math-eval/TAL-SCQ5K huggingface dataset:
Problem: Compare these fractions.  $$\frac{8}{31}$$~\uline{~~~~~~~~~~}~$$\frac{4}{15}$$, ~$$\frac{9}{61}$$~\uline{~~~~~~~~~~}~$$\frac{3}{22}$$ 
Options:
aoVal: A
content: $$\textgreater$$, $$\textgreater$$ 
aoVal: B
content: $$\textgreater$$, $$\textless$$ 
aoVal: C
content: $$\textless$$, $$\textgreater$$ 
aoVal: D
content: $$\textless$$, $$\textless$$ 
Answer: C
Analysis: $$\frac{8}{31}\textless\frac{8}{30}$$;~$$\frac{9}{61}\textgreater\frac{9}{66}$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Comparing, Ordering and Estimating->Comparing and Ordering
Problem: Compare these fractions.  $$\frac{3}{7}$$~\uline{~~~~~~~~~~}~$$\frac{5}{9}$$, $$\frac{5}{8}$$~\uline{~~~~~~~~~~}~$$\frac{7}{11}$$ 
Options:
aoVal: A
content: $$\textgreater$$, $$\textgreater$$ 
aoVal: B
content: $$\textgreater$$, $$\textless$$ 
aoVal: C
content: $$\textless$$, $$\textgreater$$ 
aoVal: D
content: $$\textless$$, $$\textless$$ 
Answer: D
Analysis: $$\frac{27}{63}\textless\frac{35}{63}$$;~$$\frac{55}{88}\textless\frac{56}{88}$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Comparing, Ordering and Estimating->Comparing and Ordering
Problem: If $ a◆b$ means $(a\times b)+b$ , then $（2◆3）◆4$ has the value． 
Options:
aoVal: A
content: $$45$$ 
aoVal: B
content: $$40$$ 
aoVal: C
content: $$16$$ 
aoVal: D
content: $$9$$ 
aoVal: E
content: $$8$$ 
Answer: B
Analysis: If $a◆b$~ represents $(a\times b)+b$ , $2◆3=(2\times3)+3=9,9◆4=(9\times4)+4=40$ . 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Operations with New Definition->Operating Directly
Problem: (2) Eddie had 120 dollars as his pocket money, and spent $$ \frac{3}{4} $$of it. How much money is left? 
Options:
aoVal: A
content: $$30$$ 
aoVal: B
content: $$60$$ 
aoVal: C
content: $$90$$ 
aoVal: D
content: $$120$$ 
Answer: A
Analysis: left！！！not spent! So you need to calculate the fraction of left money. Then use T*F=C 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Fractions
Source: 其它
Problem: Divide $$2$$、$$3$$、$$24$$、$$33$$、$$55$$ and $$60$$ into two groups with 3 numbers in each group to make the product of numbers in each group the same, so the product is． 
Options:
aoVal: A
content: $$3630$$ 
aoVal: B
content: $$1584$$ 
aoVal: C
content: $$3960$$ 
aoVal: D
content: $$2880$$ 
Answer: C
Analysis: $$2=2$$，  $$3=3$$，  $$24=2\times 2\times 2\times 3$$，  $$33=11\times 3$$，  $$55=11\times 5$$，  $$60=2\times 2\times 3\times 5$$，  $$2\times 60\times 33=24\times 55\times 3$$，  $$\textasciitilde\textasciitilde\textasciitilde\textasciitilde2\times 60\times 33$$  $$=(2\times 33)\times 60$$  $$=66\times 60$$  $$=3960$$．  $$\text{C}$$． 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Power->Computing Powers->Power of Products
Problem: Which of the following integers cannot be written as the sum of four consecutive odd integers? (2015 AMC 8 Problems, Question \#14) 
Options:
aoVal: A
content: $$16$$ 
aoVal: B
content: $$40$$ 
aoVal: C
content: $$72$$ 
aoVal: D
content: $$100$$ 
aoVal: E
content: $$200$$ 
Answer: D
Analysis: Let our $4$ numbers be $n, n+2, n+4, n+6$, where $n$ is odd. Then our sum is $4 n+12$. The only answer choice that cannot be written as $4 n+12$, where $n$ is odd, is (D) 100 . 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Sequences and Number Tables
Source: 其它
Problem: $$100-99+98-97+96-95+\cdots +4-3+2-1$$. 
