math-eval/TAL-SCQ5K huggingface dataset:
Problem: Find the next number in the following sequence  $$21, 24, 19, 26, 17, \cdots $$ 
Options:
aoVal: A
content: $$18$$ 
aoVal: B
content: $$19$$ 
aoVal: C
content: $$27$$ 
aoVal: D
content: $$28$$ 
aoVal: E
content: None of the above 
Answer: D
Analysis: $$21, 19, 17, \cdots $$ (pattern of $$-2$$)  $$24, 26, ?, \cdots $$ (pattern of $$+2$$)  Thus, the "?" is equal to $$26+2=28$$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Sequences and Number Tables->Patterns in Number Sequences
Source: 其它
Problem: The ones digit of $$1!+2!+3!+···+100!$$ is~\uline{~~~~~~~~~~}~. 
Options:
aoVal: A
content: $$1$$ 
aoVal: B
content: $$3$$ 
aoVal: C
content: $$5$$ 
aoVal: D
content: $$7$$ 
aoVal: E
content: $$9$$ 
Answer: B
Analysis: $$\textasciitilde\textasciitilde\textasciitilde\textasciitilde1+1\times 2+1\times 2\times 3+1\times 2\times 3\times 4$$  $$=1+2+6+24$$  $$=33$$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Operations through Formulas
Source: 其它
Problem: There are two girls and two cats in the room. How many legs are there in this room? (Adapted from 2011 Math Kangaroo Problem, Level 1-2, Question \#4) 
Options:
aoVal: A
content: $$2$$ 
aoVal: B
content: $$4$$ 
aoVal: C
content: $$8$$ 
aoVal: D
content: $$12$$ 
aoVal: E
content: $$16$$ 
Answer: D
Analysis: $$2+2+4+4=12$$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: Peter Rabbit eats cabbages and carrots. Each day he eats either $$10$$ carrots or $$2$$ cabbages. Last week Peter ate $$6$$ cabbages. How many carrots did he eat last week? (2014 Math Kangaroo Problem, Level 1-2, Question \#18) 
Options:
aoVal: A
content: $$20$$ 
aoVal: B
content: $$30$$ 
aoVal: C
content: $$34$$ 
aoVal: D
content: $$40$$ 
aoVal: E
content: $$50$$ 
Answer: D
Analysis: Peter ate $6$ cabbages in $6\div2=3$ days. In $7-3=4$ days, Peter ate $4\times10=10+10+10+10=40$ carrots. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Mixed Operations
Source: 其它
Problem: An amusement park has a collection of scale models, with ratio $1: 20$, of buildings and other sights from around the country. The height of the United States Capitol is 289 feet. What is the height in feet of its replica to the nearest whole number? (2018 AMC 8, Question 1) 
Options:
aoVal: A
content: $$14$$ 
aoVal: B
content: $$15$$ 
aoVal: C
content: $$16$$ 
aoVal: D
content: $$18$$ 
aoVal: E
content: $$20$$ 
Answer: A
Analysis: You can see that since the ratio of real building\textquotesingle s heights to the model building\textquotesingle s height is $1: 20$. We also know that the U.S Capitol is 289 feet in real life, so to find the height of the model, we divide by 20 . That gives us $14.45$ which rounds to 14 . Therefore, to the nearest whole number, the duplicate is (A) 14 feet. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Ratios and Proportions
Source: 其它
Problem: Seventy $$-$$ seven hundred is equal to． 
Options:
aoVal: A
content: $$70\times 70$$ 
aoVal: B
content: $$7\times 7000$$ 
aoVal: C
content: $$7\times 10$$ 
aoVal: D
content: $$11 \times 700$$ 
Answer: D
Analysis: Seventy $$-$$ seven hundred $$= 7700 = 11\times700$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Multiplication and Division
Problem: $$180\div 6=6\times $$． 
Options:
aoVal: A
content: $$180$$ 
aoVal: B
content: $$36$$ 
aoVal: C
content: $$30$$ 
aoVal: D
content: $$5$$ 
Answer: D
Analysis: $$180\div 6=30=6\times 5$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Multiplication and Division
Problem: Alen and Austin share $60$ apples between them.  Alen has twice as much as Austin.  How many apples does Alen have? 
