math-eval/TAL-SCQ5K huggingface dataset:
Problem: Some students are lining up at the cafeteria. Ellie is the third from the back and the seventh from the front. How many students are there lining up in total? 
Options:
aoVal: A
content: $$4$$ 
aoVal: B
content: $$8$$ 
aoVal: C
content: $$9$$ 
aoVal: D
content: $$10$$ 
aoVal: E
content: $$11$$ 
Answer: C
Analysis: $$3+7-1=9$$ 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules
Source: 其它
Problem: A bridge is built across a river. One quarter of the bridge is over the left bank of the river and one third of the bridge is over the right bank. The river is $$120\text{m}$$ wide. How long is the bridge? 
Options:
aoVal: A
content: $$150\text{m}$$ 
aoVal: B
content: $$190\text{m}$$ 
aoVal: C
content: $$240\text{m}$$ 
aoVal: D
content: $$288\text{m}$$ 
aoVal: E
content: $$324\text{m}$$ 
Answer: D
Analysis: The river is $$120\text{m}$$ wide and represents $$\left( 1-\frac{1}{4}-\frac{1}{3} \right)=\frac{5}{12}$$ of the length of the bridge. Therefore $$\frac{1}{12}$$ of the length of the bridge is $$24\text{m}$$. Hence the total length of the bridge is $$12\times 24\text{m}=288\text{m}$$. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Solving Problems Involving Fractions and Percentages->Finding the Base
Problem: There are $$40$$ guests queueing to enter a party. Every $$4^{\rm th}$$ guest in the queue receives a balloon and every $$6^{\rm th}$$ guest in the queue receives a mask. How many guests receive both a balloon and a mask?~\uline{~~~~~~~~~~}~ 
Options:
aoVal: A
content: $$2$$ 
aoVal: B
content: $$3$$ 
aoVal: C
content: $$4$$ 
aoVal: D
content: $$5$$ 
Answer: B
Analysis: Every 12th guest receives both a mask and a balloon. Hence, the 12th, 24th and 36th guest receive both. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Distribution Problems->Basic Distribution Problems
Problem: Daniel raises some ducks and dogs. All the ducks and dogs have $18$ legs and $5$ pairs of wings in total. How many ducks and dogs are there in total? 
Options:
aoVal: A
content: $$18$$ 
aoVal: B
content: $$13$$ 
aoVal: C
content: $$9$$ 
aoVal: D
content: $$7$$ 
aoVal: E
content: $$5$$ 
Answer: D
Analysis: $5$ pairs of wings means there are $5$ ducks, so there are $18-5\times2=8$ legs for dogs, which is $8\div4=2$. Thus, there are $2+5=7$ animals. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Chicken-Rabbit Problems
Source: 其它
Problem: The number of ounces of water needed to reduce $$9$$ ounces of shaving lotion containing $$50\textbackslash\%$$ alcohol to a lotion containing $$30\textbackslash\%$$ alcohol is: ($$1953 $$ AHSME Problem, Question \#$$9$$) 
Options:
aoVal: A
content: $$3$$ 
aoVal: B
content: $$4$$ 
aoVal: C
content: $$5$$ 
aoVal: D
content: $$6$$ 
aoVal: E
content: $$7$$ 
Answer: D
Analysis: Solution $$1$$.  Say we add $$N$$ ounces of water to the shaving lotion. Since half of an $$9$$ ounce bottle of shaving lotion is alcohol, we know that we have $$\dfrac{9}{2}$$ ounces of alcohol. We want $$\dfrac{9}{2}=0.3(9+N)$$ (because we want the amount of alcohol, $$\dfrac{9}{2}$$, to be $$30\textbackslash\%$$, or $$0.3$$, of the total amount of shaving lotion, $$9+N$$). Solving this, we find that $$9=0.6(9+N)\Rightarrow9=5.4+0.6N\hspace{0pt}\Rightarrow3.6=0.6N\hspace{0pt}\Rightarrow6=N$$. So, the total amount of water we need to add is $$6$$.  Solution $$2$$.  The concentration of alcohol after adding $$n$$ ounces of water is $$\dfrac{4.5}{9+n}$$.  To get a solution of $$30\textbackslash\%$$ alcohol, we solve $$\dfrac{4.5}{9+n}=\dfrac{3}{10}$$  $$45=27+3n$$  $$18=3n$$  $$6=n\Rightarrow6$$. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Concentration Problems->Solving Concentration Problems with Equations
Problem: In the Adventure Park, 30 children took part in two of the adventures. 15 of them participated in the "moving bridge" contest, and 20 of them went down the zip-wire. How many of the children took part in both adventures? 