Options:
aoVal: A
content: $$20$$ 
aoVal: B
content: $$40$$ 
aoVal: C
content: $$50$$ 
aoVal: D
content: $$80$$ 
aoVal: E
content: $$100$$ 
Answer: C
Analysis: $$\begin{eqnarray}\&\&\left( 100-99 \right)+\left( 98-97 \right)\cdots +\left( 4-3 \right)+\left( 2-1 \right)\textbackslash\textbackslash{} \&=\&50\times 1\textbackslash\textbackslash{} \&=\&50．\end{eqnarray}$$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Addition and Subtraction
Problem: Of the following, which has a value different from the others? 
Options:
aoVal: A
content: $$40\times 50$$ 
aoVal: B
content: $$4\times 5000$$ 
aoVal: C
content: $$50\times 400$$ 
aoVal: D
content: $$40\times 500$$ 
Answer: A
Analysis: To see which value is different, count the total number of $$0$$\textquotesingle s in each product. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Multiplication and Division
Problem: $$12 \times \left( \frac{1}{2} \times \frac{1}{3} \times \frac{1}{4}\right)=$$． 
Options:
aoVal: A
content: $$72$$ 
aoVal: B
content: $$13$$ 
aoVal: C
content: $$\frac{4}{3}$$ 
aoVal: D
content: $$\frac{1}{2}$$ 
Answer: D
Analysis: $$12 \times \left( \frac{1}{2} \times \frac{1}{3} \times \frac{1}{4}\right)$$  $$=\left(12 \times \frac{1}{2}\right) \times \frac{1}{3} \times \frac{1}{4}$$  $$=6 \times \frac{1}{3} \times \frac{1}{4}$$  $$=2 \times \frac{1}{4}$$  $$= \frac{2}{4}$$  $$=\frac{1}{2}$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Fractions->Operations of Fractions
Problem: For $\triangle ABC$, all its side lengths are integers. The primeter of $\triangle ABC$ with a side of length $3$ and a side length of $4$ is at least ． 
Options:
aoVal: A
content: $$7$$ 
aoVal: B
content: $$8$$ 
aoVal: C
content: $$9$$ 
aoVal: D
content: $$12$$ 
Answer: C
Analysis: We know from the triangle inequality that the last side, $s$, fulfills $s+3\textgreater4$. Therefore, $P\textgreater4+4$. The least integer value of $P$ is $9$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Inequalities
Source: 其它
Problem: $$(60 \div 5)\times4 =$$.　 
Options:
aoVal: A
content: $$3$$ 
aoVal: B
content: $$16$$ 
aoVal: C
content: $$48$$ 
aoVal: D
content: $$96$$ 
Answer: C
Analysis: $$(60\div5)\times4 = 12 \times4 =48$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Multiplication and Division
Problem: The fraction $$\dfrac{214}{263}$$ keeps the same value when both its numerator and denominator are~\uline{~~~~~~~~~~}~. 
