math-eval/TAL-SCQ5K huggingface dataset:
Problem: In a middle-school mentoring program, a number of the sixth graders are paired with a ninth-grade student as a buddy. No ninth grader is assigned more than one sixth-grade buddy. If $\frac{1}{3}$ of all the ninth graders are paired with $\frac{2}{5}$ of all the sixth graders, what fraction of the total number of sixth and ninth graders have a buddy? (2015 AMC 8 Problems, Question \#16) 
Options:
aoVal: A
content: $\frac{2}{15}$ 
aoVal: B
content: $\frac{4}{11}$ 
aoVal: C
content: $\frac{11}{30}$ 
aoVal: D
content: $\frac{3}{8}$ 
aoVal: E
content: $\frac{11}{15}$ 
Answer: B
Analysis: Let the number of sixth graders be $s$, and the number of ninth graders be $n$. Thus, $\frac{n}{3}=\frac{2 s}{5}$, which simplifies to $n=\frac{6 s}{5}$. Since we are trying to find the value of $\frac{\frac{n}{3}+\frac{2 s}{5}}{n+s}$, we can just substitute $\frac{6 s}{5}$ for $n$ into the equation. We then get a value of $\frac{\frac{6 s}{5}\cdot\frac13+\frac{2 s}{5}}{\frac{6 s}{5}+s}=\frac{\frac{6 s+6 s}{15}}{\frac{11 s}{5}}=\frac{\frac{4 s}{5}}{\frac{11 s}{5}}=\frac{4}{11}$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: Line $l\_1$ has equation $3 x-2 y=1$ and goes through $A=(-1,-2)$. Line $l\_2$ has equation $y=1$ and meets line $l\_1$ at point $B$. Line $l\_3$ has positive slope, goes through point $A$, and meets $l\_2$ at point $C$. The area of $\triangle A B C$ is $3$ . What is the slope of $l\_3$? (2013 AMC 12B Problems, Question \#8) 
Options:
aoVal: A
content: $$\frac23$$ 
aoVal: B
content: $$\frac34$$ 
aoVal: C
content: $$1$$ 
aoVal: D
content: $$\frac43$$ 
aoVal: E
content: $$\frac32$$ 
Answer: B
Analysis: Line $l\_1$ has the equation $y=\frac{3x}{2}-\frac12$ when rearranged. Substituting $1$ for $y$, we find that line $l\_2$ will meet this line at point $(1,1)$, which is point $B$. We call $\overline{B C}$ the base and the altitude from $A$ to the line connecting $B$ and $C, y=1$, the height. The altitude has length $\textbar-2-1\textbar=3$, and the area of $\triangle A B C=3$. Since $A=\frac{bh}{2}, b=2$. Because $l\_3$ has positive slope, it will meet $l\_2$ to the right of $B$, and the point that is $2$ to the right of $B$ is $(3,1)$. $l\_3$ passes through $(-1,-2)$ and $(3,1)$, and thus has slope $\frac{\textbar1-(-2)\textbar}{\textbar3-(-1)\textbar}=(\mathbf{B}) \frac{3}{4}$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: $$1+2+3+4+996 +997+998 + 999 =$$． 
Options:
aoVal: A
content: $$3998$$ 
aoVal: B
content: $$3999$$ 
aoVal: C
content: $$4000$$ 
aoVal: D
content: $$4001$$ 
Answer: C
Analysis: This is $$(1+999)+(2+998)+(3+997)+(4 +996) = 4000$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Addition and Subtraction
Problem: Which of the following equations are not equivalent to $x+3=11$ 
Options:
aoVal: A
content: $x+3+5=11+5$ 
aoVal: B
content: $2x=28$ 
aoVal: C
content: $x+3-11=11-11$ 
aoVal: D
content: $x+3-11=0$ 
Answer: B
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: There are $18$ bottles of water in each of $10$ boxes. Six customers want to buy some botlles of water. Four of them buy $30$ bottles of water, and two of them buy $20$ bottles of water. How many bottles of water are left in the boxes 
Options:
aoVal: A
content: $$10$$ 
aoVal: B
content: $$20$$ 
aoVal: C
content: $$30$$ 
aoVal: D
content: $$35$$ 
aoVal: E
content: $$40$$ 
Answer: B
Analysis: $18 \times 10 - 30 \times 4 - 20 \times 2 = 20$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: Erik and Ivy each writes down a fraction. They are surprised to find that the fractions they write are very similar: the denominator of each fraction is exactly the same as the other\textquotesingle s numerator. What is the product of the two fractions? 