Options:
aoVal: A
content: $$20$$ 
aoVal: B
content: $$30$$ 
aoVal: C
content: $$40$$ 
aoVal: D
content: $$45$$ 
aoVal: E
content: $$50$$ 
Answer: C
Analysis: $60 \div (2 + 1) \times 2 = 40$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: Fifteen integers $a\_1, a\_2, a\_3, \cdots , a\_{15}$ are arranged in order on a number line. The integers are equally spaced and have the property that  $$ 1 \leq a\_1 \leq 10$$,~~$$13 \leq a\_2 \leq 20$$, and $$241 \leq a\_{15} \leq 250$$.  What is the sum of the digits of $a\_{14}$? 
Options:
aoVal: A
content: $$8$$ 
aoVal: B
content: $$9$$ 
aoVal: C
content: $$10$$ 
aoVal: D
content: $$11$$ 
aoVal: E
content: $$12$$ 
Answer: A
Analysis: A 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: Given a twelve-digit number $$370592318436$$. Brandon wants to delete $4$ odd digits from the number to create an eight-digit number. What is the smallest eight-digit number that Brandon can form? 
Options:
aoVal: A
content: $$30214836$$ 
aoVal: B
content: $$30128436$$ 
aoVal: C
content: $$30231436$$ 
aoVal: D
content: $$30523136$$ 
aoVal: E
content: $$30218436$$ 
Answer: E
Analysis: NA 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Problem: How many inequalities are there in the following options?  1. $x = 1$  2. $y \textless{} 2$  3. $1 + a \geq g + p$  4. $1 \textgreater{} 2$ 
Options:
aoVal: A
content: $$0$$ 
aoVal: B
content: $$1$$ 
aoVal: C
content: $$2$$ 
aoVal: D
content: $$3$$ 
aoVal: E
content: $$4$$ 
Answer: D
Analysis: $23$ and $4$ are inequalities. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Inequalities
Source: 其它
Problem: Express $108:9$ in simplest form:． 
Options:
aoVal: A
content: $12:1$ 
aoVal: B
content: $24:6$ 
aoVal: C
content: $16:4$ 
aoVal: D
content: $4:1$ 
Answer: A
Analysis: $(108\div9):(9\div9)=12:1$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Ratios and Proportions->Ratio
Problem: Calculate: $$5 -- 4 + 3 -- 2 + 1 =$$． 
Options:
aoVal: A
content: $$1$$ 
aoVal: B
content: $$2$$ 
aoVal: C
content: $$3$$ 
aoVal: D
content: $$4$$ 
aoVal: E
content: $$5$$ 
Answer: C
Analysis: $$= (5 -- 4) + (3 -- 2) + 1 = 1 + 1 + 1 = 3$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Addition and Subtraction
Problem: Devi is $$9$$ years old and Valli is $$13$$ years old now. What is the ratio of Devi\textquotesingle s age to Valli\textquotesingle s age in $$5$$ years\textquotesingle{} time? 
Options:
aoVal: A
content: $$3:7$$ 
aoVal: B
content: $$7:9$$ 
aoVal: C
content: $$13:14$$ 
aoVal: D
content: $$9:18$$ 
Answer: B
Analysis: $5$ years later,  Devi is $9+5=14$ years old and Valli is $13+5=18$ years old.  $D:V$ $=$ $14:18$ $\to$ $7:9$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Ratios and Proportions->Ratio
Problem: Four fifths of a pitcher is filled with pineapple juice. The pitcher is emptied by pouring an equal amount of juice into each of 10 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: $$4$$ 
aoVal: B
content: $$6$$ 
aoVal: C
content: $$8$$ 
aoVal: D
content: $$12$$ 
aoVal: E
content: $$20$$ 
Answer: C
Analysis: The pitcher is $\frac{4}{5}$ full, i.e. $80 \textbackslash\%$ full. Therefore each cup receives $\frac{80}{10}=(\mathbf{C}) 8$ percent of the total capacity. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Percentage Calculation
Source: 其它
Problem: If Anna takes one step forward, it means she walks $"+1"$ step; If Anna takes one step backward, it means she walks $"-1"$ step. Now, Anna walks $"-5"$ steps, then she walks $"+4"$ steps, then she walks another $"-1"$ step. Anna\textquotesingle s position is from her original location. 
