math-eval/TAL-SCQ5K huggingface dataset:
Problem: When one ounce of water is added to a mixture of acid and water, the new mixture is $$20\textbackslash\%$$ acid. When one ounce of acid is added to the new mixture, the result is $$33\frac{1}{3}\textbackslash\%$$ acid. The percentage of acid in the original mixture is ($$1973$$ AHSME Problem, Question \#$$33$$) 
Options:
aoVal: A
content: $$22\textbackslash\%$$ 
aoVal: B
content: $$24\textbackslash\%$$ 
aoVal: C
content: $$25\textbackslash\%$$ 
aoVal: D
content: $$30\textbackslash\%$$ 
aoVal: E
content: $$33\frac{1}{3}\textbackslash\%$$ 
Answer: C
Analysis: Let $$a$$ be the original number of ounces of acid and $$w$$ be the original number of ounces of water. We can write two equations $$\frac{a}{a+w+1}=\frac{1}{5}$$ and $$\frac{a+1}{a+w+2}=\frac{1}{3}$$. Then cross-multiply to get rid of the fractions. $$5a=a+w+1$$, $$3a+3=a+w+1$$ Solve the system to get $$a=1$$ and $$w=3$$. The percentage of acid in the original mixture is $$\frac{1}{1+3}=25 \textbackslash\% $$. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Concentration Problems->Basic Concentration Problems
Problem: Brandon is engaging sprint race on the track. He is in the fifth place now, and there are $8$ players behind him. There are $5$ extra groups signing up for sprint race, and the number of players of each group are idential. How many sprint players are there in total? 
Options:
aoVal: A
content: $$65$$ 
aoVal: B
content: $$70$$ 
aoVal: C
content: $$78$$ 
aoVal: D
content: $$84$$ 
aoVal: E
content: $$96$$ 
Answer: D
Analysis: $(5 + 1 + 8) \times (5 +1) = 84$ 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules
Source: 其它
Problem: A shop advertises everything is "half price in today\textquotesingle s sale." In addition, a coupon gives a $25 \textbackslash\%$ discount on sale prices. Using the coupon, the price today represents what percentage off the original price? (adapted from 2012 AMC 8, Question \#8) 
Options:
aoVal: A
content: $$40\textbackslash\%$$ 
aoVal: B
content: $$42.5\textbackslash\%$$ 
aoVal: C
content: $$57.5\textbackslash\%$$ 
aoVal: D
content: $$62.5\textbackslash\%$$ 
aoVal: E
content: $$70\textbackslash\%$$ 
Answer: D
Analysis: $75 \textbackslash\% \cdot 50\textbackslash\% =37.5\textbackslash\%$  $1 -37.5\textbackslash\% = 62.5\textbackslash\%$ 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Solving Problems Involving Fractions and Percentages->Percentages Word Problems
Source: 其它
Problem: James can frost a cookie every $$12$$ seconds and Lacey can frost a cookie every $$15$$ seconds. Working together, how many cookies can they frost in $$10$$ minutes? (Adapted from $$2012$$ AMC $$10\rm A$$ Problem, Question \#$$11$$) 
Options:
aoVal: A
content: $$50$$ 
aoVal: B
content: $$60$$ 
aoVal: C
content: $$80$$ 
aoVal: D
content: $$90$$ 
aoVal: E
content: $$100$$ 
Answer: D
Analysis: In $$600$$ seconds ($$10$$ minutes), James will frost $$\dfrac{600}{12}=50$$ cookies, and Lacey will frost $$\dfrac{600}{15}=40$$ cookies. Therefore, working together they will frost $$50+40=\boxed{(\text{D})90}$$ cookies. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Work Word Problems->Collaborative Work Word Problems
Problem: Judy ran for $$16$$ minutes. If her average speed was $$80$$ meters per minute, then how many meters did Judy run? 