Options:
aoVal: A
content: $$0$$ 
aoVal: B
content: $\frac12$ 
aoVal: C
content: $$1$$ 
aoVal: D
content: $$2$$ 
aoVal: E
content: It cannot be determined. 
Answer: C
Analysis: Their fractions are just reciprocal of the other one. Thus, the product of the two fractions should be $1.$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Fractions
Source: 其它
Problem: $$100001^{2}$$ exceeds $$100000^{2}$$ by． 
Options:
aoVal: A
content: $$200001$$ 
aoVal: B
content: $$100001$$ 
aoVal: C
content: $$200001\times 10^{6}$$ 
aoVal: D
content: $$100001\times 10^{6}$$ 
Answer: A
Analysis: $$100001^{2}-100000^{2}=10000200001-10000000000=200001$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Operations through Formulas-> Difference of Two Squares Formula
Problem: Given that $$1^{2}+2^{2}+3^{2}+\cdots +n^{2}=\frac{n(n+1)(2n+1)}{6}$$, then $$1^{2}+2^{2}+3^{2}+\cdots +18^{2}=$$． 
Options:
aoVal: A
content: $$2016$$ 
aoVal: B
content: $$2107$$ 
aoVal: C
content: $$2018$$ 
aoVal: D
content: $$2109$$ 
Answer: D
Analysis: Directly substitute $$n$$ with $$18$$ into the formula to get $$1^{2}+2^{2}+3^{2}+\cdots +18^{2}=\frac{18\times 19\times 37}{6}=2109$$. So the answer is $$\text{D}$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Operations through Formulas->1²+2²+3²+......+n²=1/6n(n+1)(n+2)
Problem: For how many integers $x$ is the number $x^{4}-25 x^{2}+24$ negative? (  2014 AMC 10B Problems, Question \#20) 
Options:
aoVal: A
content: $$3$$ 
aoVal: B
content: $$6$$ 
aoVal: C
content: $$12$$ 
aoVal: D
content: $$14$$ 
aoVal: E
content: $$16$$ 
Answer: B
Analysis: First, note that $24+1=25$, which motivates us to factor the polynomial as $\left(x^{2}-24\right)\left(x^{2}-1\right)$. Since this expression is negative, one term must be negative and the other positive. Also, the first term is obviously smaller than the second, so $x^{2}-24\textless0\textless x^{2}-1$. Solving this inequality, we find $1\textless x^{2}\textless24$. There are exactly $6$ integers $x$ that satisfy this inequality, $\pm\textbackslash{2,3,4\textbackslash}$. Thus our answer is $(\mathbf{B}) 6$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: In a class of $$40$$ pupils, there are $$10$$ more boys than girls.  What is the ratio of the number of girls to the number of boys? 
Options:
aoVal: A
content: $$5:3$$ 
aoVal: B
content: $$3:1$$ 
aoVal: C
content: $$1:3$$ 
aoVal: D
content: $$3:5$$ 
Answer: D
Analysis: Total people $=2u+10=40$  $2u=40-10=30$  Girls $=1u=30\div2=15$  Boys $=1u+10=15+10=25$  Girls $:$ Boys $\to$ $15:25$ $\to$ $3:5$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Ratios and Proportions->Ratio
Problem: What is the largest factor of $$2^{2}\times3^{3} \times5^{5} \times7^{7}\times11^{11}$$ less than $$100$$? 
