math-eval/TAL-SCQ5K huggingface dataset:
Problem: There are $$2$$ inlets (namely $$\text{A}$$ and $$\text{B}$$) in a pool. If only inlet $$\text{A}$$ is open, it takes $$36$$ minutes to fill up the empty pool. If only inlet $$\text{B}$$ is open, it takes $$48$$ minutes to fill up the empty pool. Now, inlets $$\text{A}$$ and $$\text{B}$$ will be open in turns, according to such an order, open inlet $$\text{A}$$ for $$1$$ minute, $$\text{B}$$ for $$2$$ minutes, $$\text{A}$$ for $$2$$ minutes, $$\text{B}$$ for $$1$$ minute, $$\text{A}$$ for $$1$$ minute, $$\text{B}$$ for $$2$$ minutes $$\cdots$$ So on and so forth. How long, in the nearest minutes, does it take to fill up the empty pool? 
Options:
aoVal: A
content: $$36$$ 
aoVal: B
content: $$39$$ 
aoVal: C
content: $$41$$ 
aoVal: D
content: $$44$$ 
aoVal: E
content: $$48$$ 
aoVal: F
content: None of the above 
Answer: C
Analysis: C 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Work Word Problems
Source: 其它
Problem: Joe lives on the $3$\textsuperscript{rd}~floor. He needs to spend $12$ minutes to move from the $4$\textsuperscript{th} floor to the $5$\textsuperscript{th} floor. How many minutes does he need to climb from the $1$\textsuperscript{st} floor to the $3$\textsuperscript{rd} floor? 
Options:
aoVal: A
content: $$12$$ 
aoVal: B
content: $$24$$ 
aoVal: C
content: $$36$$ 
aoVal: D
content: $$48$$ 
aoVal: E
content: $$60$$ 
Answer: B
Analysis: $12 \times (3 - 1) = 24$ 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Interval Problems
Source: 其它
Problem: Three people $$A$$, $$B$$, and $$C$$ have a total of $$$200$$. Given that the ratio of $${A}'$$ s total to $${B}'$$s total is $$4:3$$, and $$A$$ has $$$20$$ more than $$C$$, how much does $$C$$ have? 
Options:
aoVal: A
content: $$24$$ 
aoVal: B
content: $$36$$ 
aoVal: C
content: $$48$$ 
aoVal: D
content: $$60$$ 
aoVal: E
content: $$72$$ 
Answer: D
Analysis: Assume that $$C$$ borrows $$$20$$. Therefore, $$A:B:C = 4:3:4$$, for a total sum of $$11$$ increments, or $$$220$$. One increment is $$220\div 11=20$$ dollars. Therefore, $$C$$ has a total of $$4\times 20-20=60$$ dollars. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Solving Problems Involving Proportions->Dividing Quantities Based on Ratio
Problem: If $$29^{}\text{th}$$May is Monday, what day of the week is $$1$$\textsuperscript{st~}May ? 
Options:
aoVal: A
content: Sunday　 
aoVal: B
content: Monday　 
aoVal: C
content: Wednesday　 
aoVal: D
content: Tuesday　 
aoVal: E
content: Saturday　 
Answer: B
Analysis: There are 29 days between $$29^{}\text{th}$$May and~ $$1$$\textsuperscript{st~}May.  Apart from $29$\textsuperscript{th} May itself, there are 28 days = 4 week.  Therefore,~~$$1$$\textsuperscript{st~}March is Monday. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Periodic Problems->Periodic Problems of Time
Problem: Linda paid $$8$$ dollars for $$4$$ potatoes. How many dollars did Linda pay for the same kind of potatoes if she bought $$6$$ more of them? (adapted from $$2008$$ Math kangaroo Problems, Level $$1-2$$, Question \#$$13$$) 
Options:
aoVal: A
content: $$18$$ 
aoVal: B
content: $$20$$ 
aoVal: C
content: $$24$$ 
aoVal: D
content: $$22$$ 
Answer: B
Analysis: Linda paid ＄$$8$$ for $$4$$ potatoes, $$8\div4=2$$, so we can know $$1$$ potatoes cost $$2$$ dollars. In total Linda should pay $$2\times10=20$$ dollars for $$10$$ potatoes. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Unitary Method Problems->Unitary Method with One Variable
Problem: There are $5$ people in Harry\textquotesingle s family with an average height of $168$ cm. The heights of Harry, Harry\textquotesingle s mom, Harry\textquotesingle s sister, and Harry\textquotesingle s brother are $170$ cm, $160$ cm, $162$ cm, and $175$ cm, respectively. What is the height of Harry\textquotesingle s dad? 