Options:
aoVal: A
content: $$800$$ 
aoVal: B
content: $$5$$ 
aoVal: C
content: $$1280$$ 
aoVal: D
content: $$96$$ 
Answer: C
Analysis: $$80\times 16 = 1280$$ 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules
Problem: The length of a rectangle is increased by $10 \textbackslash\%$ and the width is decreased by $10 \textbackslash\%$. What percent of the old area is the new area? (2009 AMC 8, Question \#8) 
Options:
aoVal: A
content: $$90$$ 
aoVal: B
content: $$99$$ 
aoVal: C
content: $$100$$ 
aoVal: D
content: $$101$$ 
aoVal: E
content: $$110$$ 
Answer: B
Analysis: In a rectangle with dimensions $10 \times 10$, the new rectangle would have dimensions $11 \times 9$. The ratio of the new area to the old area is $99 / 100=(\text{B}) 99$ 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Solving Problems Involving Fractions and Percentages->Percentages Word Problems
Source: 其它
Problem: The price ratio of goods $$A$$ to $$B$$ is $$7:2$$. If their prices were increased by $$\textbackslash$60$$ respectively, the price ratio would be $$5:2$$. What are the prices of these two goods? 
Options:
aoVal: A
content: $$\textbackslash$315$$, $$\textbackslash$90$$ 
aoVal: B
content: $$\textbackslash$320$$, $$\textbackslash$100$$ 
aoVal: C
content: $$\textbackslash$330$$, $$\textbackslash$120$$ 
aoVal: D
content: $$\textbackslash$350$$, $$\textbackslash$200$$ 
Answer: A
Analysis: The price difference between the two goods did not change.  Before $\to$ $$A:B:Difference$$ is $$7:2:5$$  After $\to$ $$A:B:Difference$$ is $$5:2:3$$  $\textasciitilde$  Make difference the same number of units in both ratios by finding LCM.  Before $\to$ $$A:B:Difference$$ is $$21:6:15$$  After $\to$ $$A:B:Difference$$ is $$25:10:15$$  $\textasciitilde$  Therefore,  $$4u=\textbackslash$60$$  $$1u=$$ $$\textbackslash$60\div4=\textbackslash$15$$  $\textasciitilde$  The original prices of these two goods were  $$A=21u=21\times15=\textbackslash$315$$ and  $$B=6u=6\times15=\textbackslash$90$$. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Solving Problems Involving Proportions->Complex Ratio Problems
Problem: Ela and Kasia boarded a super-train. Ela took a seat in the seventeenth car counting from the front of the train, and Kasia was seated in the thirty-fourth car counting from the end. The girls were sitting in the same car. How many cars did the super-train have? (2001 Math Kangaroo Problem, Level 3-4, Question \#13) 
Options:
aoVal: A
content: $$48$$ 
aoVal: B
content: $$49$$ 
aoVal: C
content: $$50$$ 
aoVal: D
content: $$51$$ 
aoVal: E
content: $$52$$ 
Answer: C
Analysis: $17 + 34 - 1 = 50$ 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Problems Involving Queuing->Finding the Total Number According to the Position of Two Characters in a Line
Problem: James had $$10000$$ dollars, and he deposited ＄$$7000$$ in Bank $$A$$ and ＄$$3000$$ in Bank $$B$$. At the end of the year, James received an interest of＄$$460$$ in total. Given that the sum of the annual interest rates of the two banks is $$10\textbackslash\%$$, find the interest rates of the two banks, respectively. 
Options:
aoVal: A
content: $$2\textbackslash\%$$; $$8\textbackslash\%$$. 
aoVal: B
content: $$3\textbackslash\%$$; $$7\textbackslash\%$$. 
aoVal: C
content: $$4\textbackslash\%$$; $$6\textbackslash\%$$. 
aoVal: D
content: $$5\textbackslash\%$$; $$5\textbackslash\%$$. 
Answer: C
Analysis: Let $$x$$ be the interest rate of bank $$A$$:  $$7000x+3000(10\textbackslash\%-x)=460$$  $$x=4\textbackslash\%$$. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Word Problems in Economics->Interest Problems
Problem: Peter has a rope, and he ties some knots, each using $10$-cm rope. He ties $25$ knots in total. After he ties the last knot, the knots divide the rope into pieces~ equally, and the length of each piece is $7$ cm. What is the length of the entire rope? 