Options:
aoVal: A
content: $$66$$ 
aoVal: B
content: $$77$$ 
aoVal: C
content: $$88$$ 
aoVal: D
content: $$99$$ 
Answer: D
Analysis: The largest two-digit factor is $$3^{2} \times11 = 99$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Calculation of Multi-digit Numbers
Problem: The following are the weights (in pounds) of seven people: $100, 115, 135, 140, 180, 197, 230$. Find the $36$-th percentile. 
Options:
aoVal: A
content: $$100$$ 
aoVal: B
content: $$115$$ 
aoVal: C
content: $$125$$ 
aoVal: D
content: $$135$$ 
Answer: D
Analysis: $np=7(0.36)=2.52 \uparrow 3$. The $36$-th percentile is $135$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: Given that $\frac{2}{5}(2x-3)+\frac{4}{11}x-\frac{6}{11}=0$. The solution is $x=$~\uline{~~~~~~~~~~}~. 
Options:
aoVal: A
content: $\frac{1}{2}$ 
aoVal: B
content: $\frac{3}{2}$ 
aoVal: C
content: $\frac{1}{5}$ 
aoVal: D
content: $\frac{7}{55}$ 
Answer: B
Analysis: $\frac{2}{5}(2x-3)+\frac{4}{11}x-\frac{6}{11}=0$  $\frac{2}{5}(2x-3)+\frac{2}{11}(2x-3)=0$  $(\frac{2}{5}+\frac{2}{11})(2x-3)=0$  $2x-3=0$  $x=\frac{3}{2}$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: Which of the following pairs is a solution of $$\begin{cases}2x-4=0 \textbackslash\textbackslash{} 4x-y=7 \end{cases}$$? 
Options:
aoVal: A
content: ($x$,$y$)=($2$,$-1$) 
aoVal: B
content: ($x$,$y$)=($2$,$1$) 
aoVal: C
content: ($x$,$y$)=($-2$,$1$) 
aoVal: D
content: ($x$,$y$)=($-2$,$-1$) 
Answer: B
Analysis: $x=2$  $8-y=7$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Linear Equations with Multiple Variables
Source: 其它
Problem: Which one is an equivalent fraction of $$\frac{16}{24}$$? 
Options:
aoVal: A
content: $$\frac{8}{16}$$ 
aoVal: B
content: $$\frac{2}{3}$$ 
aoVal: C
content: $$\frac{4}{20}$$ 
Answer: B
Analysis: $$\frac{16}{24}=\frac{8\times 2}{8\times 3}=\frac{2}{3}$$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Fractions->Basic Understanding of Fractions->Properties of Fractions
Problem: Find the product of $$272$$ and $$3$$. 
Options:
aoVal: A
content: $$68$$ 
aoVal: B
content: $$168$$ 
aoVal: C
content: $$270$$ 
aoVal: D
content: $$816$$ 
Answer: D
Analysis: Stack the two numbers as shown below ,~ lining up the unit digits  Remember that the 7 in 272 stands for 7 tens(70) ,the first 2 in the 272 stands for 2 hundreds(200)  First, multiply the ones~ $2\times3=6$~, regroup the 0 tens to the tens column  Write ~6 in the ones place.  Then,~ Multiply and add the tens .~$3\times7+0=21$  Write 1 in the tens place and regroup the 2 hundreds.  Last, multiply and add the hundreds.~$3\times2+2=8$  Write 8 in the hundreds place 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Multiplication and Division
Overseas In-curriculum->Knowledge Point->Operations of Numbers ->Multiplication of Whole Numbers->Multiplication of Multi-Digit Numbers and 1-Digit Numbers->Multiplication of 3-Digit and 1-Digit (with regrouping for more than once)
Problem: $$\dfrac{5}{14}\div \dfrac{10}{21}=$$~\uline{~~~~~~~~~~}~，$$\dfrac{4}{15}\div \dfrac{28}{45}=$$~\uline{~~~~~~~~~~}~． 
Options:
aoVal: A
content: $$\dfrac{3}{4}$$，$$\dfrac{4}{7}$$． 
aoVal: B
content: $$\dfrac{3}{5}$$，$$\dfrac{3}{7}$$． 
aoVal: C
content: $$\dfrac{3}{4}$$，$$\dfrac{3}{7}$$． 
aoVal: D
content: $$\dfrac{3}{5}$$，$$\dfrac{4}{7}$$． 
Answer: C
Analysis: $$\dfrac{5}{14}\div \dfrac{10}{21}=\dfrac{5}{7\times 2}\times \dfrac{3\times 7}{2\times 5}=\dfrac{3}{4}$$．  $$\dfrac{4}{15}\div \dfrac{28}{45}=\dfrac{4}{15}\times \dfrac{3\times 15}{7\times 4}=\dfrac{3}{7}$$． 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Fractions
Source: 其它
Problem: Teacher Jason leaves home every day at $8:00\text{AM}$ to go to work. If he drives at an average speed of $40$ km/h, he will be late by $3$ minutes. If he drives at an average speed of $60$ km/h, he will be early by $3$ minutes. How many km/h does Teacher Jason need to drive to get to work exactly on time? 
