math-eval/TAL-SCQ5K huggingface dataset:
Problem: Suppose $a, b$, and $c$ are nonzero real numbers, and $a+b+c=0$. What are the possible value(s) for $\frac{a}{\textbar a\textbar}+\frac{b}{\textbar b\textbar}+\frac{c}{\textbar c\textbar}+\frac{a b c}{\textbar a b c\textbar} ?$ (2017 AMC 8 Problem, Question \#21) 
Options:
aoVal: A
content: $$0$$ 
aoVal: B
content: $1$ and $-1$ 
aoVal: C
content: $2$ and $-2$ 
aoVal: D
content: $0,2$ and $-2$ 
aoVal: E
content: $0,1$ and $-1$ 
Answer: A
Analysis: There are 2 cases to consider:  \textbf{Case 1:} 2 of $a, b$, and $c$ are positive and the other is negative. WLOG, we can assume that $a$ and $b$ are positive and $c$ is negative. In this case, we have that $$ \frac{a}{\textbar a\textbar}+\frac{b}{\textbar b\textbar}+\frac{c}{\textbar c\textbar}+\frac{a b c}{\textbar a b c\textbar}=1+1-1-1=0 \text {. } $$  \textbf{Case 2}: 2 of $a, b$, and $c$ are negative and the other is positive. Without loss of generality, we can assume that $a$ and $b$ are negative and $c$ is positive. In this case, we have that $$ \frac{a}{\textbar a\textbar}+\frac{b}{\textbar b\textbar}+\frac{c}{\textbar c\textbar}+\frac{a b c}{\textbar a b c\textbar}=-1-1+1+1=0 \text {. } $$  Note these are the only valid cases, for neither 3 negatives nor 3 positives would work, as they cannot sum up to 0 . In both cases, we get that the given expression equals $(\mathbf{A}) 0$. 
Knowledge points: Overseas Competition->Knowledge Point->Combinatorics->Mathematical Thoughts->Absolute Value
Source: 其它
Problem: The numbers $$2$$, $$4$$, $$6$$, $$8$$, $$9$$ and $$11$$ are divided into groups of one or more numbers. The sum of the numbers in each group is the same. Which number must be in the group with the number $$11$$? 
Options:
aoVal: A
content: $$2$$ 
aoVal: B
content: $$4$$ 
aoVal: C
content: $$6$$ 
aoVal: D
content: $$8$$ 
aoVal: E
content: $$9$$ 
Answer: E
Analysis: $2+4+6+8+9+11=40$  $2$ groups: sum of each group is $20$, $20=9+11=2+4+6+8$ 
Knowledge points: Overseas Competition->Knowledge Point->Combinatorics->Combinatorics Involving Extreme Values->Extreme Value with Fixed Sums
Source: 其它
Problem: Joy writes a four-digit number on a piece of paper and let Sandy guess.  $\textasciitilde$  Sandy asks: "Is it $$6031$$?"  Joy says: "You guess $1$ digit correct and it\textquotesingle s in the correct place."  Sandy asks: "Is it $$5672$$?"  Joy says: "You guess $2$ digits correct and they\textquotesingle re in the wrong places."  Sandy asks: "Is it $$4796$$?"  Joy says: "You guess $4$ digits correct and they\textquotesingle re in the wrong places."  $\textasciitilde$  According to the conversation above, what number does Joy write? 
Options:
aoVal: A
content: $$4967$$ 
aoVal: B
content: $$9764$$ 
aoVal: C
content: $$9647$$ 
aoVal: D
content: $$6497$$ 
aoVal: E
content: $$6947$$ 
Answer: E
Analysis: The number is formed by $4,7,9,6$ according to last question. Thus, according to the first question, $6$ should be in the thousands place. From the second and last question, $7$ is in ones place, $4$ is in tens place, and $9$ is in hundreds place. 