Options:
aoVal: A
content: $$25$$ 
aoVal: B
content: $$15$$ 
aoVal: C
content: $$30$$ 
aoVal: D
content: $$10$$ 
aoVal: E
content: $$5$$ 
Answer: E
Analysis: There are 30 children.  15 of them participated in the "moving bridge" contest.  This means that the other 15, who did not participate in the "moving bridge" surely went down the zip-wire.  In fact 20 went down the zip-wire; therefore, 20-15=5 of the 15 "moving bridges\textquotesingle{} " participants must have also gone down the zip-wire. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Applying Addition and Subtraction
Source: 其它
Problem: When Lucy was born, Cathy was $23$ years old. The sum of their ages $3$ years later will be $45$. How old will be Cathy $3$ years later? 
Options:
aoVal: A
content: $$11$$ 
aoVal: B
content: $$15$$ 
aoVal: C
content: $$28$$ 
aoVal: D
content: $$33$$ 
aoVal: E
content: $$34$$ 
Answer: E
Analysis: Suppose that Lucy will be $x$ years old $3$ years later, Cathy will be $$(x+23)$$ years old. $x+(x+23)=45$, so $x=11$. Thus, Cathy will be $11+23=34$ years old $3$ years later. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Age Problems
Source: 其它
Problem: When Kerr was born, Barry was $13$ years old. The sum of their ages $5$ years later will be $51$. How old will be Barry $5$ years later? 
Options:
aoVal: A
content: $$11$$ 
aoVal: B
content: $$15$$ 
aoVal: C
content: $$19$$ 
aoVal: D
content: $$33$$ 
aoVal: E
content: $$34$$ 
Answer: C
Analysis: Suppose that Lucy will be $x$ years old $5$ years later, Cathy will be $$(x+13)$$ years old. $x+(x+13)=51$, so $x=19$. Thus, Cathy will be $13+19=32$ years old $5$ years later. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Age Problems
Source: 其它
Problem: A store sells cakes for $10 each, at the following discounts.  \textbf{① $1 off each every Wednesday.}  \textbf{② Buy three or more, $1 off each.}  If both rules are met at same time, the price can be reduced together.  How much will Tracy pay for 4 cakes at this store on Tuesday? 
Options:
aoVal: A
content: $36 
aoVal: B
content: $40 
aoVal: C
content: $34 
aoVal: D
content: $44 
aoVal: E
content: $30 
Answer: A
Analysis: 40-1*4=36 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Word Problems in Economics->Money Calculation
Source: 其它
Problem: If $2\text{cm}$ represents $50\text{km}$ on a map and the distance between two towns on this map is $7.5\text{cm}$, then their actual distance apart is ． 
Options:
aoVal: A
content: $375\text{km}$ 
aoVal: B
content: $275\text{km}$ 
aoVal: C
content: $187.5\text{km}$ 
aoVal: D
content: $75\text{km}$ 
Answer: C
Analysis: If $$2\textasciitilde\text{cm}$$ represents $$50\textasciitilde\text{km}$$, $$1\textasciitilde\text{cm}$$ represents $$25\textasciitilde\text{km}$$. So, their actual distance apart is $$7.5\times 25=187.5\textasciitilde\text{km}$$. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Solving Problems Involving Proportions->Basic Units
Problem: There are $10$ benches on one side of a road. Brown wants to put $1$ pot of flowers between every two adjacent benches. How many pots of flowers does he need to prepare? 