Options:
aoVal: A
content: $$45$$ 
aoVal: B
content: $$48$$ 
aoVal: C
content: $$50$$ 
aoVal: D
content: $$55$$ 
aoVal: E
content: $$58$$ 
Answer: B
Analysis: B 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: Which of these numbers is largest? 
Options:
aoVal: A
content: $$1.1$$ 
aoVal: B
content: $$0.98$$ 
aoVal: C
content: $$0.9$$ 
aoVal: D
content: $$1.09$$ 
aoVal: E
content: $$1.9$$ 
Answer: E
Analysis: Nil 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Decimals->Basic Understanding of Decimals->Comparing Decimals
Problem: Fill ``$$+$$'' or ``$$-$$'' between neighbouring numbers to make the number statement correct.  $$5$$ 　$$5$$ 　$$5$$ 　$$5$$　$$5$$ 　$$5=0$$ 
Options:
aoVal: A
content: $$+$$；$$-$$；$$+$$；$$-$$；$$+$$ 
aoVal: B
content: $$-$$；$$+$$；$$-$$；$$+$$；$$-$$ 
aoVal: C
content: $$+$$；$$+$$；$$+$$；$$-$$；$$-$$ 
Answer: B
Analysis: $$5-5+5-5+5-5=0$$． 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers
Problem: Find the value of the expression:  $$3- \frac{6}{1 \times \left (1+2\right )}- \frac{9}{\left (1+2\right ) \times \left (1+2+3\right )}-\frac{12}{\left (1+2+3\right ) \times \left (1+2+3+4\right )}- \cdots$$  $$- \frac{60}{\left (1+2+ \cdots +19\right ) \times \left (1+2+ \cdots +20\right )}$$ 
Options:
aoVal: A
content: $\frac{1}{60}$ 
aoVal: B
content: $\frac{1}{70}$ 
aoVal: C
content: $\frac{1}{210}$ 
aoVal: D
content: $\frac{1}{380}$ 
aoVal: E
content: None of the above 
Answer: B
Analysis: B 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Fractions
Source: 其它
Problem: Find the missing number: $$64 \div 2 = 2 \times $$． 
Options:
aoVal: A
content: $$16$$ 
aoVal: B
content: $$32$$ 
aoVal: C
content: $$64$$ 
aoVal: D
content: $$128$$ 
Answer: A
Analysis: $$64\div2=32=2\times16$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Multiplication and Division
Problem: How many real numbers $x$ satisfy the following equation: $$ (x^{2}+7)^{2} = 289$$ 
Options:
aoVal: A
content: $$0$$ 
aoVal: B
content: $$2$$ 
aoVal: C
content: $$3$$ 
aoVal: D
content: $$4$$ 
aoVal: E
content: $$8$$ 
Answer: B
Analysis: B 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: Aline writes a correct calculation. Then she covers two digits which are the same with a sticker:  [insert pic]  What digit is under the stickers? 