Knowledge points: Overseas Competition->Knowledge Point->Combinatorics->Logical Reasoning
Problem: Thirty apples are distributed among 4 children and each of them can get at least one apple. For the kid who gets the most apples, what\textquotesingle s the least possible number of apples he or she can get? ． 
Options:
aoVal: A
content: $7$ 
aoVal: B
content: $8$ 
aoVal: C
content: $9$ 
aoVal: D
content: $10$ 
aoVal: E
content: $11$ 
Answer: B
Analysis: $$30\div 4=7 \textbackslash{} \text{R}2$$, thus for the kid who gets the most apples, he or she can get $$7+1=8$$ apples at least. 
Knowledge points: Overseas Competition->Knowledge Point->Combinatorics->Pigeonhole Principle->Simple Pigeonhole Principle Problems
Source: 其它
Problem: Let\textquotesingle s convert!  $$1$$ hour $$45$$ minutes $$=$$~\uline{~~~~~~~~~~}~minutes 
Options:
aoVal: A
content: $$145$$ 
aoVal: B
content: $$105$$ 
aoVal: C
content: $$95$$ 
aoVal: D
content: $$75$$ 
aoVal: E
content: $$65$$ 
Answer: B
Analysis: $60+45=105$ minutes. 
Knowledge points: Overseas Competition->Knowledge Point->Combinatorics->Time Problem
Problem: Which of the following calculations shows how to work out the number of seconds in~~24 days? 
Options:
aoVal: A
content: 24 x~24~ 
aoVal: B
content: 24 x~60 
aoVal: C
content: 24 x 24 x 60 
aoVal: D
content: 24 x 60 x 60 
aoVal: E
content: 24 x~24 x~60 x~60 
Answer: E
Analysis: 1 day = 24 hours，$$1$$ hour = 60 minutes，$$1$$ minute = 60 seconds，$$24$$ days = (24 \times 24) hours = [(24\times 24) \times 60] minutes = 24\times 24\times 60\times 60 seconds 
Knowledge points: Overseas Competition->Knowledge Point->Combinatorics->Time Problem
Source: 其它
Problem: For any two numbers $$x$$ and $$y$$, the special operation $$^{}\otimes$$ is defined as follow $$x^{}\otimes y=x+y-\frac {2012}{335}$$.  Calculate $$\frac {2^{}\otimes 4^{}\otimes 6^{}\otimes \cdots ^{}\otimes 2010^{}\otimes 2012}{2012}$$. 
Options:
aoVal: A
content: $$1006$$ 
aoVal: B
content: $$1005$$ 
aoVal: C
content: $$503$$ 
aoVal: D
content: $$502$$ 
aoVal: E
content: None of the above 
Answer: E
Analysis: E 
Knowledge points: Overseas Competition->Knowledge Point->Combinatorics->Patterns of Figures->Special Changes
Source: 其它
Problem: A train started from the starting station at 7:30, it takes 8 hours and 37 minutes to arrive at the terminal station. When will it arrive? 
Options:
aoVal: A
content: $$3:30$$ A.M. 
aoVal: B
content: $$3:37$$ P.M. 
aoVal: C
content: $$4:37$$ P.M. 
aoVal: D
content: $$4:07$$ P.M. 
aoVal: E
content: $$4:37$$ A.M. 
Answer: D
Analysis: $$7:30$$+$$8$$h$$37$$m=$$16:07$$ 
Knowledge points: Overseas Competition->Knowledge Point->Combinatorics->Time Problem->Time Calculation
Problem: The weights of four boys are $45\rm{kg}$, $48\rm{kg}$, $52\rm{kg}$ and $53\rm{kg}$. Mason\textquotesingle s weight is an even number. Joshua\textquotesingle s weight is a multiple of $5$. Christopher is not the heaviest and Mateo is not the lightest. Who is the heaviest among the four boys? 
Options:
aoVal: A
content: Mason 
aoVal: B
content: Joshua 
aoVal: C
content: Christopher 
aoVal: D
content: Mateo 
aoVal: E
content: Impossible to determine 
Answer: D
Analysis: Mason\textquotesingle s weight is an even number = 48kg / 52kg.  Joshua\textquotesingle s weight is a multiple of 5 = 45kg.  Christopher is not the heaviest = x 53kg.  Mateo is not the lightest = x 45kg  Jason - \textgreater{} either Mason or Christpoher is 28/52kg -\textgreater{} Mateo. 