Options:
aoVal: A
content: $1$ step forward 
aoVal: B
content: $1$ step backward 
aoVal: C
content: $2$ step forward 
aoVal: D
content: $2$ step backward 
Answer: D
Analysis: Since walking one step forward and one step backward gets you back to your original location. Anna takes $4$ steps foward and $1+5=6$ steps backward.  $6-4=2$(steps)  Therefore, Anna is now $2$ steps backward from her original location. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Negative Numbers
Source: 其它
Problem: $$(12 \times7)-(12 + 12 + 12 + 12) =$$． 
Options:
aoVal: A
content: $$36$$ 
aoVal: B
content: $$48$$ 
aoVal: C
content: $$72$$ 
aoVal: D
content: $$108$$ 
Answer: A
Analysis: $$(12 \times7)-(12 + 12 + 12 + 12) =83-48=36$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Mixed Operations
Problem: The last five digits of Zoe\textquotesingle s phone number are $3,4,8,5,7.$ If these $5$ digits are filled in the square~$$\huge\square+\square =\square +\square $$, which number is not used?~(adapted from $$2017$$ Math kangaroo Problems, Level $$1-2$$, Question \#$$8$$) 
Options:
aoVal: A
content: $$3$$ 
aoVal: B
content: $$4$$ 
aoVal: C
content: $$5$$ 
aoVal: D
content: $$7$$ 
aoVal: E
content: $$8$$ 
Answer: C
Analysis: $4+7=3+8$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Addition and Subtraction ->Addition of Whole Numbers->Addition in Horizontal Form
Source: 其它
Problem: When $$115$$ is divided by~ \uline{?~} , the quotient is 4 and remainder is $$3$$. 
Options:
aoVal: A
content: $$27$$ 
aoVal: B
content: $$28$$ 
aoVal: C
content: $$460$$ 
aoVal: D
content: $$463$$ 
Answer: B
Analysis: $115 - 3 = 112$  $? \times 4 = 112$  $? = 112 \div ~4 = 28$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Problem: $$\left( {} \right.$$The number of seconds in a week$$\left. {} \right)$$$$\div$$$$\left( {} \right.$$the number of minutes in a week$$\left. {} \right)=$$． 
Options:
aoVal: A
content: $$60$$ 
aoVal: B
content: $$420$$ 
aoVal: C
content: $$3600$$ 
aoVal: D
content: $$7200$$ 
Answer: A
Analysis: There are $$60$$ seconds in each minute, so the quotient of the two quantities is $$60$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Ratios and Proportions
Problem: There are $12$ boys in the class. The ratio of boys to girls is $1:2$. How many students are there in the class? 