Options:
aoVal: A
content: $$9$$ 
aoVal: B
content: $$10$$ 
aoVal: C
content: $$11$$ 
aoVal: D
content: $$12$$ 
aoVal: E
content: $$13$$ 
Answer: A
Analysis: $10 - 1 = 9$ 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Interval Problems
Source: 其它
Problem: Cagney can frost a cupcake every $$30$$ seconds and Lacey can frost a cupcake every $$60$$ seconds. Working together, how many cupcakes can they frost in $$5$$ minutes? (Adapted from $$2012$$ AMC $$10\rm A$$ Problem, Question \#$$11$$) 
Options:
aoVal: A
content: $$10$$ 
aoVal: B
content: $$15$$ 
aoVal: C
content: $$20$$ 
aoVal: D
content: $$25$$ 
aoVal: E
content: $$30$$ 
Answer: B
Analysis: In $$300$$ seconds ($$5$$ minutes), Cagney will frost $$\dfrac{300}{30}=10$$ cupcakes, and Lacey will frost $$\dfrac{300}{60}=5$$ cupcakes. Therefore, working together they will frost $$10+5=\boxed{(\text{B})15}$$ cupcakes. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Work Word Problems->Collaborative Work Word Problems
Problem: Owen attends a basketball shooting game. He gets $70 \textbackslash\%$ on $10$ one-point shots , $80 \textbackslash\%$ on $10$ two-points shots and $90 \textbackslash\%$ on $10$ three-points shots. Compare Owen\textquotesingle s score with the total points possible, which percent is closest to his overall score? (Adapted from 2006 AMC 8, Question \#12) 
Options:
aoVal: A
content: $$40$$ 
aoVal: B
content: $$77$$ 
aoVal: C
content: $$80$$ 
aoVal: D
content: $$83$$ 
aoVal: E
content: $$87$$ 
Answer: D
Analysis: $70 \textbackslash\% \cdot 10=7$ $80 \textbackslash\% \cdot 20=16$ $90 \textbackslash\% \cdot 30=27$ Adding them up gets $7+16+27=50$. The overall percentage correct would be $\frac{50}{60}=\frac{5}{6}=5 \cdot 16 . \overline{6}=83 . \overline{3} \approx(\mathbf{D}) 83$. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Solving Problems Involving Fractions and Percentages->Percentages Word Problems
Source: 其它
Problem: There are five kangaroo mothers, six kangaroo babies, and one kangaroo father. Each kangaroo mother should take care of the same number of kangaroo babies. How many kangaroo babies should kangaroo father take care of? (adapted from $$2011$$ Math kangaroo Problems, Level $$1-2$$, Question \#$$6$$) 
Options:
aoVal: A
content: $$0$$ 
aoVal: B
content: $$1$$ 
aoVal: C
content: $$2$$ 
aoVal: D
content: $$3$$ 
aoVal: E
content: $$4$$ 
Answer: B
Analysis: The key point is that each kangaroo mother needs to take care of the same number of kangaroo babies. So $6$ (number of kangaroo baby) $$\div$$ $5$ (number of kangaroo mother)$=1\ldots\ldots1$, and the remaining one is taken care of by kangaroo father. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Applying Addition and Subtraction->Including and Excluding
Problem: Pamela, Pearl, and Polly went to buy some toys. Pamela paid $$2$$ dollars and $$70$$ cents for three identical dolls. Pearl paid $$3$$ dollars and $$40$$ cents for two identical toy cars. How much did Polly pay for one doll and one toy car? (Adapted from 2011 Math Kangaroo Problem, Level 3-4 , Question \#6) 
Options:
aoVal: A
content: $2$ dollars and $80$ cents 
aoVal: B
content: $2$ dollars and $70$ cents 
aoVal: C
content: $2$ dollars and $60$ cents 
aoVal: D
content: $2$ dollars and $50$ cents 
aoVal: E
content: $2$ dollars and $40$ cents 
Answer: C
Analysis: $$2$$ dollars and $$70$$ cents=$270$ cents  $$3$$ dollars and $$40$$ cents=$340$ cents  One doll costs $270\div3=90$ cents.  One toy car costs $340\div2=170$ cents.  $170+90=260$ cents = $2$ dollars $60$ cents. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Distribution Problems->Basic Distribution Problems->Basic Problems of Distribution
Problem: A police spotted a burglar from $$100\text{m}$$ apart. The burglar immdiately runs away at a speed of $$4\text{m/s}$$ and the police starts chasing him at $$8\text{m/s}$$ at the same time. At this rate, how long will it take the police to catch the burglar? 
Options:
aoVal: A
content: $$1 $$ minute 
aoVal: B
content: $$25 $$ seconds 
aoVal: C
content: $$30$$ seconds 
aoVal: D
content: $$15$$ seconds 
Answer: B
Analysis: The distance between them is $$100$$ m, and the difference between their speeds is $$12-8=4000$$ m/h. It takes $$100\div4000\times60=1.5$$minutes to catch the burglar, which is $$90$$ seconds 
Knowledge points: Overseas Competition->Knowledge Point->Distance Word Problems->Distance Word Problems on Straight Road
Problem: A sugar solution is made by mixing $$7$$ grams of sugar and $$21$$ grams of water. Find the percent concentration of the solution. 
