math-eval/TAL-SCQ5K huggingface dataset:
Problem: Peter bought three apples of $5$ dollars, and $4$ bananas of $4$ dollars. How much did he spend in total?~(adapted from $$2011$$ Math kangaroo Problems, Level $$1-2$$, Question \#$$7$$) 
Options:
aoVal: A
content: $$6$$ dollars 
aoVal: B
content: $$7$$ dollars 
aoVal: C
content: $$8$$ dollars 
aoVal: D
content: $$9$$ dollars 
aoVal: E
content: $$10$$ dollars 
Answer: E
Analysis: $5+4=9$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Addition and Subtraction ->Addition of Whole Numbers->Addition in Horizontal Form
Problem: A class had more boys than girls. After winter transfer, $3$ boys and $3$ girls joined the class. What is the relationship between boys and girls in the class now? 
Options:
aoVal: A
content: boys $\textgreater$ girls 
aoVal: B
content: boys $=$ girls 
aoVal: C
content: boys $\textless$ girls 
Answer: A
Analysis: When add the same number to both sides of the inequality, the equation is still true 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Inequalities
Source: 其它
Problem: If $$6$$ hoots $$=3$$ hollers, then $$10$$ hollers $$=$$~\uline{~~~~~~~~~~}~hoots. 
Options:
aoVal: A
content: $$5$$ 
aoVal: B
content: $$13$$ 
aoVal: C
content: $$18$$ 
aoVal: D
content: $$20$$ 
Answer: D
Analysis: If $$2$$ hoots $$=1$$ holler, then $$(10\times 1)$$ hollers $$=(10\times 2)$$ hoots. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Equivalent Substitution->Algebraic Expressions
Problem: Which three numbers can be put in the blanks to make the statement correct?  $6 \times$~\uline{~~~~~~~~~~}~$-$~\uline{~~~~~~~~~~}~$=$ $21 +$~\uline{~~~~~~~~~~}~$\times 2$ 
Options:
aoVal: A
content: $5$, $4$ and $2$ 
aoVal: B
content: $6$, $2$ and $7$ 
aoVal: C
content: $7$, $5$ and $8$ 
aoVal: D
content: $8$, $7$ and $11$ 
aoVal: E
content: $9$, $7$ and $12$ 
Answer: C
Analysis: $6 \times 7 - 5 = 21 + 8 \times 2$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: The original price of a product was $$80$$ dollars, and it\textquotesingle s on sale for 30\% off, this product isdollars cheaper than before． 
Options:
aoVal: A
content: $$7$$ 
aoVal: B
content: $$24$$ 
aoVal: C
content: $$30$$ 
aoVal: D
content: $$56$$ 
Answer: B
Analysis: the new price is $$70\textbackslash\%$$ of the original price, so the new price is $$80\times 70\textbackslash\%=56$$；  $$80-56=24$$．  so choose $$\text{B}$$． 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: A recipe that makes~$5$~servings of hot chocolate requires~$2$~squares of chocolate,~$\dfrac{1}{4}$~cup sugar,~$1$~cup water and~$4$~cups milk. Jordan has~$5$~squares of chocolate,~$2$~cups of sugar, lots of water, and~$7$~cups of milk. If he maintains the same ratio of ingredients, what is the greatest number of servings of hot chocolate he can make? 
Options:
aoVal: A
content: $5\dfrac{1}{8}$ 
aoVal: B
content: $6\dfrac{1}{4}$ 
aoVal: C
content: $7\dfrac{1}{2}$ 
aoVal: D
content: $8\dfrac{3}{4}$ 
aoVal: E
content: $9\dfrac{7}{8}$ 
Answer: D
Analysis: Assuming excesses of the other ingredients, the chocolate can make~$\dfrac{5}{2}\cdot5=12.5$~servings, the sugar can make~$\dfrac{2}{1/4}\cdot5=40$~servings, the water can make unlimited servings, and the milk can make~$\dfrac{7}{4}\cdot5=8.75$~servings. Limited by the amount of milk, Jordan can make at most~$\boxed{\left( D\right)\textbackslash{} 8\dfrac{3}{4}}$~servings. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Ratios and Proportions
Source: 其它
Problem: Bob is standing in the sixth place from the front to the back. The teacher distributes the apples from front to back according to the pattern as $1$, $3$, $5$ and so on, which means the first student has $1$ apple, and the second student has $3$ apples\ldots How many apples can Linda get if Linda is standing behind to Bob?~(adapted from $$2006$$ Math kangaroo Problems, Level $$1-2$$, Question \#$$1$$) 
Options:
aoVal: A
content: $$7$$ 
aoVal: B
content: $$9$$ 
aoVal: C
content: $$10$$ 
aoVal: D
content: $$11$$ 
aoVal: E
content: $$12$$ 
Answer: D
Analysis: $1,3,5,7,9,11$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Sequences and Number Tables->Patterns in Number Sequences
Problem: In a sequence of positive integers, each term after the second is the product of the previous two terms. The sixth term in the sequence is $4000$. What is the first term? 