Options:
aoVal: A
content: multiplied by $$2~ $$ 
aoVal: B
content: increased by $$2$$ 
aoVal: C
content: decreased by $$2$$ 
aoVal: D
content: $$ $$squared$$ $$ 
Answer: A
Analysis: The problem is about the basic property of fractions, namely, multiplying or dividing the numerator and denominator by the same number (except $$0$$), the value of the fraction is unchanged. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Fractions->Basic Understanding of Fractions
Problem: For $\triangle ABC$, all its side lengths are integer. The primeter of $\triangle ABC$ with a side of length $12$ and a side length of $7$ is at least ． 
Options:
aoVal: A
content: $$24$$ 
aoVal: B
content: $$25$$ 
aoVal: C
content: $$26$$ 
aoVal: D
content: $$27$$ 
aoVal: E
content: $$28$$ 
Answer: B
Analysis: We know from the triangle inequality that the last side, $s$, fulfills $s+7\textgreater12$. $P=s+7+12\textgreater12+12$. Therefore, $P\textgreater24+1=25$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Inequalities
Source: 其它
Problem: $$10\times20\times30\times40=24\times$$． 
Options:
aoVal: A
content: $$10^{3}$$ 
aoVal: B
content: $$10^{4}$$ 
aoVal: C
content: $$10^{5}$$ 
aoVal: D
content: $$10^{6}$$ 
Answer: B
Analysis: $$10\times20\times30\times40=(1\times2\times3\times4)\times10^{4}=24\times10^{4}$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Multiplication and Division
Problem: What is the correct ordering of the three answers from $\frac5{19}\div \frac{25}{38}$, $1\frac12 \div \frac{15}8$, and $\frac74 \div \frac{35}{12}$, in increasing order? (adapted from 2012 AMC 8 Problems, Question \#20) 
Options:
aoVal: A
content: $\frac5{19}\div \frac{25}{38}$ \textless~$1\frac12 \div \frac{15}8$ \textless{} $\frac74 \div \frac{35}{12}$ 
aoVal: B
content: $\frac74 \div \frac{35}{12}$ \textless~$1\frac12 \div \frac{15}8$ \textless~$\frac5{19}\div \frac{25}{38}$ 
aoVal: C
content: $1\frac12 \div \frac{15}8$ \textless{} $\frac74 \div \frac{35}{12}$ \textless~$\frac5{19}\div \frac{25}{38}$ 
aoVal: D
content: $\frac5{19}\div \frac{25}{38}$ \textless{} $\frac74 \div \frac{35}{12}$ \textless~$1\frac12 \div \frac{15}8$ 
aoVal: E
content: $\frac74 \div \frac{35}{12}$ \textless~$\frac5{19}\div \frac{25}{38}$ \textless~$1\frac12 \div \frac{15}8$ 
Answer: D
Analysis: $\frac5{19}\div \frac{25}{38}=\frac25$ ~  $\frac74 \div \frac{35}{12}=\frac35$  $1\frac12 \div \frac{15}8=\frac45$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Fractions->Operations of Fractions->Division of Fractions
Source: 其它
Problem: Choose the answer in the simplest form.  $$\frac{9}{4}\times \frac{8}{27}=$$． 
Options:
aoVal: A
content: $$\frac{72}{108}$$ 
aoVal: B
content: $$\frac{18}{27}$$ 
aoVal: C
content: $$\frac{8}{12}$$ 
aoVal: D
content: $$\frac{2}{3}$$ 
Answer: D
Analysis: $$\frac{9}{4}\times \frac{8}{27}=\frac{2}{3}$$． 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Fractions
Problem: In the arithmetic sequence $3$, $7$, $11$, $15$, $\cdots$ , the $26$\textsuperscript{th} number is． 
Options:
aoVal: A
content: $$103$$ 
aoVal: B
content: $$107$$ 
aoVal: C
content: $$111$$ 
aoVal: D
content: $$115$$ 
aoVal: E
content: $$119$$ 
Answer: A
Analysis: $3+4\times(26-1)=103$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Sequences and Number Tables->Arithmetic Sequences
Problem: If a whole number is multiplied by itself, the ones\textquotesingle{} digit of the product \emph{cannot} be． 
Options:
aoVal: A
content: $$1$$ 
aoVal: B
content: $$5$$ 
aoVal: C
content: $$7$$ 
aoVal: D
content: $$9$$ 
Answer: C
Analysis: If a whole number is multiplied by itself, the ones\textquotesingle{} digit of the product could be $$1 (1\times1)$$ or $$5 (5\times5)$$ or $$9 (3\times3)$$, but not $$7$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Power->Computing Powers
Problem: Bob: " Hi, Stanley. What is the coefficient of the variable in this algebraic expression $3x^{2}-4$?"  Stanley:" I can give you a hint. The value of the coefficient is $7$ more than the constant."  Stanley\textquotesingle s hint is~\uline{~~~~~~~~~~}~and the coefficient of the variable is~\uline{~~~~~~~~~~}~. 