Options:
aoVal: A
content: $$20$$\% 
aoVal: B
content: $25$\% 
aoVal: C
content: $30$\% 
aoVal: D
content: $33.3$\% 
Answer: B
Analysis: $$7\div(7+21)=25\textbackslash\%$$. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Concentration Problems->Basic Concentration Problems
Problem: Two plums and one cherry weigh $$80\text{g}$$. Two cherries and one plum weigh $$70\text{g}$$. What is the weight of four plums? 
Options:
aoVal: A
content: $$30\text{g}$$ 
aoVal: B
content: $$40\text{g}$$ 
aoVal: C
content: $$50\text{g}$$ 
aoVal: D
content: $$120\text{g}$$ 
aoVal: E
content: $$160\text{g}$$ 
Answer: D
Analysis: Let the mass of one plum and the mass of one cherry be $$p \text{g}$$ and $$c\text{g}$$ respectively.  Then $$2p +c=80$$ and $$2c+p=70$$. Hence $$3p +3c=150$$, and $$p+c=50$$. But since $$2p+c=80$$, we can deduce that $$p=80-50= 30$$, and so four plums weigh $$120 \text{g}$$. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Equation Word Problems->Equivalent Substitution in Equation Word Problems
Problem: A total of $$32$$ chickens and rabbits are caged together. If there are $$90$$ legs in total, how many chickens and rabbits are in the cage, respectively? 
Options:
aoVal: A
content: $20$; $12$ 
aoVal: B
content: $13$; $19$ 
aoVal: C
content: $18$; $14$ 
aoVal: D
content: $19$; $13$ 
Answer: D
Analysis: Suppose all the animals in the cage are chickens: there should be $$2\times 32=64$$ legs. However, $$90-64=26$$ legs are missing because we counted $$26\div2=13$$ rabbits as chickens. Hence, there are $$13$$ rabbits and $$32-13=19$$ chickens. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Chicken-Rabbit Problems->Solving Chicken-Rabbit Problems by Using Hypothesis
Problem: Anna is $$10$$ years old and the age of her mother is $$4$$ times that of Anna. How old will Anna\textquotesingle s mother be when Anna\textquotesingle s mother is twice as old as Anna? (Adapted from $$2007$$ Math kangaroo Problem, Level $$5-6$$, Question \#$$20$$) 
Options:
aoVal: A
content: $$40$$ 
aoVal: B
content: $$50$$ 
aoVal: C
content: $$60$$ 
aoVal: D
content: $$70$$ 
aoVal: E
content: $$80$$ 
Answer: C
Analysis: The age difference between them is $(4-1)\times10=30$, which stays the same forever. Then, when Anna\textquotesingle s mother is twice as old as Anna, Anna should be $30$ years old, then her mom should be $30\times2=60$. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Age Problems->Varying Multiples in Age Problems
Source: 其它
Problem: Some students are buying a boardgame together. If each person pays $$$60$$, they need another $$$60$$ for payment. If each person pays $$$70$$, they will have $$$10$$ left. How much is the boardgame? 