Options:
aoVal: A
content: $$2$$ 
aoVal: B
content: $$4$$ 
aoVal: C
content: $$5$$ 
aoVal: D
content: $$7$$ 
aoVal: E
content: $$8$$ 
Answer: D
Analysis: Two same digits that add up to "4" in the last digit, hence it must be either 2 or 7.  If we try digit 2, 42+52=94, wrong.  If we try digit 7, 27+57=104. correct. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Operations with New Definition->Finding Patterns->Encryption and Decryption
Source: 其它
Problem: The number that is 8.9 larger than 1.2 is． 
Options:
aoVal: A
content: $$7.7$$ 
aoVal: B
content: $$9.1$$ 
aoVal: C
content: $$9.7$$ 
aoVal: D
content: $$10.1$$ 
aoVal: E
content: $$9.9$$ 
Answer: D
Analysis: $$1.2+8.9=10.1$$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Decimals->Addition and Subtraction of Decimals
Source: 其它
Problem: What is the value of $$\left( 330+22 \right)\div 11$$ ? 
Options:
aoVal: A
content: $$23$$ 
aoVal: B
content: $$32$$ 
aoVal: C
content: $$34$$ 
aoVal: D
content: $$52$$ 
aoVal: E
content: $$54$$ 
Answer: B
Analysis: $$\left( 330+22 \right)\div 11=352\div 11=32$$.  Alternatively, $$\left(330 + 22 \right)\div 11 = 330 \div 11 + 22 \div 11= 30 + 2 = 32$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Mixed Operations
Problem: Let $p, q$, and $r$ be the distinct roots of the polynomial $x^{3}-22 x^{2}+80 x-67$. It is given that there exist real numbers $A, B$, and $C$ such that $$ \frac{1}{s^{3}-22 s^{2}+80 s-67}=\frac{A}{s-p}+\frac{B}{s-q}+\frac{C}{s-r} $$ for all $s \notin\textbackslash{p, q, r\textbackslash}$. What is $\frac{1}{A}+\frac{1}{B}+\frac{1}{C}$? 
Options:
aoVal: A
content: $$243$$ 
aoVal: B
content: $$244$$ 
aoVal: C
content: $$245$$ 
aoVal: D
content: $$246$$ 
aoVal: E
content: $$247$$ 
Answer: B
Analysis: Multiplying both sides by $(s-p)(s-q)(s-r)$ yields $$ 1=A(s-q)(s-r)+B(s-p)(s-r)+C(s-p)(s-q) $$ As this is a polynomial identity, and it is true for infinitely many $s$, it must be true for all $s$ (since a polynomial with infinitely many roots must in fact be the constant polynomial 0$)$. This means we can plug in $s=p$ to find that $\frac{1}{A}=(p-q)(p-r)$. Similarly, we can find $\frac{1}{B}=(q-p)(q-r)$ and $\frac{1}{C}=(r-p)(r-q)$. Summing them up, we get that $$ \frac{1}{A}+\frac{1}{B}+\frac{1}{C}=p^{2}+q^{2}+r^{2}-p q-q r-p r $$ By Vieta\textquotesingle s Formulas, we know that $p^{2}+q^{2}+r^{2}=(p+q+r)^{2}-2(p q+q r+p r)=324$ and $p q+q r+p r=80$. Thus the answer is $324-80=$ 244 . 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Proportional Equations
Source: 其它
Problem: There were four piles of strawberries and a rabbit wanted to eat some. Each pile had $24$ strawberries. The rabbit ate a few strawberries from the first pile and then ate as many strawberries from the third pile as were left in the first pile. After that the rabbit ate a few strawberries from the second pile and then ate as many strawberries from the fourth pile as were left in the second pile. How many strawberries in total did the rabbit eat? 