Options:
aoVal: A
content: $$163$$ 
aoVal: B
content: $$165$$ 
aoVal: C
content: $$171$$ 
aoVal: D
content: $$173$$ 
aoVal: E
content: $$178$$ 
Answer: D
Analysis: $(170-168)-(168-160)-(168-162)+(175-168)=-5$.  $168+5=173$. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Average Problems 
Source: 其它
Problem: Vera invited $$13$$ guests to her birthday party. She had $$2$$ pizzas, and each of them was cut into $$8$$ slices. Each person at the party ate one slice of pizza. How many slices of pizza were left over? (2015 Math Kangaroo Problem, Level 1-2, Question \#14) 
Options:
aoVal: A
content: $$5$$ 
aoVal: B
content: $$4$$ 
aoVal: C
content: $$3$$ 
aoVal: D
content: $$2$$ 
aoVal: E
content: $$1$$ 
Answer: D
Analysis: There were $8\times2=16$ slices. $13+1=14$ slices of pizza were ate, so there were $16-14=2$ slices left. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Unitary Method Problems->Applying Multiplication and Division->Simple Multiplication Applications
Source: 其它
Problem: If this is March, what month will it be $$1993$$ months from today? 
Options:
aoVal: A
content: January　 
aoVal: B
content: February　 
aoVal: C
content: March　 
aoVal: D
content: April　 
Answer: D
Analysis: Since $$12\times166 +1= 1992 + 1= 1993$$, the month will be April. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Periodic Problems->Periodic Problems of Time
Problem: In $$1975$$ it was estimated that there were $$250000$$ lions in Africa.  Over $$40$$ years this figure has decreased by $$90\textbackslash\%$$.  What is the current estimate for the number of lions in Africa ? 
Options:
aoVal: A
content: $$25000$$ 
aoVal: B
content: $$100000$$ 
aoVal: C
content: $$160000$$ 
aoVal: D
content: $$225000$$ 
aoVal: E
content: $$275000$$ 
Answer: A
Analysis: A $$90\textbackslash\%$$ decrease means $$10\textbackslash\%$$ of $$250000 = 250000 \div 10 = 25000$$ of the lions remain. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Solving Problems Involving Fractions and Percentages->Finding the Base
Problem: Annie usually leaves her cell phone on. If her cell phone is on but she is not actually using it, the battery will last for $$36$$ hours. If she is using it constantly, the battery will last for only $$10$$ hours. Since the last recharge, her phone has been on $$20$$ hours, and during that time she has used it for $$120$$ minutes. If she only keeps using the phone $3$ hours more and then leaves the phone on, how many more hours will the battery last after she leaves the phone on? (Adapted from $$2004$$ AMC $$8$$ Problem, Question \#$$12$$) 
Options:
aoVal: A
content: $$0$$ 
aoVal: B
content: $$1$$ 
aoVal: C
content: $$3$$ 
aoVal: D
content: $$4$$ 
aoVal: E
content: $$5$$ 
Answer: A
Analysis: When not being used, the cell phone uses up $$\dfrac{1}{36}$$ of its battery per hour. When being used, the cell phone uses up $$\dfrac{1}{10}$$ of its battery per hour. Since Annie\textquotesingle s phone has been on for $$20$$ hours, of those $$18$$ simply on and $2$ being used, $$18\times\left(\dfrac{1}{36}\right)+2\times\left(\dfrac{1}{10}\right)=\dfrac{7}{10}$$ of its battery has been used up. To drain the remaining $$\dfrac{3}{10}$$, the phone can last for $$(\frac{3}{10}-3\times\frac1{10})\div \frac1{36}=0$$ more hours without being used. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Work Word Problems->Simple Work Word Problems
Problem: A cow gives $24$ liters of milk each day. If the milkman sells $80$\% of the milk, how many liters of milk is left with him? 