Options:
aoVal: A
content: $$$600$$ 
aoVal: B
content: $$$480$$ 
aoVal: C
content: $$$690$$ 
aoVal: D
content: $$$550$$ 
aoVal: E
content: $$$720$$ 
Answer: B
Analysis: Number of students: $$(60+10)\div (70-60)=7$$  Price of boardgame: $$60\times7+60=480$$． 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules
Problem: In a speed skating competition, $$10$$ skaters reached the finish line. The number of skaters that came in before Tom was $$3$$ less than the number of skaters who came in after him. Which place did Tom end up in? (2015 Math Kangaroo Problem, Level 3 - 4, Question \#13) 
Options:
aoVal: A
content: $$1$$ 
aoVal: B
content: $$3$$ 
aoVal: C
content: $$4$$ 
aoVal: D
content: $$6$$ 
aoVal: E
content: $$7$$ 
Answer: C
Analysis: The sum of the number of skaters that came in before Tom and the number of skaters that came in after him is $10 - 1 = 9$. The number of skaters that came in before Tom was $(9 - 3) \div 2 = 3$, so Tom was the fourth one. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Problems Involving Queuing->Finding the Order of one Character
Source: 其它
Problem: There is a tournament at the pool, First, $$13$$ children signed up and then another $$19$$ children signed up, Six teams with an equal number of members each are needed for the tournament, At least how many more children need to sign up so that,the six teams can be formed． 
Options:
aoVal: A
content: $$1$$ 
aoVal: B
content: $$2$$ 
aoVal: C
content: $$3$$ 
aoVal: D
content: $$4$$ 
aoVal: E
content: $$5$$ 
Answer: D
Analysis: The total number of children is $$13+19=32$$, To form six teams with the same number of children in each team, we divide $$32$$ by $$6$$, $$32\div6=5$$$$\rm R$$$$2$$, The remaining $$2$$ children will need $$4$$ more children to form the sixth team. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Unitary Method Problems->Unitary Method with One Variable
Problem: Bob and Wilson are standing in line. Bob knows that there are $$5$$ people in front of him. Wilson knows that there is a total of $$12$$ people in the line. If Bob is just in front of Wilson, how many of the people in the line are behind Wilson? (Adapted from 2017 Math Kangaroo Problem, Level 1-2, Question \#14) 
Options:
aoVal: A
content: $$2$$ 
aoVal: B
content: $$3$$ 
aoVal: C
content: $$4$$ 
aoVal: D
content: $$5$$ 
aoVal: E
content: $$6$$ 
Answer: D
Analysis: $$12-5-2=5$$ 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules
Source: 其它
Problem: Mother\textquotesingle s Day in $$2020$$ was May $$10^{\rm th}$$, which was a Sunday. Father\textquotesingle s Day in $$2020$$ was June $$21^{\rm st}$$. On what day of the week did Father\textquotesingle s Day fall? 
Options:
aoVal: A
content: Monday 
aoVal: B
content: Tuesday 
aoVal: C
content: Wednesday 
aoVal: D
content: Sunday 
Answer: D
Analysis: There are in total $$21 + 21 = 42$$ days from May $$11$$ to June $$21$$. Since $$42 \div 7 = 6$$, with no remainder, the Father\textquotesingle s Day falls on Sunday as well. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Periodic Problems->Periodic Problems of Time
Problem: If today is Monday, then $63$ days from today will be． 
Options:
aoVal: A
content: $$$$Monday 
aoVal: B
content: $$$$Tuesday 
aoVal: C
content: $$$$Friday 
aoVal: D
content: $$$$Sunday 
Answer: A
Analysis: If today is Monday, every $$7$$ days is another Monday. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Periodic Problems->Periodic Problems of Time
Problem: A store sells cakes for $10 each, at the following discounts.  \textbf{① $1 off each every Wednesday.}  \textbf{② Buy three or more, $1 off each.}  If both conditions are met at same time, the price can be reduced together.  How much will Tracy pay for 4 cakes at this store on Tuesday? 
Options:
aoVal: A
content: $36 
aoVal: B
content: $40 
aoVal: C
content: $34 
aoVal: D
content: $44 
aoVal: E
content: $30 
Answer: A
Analysis: 40-1*4=36 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Word Problems in Economics->Money Calculation
Source: 其它
Problem: A balloon ride can take at most $$3$$ people at a time. If $$41$$ people want to fly in the balloon, the least number of rides needed is． 
Options:
aoVal: A
content: $$12$$ 
aoVal: B
content: $$13$$ 
aoVal: C
content: $$14$$ 
aoVal: D
content: $$15$$ 
Answer: C
Analysis: A balloon ride can take \emph{at most}~$$3$$ people at a time. If $$41$$ people want to fly in the balloon, the least number of rides needed is $$41 \div3$$, rounded \emph{up} to $$14$$. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Unitary Method Problems->Finding the Total Value using Unitary Method
Problem: Andy spent $$5$$ dollars on $$15$$ biscuits. How much would it cost if Andy bought six more of them? (adapted from $$2008$$ Math kangaroo Problems, Level $$1-2$$, Question \#$$13$$) 
Options:
aoVal: A
content: $$6$$ 
aoVal: B
content: $$7$$ 
aoVal: C
content: $$8$$ 
aoVal: D
content: $$9$$ 
Answer: B
Analysis: Fifteen cookies for five dollars means three cookies for one dollar. Six more cookies cost $$2$$ dollars. $$5+2=7$$~ dollars in total. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Unitary Method Problems->Unitary Method with One Variable
Problem: Amy mixes $$10$$ grams of a $$20\textbackslash\%$$ sugar solution and $$40$$ grams of a $$25\textbackslash\%$$ sugar solution together. After~\uline{~~~~~~~~~~}~of water evaporates, the percent concentration of solution is $40\textbackslash\%$. 