Options:
aoVal: A
content: $$425$$ cm 
aoVal: B
content: $$432$$ cm 
aoVal: C
content: $$456$$ cm 
aoVal: D
content: $$478$$ cm 
aoVal: E
content: $$480$$ cm 
Answer: B
Analysis: $25\times10+(25+1)\times7=432$ 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Interval Problems
Source: 其它
Problem: The scale of a map is:~$\dfrac{3}{4}$ of an inch $$=10$$ miles. If the distance on the map between two towns is $$12$$ inches, the actual distance between the towns is． 
Options:
aoVal: A
content: $$90$$ miles 
aoVal: B
content: $$120$$ miles 
aoVal: C
content: $$150$$ miles 
aoVal: D
content: $$160$$ miles 
Answer: D
Analysis: Since $$12 \div \frac{3}{4}=16$$, the actual distance between the towns is $$16\times10 = 160$$ miles. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Solving Problems Involving Proportions->Ratio Word Problems with an Invariant Part
Problem: Gawel lives on the second floor, and Pawel lives in the same building but has to walk up twice as many stairs as Gawel. There are no stairs to the entrance of the building. On which floor does Pawel live? (2000 Math Kangaroo Problem, Level 3-4, Question \#4) 
Options:
aoVal: A
content: on the $2$\textsuperscript{nd} floor 
aoVal: B
content: on the $3$\textsuperscript{rd} floor 
aoVal: C
content: on the $4$\textsuperscript{th} floor 
aoVal: D
content: on the $5$\textsuperscript{th} floor 
aoVal: E
content: on the $6$\textsuperscript{th} floor 
Answer: B
Analysis: Gawel lives on the second floor, so he only needs to walk up one floor.  Pawel has to walk up twice as many stairs as Gawel, so he needs to walk up two floors. So, Pawel lives on the $3$\textsuperscript{rd} floor. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Interval Problems
Source: 其它
Problem: I ate half an apple pie on Saturday and two thirds of the remainder on Sunday. What fraction of the pie was left for Monday?　 
Options:
aoVal: A
content: None　 
aoVal: B
content: $$\frac{1}{2}$$ 
aoVal: C
content: $$\frac{1}{3}$$ 
aoVal: D
content: $$\frac{2}{3}$$ 
aoVal: E
content: $$\frac{1}{6}$$ 
Answer: E
Analysis: After I eat one half, half of the apple pie is left. Eating two thirds of this half leaves one third of one half of the pie, which is one sixth, for Monday. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Solving Problems Involving Fractions and Percentages->Understanding the Base
Problem: The length of Amy\textquotesingle s string is $$35\text{cm}$$. The length of David\textquotesingle s string is $$29\text{cm}$$. What is the total length of both strings? 
Options:
aoVal: A
content: $$54\text{cm}$$ 
aoVal: B
content: $$49\text{cm}$$ 
aoVal: C
content: $$64\text{cm}$$ 
aoVal: D
content: $$65\text{cm}$$ 
aoVal: E
content: $$66\text{cm}$$ 
Answer: C
Analysis: $$35+29=64$$ 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Applying Addition and Subtraction->Simple Addition and Subtraction
Problem: Cici bought $$12$$ pencils and notebooks in total for $$66$$. Each pencil cost $$2$$. Each notebook costs $8$. How many notebooks did Cici buy? 
Options:
aoVal: A
content: $$5$$ 
aoVal: B
content: $$6$$ 
aoVal: C
content: $$7$$ 
aoVal: D
content: $8$ 
aoVal: E
content: $$10$$ 
Answer: C
Analysis: $$12\times2=24$$  $$8-2=6$$  $$66-24=42$$  $$42\div6=7$$ 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Chicken-Rabbit Problems->Solving Chicken-Rabbit Problems by Using Hypothesis
Problem: There are oranges, apricots and peaches in a big basket. How many fruits are there in the basket if the peaches and the apricots together are 18, the oranges and the apricots together are 28 and 30 fruits are not apricots? 
Options:
aoVal: A
content: $$46$$ 
aoVal: B
content: $$20$$ 
aoVal: C
content: $$40$$ 
aoVal: D
content: $$38$$ 
aoVal: E
content: $$29$$ 
Answer: D
Analysis: If we sum up the number of A and P, the result is 18.  But if we sum up A and O, result is 28 instead.  Hence, there must be 10 more O than P.  Note that there are a total of 30 I and P.  Therefore, there are (30+10)/2 = 20 Oranges.  Total = 20+18=38. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Unitary Method Problems->Unitary Method with Two Variables
Source: 其它
Problem: Suppose $15 \textbackslash\%$ of $x$ equals $20 \textbackslash\%$ of $y$. What percentage of $x$ is $y$ ? (2020 AMC 8, Question 15) 
Options:
aoVal: A
content: $$5$$ 
aoVal: B
content: $$35$$ 
aoVal: C
content: $$75$$ 
aoVal: D
content: $133\frac{1}{3}$ 
aoVal: E
content: $$300$$ 
Answer: C
Analysis: Since $20 \textbackslash\%=\frac{1}{5}$, multiplying the given condition by 5 shows that $y$ is $15 \cdot 5=(\mathbf{C}) 75$ percent of $x$. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Solving Problems Involving Fractions and Percentages->Percentages Word Problems
Source: 其它
Problem: Five people took the same test and got an average score of $80$. Four of the people scored $80$, $85$, $82$, and $90$, respectively. What is the score of the remaining person? 