Options:
aoVal: A
content: $$24$$ 
aoVal: B
content: $$36$$ 
aoVal: C
content: $$40$$ 
aoVal: D
content: $$48$$ 
aoVal: E
content: $$60$$ 
Answer: D
Analysis: $24 \times 4 \div 2 = 48$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: Consider the set of all fractions $\frac{x}{y}$, where $x$ and $y$ are relatively prime positive integers. How many of these fractions have the property that if both numerator and denominator are increased by $1$, the value of the fraction is increased by $10 \textbackslash\%$? 
Options:
aoVal: A
content: $$0$$ 
aoVal: B
content: $$1$$ 
aoVal: C
content: $$2$$ 
aoVal: D
content: $$3$$ 
aoVal: E
content: infinitely many 
Answer: B
Analysis: You can create the equation $\frac{x+1}{y+1}=\frac{11 x}{10 y}$.  Cross multiplying and combining like terms gives $x y+11 x-10 y=0$.  This can be factored into $(x-10)(y+11)=-110$.  $x$ and $y$ must be positive, so $x\textgreater0$ and $y\textgreater0$, so $x-10\textgreater-10$ and $y+11\textgreater11$.  Using the factors of 110 , we can get the factor pairs: $(-1,110),(-2,55)$, and $(-5,22)$.  But we can\textquotesingle t stop here because $x$ and $y$ must be relatively prime.  $(-1,110)$ gives $x=9$ and $y=99.9$ and 99 are not relatively prime, so this doesn\textquotesingle t work.  $(-2,55)$ gives $x=8$ and $y=44$. This doesn\textquotesingle t work.  $(-5,22)$ gives $x=5$ and $y=11$. This does work. We found one valid solution so the answer is (B)$1$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: On her first day of work, Janabel sold one widget. On day two, she sold three widgets. On day three, she sold five widgets, and on each succeeding day, she sold two more widgets than she had sold on the previous day. How many widgets in total had sabing $20$ days? (2015 AMC 8 Problems, Question \#9) 
Options:
aoVal: A
content: $$39$$ 
aoVal: B
content: $$40$$ 
aoVal: C
content: $$210$$ 
aoVal: D
content: $$400$$ 
aoVal: E
content: $$401$$ 
Answer: D
Analysis: We can easily find out she makes $2 \cdot 20-1=39$ widgets on Day $20$ . Then, we make the sum of $1,3,5, \ldots \ldots, 35,37,39$ by adding in this way: $(1+39)+(3+37)+(5+35)+\ldots+(19+21)$, which include $10$ pairs of $40$ . So the sum of $1,3,5, \ldots \ldots \ldots 39$ is $(40 \cdot 10)=(\text{D}) 400$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Sequences and Number Tables
Source: 其它
Problem: Three fifths of a pitcher is filled with pineapple juice. The pitcher is emptied by pouring an equal amount of juice into each of 6 cups. What percent of the total capacity of the pitcher did each cup receive?~ (adapted from 2020 AMC 8, Question \#5) 
Options:
aoVal: A
content: $$5$$ 
aoVal: B
content: $$10$$ 
aoVal: C
content: $$15$$ 
aoVal: D
content: $$20$$ 
aoVal: E
content: $$25$$ 
Answer: B
Analysis: The pitcher is $\frac{3}{5}$ full, i.e. $60 \textbackslash\%$ full. Therefore each cup receives $\frac{60}{6}=(\mathbf{C}) 10$ percent of the total capacity. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Percentage Calculation
Source: 其它
Problem: Observe the characteristics of the numbers given and write the numbers underlined below.~(adapted from $$2006$$ Math kangaroo Problems, Level $$1-2$$, Question \#$$8$$)  $$3$$~ ~ $$4$$~ ~ $$7$$~ ~ $$11$$~ ~$18$~\uline{~~~~~~~~~~}~$47$~ ~$76$~ ~$123$~ $$\cdots$$ 
Options:
aoVal: A
content: $$21$$ 
aoVal: B
content: $$24$$ 
aoVal: C
content: $$27$$ 
aoVal: D
content: $$28$$ 
aoVal: E
content: $$29$$ 
Answer: E
