math-eval/TAL-SCQ5K huggingface dataset:
Problem: After a test, teacher Valeria collects the data of the scores. Given that:  The average score of class $A$ is $76$.  The average score of class $B$ is $84$.  The average score of class $C$ is $89$.  The average score of class $A$ and $B$ is $79$.  The average score of all the three classes is $81$.  There are $40$ students in class $A.$ What is the ratio of number of students of class $B$ to that of $C$? 
Options:
aoVal: A
content: $3:2$ 
aoVal: B
content: $1:1$ 
aoVal: C
content: $4:1$ 
aoVal: D
content: $5:2$ 
aoVal: E
content: $2:3$ 
Answer: A
Analysis: There are $40\div\frac{84-79}{79-76}=24$ students in class $B.$  There are $(40+24)\div\frac{89-81}{81-79}=16$ students in class $C.$  $24:16=3:2$ 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Concentration Problems
Source: 其它
Problem: Ala, Lenka and Miso went out for dessert. Lenka paid $$4$$ dollars and $$50$$ cents for three scoops of ice cream. Miso paid $$3$$ dollars and $$60$$ cents for two cookies. How much did Ala pay for one scoop of ice cream and one cookie? (2011 Math Kangaroo Problem, Level 3 - 4, Question \#6) 
Options:
aoVal: A
content: $3$~dollars and~$30$~cents 
aoVal: B
content: $4$~dollars and~$80$~cents 
aoVal: C
content: $5$~dollars and~$10$~cents 
aoVal: D
content: $6$~dollars and~$30$~cents 
aoVal: E
content: $8$~dollars and~$10$~cents 
Answer: A
Analysis: $4$ dollars and $50$ cents are equal to $450$ cents, so one scoop of ice cream costs $450 \div 3 = 150$ cents. $$3$$ dollars and $$60$$ cents are equal to $360$ cents, so one cookie costs $360 \div 2 = 180$ cents. Thus, Ala should pay $ 150 + 180 = 330$ cents for one scoop of ice cream and one cookie. $330$ cents are equal to $3$ dollars and $30$ cents, so the answer is $A$. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Distribution Problems->Basic Distribution Problems->Basic Problems of Distribution
Problem: Danni bought a painting for £$$42$$ last year. She sold it this year for £$$55$$. How much did she earn? 
Options:
aoVal: A
content: $$5$$ 
aoVal: B
content: $$10$$ 
aoVal: C
content: $$13$$ 
aoVal: D
content: $$23$$ 
Answer: C
Analysis: $$55-42=13$$. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Applying Addition and Subtraction->Simple Addition and Subtraction
Problem: There are $38$ chickens and rabbits in a farm in total. The number of ducks is $2$ more than five times that of rabbits. How many ducks are there in the farm? 
Options:
aoVal: A
content: $$32$$ 
aoVal: B
content: $$33$$ 
aoVal: C
content: $$29$$ 
aoVal: D
content: $$26$$ 
aoVal: E
content: $$24$$ 
Answer: A
Analysis: $$(38-2)\div (1+5)=36\div 6=6$$  $$6\times 5+2=32$$ 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Questions Involving Sum, Difference and Multiples->Problems of Sum and Multiple
Problem: A fruit shop brought in some fruit. A quarter of them were sold last week. This week another $120$ kilograms were sold. One third of the original fruit is left now. How many kilos of fruit did the fruit shop buy? 
Options:
aoVal: A
content: $$150\text{kg}$$ 
aoVal: B
content: $$190\text{kg}$$ 
aoVal: C
content: $$240\text{kg}$$ 
aoVal: D
content: $$288\text{kg}$$ 
aoVal: E
content: $$324\text{kg}$$ 
Answer: D
Analysis: $$1-\frac{1}{4}-\frac{1}{3} =\frac{5}{12}$$, $$120\div\frac{5}{12}=288\text{kg}$$. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Solving Problems Involving Fractions and Percentages->Finding the Base
Problem: Nicole has a $$500$$ ml bottle of mouthwash. Every morning, she uses two capfuls of mouthwash. Each capful contains $$4$$ ml of mouthwash.  If Nicole open a new bottle of mouthwash on $$12$$ April, on which of these dates will she use up the whole bottle of mouthwash? 