Options:
aoVal: A
content: $$80$$ 
aoVal: B
content: $$76$$ 
aoVal: C
content: $$73$$ 
aoVal: D
content: $$63$$ 
aoVal: E
content: $$56$$ 
Answer: D
Analysis: There are $0+5+2+10=17$ points more than the average, so the last score is $80-17=63$. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Average Problems 
Source: 其它
Problem: It is $4:00$ now. How many minutes does it take for the clock\textquotesingle s hour and minute hands to overlap for the first time? 
Options:
aoVal: A
content: $43\frac 7{11}$ minutes 
aoVal: B
content: $10\frac {10}{11}$ minutes 
aoVal: C
content: $21\frac 9{11}$ minutes 
aoVal: D
content: $22\frac 9{11}$ minutes 
aoVal: E
content: $42\frac 7{11}$ minutes 
Answer: C
Analysis: $(30\times 4)\div (6-0.5)=120\div 5.5=\frac {240}{11}=21\frac 9{11}$ minutes. 
Knowledge points: Overseas Competition->Knowledge Point->Distance Word Problems->Clock Problems
Problem: Chloe and Zoe are both students in Ms. Demeanor\textquotesingle s math class. Last night they each solved half of the problems in their homework assignment alone and then solved the other half together. Chloe had correct answers to only $80 \textbackslash\%$ of the problems she solved alone, but overall $88 \textbackslash\%$ of her answers were correct. Zoe had correct answers to $90 \textbackslash\%$ of the problems she solved alone. What was Zoe\textquotesingle s overall percentage of correct answers? (2017 AMC 8, Question \#14) 
Options:
aoVal: A
content: $$89$$ 
aoVal: B
content: $$92$$ 
aoVal: C
content: $$93$$ 
aoVal: D
content: $$96$$ 
aoVal: E
content: $$98$$ 
Answer: C
Analysis: Assume the total amount of problems is 100 per half homework assignment, since we are dealing with percentages, and not values. Then, we know that Chloe got 80 problems correct by herself, and got 176 problems correct overall. We also know that Zoe had 90 problems she did correct alone. We can see that the total amount of correct problems Chloe and Zoe did was $176-80=96$. Therefore Zoe has $96+90=186$ problems out of 200 problems correct. This is (C) $93$ percent. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Solving Problems Involving Fractions and Percentages->Percentages Word Problems
Source: 其它
Problem: Three boxes under my stairs contain apples or pears or both. Each box contains the same number of pieces of fruit. The first box contains all twelve of the apples and one-ninth of the pears. How many pieces of fruit are there in each box? 
Options:
aoVal: A
content: $$14$$ 
aoVal: B
content: $$16$$ 
aoVal: C
content: $$18$$ 
aoVal: D
content: $$20$$ 
aoVal: E
content: $$36$$ 
Answer: B
Analysis: $(1-\frac19)\div2=\frac49$, $\frac49-\frac19=\frac13$, $12\div\frac13=36$, $36\times\frac49=16$. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Equation Word Problems->Word Problems for Linear Equations with One Variable
Problem: Duckings are walking in a queue when they are following the mother duck. Joey the duckling is the 6th from the front and 8th from the back (including the mother duck). Some of the ducklings at the back moved away from the queue making Joey the middle duckling in the queue. How many ducklings moved away from the queue? 
Options:
aoVal: A
content: $$2$$ 
aoVal: B
content: $$3$$ 
aoVal: C
content: $$4$$ 
aoVal: D
content: $$5$$ 
aoVal: E
content: None of the above. 