Analysis: The sum of the first digit and the second digit is the third digit, and so on. The horizontal line is $11+18=29$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Sequences and Number Tables->Arithmetic Sequences->Concepts of Arithmetic Sequences
Problem: A number of students from Fibonacci Middle School are taking part in a community service project. The ratio of $8^{\text {th }}$-graders to $6^{\text {th }}$-graders is $11: 6$, and the the ratio of $8^{\text {th }}$-graders to $7^{\text {th }}$ graders is $8: 13$. What is the smallest number of students that could be participating in the project? (2013 AMC 8, Question 16) 
Options:
aoVal: A
content: $$44$$ 
aoVal: B
content: $$84$$ 
aoVal: C
content: $$107$$ 
aoVal: D
content: $$150$$ 
aoVal: E
content: $$214$$ 
Answer: C
Analysis: We multiply the first ratio by 4 on both sides, and the second ratio by 3 to get the same number for 8 th graders, in order that we can put the two ratios together: $$ \begin{aligned} \&11: 6=11(4): 6(4)=44: 24 \textbackslash\textbackslash{} \&8: 13=8(3): 13(3)=24: 39 \end{aligned} $$ Therefore, the ratio of 8th graders to 7th graders to 6th graders is $44: 24: 39$. Since the ratio is in lowest terms, the smallest number of students participating in the project is $$ 44+24+39=\text { (C) } 107 $$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Ratios and Proportions
Source: 其它
Problem: Let $f$ be a linear function for which $f(5)-f(2)=10$. What is $f(8)-f(2)$? (  Adapted From 2003 AMC 12B Problems, Question \#9) 
Options:
aoVal: A
content: $$10$$ 
aoVal: B
content: $$14$$ 
aoVal: C
content: $$20$$ 
aoVal: D
content: $$22$$ 
aoVal: E
content: $$24$$ 
Answer: C
Analysis: Let $f$ be a linear function with slope $m$.  $$ \begin{gathered} m=\frac{f(5)-f(2)}{\Delta x}=\frac{10}{5-2}=\frac{10}{3} \textbackslash\textbackslash{} f(8)-f(2)=m \Delta x=\frac{10}{3}(8-2)=20 \Rightarrow(C) \end{gathered}$$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: How many odd numbers are there?  1, 3, 4, 6, 7, 9, 5, 8. 
Options:
aoVal: A
content: $$3$$ 
aoVal: B
content: $$4$$ 
aoVal: C
content: $$5$$ 
Answer: C
Analysis: $$Omitted.$$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers
Source: 其它
Problem: For how many integers $x$ is the number $x^{4}-51 x^{2}+50$ negative? 
Options:
aoVal: A
content: $$8$$ 
aoVal: B
content: $$10$$ 
aoVal: C
content: $$12$$ 
aoVal: D
content: $$14$$ 
aoVal: E
content: $$16$$ 
Answer: C
Analysis: First, note that $50+1=51$, which motivates us to factor the polynomial as $\left(x^{2}-50\right)\left(x^{2}-1\right)$. Since this expression is negative, one term must be negative and the other positive. Also, the first term is obviously smaller than the second, so $x^{2}-50\textless0\textless x^{2}-1$. Solving this inequality, we find $1\textless x^{2}\textless50$. There are exactly 12 integers $x$ that satisfy this inequality, $\pm\textbackslash{2,3,4,5,6,7\textbackslash}$. Thus our answer is $(\mathbf{C}) 12$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation
Source: 其它
Problem: The following are the heights (in cm) of eight people: $157$, $175$, $165$, $184$, $180$, $197$, $160$, $177$. Find the $42$-th percentile. 
Options:
aoVal: A
content: $$165$$ 
aoVal: B
content: $$175$$ 
aoVal: C
content: $$177$$ 
aoVal: D
content: $$180$$ 
Answer: B
Analysis: Arrange the data from the least to the largest: $157$, $160$, $165$, $175$, $177$, $180$, $184$, $197$.  $np=8(0.42)=3.36 \uparrow 4$. The $42$-th percentile is $175$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它