Options:
aoVal: A
content: $$18$$ 
aoVal: B
content: $$24$$ 
aoVal: C
content: $$30$$ 
aoVal: D
content: $$36$$ 
Answer: D
Analysis: $12-\textgreater1$  $24-\textgreater2$  $12+24=36$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Ratios and Proportions
Source: 其它
Problem: \textbf{A news organization conducted a survey about preferred methods for obtaining the news. A random sample of 1,605 adults living in a certain state was selected, and 16.2 percent of the adults in the sample reported that television was their preferred method. Which of the following is an appropriate margin of error for a 90 percent confidence interval to estimate the population proportion of all adults living in the state who would report that television is their preferred method for obtaining the news?} 
Options:
aoVal: A
content: \textbf{$1.645 \sqrt{\frac{(0.162)(1-0.162)}{1605}}$} 
aoVal: B
content: \textbf{$1.645 \sqrt{\frac{(0.5)(1-0.5)}{1605}}$} 
aoVal: C
content: \textbf{$1.96 \sqrt{\frac{(0.162)(1-0.162)}{1605}}$} 
aoVal: D
content: \textbf{$1.96 \sqrt{\frac{(0.5)(1-0.5)}{1605}}$} 
aoVal: E
content: \textbf{$1.83 \sqrt{\frac{(0.162)(1-0.162)}{1605}}$} 
Answer: A
Analysis: \textbf{$\hat{p} = 0.162$}  \textbf{MOE = $Z\_{0.05}\sqrt{\frac{0.162*(1-0.162)}{1605}}$ = 1.65*$\sqrt{\frac{0.162*(1-0.162)}{1605}}$} 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: Cassandra sets her watch to the correct time at noon. At the actual time of $1: 00$ PM, she notices that her watch reads $12: 57$ and $36$ seconds. Assuming that her watch loses time at a constant rate, what will be the actual time when her watch first reads $$10:00$$ PM? (2003 AMC 12B Problems, Question \#11) 
Options:
aoVal: A
content: $10:22$ PM and $24$ seconds 
aoVal: B
content: $10:24$ PM 
aoVal: C
content: $10:25$ PM 
aoVal: D
content: $10:27$ PM 
aoVal: E
content: $10:30$ PM 
Answer: C
Analysis: For every $60$ minutes that pass by in actual time, $57+\frac{36}{60}=57.6$ minutes pass by on Cassandra\textquotesingle s watch. When her watch first reads, $10: 00$ PM, $10(60)=600$ minutes have passed by on her watch. Setting up a proportion, $$ \frac{57.6}{60}=\frac{600}{x} $$ where $x$ is the number of minutes that have passed by in actual time. Solve for $x$ to get $625$ minutes, or $10$ hours and $25$ minutes $$ \Rightarrow \text { (C) 10:25} $$ PM. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: Yesterday is Saturday, what day is the day after tomorrow? (Adapted from 2019 Math Kangaroo Problem, Level 3-4, Question \#3) 
Options:
aoVal: A
content: Monday 
aoVal: B
content: Tuesday 
aoVal: C
content: Wednesday 
aoVal: D
content: Thursday 
aoVal: E
content: Sunday 
Answer: B
Analysis: Yesterday: Saturday  Today: Sunday  Tomorrow: Monday  The day after tomorrow: Tuesday 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: \textbf{Free-response questions on the AP Statistics Exam are graded 4, 3, 2, 1, or~ 0. Question 2 on the exam was of moderate difficulty. The average score on question 2 was 2.05 with a standard deviation of 1. To the nearest tenth, what score was achieved by a student who was at the 90th percentile of all students on the test? You may assume that the scores on the question were approximately normally distributed.} 
Options:
aoVal: A
content: $$3.5$$ 
aoVal: B
content: $$3.3$$ 
aoVal: C
content: $$2.9$$ 
aoVal: D
content: $$3.7$$ 
aoVal: E
content: $$3.1$$ 
Answer: B
Analysis: \textbf{Z = 1.28}  \textbf{$$\frac{x-2.05}{1}$$ = 1.28}  \textbf{x= 3.33} 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Fractions
Source: 其它
Problem: What is the simplest form of $\frac{15}{20}$? 
Options:
aoVal: A
content: $\frac{4}{5}$ 
aoVal: B
content: $\frac{3}{4}$ 
aoVal: C
content: $\frac{2}{3}$ 
aoVal: D
content: $\frac{15}{20}$ 
Answer: B
Analysis: Factors of numerator $15$ are $1, 3, 5, 15$ and factors of denominator $20$ are $1, 2, 4, 5, 10, 20$.  The highest common factor(HCF) is $5$.  Divide $15$ and $20$ by the HCF:  $\frac{15 \div 5}{20 \div 5}$=$\frac{3}{4}$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Fractions->Basic Understanding of Fractions->Using Common Factors to Simplify Fractions
Source: 其它
Problem: Which of the following sequences is different from the rest? 