Answer: A
Analysis: total number of duck and ducklings in the queue is 6+(8-1) = 13.  if 2 ducklings at the back moved away, it would be 13-2=11 in the queue, and the middle ducking would be 6th from the front, which is Joey\textquotesingle s position. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Applying Addition and Subtraction->Simple Addition and Subtraction
Source: 其它
Problem: Every week, Robyn swims while Sam sings.  For every $$2$$ of Sam\textquotesingle s songs, Robyn swims $$9$$ lengths.  If Robyn swims $$54$$ lengths, how many songs does Sam sing? 
Options:
aoVal: A
content: $$6$$ 
aoVal: B
content: $$8$$ 
aoVal: C
content: $$10$$ 
aoVal: D
content: $$12$$ 
aoVal: E
content: $$18$$ 
Answer: D
Analysis: Sam sings $$2$$ songs in the time that Robyn swims $$9$$ lengths. The time that she takes for $$54$$ lengths is $$54\div9 =6$$ times longer, so Sam will sing $$2\times6 =12$$ songs. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Unitary Method Problems->Applying Multiplication and Division
Problem: If today is a Wednesday, then $$22$$ days ago was a． 
Options:
aoVal: A
content: Tuesday 
aoVal: B
content: Wednesday 
aoVal: C
content: Thursday 
aoVal: D
content: Friday 
Answer: A
Analysis: Today is Wednesday. $$21$$ days ago was Wednesday. $$22$$ days ago was Tuesday.  $\textasciitilde$  or  $\textasciitilde$  $$22\div7=3$$ weeks $$1$$ day.  $$1$$ day before Wednesday was a Tuesday. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Periodic Problems->Periodic Problems of Time
Problem: John drinks $x$ cups of boba everyday on weekdays and $1$ cups of boba everyday on the weekends. Which of the following equations represents how many cups of boba John drinks every week? 
Options:
aoVal: A
content: $x+2$ 
aoVal: B
content: $5x+2$ 
aoVal: C
content: $5x+1$ 
aoVal: D
content: $5x-1$ 
Answer: B
Analysis: The boba that John drinks during weekdays: $$5x$$  The boba that John drinks during weekends: $$2\times 1 =2$$  The total boba that John drinks during a week: $$5x+2$$ 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Equation Word Problems
Source: 其它
Problem: This year Gareth is seven times the age of his son Herbie. In six years\textquotesingle{} time Gareth will be four times as old as Herbie. What is the difference between Gareth\textquotesingle s and Herbie\textquotesingle s ages this year? 
Options:
aoVal: A
content: $$16$$ years 
aoVal: B
content: $$20$$ years 
aoVal: C
content: $$24$$ years 
aoVal: D
content: $$36$$ years 
aoVal: E
content: $$42$$ years 
Answer: D
Analysis: Let Herbie\textquotesingle s age this year be $$h$$ years. Then Gareth\textquotesingle s age this year is $$7h$$ years. In $$6$$ years\textquotesingle{} time, their ages will be $$h+6$$ years and $$7h+6$$ years respectively. We are told that $$7h+6=$$$$4$$$$\left(h+6\right)$$. This gives $$7h+6=4h+24$$ so that $$3h=18$$ and $$h=6$$.  So the present ages are $$6$$ and $$7\times 6=42$$, giving a difference of $$36$$ years. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Age Problems->Basic Relationships in Age Problems
Problem: A piece of pasture can feed $$17$$ cow for $$30$$ days or $$19$$ cows for $$24$$ days. If the grass grows at the same speed every day, how many cows can it feed for $$6$$ days? ． 
Options:
aoVal: A
content: $$38$$ 
aoVal: B
content: $$45$$ 
aoVal: C
content: $$49$$ 
aoVal: D
content: $$52$$ 
Answer: C
Analysis: $17\times30=510$ $19\times 24=456$ $510-456=54$ $54\div(30-24)=9$  $510-30\times9=240$ $240+6\times9=294$ $294\div 6=49$ 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Newton's Problem of Cows and Fields
Source: 其它
Problem: The month of February of a given year has five Fridays. What day of the week is January $$31$$ of that year? 