Options:
aoVal: A
content: $$4$$ 
aoVal: B
content: $$4.8$$ 
aoVal: C
content: $$19.2$$ 
aoVal: D
content: $$20$$ 
Answer: B
Analysis: $24 \times 80$\%=$19.2$,  $24-19.2=4.8$,  $4.8$ liters of milk is left with him. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Solving Problems Involving Fractions and Percentages->Percentages Word Problems
Source: 其它
Problem: Patrick got into a lift. He went down six floors, up seven floors and then down eight floors. He was finally on the third floor. Which floor did Patrick get into the lift? 
Options:
aoVal: A
content: $$3^{rd}$$ 
aoVal: B
content: $$4^{th}$$ 
aoVal: C
content: $$10^{th}$$ 
aoVal: D
content: $$11^{th}$$ 
Answer: C
Analysis: $3+8-7+6=10$ 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Inverse Operation Problems->Inverse Operations
Problem: Together, Bob, Colin and Dave weigh $$195\text{kg}$$. Bob weighs $$9\text{kg}$$ more than Colin and $$6\text{kg}$$ more than Dave. How much does Bob weigh? 
Options:
aoVal: A
content: $$60\text{kg}$$ 
aoVal: B
content: $$66\text{kg}$$ 
aoVal: C
content: $$69\text{kg}$$ 
aoVal: D
content: $$70\text{kg}$$ 
Answer: D
Analysis: Nil 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Questions Involving Sum, Difference and Multiples->Problems Involving Sum and Difference->Sum and Differences Problems with multiple Variables
Problem: Helen said to Mary, "If you give me $$4$$ apples, I will have exactly as many apples as you." How many more apples does Mary have than Helen? 
Options:
aoVal: A
content: $$3$$ 
aoVal: B
content: $$6$$ 
aoVal: C
content: $$8$$ 
aoVal: D
content: $$12$$ 
aoVal: E
content: $$15$$ 
Answer: C
Analysis: Difference: $4+4=8$ 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Questions Involving Sum, Difference and Multiples->Problems of Difference and Multiple->Differences and Multiples of Two Variables->Differences and Integer Multiples
Problem: To make a pudding, you need $$2$$ ounces of flour and $$10$$ ounces of milk. You have $$10$$ ounces of flour and want to make as many puddings as possible. How much milk do you need? 
Options:
aoVal: A
content: $$20$$ ounces 
aoVal: B
content: $$30$$ ounces 
aoVal: C
content: $$40$$ ounces 
aoVal: D
content: $$50$$ ounces 
aoVal: E
content: $$60$$ ounces 
Answer: D
Analysis: The ratio of flour to milk is $$2:10$$ or $$1:5$$. With $$10$$ ounces of flour, you could make as many as $$5$$ puddings. At the same time, you need $$50$$ more ounces of milk. The ratio of flour to milk is $$10:50$$, or $$1:5$$. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Solving Problems Involving Proportions->Dividing Quantities Based on Ratio
Problem: NBA playoffs are held annually. The $62$\textsuperscript{nd~}NBA playoffs were held in $2008$. When Judy was $10$ years old, the $58$\textsuperscript{th}~NBA playoffs were held. When was Judy born? 
Options:
aoVal: A
content: $$1991$$ 
aoVal: B
content: $$1992$$ 
aoVal: C
content: $$1994$$ 
aoVal: D
content: $$1996$$ 
aoVal: E
content: $$1998$$ 
Answer: C
Analysis: $62-58=4$  $2008-4-10=1994$ 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Interval Problems
Source: 其它
Problem: Frank stands in line and has $$76$$ people behind him, If there are a total of $$110$$ people in line, how many people are there in front of Frank? 
Options:
aoVal: A
content: $$33$$ 
aoVal: B
content: $$34$$ 
aoVal: C
content: $$35$$ 
aoVal: D
content: $$36$$ 
Answer: A
Analysis: If there are a total of $$110$$ people in line, subtract those behind Frank and Frank himself: $$110-76-1=33$$, the number in front of Frank. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Problems Involving Queuing->Finding the Total Number According to the Position of one Character in a Line
Problem: A teacher distributes scorecards to students, if everyone gets $3$ cards, there will be a shortage of $12$ cards. If everyone gets $2$ cards, all these cards will just be divided. Then, how many students are there in total? 