Options:
aoVal: A
content: $$12$$ June 
aoVal: B
content: $$13$$ June 
aoVal: C
content: $$14$$ August 
aoVal: D
content: $$15$$ August 
Answer: B
Analysis: Nicole uses $$4\times2=8$$ ml of mouthwash every day.  $$500\div8=62$$ R $$4$$  So, $$62$$ days after $$12$$ April, the bottle of mouthwash will become empty.  There are $$18$$ remaining days in April and $$31$$ days in May.  $$62-18-31=13$$ 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules
Source: 其它
Problem: A bag of toffee is $5$ dollars, a bag of cotton candy is $3$ dollars, and a bag of orange candy is $12$ dollars. Now, the candy shop decides to mix $25$ bags of toffee, $60$ bags of cotton candy, and $15$ bags of orange candy for $100$ bags of assorted candy. What should be the price of the assorted candy in dollars to keep the total revenue unchanged? 
Options:
aoVal: A
content: $$3$$ 
aoVal: B
content: $$3.5$$ 
aoVal: C
content: $$4$$ 
aoVal: D
content: $$4.85$$ 
aoVal: E
content: $$5$$ 
Answer: D
Analysis: Total revenue: $25\times 5+60\times 3+15\times 12=485$ dollars  A bag of assorted candy: $$485\div 100=4.85$$ dollars 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Average Problems
Problem: Jill is now $$14$$ years old. Jack is now $$6$$ years older than Jill was $$2$$ years ago. How old is Jack now? 
Options:
aoVal: A
content: $$10$$ 
aoVal: B
content: $$18$$ 
aoVal: C
content: $$20$$ 
aoVal: D
content: $$22$$ 
Answer: B
Analysis: Jill is now $$14$$. Two years ago, she was $$12$$. Since Jack is now $$6$$ years older than Jill was $$2$$ years ago, Jack is now $$12 + 6= 18$$. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Age Problems->Basic Relationships in Age Problems
Problem: Mr. Ronald sold $$300$$ burgers on Monday. He sold $$133$$ fewer burgers on Tuesday.  How many burgers did he sell on Tuesday? 
Options:
aoVal: A
content: $$167$$ 
aoVal: B
content: $$133$$ 
aoVal: C
content: $$166$$ 
aoVal: D
content: $$433$$ 
aoVal: E
content: $$177$$ 
Answer: A
Analysis: $$300-133=167$$ 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Applying Addition and Subtraction->Simple Addition and Subtraction
Problem: To make coleslaw Cathy uses twice as much carrot (by weight) as cabbage. She then adds half as much yoghurt as cabbage. A pot of Cathy\textquotesingle s coleslaw weighs $$175\text{g}$$. How many pots of coleslaw can she make with a $$2 \text{kg}$$ cabbage? 
Options:
aoVal: A
content: $$30$$ 
aoVal: B
content: $$40$$ 
aoVal: C
content: $$50$$ 
aoVal: D
content: $$60$$ 
aoVal: E
content: $$80$$ 
Answer: B
Analysis: The ratio by weight of carrot: cabbage:yoghurt$$=2:1:0.5$$ which we can double to give $$4:2:1$$. We can see that here $$2$$ represents the amount of cabbage and we want $$2 \text{kg}$$ of cabbage, so there will be $$4+2+1=7\text{kg}$$ of coleslaw altogether. Therefore the number of pots is $$7000\div 175 =40$$. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Solving Problems Involving Proportions->Basic Units
Problem: In a traditional Chinese novel, there are $108$ heroes, three of whom are women. How many male heroes are there in this novel？(adapted from $$2007$$ Math kangaroo Problems, Level $$1-2$$, Question \#$$6$$) 
Options:
aoVal: A
content: $$105$$ 
aoVal: B
content: $$98$$ 
aoVal: C
content: $$96$$ 
aoVal: D
content: $$94$$ 
aoVal: E
content: $$90$$ 
Answer: A
Analysis: There are $108$ heroes in total. Subtract three heroines to get the number of male heroes. That is $108-3=105$. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Applying Addition and Subtraction->Simple Addition and Subtraction
Problem: A total of $$46$$ bicycles and tricycles are in the garage. If there are $$100$$ wheels in total, there should be~\uline{~~~~~~~~~~}~tricycles in the garage. 