Options:
aoVal: A
content: $$1$$ 
aoVal: B
content: $$2$$ 
aoVal: C
content: $$4$$ 
aoVal: D
content: $$5$$ 
aoVal: E
content: $$10$$ 
Answer: D
Analysis: D 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: Compare the following pair of fractions using an inequality sign.  $$A=\frac{773}{778}$$, $$B=\frac{884}{889}$$ 
Options:
aoVal: A
content: $$\frac{773}{778}\textgreater\frac{884}{889}$$ 
aoVal: B
content: $$\frac{773}{778}\textless{}\frac{884}{889}$$ 
aoVal: C
content: $$\frac{773}{778}=\frac{884}{889}$$ 
Answer: B
Analysis: $$\frac{773}{778}\textless{}\frac{(773+111)}{(778+111)}=\frac{884}{889}$$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Fractions
Problem: $$9 + 99 + 99 + 101 + 101 + 101 =$$． 
Options:
aoVal: A
content: $$200$$ 
aoVal: B
content: $$300$$ 
aoVal: C
content: $$600$$ 
aoVal: D
content: $$919$$ 
Answer: C
Analysis: $$99+101 + 99+101 + 99+101 = 200 + 200 + 200 = 600$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Addition and Subtraction
Problem: Alysha and Julia have some biscuits. Altogether they have $28$ biscuits.  Alysha has $4$ more biscuits than Julia.  How many biscuits does Alysha have? 
Options:
aoVal: A
content: $$12$$ 
aoVal: B
content: $$13$$ 
aoVal: C
content: $$14$$ 
aoVal: D
content: $$15$$ 
aoVal: E
content: $$16$$ 
Answer: E
Analysis: $(28 + 4) \div 2 = 16$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: Supposed that $x$ and $y$ are nonzero real numbers such that $\frac{3 x+y}{x-3 y}=-2$. What is the value of $\frac{x+3 y}{3 x-y}$? (2017 AMC 10B Problems, Question \#4) 
Options:
aoVal: A
content: $$-3$$ 
aoVal: B
content: $$-1$$ 
aoVal: C
content: $$1$$ 
aoVal: D
content: $$2$$ 
aoVal: E
content: $$3$$ 
Answer: D
Analysis: Rearranging, we find $3 x+y=-2 x+6 y$, or $5 x=5 y \Longrightarrow x=y$. Substituting, we can convert the second equation into $$\frac{x+3 x}{3 x-x}=\frac{4 x}{2 x}=\text { (D) } 2$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: Which set of numbers go into the missing blanks respectively to make the equation true?  $7=14$ $\div$~\uline{~~~~~~~~~~}~$=\frac{14}{\square}=\frac{14}{1}\times\frac{1}{\square}$ 
Options:
aoVal: A
content: $2, 1, 2$ 
aoVal: B
content: $1, 2, 2$ 
aoVal: C
content: $2, 2, 1$ 
aoVal: D
content: $2, 2, 2$ 
Answer: D
Analysis: Dividing a term by $2$ is the same as multiplying it by half ($\frac{1}{2}$)! 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: The sixteenth term of an arithmetic progression is 40. The fifty-fifth term of this arithmetic progression is 157. Find the eighty-first term of this arithmetic progression. 