Options:
aoVal: A
content: $$1$$，$$1$$，$$2$$，$$3$$，$$5$$，$$8$$，$$13$$，$$\cdots $$ 
aoVal: B
content: $$0$$，$$2$$，$$2$$，$$4$$，$$6$$，$$10$$，$$16$$，$$\cdots $$ 
aoVal: C
content: $$1$$，$$3$$，$$4$$，$$7$$，$$11$$，$$18$$，$$29$$，$$\cdots $$ 
aoVal: D
content: $$1$$，$$2$$，$$3$$，$$6$$，$$11$$，$$20$$，$$37$$，$$\cdots $$ 
Answer: D
Analysis: All sequences are Fibonacci numbers except for ($$\text{D}$$).  （Maths Olympiad 《Looking for a Pattern》 Pr$$3$$\&$$4$$ Question \#$$18$$） 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Sequences and Number Tables->Patterns in Number Sequences
Problem: What is the product of $286$ and $4$? 
Options:
aoVal: A
content: $$1144$$ 
aoVal: B
content: $$1124$$ 
aoVal: C
content: $$844$$ 
aoVal: D
content: $$824$$ 
Answer: A
Analysis: omitted 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Multiplication and Division
Problem: Let $ a◆b=a+(2\times b)$, then $1◆(2◆3) =$~\uline{~~~~~~~~~~}~. 
Options:
aoVal: A
content: $$9$$ 
aoVal: B
content: $$11$$ 
aoVal: C
content: $$17$$ 
aoVal: D
content: $$36$$ 
Answer: C
Analysis: $1◆(2◆3)=1◆[2+(2\times3)]=1◆8=1+16=17$.  So the answer is $\rm C$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Operations with New Definition->Operating Directly
Problem: There is a total of $3567$ dollars in Thompson\textquotesingle s wallet. What is the sum of the digits of his money? (adapted from $$2011$$ Math kangaroo Problems, Level $$1-2$$, Question \#$$7$$) 
Options:
aoVal: A
content: $$356$$ 
aoVal: B
content: $$57$$ 
aoVal: C
content: $$22$$ 
aoVal: D
content: $$21$$ 
aoVal: E
content: $$24$$ 
Answer: D
Analysis: $3+5+6+7=21$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Addition and Subtraction ->Addition of Whole Numbers->Addition in Horizontal Form
Problem: Sreshtha needs to estimate the quantity $\frac{a}{b}-c$, where $a, b$, and $c$ are large positive integers. She rounds each of the integers so that the calculation will be easier to do mentally. In which of these situations will her answer necessarily be greater than the exact value of $\frac{a}{b}-c ?$ ．(2015 AMC 12A Problem, Question \#5) 
Options:
aoVal: A
content: She rounds all three numbers up. 
aoVal: B
content: She rounds $a$ and $b$ up, and she rounds $c$ down. 
aoVal: C
content: She rounds $a$ and $c$ up, and she rounds $b$ down. 
aoVal: D
content: She rounds $a$ up, and she rounds $b$ and $c$ down. 
aoVal: E
content: She rounds $c$ up, and she rounds $a$ and $b$ down. 
Answer: D
Analysis: To maximize our estimate, we want to maximize $\frac{a}{b}$ and minimize $c$, because both terms are positive values. Therefore we round $c$ down. To maximize $\frac{a}{b}$, round $a$ up and $b$ down. $\Rightarrow(\mathbf{D})$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: Let $a, b$, and $c$ be three distinct one-digit numbers. What is the maximum value of the sum of the roots of the equation $(x-a)(x-b)+(x-b)(x-c)=0$? 