Options:
aoVal: A
content: correct, $3$ 
aoVal: B
content: correct, $-4$ 
aoVal: C
content: incorrect, $3$ 
aoVal: D
content: incorrect, $4$ 
Answer: A
Analysis: Constant: $-4$  Coefficient of the variable: $3$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Equivalent Substitution->Algebraic Expressions
Source: 其它
Problem: Gilda has a bag of marbles. She gives $20 \textbackslash\%$ of them to her friend Pedro. Then Gilda gives $25 \textbackslash\%$ of what is left to another friend, Ebony. Finally, Gilda gives $5 \textbackslash\%$ of what is now left in the bag to her brother Jimmy. What percentage of her original bag of marbles does Gilda have left for herself? (adapted 2019 AMC 8, Question \#8) 
Options:
aoVal: A
content: $$95\textbackslash\%$$ 
aoVal: B
content: $33\textbackslash\%$ 
aoVal: C
content: $$45\textbackslash\%$$ 
aoVal: D
content: $$57\textbackslash\%$$ 
aoVal: E
content: $$63\textbackslash\%$$ 
Answer: D
Analysis: After Gilda gives $20 \textbackslash\%$ of the marbles to Pedro, she has $80 \textbackslash\%$ of the marbles left. If she then gives $25 \textbackslash\%$ of what\textquotesingle s left to Ebony, she has $(0.75 * 0.8)=60 \textbackslash\%$ of what she had at the beginning. Finally, she gives $5 \textbackslash\%$ of what\textquotesingle s left to her brother, so she has $(0.6 * 0.95)$ (D) $57\textbackslash\%$~ of what she had in the beginning left. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Percentage Calculation
Source: 其它
Problem: Given that $$x\otimes y=6\times x-5\times y$$, find $$7\otimes 6$$. 
Options:
aoVal: A
content: $$12$$ 
aoVal: B
content: $$13$$ 
aoVal: C
content: $$42$$ 
aoVal: D
content: $$30$$ 
Answer: A
Analysis: $$7\otimes 6=6\times 7-5\times 6=12$$, so $$\text{A}$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Operations with New Definition->Operating Directly
Problem: For $\triangle ABC$, all its side lengths are integer. The primeter of $\triangle ABC$ with a side of length $12$ and a side length of $7$ is at least ． 
Options:
aoVal: A
content: $$24$$ 
aoVal: B
content: $$25$$ 
aoVal: C
content: $$26$$ 
aoVal: D
content: $$27$$ 
Answer: B
Analysis: We know from the triangle inequality that the last side, $s$, fulfills $s+7\textgreater12$. $P=s+7+12\textgreater12+12$. Therefore, $P\textgreater24+1=25$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Inequalities
Source: 其它
Problem: A slug called Glug eats $$2$$ tomatoes for every $$3$$ strawberries. Yesterday it had eaten $$35$$ tomatoes and strawberries altogether. How many tomatoes did it eat? 