Options:
aoVal: A
content: $$235$$ 
aoVal: B
content: $$241$$ 
aoVal: C
content: $$299$$ 
aoVal: D
content: $$274$$ 
aoVal: E
content: $$171$$ 
Answer: A
Analysis: By the formula~$d=\dfrac{a\_{m}-a\_{n}}{m-n}\textbackslash{} \Rightarrow\textbackslash{} d=\dfrac{a\_{55}-a\_{16}}{55-16}=\dfrac{157-40}{39}=3.$  By the formula~$a\_{n}=a\_{m}+\left( n-m\right)d\textbackslash{} \textbackslash{} \textbackslash{} \Rightarrow\textbackslash{} \textbackslash{} a\_{81}=a\_{55}+\left( 81-55\right)\times3=235.$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Sequences and Number Tables->Arithmetic Sequences
Source: 其它
Problem: Fido the Dog, Philemon the Cat and $4$ monkeys together weigh $24$ lbs. Fido and one monkey together weigh $11$ lbs. Philemon and $2$ monkeys together weigh $1$ lb less than Fido and one monkey weigh together. Each monkey weighs the same. How much does Philemon weigh? (2011 Math Kangaroo Problem, Level 1-2, Question \#21) 
Options:
aoVal: A
content: $3$ lbs 
aoVal: B
content: $4$ lbs 
aoVal: C
content: $5$ lbs 
aoVal: D
content: $6$ lbs 
Answer: B
Analysis: We can write their relationships as the equations below:  $$F+P+M+M+M+M=24$$  $$F+M=11$$  $$P+M+M=F+M-1$$  $ $  $$P+M+M=F+M-1$$, so $P+M+M=10$  $$F+P+M+M+M+M=24$$, so $11+10+M=24$, $M=3$, $P=4$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Equivalent Substitution
Source: 其它
Problem: $$9.25\times 0.8+9\frac{1}{4}\times 0.2=$$． 
Options:
aoVal: A
content: $$9.25$$ 
aoVal: B
content: $$92.5$$ 
aoVal: C
content: $$925$$ 
Answer: A
Analysis: $$9.25\times 0.8+9\frac{1}{4}\times 0.2$$  $$=9.25\times 0.8+9.25\times 0.2$$  $$=9.25\times (0.8+0.2)$$  $$=9.25\times 1$$  $$=9.25$$  So, $$\text{A}$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Problem: Which of the following is not an expression? 
Options:
aoVal: A
content: $$a-b=2c$$ 
aoVal: B
content: $$z$$ 
aoVal: C
content: $$a+b-5$$ 
aoVal: D
content: $$3$$ 
Answer: A
Analysis: Equations are not expressions. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Equivalent Substitution->Algebraic Expressions
Problem: David drives from his home to the airport to catch a flight. He drives $35$ miles in the first hour, but realizes that he will be $1$ hour late if he continues at this speed. He increases his speed by $15$ miles per hour for the rest of the way to the airport and arrives $30$ minutes early. How many miles is the airport from his home? (2014 AMC 10A Problems, Question \#15) 
Options:
aoVal: A
content: $$140$$ 
aoVal: B
content: $$175$$ 
aoVal: C
content: $$210$$ 
aoVal: D
content: $$245$$ 
aoVal: E
content: $$280$$ 
Answer: C
Analysis: Note that he drives at $50$ miles per hour after the first hour and continues doing so until he arrives. Let $d$ be the distance still needed to travel after $1$ hour. We have that $\frac{d}{50}+1.5=\frac{d}{35}$, where the $1.5$ comes from $1$ hour late decreased to $0.5$ hours early. Simplifying gives $7 d+525=10 d$, or $d=175$. Now, we must add an extra $35$ miles traveled in the first hour, giving a total of (C) $210$ miles. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: \textbf{A grocer would like to determine the proportion of milk cartons that have expired within 0.05 of the true proportion with a 95 percent confidence interval. What is the minimum required sample size?} 
Options:
aoVal: A
content: $$300$$ 
aoVal: B
content: $$383$$ 
aoVal: C
content: $$384$$ 
aoVal: D
content: $$385$$ 
aoVal: E
content: $$400$$ 
Answer: D
Analysis: \textbf{$n \geq (\frac{Z\_{\alpha/2}}{ME})^{2} {p(1-p)}$}  \textbf{Since p is unknown, then use p = ½. $n \geq (\frac{Z\_{\alpha/2}}{2ME})^{2}$}  \textbf{$Z\_{\alpha/2}=1.96$ The ME is how far off from the true proportion you are willing to be, in this case, 0.05}  \textbf{$n \geq (\frac{1.96}{2*0.05})^{2} = 384.16$ round up} 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: $$(101+100+\cdots +3+2)-(100+99+\cdots +2+1)=$$． 
Options:
aoVal: A
content: $$99$$ 
aoVal: B
content: $$100$$ 
aoVal: C
content: $$101$$ 
aoVal: D
content: $$102$$ 
Answer: B
Analysis: $$(101+100+\cdots +3+2)-(100+99+\cdots +2+1)=(101-1)=100$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Addition and Subtraction
Problem: $$\sqrt{9+16+144}=\sqrt{9}+\sqrt{16}+$$． 
Options:
aoVal: A
content: $$\sqrt{36}$$ 
aoVal: B
content: $$\sqrt{100}$$ 
aoVal: C
content: $$\sqrt{144}$$ 
aoVal: D
content: $$\sqrt{169}$$ 
Answer: A
Analysis: $$\sqrt{9+16+144}=\sqrt{169}=13=3+4+6=\sqrt{9}+\sqrt{16}+\sqrt{36}$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Power->Computing Powers
Problem: Convert the decimal $$0.65$$ to a fraction in its simplest form. 
