
aoVal: D
content: $$90$$ 
aoVal: E
content: None of the above 
Answer: C
Analysis: 75 is divisible by 1, 3, 5,15, 25 and 75.  30 is divisible by 1, 2, 3, 5\textsubscript{Z} 6,10,15 and 30.  15 is the highest common divisor for both 75 and 30.  $\textasciitilde$  To get the least number of squares, the length of the squares must be the largest possible. Hence the length must be 15 and there must be 5 x 2 = \textbf{10} squares. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Extracting Common Factors from Whole Numbers
Source: 其它
Problem: Seven numbers add up to $2022$ and one of the numbers is $522$. Replace $522$ with $321$. What is the new sum of the seven numbers? 
Options:
aoVal: A
content: $$2223$$ 
aoVal: B
content: $$2022$$ 
aoVal: C
content: $$1921$$ 
aoVal: D
content: $$1821$$ 
aoVal: E
content: $$1793$$ 
Answer: D
Analysis: $522-321=201$, so the sum will decrease by $201$. $2022-201=1821$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Addition and Subtraction 
Source: 其它
Problem: There is a rule that the symbol "\&" represents an operation of producing the larger one of the two numbers (for example, $7$ \&~$15 = 15$). Calculate: ($13$ \&~$22$ ) $\times$ ($3$ \&~$6$)=~\uline{~~~~~~~~~~}~. 
Options:
aoVal: A
content: $$39$$ 
aoVal: B
content: $$66$$ 
aoVal: C
content: $$78$$ 
aoVal: D
content: $$132$$ 
Answer: D
Analysis: ($13$ \&~$22$) $\times$ ($3$ \&~$6$) = $22\times6=132$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Operations with New Definition->Operating Directly
Problem: If $3^{}p+3^{4}=90$, and $2^{}r+44=76$, what is the product of $p$ and $r$? (Adapted from 2013 AMC 8 Problem, Question \#15) 
Options:
aoVal: A
content: $$8$$ 
aoVal: B
content: $$10$$ 
aoVal: C
content: $$12$$ 
aoVal: D
content: $$15$$ 
aoVal: E
content: $$18$$ 
Answer: B
Analysis: First, solve for $p$. Start with $3^{}p+3^{4}=90$. Then, change $3^{4}$ to $81$. Subtract $81$ from both sides to get $3^{}p=9$ and see that $p$ is $2$. Now, solve for $r$. Since $2^{}r+44=76$, $2^{}r$ must equal $32$, $r=5$. $pr$ equals $2\times5$ which equals $10$. So, the answer is $10$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Power->Computing Powers->Exponentiation of Powers
Source: 其它
Problem: What is the biggest digit in ones place? 
Options:
aoVal: A
content: $$6$$ 
aoVal: B
content: $$7$$ 
aoVal: C
content: $$8$$ 
aoVal: D
content: $$9$$ 
Answer: C
Analysis: NA 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: What is the least possible value of $$ (x+1)(x+2)(x+3)(x+4)+2019 $$ where $x$ is a real number? 
Options:
aoVal: A
content: $$2017$$ 
aoVal: B
content: $$2018$$ 
aoVal: C
content: $$2019$$ 
aoVal: D
content: $$2020$$ 
aoVal: E
content: $$2021$$ 
Answer: B
Analysis: Grouping the first and last terms and two middle terms gives $\left(x^{2}+5 x+4\right)\left(x^{2}+5 x+6\right)+2019$, which can be simplified to $\left(x^{2}+5 x+5\right)^{2}-1+2019$. Noting that squares are nonnegative, and verifying that $x^{2}+5 x+5=0$ for some real $x$, the answer is 2018. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation
Source: 其它
Problem: Avril\textquotesingle s father bought 14 books. Avril takes away 6 books. How many books does Avril\textquotesingle s father have left? 
Options:
aoVal: A
content: $$9$$ 
aoVal: B
content: $$8$$ 
aoVal: C
content: $$7$$ 
Answer: B
Analysis: $$14-6=8$$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: Before I began snacking, there were $$60\div 4+1\times 3$$ gumballs here. If I ate all of them, how many gumballs did I eat? 