Options:
aoVal: A
content: Tuesday 
aoVal: B
content: Thursday 
aoVal: C
content: Friday 
aoVal: D
content: Saturday 
Answer: B
Analysis: There are seven days in a week; so, in order to have five Fridays, a month has at least $$4\times7+1=29$$ days. Since February has at most $$29$$ days, the $$1^{\rm st}$$ and $$29^{\rm th}$$ days of this February must both be Fridays, which means that January $$31$$ is a Thursday. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Periodic Problems->Periodic Problems of Time
Problem: Several students are singing in a line. Amy is the fourth one from the left. From the right, Amy is the fourth too. How many students are singing? (adapted from $$2019$$ Math kangaroo Problems, Level $$1-2$$, Question \#$$5$$) 
Options:
aoVal: A
content: $5$ 
aoVal: B
content: $6$ 
aoVal: C
content: $7$ 
aoVal: D
content: $8$ 
aoVal: E
content: $9$ 
Answer: C
Analysis: It comes $4+4=8$ when we count Amy twice.  The number of all singing students is $4+4-1=7$ 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Applying Addition and Subtraction
Source: 其它
Problem: If six days ago was a Tuesday, then eight days from today is a． 
Options:
aoVal: A
content: Monday　 
aoVal: B
content: Tuesday　 
aoVal: C
content: Wednesday　 
aoVal: D
content: Thursday　 
Answer: B
Analysis: $$6$$ days ago was a Tuesday. Today is Monday. Thus, in $$8$$ days\textquotesingle{} time, it will be a Tuesday. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Periodic Problems->Periodic Problems of Time
Problem: In the sequence of letters $$\text{KANGAROOKANGAROOKANG}\cdots$$ the word $$\text{KANGAROO}$$ is repeated indefinitely. What is the $$2017\text{th}$$ letter in this sequence? 
Options:
aoVal: A
content: $$\text{K}$$ 
aoVal: B
content: $$\text{N}$$ 
aoVal: C
content: $$\text{G}$$ 
aoVal: D
content: $$\text{R}$$ 
aoVal: E
content: $$\text{O}$$ 
Answer: A
Analysis: The sequence $$\text{KANGAROOKANGAROOKANG}\cdots$$ repeats every $$8$$ letters. Since $$2017=8\times252+1$$, the $$2017\text{th}$$ letter in the sequence is the first of the repeating sequence and hence is $$\text{K}$$. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Periodic Problems->Periodic Problems of Basic Permutation
Problem: There are $$30$$ students in your class. The ratio of boys to girls is $$3:2$$. How many boys and girls are there?. 
Options:
aoVal: A
content: $$30$$ and $$20$$ 
aoVal: B
content: $$20$$ and $$30$$ 
aoVal: C
content: $$18$$ and $$12$$ 
aoVal: D
content: $$12$$ and $$18$$ 
Answer: C
Analysis: According to the ratio $$3:2$$, we could know the total is $$5$$. Then the number of boys is $$30\times \dfrac{3}{5} =18$$.The number of girls is $$30\times\dfrac{2}{5}=12$$. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Solving Problems Involving Proportions->Basic Units
Problem: A bag contains $$100$$ raffle tickets in red, white and blue. The ratio of red to white tickets is $$5:4$$ and the ratio of white to blue tickets is $$8:7$$. What is the smallest number of tickets that would have to be picked to make sure having selected at least half of \textbf{one} colour? 
Options:
aoVal: A
content: $$48$$ 
aoVal: B
content: $$49$$ 
aoVal: C
content: $$50$$ 
aoVal: D
content: $$51$$ 
aoVal: E
content: $$52$$ 
Answer: A
Analysis: First we must work out the number of tickets of each colour. Let the numbers of red, white and blue tickets be $$r$$, $$w$$ and b respectively; we have $$r : w = 5 : 4$$ and $$w : b = 8 : 7$$. Writing $$r : w$$ as $$10 : 8$$, we can combine the ratios to find $$r : w : b = 10 : 8 : 7$$. Hence for a total of $$100$$ tickets, there must be $$40$$ red, $$32$$ white and $$28$$ blue tickets. Now the largest number of red tickets one can select without having half of them is $$19$$, and likewise $$15$$ and $$13$$ for white and blue tickets. So it would be possible to select $$19 + 15 + 13 = 47$$ tickets and still not have half or more of any particular colour of ticket. However, the very next ticket, the $$48\text{th}$$, would have to be a red, a white or a blue one, and so one of the colours would now have at least half of its number selected. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Solving Problems Involving Proportions->Complex Ratio Problems