Options:
aoVal: A
content: $\dfrac{13}{20}$ 
aoVal: B
content: $\dfrac{3}{4}$ 
aoVal: C
content: $$\dfrac{17}{20}$$ 
aoVal: D
content: $$\dfrac{6}{5}$$ 
Answer: A
Analysis: First, write down the decimal "over" the number 1 :~$0.65=\dfrac{0.65}{1}$  Then multiply top and bottom by 100 since there are two numbers after the decimal point :~$\dfrac{0.65}{1}=\dfrac{0.65\times100}{1\times100}=\dfrac{65}{100}$  This makes it a correctly formed fraction. Then simplify the fraction (in this case by dividing top and bottom by 5) :~$\dfrac{65}{100}=\dfrac{13}{20}$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Decimals->Basic Understanding of Decimals->Converting Decimals into Fractions
Source: 其它
Problem: Choose the correct number to make the number statement true.  $$17-9+$$$$\textgreater15+3-8$$ 
Options:
aoVal: A
content: $$1$$ 
aoVal: B
content: $$2$$ 
aoVal: C
content: $$5$$ 
Answer: C
Analysis: $$15+3-8=10$$，$$17-9=8$$，$$8+\left( 2 \right)=10$$，$$number$$ has to be bigger than $$2$$ so $$5$$ fits the requirement. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers
Problem: A consumer is willing to pay $\textbackslash$ 12$ for a good, but is able to purchase it for $\textbackslash$ 10$. What is the consumer surplus in this scenario? 
Options:
aoVal: A
content: $\textbackslash$ 2$ 
aoVal: B
content: $\textbackslash$ 10$ 
aoVal: C
content: $\textbackslash$ 12$ 
aoVal: D
content: $\textbackslash$ 22$ 
Answer: A
Analysis: Consumer surplus is calculated as the difference between the maximum price a consumer is willing to pay for a good and the actual price they pay, which is $\textbackslash$ 12$- $\textbackslash$10$ = $\textbackslash$ 2$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: Calculate: $1^{2}+2^{2}+3^{2}+\cdots +10^{2}$ 
Options:
aoVal: A
content: $$55$$ 
aoVal: B
content: $$385$$ 
aoVal: C
content: $$1155$$ 
aoVal: D
content: $$2310$$ 
Answer: B
Analysis: Answer is $385$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Sequences and Number Tables->Arithmetic Sequences->Sum of Terms in Arithmetic Sequences
Source: 其它
Problem: Calculate: $$\frac{1}{2}\times \frac{5}{3}+\frac{11}{5}\times \frac{7}{6}=$$． 
Options:
aoVal: A
content: $$\frac{77}{30}$$ 
aoVal: B
content: $$\frac{5}{6}$$ 
aoVal: C
content: $$\frac{17}{5}$$ 
aoVal: D
content: $$\frac{41}{18}$$ 
Answer: C
Analysis: $\frac{5}{6}+\frac{77}{30}=\frac{25}{30}+\frac{77}{30}=\frac{102}{30}=\frac{17}{5}$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Fractions->Operations of Fractions
Problem: What is the next number in the sequence below?  $$3, 5, 8, 13, 21, $$~\uline{~~~~~~~~~~}~ 
Options:
aoVal: A
content: $$27$$ 
aoVal: B
content: $$29$$ 
aoVal: C
content: $$31$$ 
aoVal: D
content: $$34$$ 
aoVal: E
content: None of the above 
Answer: D
Analysis: Sum of the previous two number. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Sequences and Number Tables->Patterns in Number Sequences
Problem: Which calculation has a result of an odd number． 
Options:
aoVal: A
content: $$2\times \left( 0+2+2 \right)\times 2021$$ 
aoVal: B
content: $$2\times 2021-0-2-2$$ 
aoVal: C
content: $$2021-2\times 0\times 2\times 2$$ 
aoVal: D
content: $$2+\left( 0\times 2\times 2 \right)\times 2021$$ 
Answer: C
Analysis: C 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Mixed Operations
Problem: The $2021^{st}$ digit at the right of the decimal point in the decimal expression of $\dfrac{2}{7}$ is~\uline{~~~~~~~~~~}~. 