Options:
aoVal: A
content: $$4$$ 
aoVal: B
content: $$18$$ 
aoVal: C
content: $$36$$ 
aoVal: D
content: $$48$$ 
Answer: B
Analysis: To evaluate $$60\div 4+1\times 3$$, we first do the $$\times $$ and $$\div $$ in the order in which they appear. Do the addition last. We get $$15+3=18$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Mixed Operations
Problem: $$99\times 99=$$$$-99$$． 
Options:
aoVal: A
content: $$199\times 99$$ 
aoVal: B
content: $$198\times 99$$ 
aoVal: C
content: $$100\times 100$$ 
aoVal: D
content: $$100\times 99$$ 
Answer: D
Analysis: $$(100-1)\times 99=(100\times 99)-(1\times 99)$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Distributive Law of Whole Numbers
Problem: Calculate: $$1+2+3+\ldots\ldots+37+38+39+40+39+38+37+\ldots\ldots+13+12+11$$=~\uline{~~~~~~~~~~}~ 
Options:
aoVal: A
content: $$4000$$ 
aoVal: B
content: $$1000$$ 
aoVal: C
content: $$1545$$ 
aoVal: D
content: $$4545$$ 
Answer: C
Analysis: $$1+2+3+\ldots\ldots+37+38+39+40+39+38+37+\ldots\ldots+13+12+11$$  $$=（1+2+3+\ldots\ldots+37+38+39+40+39+38+37+\ldots\ldots+3+2+1）-（1+2+3+\ldots\ldots+8+9+10）$$  $$=40\times40-55$$  $$=1545.$$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Addition and Subtraction 
Source: 其它
Problem: Gary sold $5$ times as many tickets as Louis.They sold a total of $54$ tickets. How many tickets did Louis sell? 
Options:
aoVal: A
content: $$9$$ 
aoVal: B
content: $$18$$ 
aoVal: C
content: $$27$$ 
aoVal: D
content: $$36$$ 
aoVal: E
content: $$45$$ 
Answer: A
Analysis: $54 \div (5 + 1) = 9$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: What is the value of $\dfrac{21w}{10}-w+1$ when $w=2$? 
Options:
aoVal: A
content: $$1.2$$ 
aoVal: B
content: $$3.1$$ 
aoVal: C
content: $$3.2$$ 
aoVal: D
content: $$5.2$$ 
Answer: C
Analysis: When $w= 2$,  $$\frac{21w}{10}-w+1= \frac{21 \times 2}{10}-2+1$$  $$= \frac{42}{10}-2+1$$  $=4.2-2+1$  $= 2.2 + 1$  $= 3.2$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Equivalent Substitution->Algebraic Expressions
Problem: What is the correct ordering of the three numbers, $$10^{8}$$, $$5^{12}$$, and $$2^{24}$$? ($$2010$$ AMC $$8$$ Problem, Question \#$$24$$) 
Options:
aoVal: A
content: $$2^{24}$$\textless{} $$10^{8}$$\textless$$5^{12}$$ 
aoVal: B
content: $$2^{24}$$\textless$$5^{12}$$\textless$$10^{8}$$ 
aoVal: C
content: $$5^{12}$$\textless$$2^{24}$$\textless$$10^{8}$$ 
aoVal: D
content: $$10^{8}$$\textless$$5^{12}$$\textless$$2^{24}$$ 
aoVal: E
content: $$10^{8}$$\textless$$2^{24}$$\textless$$5^{12}$$ 
Answer: A
Analysis: $$\rm Method$$ $$1$$:  Since all of the exponents are multiples of $4$, we can simplify the problem by taking the fourth root of each number. Evaluating we get $$10^{2}=100$$, $$5^{3}=125$$, and $$2^{6}=64$$. $$64\textless100\textless125$$. So, $$2^{24} \textless$$ $$10^{8}$$ $$\textless{} 5^{12}$$.  $$\rm Method$$ $$2$$:  First, let us make all exponents equal to $$8$$. Then, it will be easy to order the  numbers without doing any computations. $$10^{8}$$ is fine as it is. We can rewrite $$2^{24}$$ as $$(2^{3})^{8}=8^{8}$$. We can rewrite $$5^{12}$$ as $$\left( 5^{\frac{3}{2}}\right)^{8}=\left( \sqrt{125}\right)^{8}$$. We take the eighth root of all of these to get $$10$$, $$8$$,~$\sqrt{125}$. Obviously, $$8\textless10\textless\sqrt{ 125}$$, so $$2^{24}\textless10^{8}\textless5^{12}$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Comparing, Ordering and Estimating->Comparing and Ordering