Options:
aoVal: A
content: $$5$$ 
aoVal: B
content: $$7$$ 
aoVal: C
content: $$14$$ 
aoVal: D
content: $$15$$ 
aoVal: E
content: $$21$$ 
Answer: C
Analysis: $$35\times \frac {2} {2+3} = 14$$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Ratios and Proportions
Source: 其它
Problem: How many perfect cubes lie between $2^{2}+1$ and $2^{8}+1$, inclusive? (Adapted from 2018 AMC 8 Problem, Question \#25) 
Options:
aoVal: A
content: $$2$$ 
aoVal: B
content: $$3$$ 
aoVal: C
content: $$4$$ 
aoVal: D
content: $$5$$ 
aoVal: E
content: $$6$$ 
Answer: D
Analysis: $2^{2}+1=5$, $2^{8}+1=257$.  $2^{3}=8$, $3^{3}=27$, $4^{3}=64$, $5^{3}=125$, $6^{3}=216$.  Thus, the answer is $5$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Power->Computing Powers->Exponentiation of Powers
Source: 其它
Problem: Steven subtracts the units digit from the tens digit for each two-digit number. He then finds the sum of all his answers.  What is the value of Steven\textquotesingle s sum? 
Options:
aoVal: A
content: $$30$$ 
aoVal: B
content: $$45$$ 
aoVal: C
content: $$55$$ 
aoVal: D
content: $$90$$ 
aoVal: E
content: $$100$$ 
Answer: B
Analysis: The sum of all the tens digits is $$\left(1+2+3+4+5 +6 +7+8+9\right)\times10$$.  The sum of all the units digits is $$\left(0 + 1+2+3 +4+5 +6 +7+8+9\right)\times9$$.  Therefore Steven\textquotesingle s sum is $$\left(1+2+3 +4+5+6+7+8+9\right)\times1=45$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Sequences and Number Tables->Arithmetic Sequences
Problem: The product of two different nonzero integers cannot be． 
Options:
aoVal: A
content: prime　 
aoVal: B
content: zero　 
aoVal: C
content: even　 
aoVal: D
content: composite　 
Answer: B
Analysis: The product of two different nonzero integers can never be $$0$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Multiplication and Division
Problem: Bob bought three kinds of meat: pork, beef and chicken with the total cost of $152. The ratio of the weight of pork, beef and chicken is~$2:4:3$. The ratio of the price per pound of pork, beef and chicken is~$6:5:2$. What is the sum of the last digits of the cost of each kind in dollars? 
Options:
aoVal: A
content: $$8$$ 
aoVal: B
content: $$10$$ 
aoVal: C
content: $$12$$ 
aoVal: D
content: $$15$$ 
aoVal: E
content: $$16$$ 
Answer: C
Analysis: Let~$x,\textbackslash{} y\textbackslash{} and\textbackslash{} z$~be the weight of pork, beef and chicken, respectively.  ~$\textbackslash{} x:y:z=2:4:3$~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ (1)  The ratio of the price will be~$\left( 6x\right):\left( 5y\right):\left( 2z\right).$  The costs of pork, beef and chicken are~$A,B\textbackslash{} and\textbackslash{} C,$~respectively.  ~$A=\dfrac{6x}{6x+5y+2z}\times152$~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~~(2)  From (1), we get:  ~$y=2x,\textbackslash{} and\textbackslash{} z=\dfrac{3}{2}x$~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~(3)  (2) becomes:~$A=\dfrac{6x}{6x+5\left( 2x\right)+2\times\dfrac{3}{2}x}\times152=\dfrac{6x}{19x}\times152=48$  Similarly,~$B=\dfrac{5y}{6x+5y+2z}\times152=\dfrac{5y}{19x}\times152=\dfrac{10x}{19x}\times152=80.$  And~$C=\dfrac{2z}{6x+5y+2z}\times152=\dfrac{2z}{19x}\times152=\dfrac{3x}{19x}\times152=24.$  Pork costs $48 per pound, beef costs $80, and chicken costs $24.  The sum of the last digits of the costs is 8+0+4=12. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Ratios and Proportions
Source: 其它
Problem: $$1000\text{m}$$ per second $$=$$$$\text{km}$$ per hour. 