Options:
aoVal: A
content: $$4$$ 
aoVal: B
content: $$6$$ 
aoVal: C
content: $$10$$ 
aoVal: D
content: $$12$$ 
Answer: D
Analysis: If $2-1=1$ less card is given to students, the situation would transfer from \textquotesingle a shortage of $12$ cards\textquotesingle~ to \textquotesingle all cards are just divided\textquotesingle{} . Therefore. the shortage of $12$ cards is equal to the number of cards that everyone gets $1$ less card. So, there are in total $12\div1=12$ students. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Distribution Problems->Basic Distribution Problems
Problem: Which date is $$100$$ days after November $$6\text{th}$$? 
Options:
aoVal: A
content: February $$14$$ 
aoVal: B
content: February $$15$$ 
aoVal: C
content: February $$16$$ 
aoVal: D
content: February $$17$$ 
Answer: A
Analysis: It\textquotesingle s $$92$$ days $$\left( 30+31+31 \right)$$ from Nov. $$6$$ to Feb. $$6$$;~$$8$$ days later is $$100$$ days. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Periodic Problems->Periodic Problems of Time
Problem: Karl Lecter has been collecting $$1\text{p}$$, $$2\text{p}$$ and $$5\text{p}$$ coins in a jar. All but $$10$$ of his coins are $$1\text{p}$$ coins, all but $$10$$ are $$2\text{p}$$ coins, and all but $$10$$ are $$5\text{p}$$ coins. How much money does he have? 
Options:
aoVal: A
content: $$8\text{p}$$ 
aoVal: B
content: $$10\text{p}$$ 
aoVal: C
content: $$25\text{p}$$ 
aoVal: D
content: $$40\text{p}$$ 
aoVal: E
content: $$80\text{p}$$ 
Answer: D
Analysis: All but $$10$$ of the coins are $$1\text{p}$$ coins, which tells us that the number of $$2\text{p}$$ and $$5\text{p}$$ coins adds up to $$10$$. Similarly, the total of the $$1\text{p}$$ and $$5\text{p}$$ coins is $$10$$ and the total of the $$1\text{p}$$ and $$2\text{p}$$ coins is $$10$$. He therefore has $$5$$ of each coin, giving a total of $$(5 \times1)+ (5 \times2)+(5\times5)=40\text{p}$$. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Unitary Method Problems->Finding the Total Value using Unitary Method
Problem: When I add together the number of sides of a quadrilateral, a trapezoid, and a parallelogram, I get a total of． 
Options:
aoVal: A
content: $$12$$ 
aoVal: B
content: $$13$$ 
aoVal: C
content: $$14$$ 
aoVal: D
content: $$15$$ 
Answer: A
Analysis: Quadrilaterals, trapezoids, and parallelograms each have $$4$$ sides each, so the total number of sides is $$4+4+4=12$$. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Applying Addition and Subtraction->Simple Addition and Subtraction
Problem: A ship was attacked by pirates. One by one the pirates climb a rope to get on the ship. The pirate captain was the $$8$$\textsuperscript{th}~pirate to climb in line, and there were as many pirates in front of him as behind him. How many pirates climbed the rope? (2015 Math Kangaroo Problem, Level 1 - 2, Question \#18)  $$\textasciitilde$$ 
Options:
aoVal: A
content: $$7$$ 
aoVal: B
content: $$8$$ 
aoVal: C
content: $$12$$ 
aoVal: D
content: $$15$$ 
aoVal: E
content: $$16$$ 
Answer: D
Analysis: The pirate captain was the $$8$$\textsuperscript{th}~pirate and there were as many pirates in front of him as behind him, which means there were $$7$$ pirates in front of him and there were also $$7$$ pirates behind him. The total number of pirates is~~$$7+7+1=15$$. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Problems Involving Queuing->Finding the Total Number According to the Position of one Character in a Line
Problem: If $$28$$ January $$2000$$ falls on a Friday, which day of the week will $$28$$ January $2001$ fall on? 
Options:
aoVal: A
content: Thursday 
aoVal: B
content: Friday 
aoVal: C
content: Saturday 
aoVal: D
content: Sunday 
aoVal: E
content: Monday 
Answer: D
Analysis: $2000$ is a leap year and the period includes $29$ Feb, so there are $366$ days.  $366\div7=52$ R $2$  Friday $+$ $2$ days $\rightarrow$ Sunday 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Periodic Problems->Periodic Problems of Time
Problem: There are $$9$$ benches in the park, which have the same length. They are placed on one side of the road every $$7$$ meters from end to end. Given that the length of the road is $$74$$ meters, how long is the bench? 