Problem: $3+\dfrac{1}{1+\dfrac{1}{5+\dfrac{1}{16}}}=\left( \textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \right).$ 
Options:
aoVal: A
content: $\dfrac{307}{97}$ 
aoVal: B
content: $\dfrac{16}{97}$ 
aoVal: C
content: $\dfrac{145}{16}$ 
aoVal: D
content: $\dfrac{372}{97}$ 
Answer: D
Analysis: $=3+\dfrac{1}{1+\dfrac{1}{\dfrac{81}{16}}}$  $=3+\dfrac{1}{1+\dfrac{16}{81}}$  $=3+\dfrac{1}{\dfrac{97}{81}}$  $=3+\dfrac{81}{97}$  $=\dfrac{372}{97}$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Fractions->Complex Fractions
Source: 其它
Problem: Which of the following is an algebraic equation with variable(s)? ． 
Options:
aoVal: A
content: $xy+91$ 
aoVal: B
content: $5=3$ 
aoVal: C
content: $x+1=y+4$ 
Answer: C
Analysis: both sides are algebraic expressions. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation
Source: 其它
Problem: Calculate:．  $$\frac{3}{4}+ \frac{1}{4}\times \frac{2}{3}- \frac{1}{3}=$$~\uline{~~~~~~~~~~}~;  $$\frac{2}{47}\times 15 \times \frac{47}{36}\times \frac{2}{15}\div \frac{1}{18}=$$~\uline{~~~~~~~~~~}~. 
Options:
aoVal: A
content: $$\frac{7}{12};2$$ 
aoVal: B
content: $$\frac{2}{3}; \frac{1}{2}$$ 
aoVal: C
content: $$\frac{1}{12};2$$ 
aoVal: D
content: $$\frac{2}{3}; \frac{1}{9}$$ 
aoVal: E
content: $$\frac{1}{12}; \frac{1}{9}$$ 
Answer: A
Analysis: $$\frac{3}{4}+ \frac{1}{4}\times \frac{2}{3}- \frac{1}{3}= \frac{3}{4}+\frac{1}{6}- \frac{1}{3}=\frac{9}{12}+\frac{2}{12}-\frac{4}{12}=\frac{7}{12}$$;  $$\frac{2}{47}\times 15 \times \frac{47}{36}\times \frac{2}{15}\div \frac{1}{18}= \frac{4}{36}\times 18=2$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Fractions->Operations of Fractions
Problem: If $n$ is an even positive integer, the double factorial notation $n!!$ represents the product of all the even integers from $2$ to $n$. For example, $8!! = 2·4·6·8$. What is the units digit of the following sum?  $2!! + 4!! + 6!! + · · · + 2018!! + 2020!! + 2022!!$ 
Options:
aoVal: A
content: $$0$$ 
aoVal: B
content: $$2$$ 
aoVal: C
content: $$4$$ 
aoVal: D
content: $$6$$ 
aoVal: E
content: $$8$$ 
Answer: B
Analysis: NA 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: Happy Hotel is offering $$40 \textbackslash\%$$ off discount for any bookings. David booked a room, the new price is $$80$$ dollars cheaper than the original price, the original price is~\uline{~~~~~~~~~~}~. 
Options:
aoVal: A
content: $$180$$ 
aoVal: B
content: $$200$$ 
aoVal: C
content: $$300$$ 
aoVal: D
content: $$480$$ 
Answer: B
Analysis: $$80\div40\textbackslash\%=200$$．  so choose $$\text{B}$$． 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: Among all the whole numbers from $$1$$ to $$30$$, how many numbers are multiples of $$3$$? 
Options:
aoVal: A
content: $$5$$ 
aoVal: B
content: $$10$$ 
aoVal: C
content: $$15$$ 
aoVal: D
content: $$20$$ 
Answer: B
Analysis: $30\div3=10$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: $$1423$$, $$1723$$, $$2123$$,~\uline{~~~~~~~~~~}~, $$3223$$  Which one of the following is the missing number? 