Options:
aoVal: A
content: $$8$$ 
aoVal: B
content: $$12$$ 
aoVal: C
content: $$24$$ 
aoVal: D
content: $$38$$ 
aoVal: E
content: $$6$$ 
Answer: A
Analysis: $(100-2\times46)\div(3-2)=8$ 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Chicken-Rabbit Problems->Solving Chicken-Rabbit Problems by Using Hypothesis
Problem: Amanda is $$7$$ years old last year. In $$2$$ years times, she is double of her sister\textquotesingle s age. What is the sum of their age this year? 
Options:
aoVal: A
content: $$11$$ 
aoVal: B
content: $$12$$ 
aoVal: C
content: $$13$$ 
aoVal: D
content: $$14$$ 
aoVal: E
content: $$15$$ 
Answer: A
Analysis: $$2$$ years time, Amanda $$10$$ and sister $$5$$. Total $$15$$.  This year, $$15-2-2=11$$. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Age Problems
Source: 其它
Problem: If I have $$2500$$ quarters, then I have． 
Options:
aoVal: A
content: ＄$$100$$ 
aoVal: B
content: ＄$$500$$ 
aoVal: C
content: ＄$$625$$ 
aoVal: D
content: ＄$$1000$$ 
Answer: C
Analysis: Every $$4$$ quarters is ＄$$1$$; the number of dollars I have is $$2500\div 4=$$＄$$625$$. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Solving Problems Involving Fractions and Percentages->Knowing the Base
Problem: Bud needs to recite 120 words for a dictation in a week (7 days). She plans to recite 15 words everyday. Can she be fully prepared for the test? 
Options:
aoVal: A
content: Yes, she can. 
aoVal: B
content: No, she can\textquotesingle t. 
Answer: B
Analysis: $$15\times7=105$$，$$105\textless120$$． 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Applying of Multiplication and Division
Problem: Two plums and one cherry weigh $$80\text{g}$$. One cherry and one plum weigh $$50\text{g}$$. 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 $$p+c=50$$. We can deduce that $$p=80-50= 30$$, and so four plums weigh $$120 \text{g}$$.  Revise from ($$2017$$ Primary Mathematics Challenge-February, Question \#$$18$$) 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Equation Word Problems->Equivalent Substitution in Equation Word Problems
Problem: In a mathematics contest with ten problems, a student gains $$5$$ points for a correct answer and loses $$2$$ points for an incorrect answer. If Olivia answered every problem and her score was $$29$$, how many correct answers did she have? 
Options:
aoVal: A
content: $$5$$ 
aoVal: B
content: $$6$$ 
aoVal: C
content: $$7$$ 
aoVal: D
content: $$8$$ 
aoVal: E
content: $$9$$ 
Answer: C
Analysis: Suppose Olivia has $$x$$ correct answers. $$5x-2(10-x)=29$$, $$x=7$$. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Equation Word Problems->Word Problems for Linear Equations with One Variable
Problem: On Kellin\textquotesingle s $$13$$th birthday, Allen was four times her age. On Kellin\textquotesingle s $$24$$st birthday, how old was Allen? 
Options:
aoVal: A
content: $$39$$ 
aoVal: B
content: $$44$$ 
aoVal: C
content: $$ 52 $$ 
aoVal: D
content: $$ 63 $$ 
aoVal: E
content: $$ 74$$ 
Answer: D
Analysis: On Kellin\textquotesingle s $$13$$th birthday, Allen was four times her age, that is $$52$$. It is then eleven years until Kellin\textquotesingle s $$24$$st birthday, so Allen\textquotesingle s age at that time was $$52 +11 =63$$. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Age Problems->Basic Relationships in Age Problems
Problem: In a jar of red, green, and blue marbles, all but $$6$$ are red marbles, all but $$8$$ are green, and all but $$4$$ are blue. How many marbles are in the jar? 