Knowledge points: Overseas Competition->Knowledge Point->Combinatorics->Logical Reasoning
Source: 其它
Problem: Math workshop starts at $5:00$ PM. Today Allan was $15$ minutes late to the workshop. What time did Allan come? (2010 Math Kangaroo Problem, Level 1--2, Question \#12) 
Options:
aoVal: A
content: $5:00$ PM 
aoVal: B
content: $5:05$ PM 
aoVal: C
content: $5:15$ PM 
aoVal: D
content: $5:20$ PM 
Answer: C
Analysis: $15$ minutes past $5:00$ PM is $5:15$ PM. 
Knowledge points: Overseas Competition->Knowledge Point->Combinatorics->Time Problem
Source: 其它
Problem: Peter can draw a cartoon in $30$ minutes. How many hours does it take to paint four cartoons?~(adapted from 2011 Math kangaroo Problems, Level 1-2, Question \#4) 
Options:
aoVal: A
content: $$2$$ 
aoVal: B
content: $$3$$ 
aoVal: C
content: $$4$$ 
aoVal: D
content: $$5$$ 
aoVal: E
content: $$6$$ 
Answer: A
Analysis: There are $60$ minutes in an hour.  It will take $120$ minutes to draw $4$ cartoons.~  $120$ minutes=$2$ hours. 
Knowledge points: Overseas Competition->Knowledge Point->Combinatorics->Time Problem->Time Calculation
Source: 其它
Problem: Bob and Linda went mountain climbing together. They climbed the mountain from $8$:$30$. Bob reached the top at $14$:$00$, and Linda was $1$ hour and $40$ minutes behind him. How long did it take Linda climb?~(adapted from 2003 Math Kangaroo Problem, Level 3 - 4, Question \#10) 
Options:
aoVal: A
content: $3$ hr $30$ min 
aoVal: B
content: $4$ hr $30$ min 
aoVal: C
content: $7$ hr $10$ min 
aoVal: D
content: $6$ hr $20$ min 
aoVal: E
content: $8$ hr $40$ min 
Answer: C
Analysis: It lasted $3$ hours and $30$ minutes from $8$:$30$ to $12$:$00$. It lasted $2$ hours from $12$:$00$ to $14$:$00$. So Bob took $5$ hours and $30$ minutes. Linda was $1$ hour and $40$ minutes behind Bob. Linda\textquotesingle s time was $7$ hours and $10$ minutes.  Bob: $3$h+$30$min+$2$h=$5$h$30$min  Linda:$3$h+$30$min+$2$h+$1$h+$40$min=$7$h$10$min 
Knowledge points: Overseas Competition->Knowledge Point->Combinatorics->Time Problem->Time Calculation
Problem: A certain play has three $30$-minute parts and two intermission among them. The play started at $8:30$ and ended at $10:15$. How many minutes long were the intermissions in total? 
Options:
aoVal: A
content: $$10$$ 
aoVal: B
content: $$15$$ 
aoVal: C
content: $$20$$ 
aoVal: D
content: $$25$$ 
Answer: B
Analysis: From $8:30$ to $10:15$ = $1$ hr $45$ min, $1$ hr $45$ min = $105$ min, three $30$-minute parts = $90$ min, $105 - 90$ = $15$ min. 
Knowledge points: Overseas Competition->Knowledge Point->Combinatorics->Time Problem
Source: 其它
Problem: There are two piles of matches. Each pile has $$6$$ sticks. Cindy and Doris take turns to pick up matches from either pile. There is no limit to how many matches they can pick up, but they must pick up at least one match each turn. The person who picks up the last match will be the winner. If Cindy starts picking up matches first,~\uline{~~~~~~~~~~}~will be the winner (has a winning strategy). 