Options:
aoVal: A
content: $$2$$ 
aoVal: B
content: $$1$$ 
aoVal: C
content: $$5$$ 
aoVal: D
content: $$8$$ 
Answer: B
Analysis: $$\frac{2}{7}=0.\overline{285714}$$, it is a decimal which repeats in cycles of $6$ digits. Every $6$$^{th}$ digit is $4$. The $2022$$$^{nd}$$ digit is $4$, so the $2021$$^{st}$ digit is $1$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Decimals->Basic Understanding of Decimals
Problem: Betty has more than $90$ toys. All her toys can be divided evenly between $2$, $3$, or $4$ children. However, they cannot be divided evenly between $9$ children because $3$ more toys would be needed. How many toys does she have at least? 
Options:
aoVal: A
content: $$87$$ 
aoVal: B
content: $$96$$ 
aoVal: C
content: $$132$$ 
aoVal: D
content: $$135$$ 
Answer: B
Analysis: After adding $3$ toys, the number of toys should be divisible by $9$, and it also should be divisible by $2, 3, $ and $4$. Thus, the answer is $B$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: $$49\div0.035=$$． 
Options:
aoVal: A
content: $$1.4$$ 
aoVal: B
content: $$14$$ 
aoVal: C
content: $$140$$ 
aoVal: D
content: $$1400$$ 
Answer: D
Analysis: $$49\div0.035=49000\div35=1400$$, so the answer is $$\rm{D}$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Decimals->Multiplication and Division of Decimals
Problem: Four fair six-sided dice are rolled. What is the probability that at least three of the four dice show the same value? 
Options:
aoVal: A
content: $\frac{1}{36}$ 
aoVal: B
content: $\frac{7}{72}$ 
aoVal: C
content: $\frac{1}{9}$ 
aoVal: D
content: $\frac{5}{36}$ 
aoVal: E
content: $\frac{1}{6}$ 
Answer: B
Analysis: B 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: Express the recurring decimal~$0.2\dot{5}$~as a fraction. ． 
Options:
aoVal: A
content: $$\frac{23}{99}$$ 
aoVal: B
content: $$\frac{23}{90}$$ 
aoVal: C
content: $$\frac{25}{99}$$ 
aoVal: D
content: $$\frac{25}{90}$$ 
Answer: B
Analysis: $$x=0.2\dot{5}$$,  $$100x=25. \dot{5}$$,  $$10x=2. \dot{5}$$,  $$90x=23$$,  $$x= \frac{23}{90}$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Decimals->Recurring Decimals
Source: 其它
Problem: Ethan is shopping for an Eiffel Tower model online. The description says the scale of the model is $1:1000$. The height of the Eiffel Tower is $1083$ feet, so the model should be~\uline{~~~~~~~~~~}~inches tall (round to the nearest integer, $12$ inches $=1$ foot ). 
Options:
aoVal: A
content: $$12$$ 
aoVal: B
content: $$13$$ 
aoVal: C
content: $$14$$ 
aoVal: D
content: $$15$$ 
aoVal: E
content: $$16$$ 
Answer: B
Analysis: $1083\div 1000\times 12=12.996$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Ratios and Proportions
Source: 其它
Problem: For $\triangle ABC$, all of its side lengths are integers. The perimeter 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: There are seven people in a bus. Four passengers get on and $6$ passengers get off at the first station. Then seven passengers get on and three get off at the second station. How many passengers are there on the bus at this time? 
Options:
aoVal: A
content: $$5$$ 
aoVal: B
content: $$7$$ 
aoVal: C
content: $$9$$ 
aoVal: D
content: $$11$$ 
aoVal: E
content: $$13$$ 
Answer: C
Analysis: $7+4-6+7-3=9$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Addition and Subtraction ->Addition and Subtraction of Whole Numbers
Source: 其它
Problem: What is the sum: $$3+7+2+8+1=?$$ (2007 Math Kangaroo Problem, Level 1-2, Question \#17) 
Options:
aoVal: A
content: $$12$$ 
aoVal: B
content: $$14$$ 
aoVal: C
content: $$17$$ 
aoVal: D
content: $$21$$ 
aoVal: E
content: $$27$$ 
Answer: D
Analysis: $3+7+2+8+1$  $=(3+7)+(2+8)+1$  $=10+10+1$  $=21$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: What is the product of $$ \frac { 3 } { 2 } \times \frac { 4 } { 3 } \times \frac { 5 } { 4 } \times \cdots \times \frac { 2 0 0 6 } { 2 0 0 5 }$$? 