Options:
aoVal: A
content: $$60$$ 
aoVal: B
content: $$360$$ 
aoVal: C
content: $$3600$$ 
aoVal: D
content: $$6000$$ 
Answer: C
Analysis: $$1000\text{m/s}=1\text{km/s}=3600\text{km/hr}$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Unit Conversion->Converting between Compound Units
Problem: Solve the equation: $$16\times 25-13\left( 3x+2 \right)=179$$． 
Options:
aoVal: A
content: $$4$$ 
aoVal: B
content: $$5$$ 
aoVal: C
content: $$6$$ 
aoVal: D
content: $$7$$ 
Answer: B
Analysis: $$39x=400-26-179$$  $$\textasciitilde\textasciitilde\textasciitilde\textasciitilde x=5$$． 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Equivalent Substitution->Algebraic Expressions
Problem: Find the missing term in the following sequence: $$1, 2, 4, 7,~\uline{~~~~~~~~~~}~, 16$$. 
Options:
aoVal: A
content: $$10$$ 
aoVal: B
content: $$11$$ 
aoVal: C
content: $$12$$ 
aoVal: D
content: $$13$$ 
aoVal: E
content: $$14$$ 
Answer: B
Analysis: The number sequence are in the pattern of $$+1, +2, +3, +4, +5, \cdots $$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Sequences and Number Tables->Patterns in Number Sequences
Source: 其它
Problem: On Kangaroo planet each kangyear has 20 kangmonths and each kangmonth has 6 kangweeks, How many kangweeks are there in one quarter of a kangyear? 
Options:
aoVal: A
content: $$9$$ 
aoVal: B
content: $$30$$ 
aoVal: C
content: $$60$$ 
aoVal: D
content: $$90$$ 
aoVal: E
content: $$120$$ 
Answer: B
Analysis: $\dfrac{1}{4}$kangyear x~$\dfrac{20\textbackslash{} kangmonths}{1\textbackslash{} kangyear}$~x~$\dfrac{6\textbackslash{} kangweeks}{1\textbackslash{} kangmonth}$~= 30 kangweeks. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Fractions
Source: 其它
Problem: A vase was being sold at the price of $250$ dollars. The store decides to sell it with a $40\textbackslash\%$ discount. If you buy the vase now, how much will you save? 
Options:
aoVal: A
content: $$150$$ 
aoVal: B
content: $$100$$ 
aoVal: C
content: $$80$$ 
aoVal: D
content: $$50$$ 
Answer: B
Analysis: $250\times 40\textbackslash\%$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Percentage Calculation
Source: 其它
Problem: Calculate: $$2.7\times0.2\div3=$$． 
Options:
aoVal: A
content: $0.16$ 
aoVal: B
content: $1.6$ 
aoVal: C
content: $0.18$ 
aoVal: D
content: $1.8$ 
aoVal: E
content: $0.14$ 
Answer: C
Analysis: $$2.7\times0.2\div3$$  $$=0.54\div3$$  $$=0.18$$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Decimals->Multiplication and Division of Decimals
Source: 其它
Problem: Mary wanted to buy $12$ lemons. The lemons were sold either for $90$ cents each or at $3$ dollars for a bag of $4$ lemons. How much money would Mary save if she buys three bags of lemons instead of buying $12$ lemons separately? 
Options:
aoVal: A
content: $1.2$ dollars 
aoVal: B
content: $1.5$ dollars 
aoVal: C
content: $1.6$ dollars 
aoVal: D
content: $1.8$ dollars 
aoVal: E
content: $2.1$ dollars 
Answer: D
Analysis: $0.9 \times 12 - 3 \times 3 = 1.8$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: The following are the weights (in pounds) of ten people: $100,115, 135, 140, 180, 197, 203, 230, x, y$ (not necessarily in increasing order). It is also given that the average weight of these ten people is $157$ pounds, and there is a unique mode of $135$. Find the $56$-th percentile. 