Options:
aoVal: A
content: $$15$$ 
aoVal: B
content: $$15.5$$ 
aoVal: C
content: $$16$$ 
aoVal: D
content: $$16.5$$ 
aoVal: E
content: $$17$$ 
Answer: D
Analysis: Expanding the equation and combining like terms results in $2 x^{2}-(a+2 b+c) x+(a b+b c)=0$. By Vieta\textquotesingle s formula the sum of the roots is $\frac{-[-(a+2 b+c)]}{2}=\frac{a+2 b+c}{2}$. To maximize this expression we want $b$ to be the largest, and from there we can assign the next highest values to $a$ and $c$. So let $b=9, a=8$, and $c=7$. Then the answer is $\frac{8+18+7}{2}=16.5$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Unary Quadratic Equations
Source: 其它
Problem: $$10+11+12+13+14=\left(30+31+32+33+34\right)-$$． 
Options:
aoVal: A
content: $$5$$ 
aoVal: B
content: $$20$$ 
aoVal: C
content: $$3\times 20$$ 
aoVal: D
content: $$5\times 20$$ 
Answer: D
Analysis: Add five $$20$$\textquotesingle s to $$10+11+12+13+14$$ to get $$30+31+32+33+34$$. The answer is $$\text{D}$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Addition and Subtraction
Problem: Which of the following is the correct order of the fractions $$\dfrac{15}{11}$$, $$\dfrac{19}{15}$$ and $$\dfrac{17}{13}$$, from least to greatest? ($$2019$$ AMC $$8$$ Problem, Question \#$$3$$) 
Options:
aoVal: A
content: $$\dfrac{15}{11}\textless{} \dfrac{17}{13}\textless{} \dfrac{19}{15}$$ 
aoVal: B
content: $$\dfrac{15}{11}\textless{} \dfrac{19}{15}\textless{} \dfrac{17}{13}$$ 
aoVal: C
content: $$\dfrac{17}{13}\textless{} \dfrac{19}{15}\textless{} \dfrac{15}{11}$$ 
aoVal: D
content: $$\dfrac{19}{15}\textless{} \dfrac{15}{11}\textless{} \dfrac{17}{13}$$ 
aoVal: E
content: $$\dfrac{19}{15}\textless{} \dfrac{17}{13}\textless{} \dfrac{15}{11}$$ 
Answer: E
Analysis: Method $$1$$:  Each one is $$x+\dfrac{4}{x}$$ so we are really comparing $$\dfrac{4}{11}$$, $$\dfrac{4}{15}$$, and $$\dfrac{4}{13}$$, where you can see $$\dfrac{4}{11}$$ $$\textgreater$$ $$\dfrac{4}{13}$$ $$\textgreater$$ $$\dfrac{4}{15}$$. So, $$\dfrac{19}{15}\textless{} \dfrac{17}{13}\textless{} \dfrac{15}{11}$$.  Method $$2$$:  We take a common denominator:  $$\dfrac{15}{11}, \dfrac{19}{15}, \dfrac{17}{13}= \dfrac{15 \cdot 15 \cdot 13}{11 \cdot 15 \cdot 13}, \dfrac{19 \cdot 11 \cdot 13}{15 \cdot 11 \cdot 13}, \dfrac{17 \cdot 11 \cdot 15}{13 \cdot 11 \cdot 15}= \dfrac{2925}{2145}, \dfrac{2717}{2145}, \dfrac{2805}{2145}$$.  $$2717 \textless{} 2805\textless{} 2925$$. So, $$\dfrac{19}{15}\textless{} \dfrac{17}{13}\textless{} \dfrac{15}{11}$$.  Method $$3$$:  When $$\frac{x}{y}\textgreater1$$, $$\frac{x+z}{y+z}\textless\frac{x}{y}$$.  Hence, $$\dfrac{19}{15}\textless{} \dfrac{17}{13}\textless{} \dfrac{15}{11}$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Comparing, Ordering and Estimating->Comparing and Ordering
Problem: $$1+3\frac{1}{6}+5\frac{1}{12}+7\frac{1}{20}+9\frac{1}{30}+11\frac{1}{42}=$$． 
Options:
aoVal: A
content: $$36\frac{5}{14}$$ 