Options:
aoVal: A
content: $$1$$ 
aoVal: B
content: $$1002$$ 
aoVal: C
content: $$1003$$ 
aoVal: D
content: $$2005$$ 
aoVal: E
content: $$2006$$ 
Answer: C
Analysis: By telescoping, it\textquotesingle s easy to see the sum becomes $$\frac {2006}2=1003$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Fractions->Fast Calculation in Fractions->Reducing Fractions by cancelling out successively
Problem: In a dining room, there are $$15$$ chairs, $$5$$ tables, and $$20$$ cups. What is the ratio of chairs to cups? 
Options:
aoVal: A
content: $$1:4$$ 
aoVal: B
content: $$15:20$$ 
aoVal: C
content: $$4:3$$ 
aoVal: D
content: $$3:4$$ 
Answer: D
Analysis: There are $$15$$ chairs and $$20$$ cups. So the ratio of chairs to cups is $$15:20$$. The simplest form is $$3:4$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Ratios and Proportions->Ratio
Problem: Which of the following expression has the maximum value? 
Options:
aoVal: A
content: $$3.2\times0.16$$ 
aoVal: B
content: $$0.32\times0.16$$ 
aoVal: C
content: $$32\times0.016$$ 
aoVal: D
content: $$0.032\times160$$ 
Answer: D
Analysis: $$3.2\times0.16=0.512$$, $$0.32\times0.16=0.0512$$,  $$32\times0.016=0.512$$, $$0.032\times160=5.12$$.  Therefore, expression $$\rm D$$ has the maximum value. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Decimals->Multiplication and Division of Decimals
Problem: Teacher Judy has $71$ stickers in total, and she gives all the stickers to her students Martin and David. Martin gets $7$ more than David. How many stickers does David get? 
Options:
aoVal: A
content: $$29$$ 
aoVal: B
content: $$30$$ 
aoVal: C
content: $$32$$ 
aoVal: D
content: $$35$$ 
aoVal: E
content: $$39$$ 
Answer: C
Analysis: $(71-7) \div 2 = 32$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: What is the last number on the $20^{th}$ row? 
Options:
aoVal: A
content: $$190$$ 
aoVal: B
content: $$200$$ 
aoVal: C
content: $$210$$ 
aoVal: D
content: $$220$$ 
Answer: C
Analysis: C 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Sequences and Number Tables->Patterns of Number Tables->Triangle Number Table
Source: 其它
Problem: Pinocchio has a magic nose, which will stretch out $5$ cm if he tells a lie and will shorten itself $1$ cm if he tells a truth. At the beginning his nose was $11$ cm long. The length of his nose changed into $13$ cm after he had said $10$ sentences. If these sentences were not true, they were definitely false. Thus, there were~\uline{~~~~~~~~~~}~truths among these $10$ sentences. 
Options:
aoVal: A
content: $$5$$ 
aoVal: B
content: $$6$$ 
aoVal: C
content: $$7$$ 
aoVal: D
content: $$8$$ 
aoVal: E
content: $$9$$ 
Answer: D
Analysis: If these $10$ sentences are true, the nose will shorten $10$ cm. There are $(13-11+10)\div(5+1)=2$ sentences which are false. Thus, $10-2=8$ sentences are true. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: $$\frac{1}{2+\dfrac{1}{2+\dfrac{1}{2}}}=$$． 
Options:
aoVal: A
content: $\dfrac{1}{3}$ 
aoVal: B
content: $\dfrac{2}{5}$ 
aoVal: C
content: $\dfrac{3}{8}$ 
aoVal: D
content: $\dfrac{2}{9}$ 
aoVal: E
content: $\dfrac{5}{12}$ 
Answer: E
Analysis: $$\frac{1}{2+\dfrac{1}{2+\dfrac{1}{2}}}=\frac{1}{2+ \dfrac{1}{ \dfrac{5}{2}}}= \frac{1}{2+ \dfrac{2}{5}}= \frac{1}{ \dfrac{12}{5}}= \frac{5}{12}$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Fractions->Complex Fractions