Options:
aoVal: A
content: $$1$$ m 
aoVal: B
content: $$2$$ m 
aoVal: C
content: $$3$$ m 
aoVal: D
content: $$4$$ m 
Answer: B
Analysis: The $$9-1=8$$ intervals are $$7\times8=56$$ meters in total.  So, the sum of length of all the benches is $$74-56=18$$ meters and each of them is $$18\div9=2$$ meters. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Interval Problems->Straight Line Type with Objects on Both Sides
Problem: Nancy did an experiment. She inserted a paper straw vertically into the bottom of a bottle of black ink. Then, she found that the length of the black part of the paper straw was exactly $12$ cm. She turned the straw upside down and inserted the other end vertically into the bottom of the bottle. She found that the length of the part not colored of the straw was exactly half of that of all the black parts. The paper straw was~\uline{~~~~~~~~~~}~cm. 
Options:
aoVal: A
content: $$12$$ 
aoVal: B
content: $$24$$ 
aoVal: C
content: $$28$$ 
aoVal: D
content: $$30$$ 
aoVal: E
content: $$36$$ 
Answer: E
Analysis: The black parts of the straw was $$12+12=24$$ cm, so the part not colored was $$24\div2=12$$ cm. Thus, the length of the straw was $$24+12=36$$ cm. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules
Source: 其它
Problem: $A$, $B$ and $C$ start from the same place at the same time and chase a cyclist in front of them along the same road. The three cars catch up with him in $6$, $10$ and $12$ minutes respectively. Given that car $A$\textquotesingle s speed is $24$ kilometers per hour, car $B$\textquotesingle s speed is $20$ kilometers per hour. Find car $C$\textquotesingle s speed. 
Options:
aoVal: A
content: $$21$$ 
aoVal: B
content: $$19$$ 
aoVal: C
content: $$18$$ 
aoVal: D
content: $$16$$ 
aoVal: E
content: $$15$$ 
Answer: B
Analysis: B 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Newton's Problem of Cows and Fields->Basic Newton's Problem of Cows and Fields->Finding the Number of Cows
Source: 其它
Problem: A $$25\textbackslash\%$$ salt solution contains $$75$$ g of water. How many g of solution are there? 
Options:
aoVal: A
content: $$100$$ g 
aoVal: B
content: $$25.5$$ g 
aoVal: C
content: $$25$$ g 
aoVal: D
content: $$102$$ g 
Answer: A
Analysis: $$75\div(1-25\textbackslash\%) = 100$$ g. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Concentration Problems->Basic Concentration Problems
Problem: At the red light, seven buses of the same length stop in a line. Given that the length of each bus is $$5$$ meters, and the distance between each two adjacent buses is $$2$$ meters, how long is the line? 
Options:
aoVal: A
content: $$35$$ m 
aoVal: B
content: $$49$$ m 
aoVal: C
content: $$47$$ m 
aoVal: D
content: $$54$$ m 
Answer: C
Analysis: Among the $$7$$ buses, there are $$7-1=6$$ intervals. So the answer is $$7\times5+2\times6=47$$ meters. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Interval Problems->Straight Line Type with Objects on Both Sides
Problem: $$37$$ sakura trees were planted along one side of the road. The trees were planted at $$4m$$ intervals. After drivers complained the road was too pink, pineapple trees were planted on the other side of the road at $$6m$$ intervals. How many pineapple trees were planted, if there were sakura and pineapple trees at both ends of the road? 
Options:
aoVal: A
content: $$24$$ 
aoVal: B
content: $$25$$ 
aoVal: C
content: $$144$$ 
aoVal: D
content: $$18$$ 
Answer: B
Analysis: Number of sakura intervals = $$37-1 = 36$$  Length of road = $$36 \times 4 = 144m$$  Number of pineapple intervals = $$144m \div 6m = 24$$  Number of pineapple trees = $$24 + 1 = 25$$ 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Interval Problems
Source: 其它
Problem: Agnĳo has half as many apps as Sam who has a third as many apps as Naomi. Altogether, they have $$180$$ apps. How many apps does Sam have? 