Options:
aoVal: A
content: Cindy 
aoVal: B
content: Doris 
Answer: B
Analysis: Doris will be the winner, since she can simply mirror the number of matches Cindy picks up every turn. 
Knowledge points: Overseas Competition->Knowledge Point->Combinatorics->Strategies and Operations->Game Strategy
Problem: Two cups with capacities of 9 liters and 3 liters, respectively, have different diameters at the top and bottom and are unmarked. If water can be taken arbitrarily, how many operations are required at least to pour out 6 liters of water? (Filling the cup is counted as one operation, pouring the water out is counted as another operation.) 
Options:
aoVal: A
content: $$1$$ 
aoVal: B
content: $$2$$ 
aoVal: C
content: $$3$$ 
aoVal: D
content: $$4$$ 
Answer: B
Analysis: 2 operations at least.  1. Fill the 9 liter cup to the top.  2. Pour water from the 9 liter cup into the 3 liter cup until the 3 liter cup is full, leaving 6 liters of water in the 9 liter cup.  The answer is B 
Knowledge points: Overseas Competition->Knowledge Point->Combinatorics->Strategies and Operations->Operational Problem->Pouring Water Problems
Problem: From $1$ to $30$, how many integers can be chosen at most to ensure that no two of chosen numbers sum to $36$? 
Options:
aoVal: A
content: $$15$$ 
aoVal: B
content: $$16$$ 
aoVal: C
content: $$17$$ 
aoVal: D
content: $$18$$ 
aoVal: E
content: $$19$$ 
Answer: D
Analysis: 1, 2, 3, 4, 5, 18→ 6  $30-5-1=24$, $24\div2=12$  $$6+12=18$$ 
Knowledge points: Overseas Competition->Knowledge Point->Combinatorics->Constructing and Proving
Source: 其它
Problem: Out of 8 bottles of sugar, 1 bottle has a lighter mass than the other 7. How many times can the balance be used to ensure that the lighter one is found? 
Options:
aoVal: A
content: $$1$$ 
aoVal: B
content: $$2$$ 
aoVal: C
content: $$3$$ 
aoVal: D
content: $$4$$ 
Answer: B
Analysis: omit 
Knowledge points: Overseas Competition->Knowledge Point->Combinatorics->Fun Problems in Math
Source: 其它
Problem: Ivy, Vivian, and Candy are playing the truth or lie game. The rule is: the person who picks the truth card can only tell the truth, and the person who picks the lie card must tell a lie.  \textbf{Ivy said: "Vivian and Candy lied."}  \textbf{Vivian said: "I did\textquotesingle t lie."}  \textbf{Candy said: "Vivian lied."}  How many of them told the truth? How many of them told a lie?~\hspace{0pt} 
Options:
aoVal: A
content: $2$; $1$ 
aoVal: B
content: $0$; $3$ 
aoVal: C
content: $1$; $2$ 
aoVal: D
content: $3$; $0$ 
Answer: C
Analysis: Vivian and Candy told contradictory information, so one of them told the truth and the other one told the lie. Therefore, Ivy has definitely told the lie. So two people told the lie, and one people told the truth. 