Options:
aoVal: A
content: $$2523$$ 
aoVal: B
content: $$2623$$ 
aoVal: C
content: $$2723$$ 
aoVal: D
content: $$2823$$ 
Answer: B
Analysis: The sequence increases in this order: $$300$$, $$400$$, $$500$$, $$600$$, $$\dots$$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Sequences and Number Tables->Patterns in Number Sequences
Problem: Starting with some gold coins and some empty treasure chests, I tried to put $15$ gold coins in each treasure chest, but that left $$1$$ treasure chests empty. So instead I put $12$ gold coins in each treasure chest, but then I had $6$ gold coins left over. How many gold coins did I have? ( adapted from 2017 AMC8, Questions \#17) 
Options:
aoVal: A
content: $$36$$ 
aoVal: B
content: $$49$$ 
aoVal: C
content: $$75$$ 
aoVal: D
content: $$84$$ 
aoVal: E
content: $$90$$ 
Answer: E
Analysis: We can represent the amount of gold with $g$ and the amount of chests with $c$. We can use the problem to make the following equations: $$ \begin{gathered} 15 c-15=g \textbackslash\textbackslash{} 12 c+6=g \end{gathered} $$  Therefore, $15 c-15=12 c+6$. This implies that $c=7$. We therefore have $g=90$. So, our answer is (E) $90$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Linear Equations with Multiple Variables
Source: 其它
Problem: $$6\times \dfrac{11}{12}=$$~\uline{~~~~~~~~~~}~，$$\dfrac{11}{24}\times 16=$$~\uline{~~~~~~~~~~}~． 
Options:
aoVal: A
content: $$5\dfrac{1}{2}$$，$$7\dfrac{1}{2}$$． 
aoVal: B
content: $$5\dfrac{1}{3}$$，$$7\dfrac{1}{3}$$． 
aoVal: C
content: $$5\dfrac{1}{3}$$，$$7\dfrac{1}{2}$$． 
aoVal: D
content: $$5\dfrac{1}{2}$$，$$7\dfrac{1}{3}$$． 
Answer: D
Analysis: $$6\times \dfrac{11}{12}=\dfrac{11}{2}=5\dfrac{1}{2}$$．  $$\dfrac{11}{24}\times 16=\dfrac{22}{3}=7\dfrac{1}{3}$$． 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Fractions
Source: 其它
Problem: From a regular octagon, a triangle is formed by connecting three randomly chosen vertices of the octagon. What is the probability that at least one of the sides of the triangle is also a side of the octagon? (2018 AMC 8 Problems, Question \#23) 
Options:
aoVal: A
content: $\frac{2}{7}$ 
aoVal: B
content: $\frac{5}{42}$ 
aoVal: C
content: $\frac{11}{14}$ 
aoVal: D
content: $\frac{5}{7}$ 
aoVal: E
content: $\frac{6}{7}$ 
Answer: D
Analysis: We will use constructive counting to solve this. There are $2$ cases: Either all $3$ points are adjacent, or exactly $2$ points are adjacent. If all $3$ points are adjacent, then we have $8$ choices. If we have exactly $2$ adjacent points, then we will have $8$ places to put the adjacent points and $4$ places to put the remaining point, so we have $8 \cdot 4$ choices. The total amount of choices is $\left(\begin{array}{l}8 \textbackslash\textbackslash{} 3\end{array}\right)=8 \cdot 7$. Thus, our answer is $\frac{8+8 \cdot 4}{8 \cdot 7}=\frac{1+4}{7}=$ (D) $\frac{5}{7}$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: Find the result of $$1\times12\times23\times34\times45\times \cdots \times78\times89$$. What is the sum of its last $2$ digits? 