Options:
aoVal: A
content: $$20$$ 
aoVal: B
content: $$30$$ 
aoVal: C
content: $$40$$ 
aoVal: D
content: $$60$$ 
aoVal: E
content: $$90$$ 
Answer: C
Analysis: Let Agnijo have $$n$$ apps. Now Sam has $$2n$$, and Naomi $$6n$$. Therefore $$n+2n+6n = 9n = 180$$, and so $$n = 180\div9 = 20$$. Hence Sam has $$2\times20 = 40$$ apps. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Questions Involving Sum, Difference and Multiples->Problems of Sum and Multiple
Problem: In the amusement park, Jessie is playing with darts. For each time she hits the bullseye, she can win two toys. At the beginning she has $$3$$ toys and at the end she has $$23$$ toys. How many times did she hit the bullseye? (Adapted from 2006 Math Kangaroo Problem, Level 3-4, Question \#5) 
Options:
aoVal: A
content: $$6$$ 
aoVal: B
content: $$8$$ 
aoVal: C
content: $$10$$ 
aoVal: D
content: $$5$$ 
aoVal: E
content: $$4$$ 
Answer: C
Analysis: $(23-3)\div2=10$ 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Unitary Method Problems->Unitary Method with One Variable
Problem: October 10th, 2021 is Sunday. What day is October $$26$$th of the same year?~\uline{~~~~~~~~~~}~ 
Options:
aoVal: A
content: Sunday 
aoVal: B
content: Monday 
aoVal: C
content: Tuesday 
aoVal: D
content: Wednesday 
Answer: C
Analysis: $26-10=16$, $16\div7=2\textbackslash{} \rm R\textasciitilde2$,~ it\textquotesingle s Tuesday. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Periodic Problems->Periodic Problems of Time->Period of Dates
Problem: $$380$$ is~\uline{~~~~~~~~~~}~more than $$254$$. 
Options:
aoVal: A
content: $$126$$ 
aoVal: B
content: $$124$$ 
aoVal: C
content: $$534$$ 
aoVal: D
content: $$634$$ 
Answer: A
Analysis: $$380-254=126$$ 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules
Source: 其它
Problem: There are $$11$$ flags on one side of a road. The distance between neighboring flags is $$5$$ meters. Grace walked from the first flag to the last flag. How many meters did she walk? (Adapted from 2007 Math Kangaroo Problem, Level 3-4, Question \#5) 
Options:
aoVal: A
content: $$45$$ 
aoVal: B
content: $$50$$ 
aoVal: C
content: $$55$$ 
aoVal: D
content: $$65$$ 
aoVal: E
content: $$75$$ 
Answer: B
Analysis: $(11-1)\times5=50$ 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Interval Problems->Straight Line Type with Objects on Both Sides->Planting Trees on Both Sides
Problem: Colin was preparing for the PE test. On the first day, he ran $1$ km. He decided that each day he would be running $100$ m more than the day before. How many meters did he run in total in the first $5$ days? 
Options:
aoVal: A
content: $$5600$$ 
aoVal: B
content: $$5700$$ 
aoVal: C
content: $$5800$$ 
aoVal: D
content: $$6000$$ 
aoVal: E
content: $$6100$$ 
Answer: D
Analysis: $1000 + 1100 + 1200 + 1300 + 1400 = 6000$ 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Practical Application of Arithmetic Progression
Source: 其它
Problem: Wendy had $$30$$ stickers. First, she gave Aiden three stickers. Then, she gave Terry $7$ stickers. Now, each of the three people had the same number of stickers. At the beginning, how many stickers did they have in total? 
Options:
aoVal: A
content: $$120$$ 
aoVal: B
content: $$90$$ 
aoVal: C
content: $$85$$ 
aoVal: D
content: $$60$$ 
aoVal: E
content: $$55$$ 
Answer: D
Analysis: Now, Wendy had $30-3-7=20$ stickers. Then, all of them had $20\times3=60$ stickers, which was equal to the total number of stickers they had at beginning. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Inverse Operation Problems
Problem: Bram bought $$60$$ beans for ＄$$3.00$$. At this price, $$100$$ beans cost． 
Options:
aoVal: A
content: ＄$$3.50$$ 
aoVal: B
content: ＄$$4.00$$ 
aoVal: C
content: ＄$$5.00$$ 
aoVal: D
content: ＄$$5.50$$ 
Answer: C
Analysis: If $$60$$ beans cost ＄$$3.00$$, then $$1$$ bean costs $$5$$￠and $$100$$ beans cost ＄$$5$$. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Unitary Method Problems->Finding the Total Value using Unitary Method