Knowledge points: Overseas Competition->Knowledge Point->Combinatorics->Logical Reasoning
Problem: If the exam lasts for two days and ends on Wednesday, what day does the exam start?~(adapted from $$2008$$ Math kangaroo Problems, Level $$1-2$$, Question \#$$4$$) 
Options:
aoVal: A
content: Monday 
aoVal: B
content: Tuesday 
aoVal: C
content: Friday 
aoVal: D
content: Saturday 
aoVal: E
content: Thursday 
Answer: B
Analysis: Tuesday to Wednesday is exactly two days. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Periodic Problems->Periodic Problems of Time->Period of Dates
Problem: The number of apples in basket $$A$$ was $$\frac{13}{19}$$ of that in basket $$B$$. After $$10$$ apples were taken out from basket $$A$$ and were put in basket $$B$$, the number of apples in basket $$A$$ became $$\frac{1}{3}$$ of that in basket $$B$$. How many apples were there in basket $$A$$ and basket $$B$$, respectively, in the beginning? ． 
Options:
aoVal: A
content: $$13$$ apples in basket $$A$$; $$19$$ apples in basket $$B$$ 
aoVal: B
content: $$26$$ apples in basket $$A$$; $$19$$ apples in basket $$B$$ 
aoVal: C
content: $$26$$ apples in basket $$A$$;~$$38$$ apples in basket $$B$$ 
aoVal: D
content: $$19$$ apples in basket $$A$$; $$13$$ apples in basket $$B$$ 
Answer: C
Analysis: The total number of apples in basket $$A$$ and basket $$B$$ is constant. In the beginning, $$A: B=13:19$$; now, $$A: B=1:3=8:24$$ (the total number of portions is still $$32$$). By comparison, the number of portions in basket $$A$$ decreases by $$5$$, and the number of portions in basket $$B$$ increases by $$5$$, so each portion is $$10\div5=2\text{kg}$$. Formerly, there were $$13\times2=26$$ apples in basket $$A$$ and $$19\times2=38$$ apples in basket $$B$$. So, the answer is $$C$$. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Solving Problems Involving Proportions->Complex Ratio Problems
Problem: The average cost of a long-distance call in the USA in 1985 was 50 cents per minute, and the average cost of a long-distance call in the USA in 2008 was 5 cents per minute. Find the approximate percent decrease in the cost per minute of a long- distance call. (adapted from 2007 AMC 8, Question \#6) 
Options:
aoVal: A
content: $$30$$ 
aoVal: B
content: $$40$$ 
aoVal: C
content: $$50$$ 
aoVal: D
content: $$80$$ 
aoVal: E
content: $$90$$ 
Answer: E
Analysis: The percent decrease is (the amount of decrease)/(original amount) the amount of decrease is $50-5=45$ so the percent decrease is $\frac{45}{50}$ which is ~$ 90 \textbackslash\%$ 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Solving Problems Involving Fractions and Percentages->Percentages Word Problems
Source: 其它
Problem: My mom\textquotesingle s birthday is on Sunday, and my dad\textquotesingle s birthday is $$55$$ days later. On what day of the week will my dad\textquotesingle s birthday be? 
Options:
aoVal: A
content: $$ $$Sunday$$ $$ 
aoVal: B
content: $$ $$Monday$$ $$ 
aoVal: C
content: $$ $$Tuesday$$ $$ 
aoVal: D
content: $$ $$Thursday$$ $$ 
aoVal: E
content: $$ $$Saturday$$ $$ 
Answer: E
Analysis: After $$1$$ day, it is Monday; after $$2$$ days, it is Tuesday; after $$3$$ days, it is Wednesday $$\cdots$$ After $$8$$ days, it is still Monday, and the period is $$7$$ days. $$55\div 7=7$$(weeks) $$\rm R$$ $$6$$ (days). The birthday is Saturday. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Periodic Problems->Periodic Problems of Time
Problem: If $$10$$ February $$2013$$ falls on Friday, which day of the week will $$10$$ February $2014$ fall on? 
Options:
aoVal: A
content: Sunday 
aoVal: B
content: Monday 
aoVal: C
content: Tuesday 
aoVal: D
content: Wednesday 
aoVal: E
content: Thursday 
aoVal: F
content: Friday 
aoVal: G
content: Saturday 
Answer: G
Analysis: No leap year, 365 days 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Periodic Problems->Periodic Problems of Time
Problem: On a map of England and Wales the distance between St Ives in Cornwall and St Ives in Cambridgeshire measures $$18\text{cm}$$. In reality the distance between the two towns is about $$450\text{km}$$.  Which of the options below is the scale of the map? 