Options:
aoVal: A
content: $$15$$ grams 
aoVal: B
content: $$18$$ grams 
aoVal: C
content: $$20$$ grams 
aoVal: D
content: $$25$$ grams 
Answer: C
Analysis: $$10\times20\textbackslash\%+40\times 25\textbackslash\%=12$$ ounces.  $$(10+40)-12\div40\textbackslash\%=20$$ ounces. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Concentration Problems->Finding and Making Use of Invariants
Problem: One of my two brothers is $$4$$ years older than the other. If the sum of their ages is $$38$$, the older brother isyears old. 
Options:
aoVal: A
content: $$17$$ 
aoVal: B
content: $$21$$ 
aoVal: C
content: $$23$$ 
aoVal: D
content: $$27$$ 
aoVal: E
content: $$42$$ 
Answer: B
Analysis: One of my two brothers is $$4$$ years older than the other. If they were the same age, they\textquotesingle d each be $$19$$. Thus, one is $$17$$ and the other is $$21$$. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Questions Involving Sum, Difference and Multiples->Problems Involving Sum and Difference
Problem: Lee is $14$ years old and Lay is $41$ years old. How many years ago, Lay\textquotesingle s age is exactaly $4$ times Lee\textquotesingle s age? 
Options:
aoVal: A
content: $$3$$ 
aoVal: B
content: $$4$$ 
aoVal: C
content: $$5$$ 
aoVal: D
content: $$6$$ 
aoVal: E
content: $$7$$ 
Answer: C
Analysis: Let the year be $x$ years ago.  $14-x=(41-x)\div 4$  $4(14-x)=41-x$  $56-4x=41-x$  $15=3x$  $3x=15$  $x=5$ 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Equation Word Problems
Problem: Which of the following equations represents the situation: John drinks $x$ cups of boba everyday on weekdays and $2$ cups of boba everyday on the weekends. 
Options:
aoVal: A
content: $x+2$ 
aoVal: B
content: $5x+2$ 
aoVal: C
content: $5x+4$ 
aoVal: D
content: $5x-4$ 
Answer: C
Analysis: The boba that John drinks during weekdays: $$5x$$  The boba that John drinks during weekends: $$2\times 2 =4$$  The total boba that John drinks during a week: $$5x+4$$ 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Equation Word Problems
Source: 其它
Problem: Granny and the triplets Cara, Cate and Chris all have their birthdays today.  The ages of all four of them total $$120$$ years.  Granny is $$5$$ times as old as each of the triplets.  When were the triplets born? 
Options:
aoVal: A
content: $$2000$$ 
aoVal: B
content: $$2002$$ 
aoVal: C
content: $$2004$$ 
aoVal: D
content: $$2007$$ 
aoVal: E
content: $$2009$$ 
Answer: D
Analysis: Given that Granny is $$5$$ times older than each of the three triplets, we know that the age of each triplet is $$120\div (5+3)= 15$$. So Cara, Cate and Chris were born in $$2007$$. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Age Problems->Sums and Multiples in Age Problems
Problem: Nick has $83$ books and Steven has $140$ books. Starting from tomorrow, Nick will buy $8$ books and Steven will buy $5$ books every day. How many days later will they have the same number of books? 
Options:
aoVal: A
content: $$100$$ 
aoVal: B
content: $$19$$ 
aoVal: C
content: $$32$$ 
aoVal: D
content: $$20$$ 
aoVal: E
content: Never 
Answer: B
Analysis: $$(140-83) \div (8-5)=19$$ 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules
Source: 其它
Problem: In a basket, there are $60$ bananas. $20$\% of them are rotten. How many bananas are in good condition? 
Options:
aoVal: A
content: $$12$$ 
aoVal: B
content: $$20$$ 
aoVal: C
content: $$48$$ 
aoVal: D
content: $$60$$ 
Answer: C
Analysis: $1-20$\%=$80$\%,  $60 \times 80$\%=$48$.  $48$ bananas are in good condition. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Solving Problems Involving Fractions and Percentages->Percentages Word Problems
Source: 其它