Options:
aoVal: A
content: $$9$$ 
aoVal: B
content: $$8$$ 
aoVal: C
content: $$4$$ 
aoVal: D
content: $$2$$ 
aoVal: E
content: $$7$$ 
Answer: D
Analysis: $$\textasciitilde\textasciitilde\textasciitilde\textasciitilde1\times12\times23\times34\times45\times \cdots \times78\times89$$  $$=1\times6\times23\times34\times9\times \cdots \times78\times89\times2\times5$$  ∴$$1\times6\times3\times4\times9\times6\times7\times8\times9$$ has the last digit of $2$  ∴ the last two digits are $20.$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Calculation of Multi-digit Numbers
Problem: Calculate: $$3.75\times 20\textbackslash\%\times \frac{3}{23}-4\times \frac{7}{23}+\frac{9}{23}\times 3\frac{1}{4}=$$． 
Options:
aoVal: A
content: $$\frac{9}{23}$$ 
aoVal: B
content: $$\frac{7}{23}$$ 
aoVal: C
content: $$\frac{7}{46}$$ 
aoVal: D
content: $$4$$ 
Answer: C
Analysis: $$3.75\times 20\textbackslash\%\times \frac{3}{23}-4\times \frac{7}{23}+\frac{9}{23}\times 3\frac{1}{4}$$  $$=\frac{15}{4}\times \frac{1}{5}\times \frac{3}{23}-\frac{4\times 7}{23}+\frac{9}{23}\times \frac{13}{4}$$  $$=\frac{3}{4}\times \frac{3}{23}-\frac{28}{23}+\frac{9\times 13}{4\times 23}$$  $$=\frac{9}{92}-\frac{28\times 4}{23\times 4}+\frac{117}{92}$$  $$=\frac{9-112+117}{92}$$  $$=\frac{14}{92}$$  $$=\frac{7}{46}$$． 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Mixed Operations
Problem: $$5+10+15+20+25=\left(1+2+3+4+5\right)\times$$． 
Options:
aoVal: A
content: $$2$$ 
aoVal: B
content: $$3$$ 
aoVal: C
content: $$4$$ 
aoVal: D
content: $$5$$ 
Answer: D
Analysis: $$5+10+15+20+25=1\times 5+2\times 5+\cdots+5\times 5$$  $$=\left(1+2+3+4+5\right)\times 5$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Extracting Common Factors from Whole Numbers
Problem: Greta Grasshopper sits on a long line of lily pads in a pond. From any lily pad, Greta can jump $5$ pads to the right or $3$ pads to the left. What is the fewest number of jumps Greta must make to reach the lily pad located $2023$ pads to the right of her starting position? 
Options:
aoVal: A
content: $$405$$ 
aoVal: B
content: $$407$$ 
aoVal: C
content: $$409$$ 
aoVal: D
content: $$411$$ 
aoVal: E
content: $$413$$ 
Answer: D
Analysis: D 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: \textbf{If P(A) = 0.34 and P(A or B) = 0.71, which of the following is false?} 
Options:
aoVal: A
content: \textbf{P(B) = 0.37 , if A and B are mutually exclusive.} 
aoVal: B
content: \textbf{P(B) = 0.561, if A and B are independent.} 
aoVal: C
content: \textbf{P(B) cannot be determined if A and B are neither mutually exclusive nor independent.} 
aoVal: D
content: \textbf{P(A and B) = 0.191 , if A and B are independent.} 
aoVal: E
content: \textbf{P(A\textbar B) = 0.34 , if A and B are mutually exclusive.} 
Answer: E
Analysis: \textbf{P(A) = 0.34, P(A ∪ B) = 0.71 =~ P(A) + P(B) -- P(A ∩ B)}  \textbf{(A) If A and B are mutually exclusive, P(A ∩ B) = 0. Then, P(A ∪ B) = P(A) + P(B) which is 0.71= 0.34+P(B). So P(B) = 0.37.}  \textbf{(B) If A and B is independent, P(A ∩ B) = P(A)*P(B). Then, P(A ∪ B) =~ P(A) + P(B) --~ P(A)*P(B) which 0.71 = 0.34 + P(B) - 0.34*P(B). So P(B) = 0.561}  \textbf{(C) P(A ∪ B) = P(A) + P(B) -- P(A ∩ B). We have to know it is mutually exclusive or independent in order to know P(A ∩ B).}  \textbf{(D) If A and B is independent, P(A ∩ B) = P(A)*P(B) =0.34*0.561=0.191}  \textbf{(E) P(A \textbar{} B)=P(A∩B)P(B) = 0} 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: $$a:b=5:6$$, $$b:c=8:3$$, $$a:c=$$~\uline{~~~~~~~~~~}~. 
Options:
aoVal: A
content: $$20:9$$ 
aoVal: B
content: $$5:3$$ 
aoVal: C
content: $$40:21$$ 
aoVal: D
content: $$5:8$$ 
Answer: A
Analysis: $$a:b=20:24$$, $$b:c=24:9$$, $$a:c=20:9$$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Ratios and Proportions