Options:
aoVal: A
content: $$1:2500000$$ 
aoVal: B
content: $$1:1000000$$ 
aoVal: C
content: $$1:750000$$ 
aoVal: D
content: $$1:500000$$ 
aoVal: E
content: $$1:250000$$ 
Answer: A
Analysis: The scale of the map is $$18\text{cm} : 450\text{km}$$; converting the actual distance to centimetres, this is equivalent to $$18 : 450 \times 1000 \times 100 = 18 : 45 000 000$$. Simplifying gives a scale of $$1 : 2 500 000$$. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Solving Problems Involving Proportions->Basic Units
Problem: A mixture of $$30$$ litres of paint is $$25\textbackslash\%$$ red tint, $$30\textbackslash\%$$ yellow tint and $$45\textbackslash\%$$ water. Five litres of yellow tint are added to the original mixture. What is the percentage of yellow tint in the new mixture? 
Options:
aoVal: A
content: $$25\textbackslash\%$$ 
aoVal: B
content: $$35\textbackslash\%$$ 
aoVal: C
content: $$40\textbackslash\%$$ 
aoVal: D
content: $$45\textbackslash\%$$ 
aoVal: E
content: $$50\textbackslash\%$$ 
Answer: C
Analysis: Since $$30\textbackslash\%$$ of the original $$30$$ liters of paint was yellow, and $$5$$ liters of yellow paint were added to make the new mixture, there are $$9+5=14$$ liters of yellow tint in the new mixture. Since only $$5$$ liters of paint were added to the original $$30$$, there are a total of $$35$$ liters of paint in the new mixture. This gives $$40\textbackslash\%$$ of yellow tint in the new mixture, which is $$40$$. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Concentration Problems->Basic Concentration Problems
Problem: Rick and Roy each stands at different ends of a straight road that is $$2021$$m long. They run toward each other. Rick\textquotesingle s speed is $$21\text{m/s}$$ and Roy\textquotesingle s speed is $$22\text{m/s}$$. They will meet in~\uline{~~~~~~~~~~}~seconds. ($$2016$$ Math League.com contest problem, $$5$$th Grade, Question \#$$26$$) 
Options:
aoVal: A
content: $$35$$ 
aoVal: B
content: $$4$$ 
aoVal: C
content: $$47$$ 
aoVal: D
content: $$50$$ 
Answer: C
Analysis: $$2021\div (21+22)=47$$. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules
Problem: If $$13^{}\text{th}$$February is Friday, what day of the week is $$1$$\textsuperscript{st~}February? 
Options:
aoVal: A
content: Sunday　 
aoVal: B
content: Monday　 
aoVal: C
content: Wednesday　 
aoVal: D
content: Thursday　 
aoVal: E
content: Saturday　 
Answer: A
Analysis: We can count backward by days or by weeks. Count a few weeks back to find that February $$6$$ is a Friday. Then count a few days back to find that February $$1$$ is a Sunday. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Periodic Problems->Periodic Problems of Time
Problem: To defeat a dragon, Matthias has to cut off all the dragon\textquotesingle s heads. If he can cut off $$3$$ of the dragon\textquotesingle s heads, one new head immediately grows. Matthias defeats the dragon by cutting off $$17$$ heads in total. How many heads did the dragon have at the beginning? 