aoVal: B
content: $$25\frac{5}{14}$$ 
aoVal: C
content: $$36\frac{1}{3}$$ 
aoVal: D
content: $$25\frac{1}{3}$$ 
Answer: A
Analysis: $$\begin{eqnarray}\&\&1+3\frac{1}{6}+5\frac{1}{12}+7\frac{1}{20}+9\frac{1}{30}+11\frac{1}{42}\textbackslash\textbackslash{} \&=\&1+3+5+7+9+11+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\textbackslash\textbackslash{} \&=\&\left[ (1+9)+(3+7)+(5+11) \right]+\left( \frac{1}{2}-\frac{1}{3} \right)+\left( \frac{1}{3}-\frac{1}{4} \right)+\left( \frac{1}{4}-\frac{1}{5} \right)+\left( \frac{1}{5}-\frac{1}{6} \right)+\left( \frac{1}{6}-\frac{1}{7} \right)\textbackslash\textbackslash{} \&=\&36+\left( \frac{1}{2}-\frac{1}{7} \right)\textbackslash\textbackslash{} \&=\&36\frac{5}{14}\end{eqnarray}$$  $$\text{A}$$． 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Fractions
Problem: For real numbers $w$ and $z$, $$ \frac{\frac{1}{w}+\frac{1}{z}}{\frac{1}{w}-\frac{1}{z}}=2014 $$ What is $\frac{w+z}{w-z} ?$ 
Options:
aoVal: A
content: - 2014 
aoVal: B
content: $\frac{-1}{2014}$ 
aoVal: C
content: $\frac{1}{2014}$ 
aoVal: D
content: $$1$$ 
aoVal: E
content: $$2014$$ 
Answer: A
Analysis: Muliply both sides by $\left(\frac{1}{w}-\frac{1}{z}\right)$ to get $\frac{1}{w}+\frac{1}{z}=2014\left(\frac{1}{w}-\frac{1}{z}\right)$. Then, add $2014 \cdot \frac{1}{z}$ to both sides and subtract $\frac{1}{w}$ from both sides to get $2015 \cdot \frac{1}{z}=2013 \cdot \frac{1}{w}$. Then, we can plug in the most simple values for $z$ and $w(2015$ and 2013 , respectively), and find $\frac{2013+2015}{2013-2015}=\frac{2(2014)}{-2}=-2014$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Proportional Equations
Source: 其它
Problem: In a dining room, there are $$15$$ chairs, $$5$$ tables, and $$20$$ cups. What is the ratio of chairs to cups in the simplest form? 
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: 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 $7: 6$, and the the ratio of $8^{\text {th }}$-graders to $7^{\text {th }}$ graders is $4: 7$. What is the smallest number of students that could be participating in the project? (2013 AMC 8, Question 16) 
Options:
aoVal: A
content: $$21$$ 
aoVal: B
content: $$47$$ 
aoVal: C
content: $$56$$ 
aoVal: D
content: $$94$$ 
aoVal: E
content: $$100$$ 
Answer: B
Analysis: We multiply the first ratio by 2 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: $$ \begin{aligned} \&7: 6=7(2): 6(2)=14: 12 \textbackslash\textbackslash{} \&4: 7=4(3): 7(3)=12: 21\end{aligned} $$ Therefore, the ratio of 8th graders to 7th graders to 6th graders is $14: 12: 21$. Since the ratio is in lowest terms, the smallest number of students participating in the project is $$ 14+12+21=\text { (B) } 47 $$  Notice if you get $$28:24:42$$, you have to simplfy to $14: 12: 21$ for the smallest number of students. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Ratios and Proportions
Source: 其它