Options:
aoVal: A
content: $$100$$ 
aoVal: B
content: $$115$$ 
aoVal: C
content: $$125$$ 
aoVal: D
content: $$135$$ 
Answer: D
Analysis: First of all, we find the values of $x$ and $y$.  Since the average is $157$, we have  $(100+115+135+140+ 180+197+203+230+x+y = 157 \times 10 = 1570$  $1300+x+y =1570$  $x+y = 270$.  There is a unique mode, $135$, then $135$ must appear at least twice. Therefore, one of $x, y$ is $135$. It is easy to deduce that both $x$ and $y$ are $135$.  List the weights in increasing order: $100,115, 135, 135, 135, 140, 180, 197, 203, 230$.  $np=10(0.56)=5.6 \uparrow 6$. The $56$-th percentile is $140$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: Which of the following equations are NOT equivalent to $x+5=13$? 
Options:
aoVal: A
content: $x+5-5=13-5$ 
aoVal: B
content: $2(x+5)=26$ 
aoVal: C
content: $x+5-13=0$ 
aoVal: D
content: $x+5+13=0$ 
Answer: D
Analysis: Equations are equivalent when you can obtain one by subtracting, adding, dividing, or multiplying the same number on the other. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation
Source: 其它
Problem: $$\frac{1}{9}+ \frac{3}{9}+ \frac{5}{9}=$$． 
Options:
aoVal: A
content: $$\frac{1}{3}$$ 
aoVal: B
content: $$1$$ 
aoVal: C
content: $$\frac{10}{9}$$ 
aoVal: D
content: $$\frac{5}{243}$$ 
Answer: B
Analysis: $$\frac{1}{9}+ \frac{3}{9}+ \frac{5}{9}= \frac{9}{9}=1$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Fractions->Operations of Fractions
Problem: What is $$9762 + 7 \times 8 \times 9 \times 4 \times 99 \times 0$$ 
Options:
aoVal: A
content: $$9762$$ 
aoVal: B
content: $$9818$$ 
aoVal: C
content: $$9889$$ 
aoVal: D
content: $$209346$$ 
aoVal: E
content: None of the above 
Answer: A
Analysis: Anything times $$0$$ equal $$0$$.  $$9762 + 0 = 9762$$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: In a triangle with integer side lengths, one side is three times as long as a second side, and the length of the third side is $15$. What is the greatest possible perimeter of the triangle? （2006 AMC10B, Question 10) 
Options:
aoVal: A
content: $$43$$ 
aoVal: B
content: $$44$$ 
aoVal: C
content: $$45$$ 
aoVal: D
content: $$46$$ 
aoVal: E
content: $$47$$ 
Answer: A
Analysis: Let $x$ be the length of the first side. The lengths of the sides are: $x, 3 x$, and 15 . By the Triangle Inequality, $$ \begin{aligned} \&3 x\textless x+15 \textbackslash\textbackslash{} \&2 x\textless15 \textbackslash\textbackslash{} \&x\textless\frac{15}{2} \end{aligned} $$ The greatest integer satisfying this inequality is 7 . So the greatest possible perimeter is $7+3 \cdot 7+15=$ (A) 43 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Inequalities
Source: 其它
Problem: $$1 + 10 + 100 + 1000 =$$． 
Options:
aoVal: A
content: $$4$$ 
aoVal: B
content: $$1111$$ 
aoVal: C
content: $$1234$$ 
aoVal: D
content: $$4000$$ 
Answer: B
Analysis: $$1 + 10 + 100 + 1000 =11+11$$ hundred $$=1111$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Addition and Subtraction
Problem: The $$2022$$th digit to the right of the decimal point in the decimal representation of $$\frac 1{54}$$ is． 
Options:
aoVal: A
content: $$0$$ 
aoVal: B
content: $$1$$ 
aoVal: C
content: $$5$$ 
aoVal: D
content: $$8$$ 
Answer: D
Analysis: The decimal is $$0.0185185\cdots $$. An "$$8$$" appears in the $$3$$rd, $$6$$th, $$9$$th, $$\cdots $$, $$2022$$th decimal place. So a "$$8$$" is in the $$2022$$th place. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Periodic Problems->Periodic Problems of Sequence Operations