Options:
aoVal: A
content: $$8$$ 
aoVal: B
content: $$9$$ 
aoVal: C
content: $$10$$ 
aoVal: D
content: $$11$$ 
aoVal: E
content: $$12$$ 
Answer: E
Analysis: $17\div3=5R2$, so Matthias has $5$ groups of $3$ cuts, which means $5$ new heads grew. Thus, there are $17-5=12$ heads originally. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Applying Addition and Subtraction->Including and Excluding 
Source: 其它
Problem: Gabriel wrote his name $$100$$ times in the sand. What was the $$100\text{th}$$ letter he wrote? ． 
Options:
aoVal: A
content: $$a$$ 
aoVal: B
content: $$b$$ 
aoVal: C
content: $$r$$ 
aoVal: D
content: $$i$$ 
Answer: A
Analysis: The $$1\text{st}$$ letter Gabriel wrote was $$G$$, and every $$7\text{th}$$ letter after was $$G$$.  The $$99\text{th}$$ letter he wrote was $$G$$, and the $$100\text{th}$$ was an"$$a$$." 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Periodic Problems->Periodic Problems of Basic Permutation
Problem: At the $2018$ Tompkins County Fair a vendor is offering a "fair special" on hats. If you buy one hat at the regular price of $\textbackslash$ 30$, you get a second hat at a $40\textbackslash\%$ discount, and a third pair at half the regular price. James took advantage of the "fair special" to buy three hats. What percentage of the $\textbackslash$ 90$ regular price did he save? (adapted from 2013 AMC 8, Question \#12) 
Options:
aoVal: A
content: $$15$$ 
aoVal: B
content: $$20$$ 
aoVal: C
content: $$25$$ 
aoVal: D
content: $$30$$ 
aoVal: E
content: $$40$$ 
Answer: D
Analysis: First, the amount of money one will pay for three hats without the discount $=\textbackslash$ 90$. Then, find the amount of money using the discount: $30+0.6 \times 30+\frac{1}{2} \times 30=\textbackslash$ 63$. Finding the percentage yields $\frac{63}{90}=70\textbackslash\%$ To find the percent saved, we have $100 \textbackslash\%-70\textbackslash\%= 30\textbackslash\%$ 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Solving Problems Involving Fractions and Percentages->Percentages Word Problems
Source: 其它
Problem: The sales tax rate in Bergville is $$6\textbackslash\%$$. During a sale at the Bergville Coat Closet, the price of a coat is discounted $$20\textbackslash\%$$ from its ＄$$90.00$$ price. Two clerks, Jack and Jill, calculate the bill independently. Jack rings up ＄$$90.00$$ and adds $$6\textbackslash\%$$ sales tax, then subtracts $$20\textbackslash\%$$ from this total. Jill rings up ＄$$90.00$$, subtracts $$20\textbackslash\%$$ of the price, then adds $$6\textbackslash\%$$ of the discounted price for sales tax. What is Jack's total minus Jill's total? 
Options:
aoVal: A
content: ＄$$-1.06$$ 
aoVal: B
content: ＄$$-0.53$$ 
aoVal: C
content: ＄$$0$$ 
aoVal: D
content: ＄$$0.53$$ 
aoVal: E
content: ＄$$1.06$$ 
Answer: C
Analysis: The price Jacks rings up is $$(90.00)(1.06)(0.80)$$. The price Jill rings up is $$(90.00)(0.80)(1.06)$$. By the commutative property of multiplication, these quantities are the same, and the difference is $$(\text{C})$$＄$$0$$. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Word Problems in Economics->Complex Money Word Problems
Problem: The distance between Exeter and London is $$175$$ miles. Sam left Exeter at $$10:00$$ on Tuesday for London. Morgan left Exeter~ for London at $$13:00$$ the same day. They travelled on the same road. Sam\textquotesingle s average speed was $$25$$ miles per hour, and Morgan\textquotesingle s average speed was $$50$$ miles an hour. At what time did Sam and Morgan meet? 
Options:
aoVal: A
content: $$16:00$$ 
aoVal: B
content: $$15:55$$ 
aoVal: C
content: $$15:30$$ 
aoVal: D
content: $$15:00$$ 
aoVal: E
content: $$14:40$$ 
Answer: A
Analysis: $25\times3=75$miles  $75\div (50-25)=3$h  $13+3=16$  They meet at $16:00$.  revise from ($$2017$$ Junior Mathematical Challenge Question \#$$25$$) 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules
Problem: There were $$31$$ runners competing in a race. The number of runners who finished before John is four times smaller than the number of runners who finished later than John. At what place did John finish? 
Options:
aoVal: A
content: $$6$$ 
aoVal: B
content: $$7$$ 
aoVal: C
content: $$8$$ 
aoVal: D
content: $$20$$ 
aoVal: E
content: $$21$$ 
Answer: B
Analysis: Before is ``$$1$$'', so after is ``$$4$$'', Before: $$(31-1) \div (4+1) = 6$$, and John: $$6+1=7$$. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Questions Involving Sum, Difference and Multiples->Problems of Sum and Multiple