Options:
aoVal: A
content: $$6$$ 
aoVal: B
content: $$8$$ 
aoVal: C
content: $$9$$ 
aoVal: D
content: $$10$$ 
aoVal: E
content: $$12$$ 
Answer: C
Analysis: Suppose there are $$x$$ red marbles, $$y$$ green marbles and $$z$$ blue marbles. Thus, we can get  $$\begin{cases} x+y=4① \textbackslash\textbackslash{} y+z=6②\textbackslash\x+z=8③\end{cases}$$,  ①$$+$$②$$+$$③: $$2x+2y+2z=18$$, $$x+y+z=9$$. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Equation Word Problems->Multivariate Linear Equation Word Problems
Problem: Hailey is $$6$$ years old and Xavier is $$7$$ years old this year. What is the sum of their ages after $$10$$ years? 
Options:
aoVal: A
content: $$23$$ 
aoVal: B
content: $$26$$ 
aoVal: C
content: $$33$$ 
aoVal: D
content: $$35$$ 
aoVal: E
content: $$43$$ 
Answer: C
Analysis: $6+7+10\times2=33$ 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Average Problems
Problem: Water from the first faucet fills the swimming pool in $20$ hours. Water from each of the three other faucets fills the same swimming pool $3$ times faster. In how many hours will the swimming pool be filled if all three faucets are opened? 
Options:
aoVal: A
content: $$2$$ 
aoVal: B
content: $$3$$ 
aoVal: C
content: $$4$$ 
aoVal: D
content: $$5$$ 
Answer: A
Analysis: The efficiency of the first faucet is $\frac1{20}$ and that of the other two is $\frac3{20}$.  Thus it takes $1\div (\frac1{20}+\frac3{20}\times3)=2$ hours to fill the pool. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Work Word Problems
Source: 其它
Problem: A red kangaroo and a gray kangaroo weigh $$139\textasciitilde\text{kg}$$ altogether. The red kangaroo weighs $$35\textasciitilde\text{kg}$$ less than the gray kangaroo. How much does the gray kangaroo weigh? 
Options:
aoVal: A
content: $$104\text{kg}$$ 
aoVal: B
content: $$52\text{kg}$$ 
aoVal: C
content: $$87\text{kg}$$ 
aoVal: D
content: $$96\text{kg}$$ 
aoVal: E
content: $$53\text{kg}$$ 
Answer: C
Analysis: $$139-35=104$$  Red kangaroo: $$104 \div2 = 52$$  Gray kangaroo: $$52+35=87$$ 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Questions Involving Sum, Difference and Multiples->Problems Involving Sum and Difference
Source: 其它
Problem: I ate half of 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: $\frac12$ 
aoVal: C
content: $\frac13$ 
aoVal: D
content: $\frac23$ 
aoVal: E
content: $\frac16$ 
Answer: E
Analysis: $(1-\frac12)\times(1-\frac23)=\frac12\times\frac13=\frac16$. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Solving Problems Involving Fractions and Percentages
Source: 其它
Problem: In a group of $$48$$ children, the ratio of boys to girls is $$3:5$$. How many boys must join the group to make the ratio of boys to girls $$5:3$$? 
Options:
aoVal: A
content: $$48$$ 
aoVal: B
content: $$40$$ 
aoVal: C
content: $$32$$ 
aoVal: D
content: $$24$$ 
aoVal: E
content: $$8$$ 
Answer: C
Analysis: Initially there are $$48$$ children of whom $$\frac{3}{8}$$ are boys and $$\frac{5}{8}$$ are girls, so there are $$18$$ boys and $$30$$ girls. When more boys join, there are still $$30$$ girls but now they form $$\frac{3}{8}$$ of the total. So the total number of pupils is now $$\frac{8}{3}\times30= 80$$, of whom $$80-30=50$$ are boys. Hence the number of boys joining is $$50-18=32$$. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Solving Problems Involving Proportions->Complex Ratio Problems
Problem: Lucy joined a math test which had the scoring rules below: for any correct answer, she got $5$ points; for any skipped or wrong answer, she lost $7$ points. There were $20$ problems in total. When she finished the test, she got only $4$ points. How many problems did she answer correct? 
Options:
aoVal: A
content: $$4$$ 
aoVal: B
content: $$8$$ 
aoVal: C
content: $$12$$ 
aoVal: D
content: $$16$$ 
aoVal: E
content: $$3$$ 
Answer: C
Analysis: The total points is $20\times5=100.$ If there was one wrong anwser, she would lose $5+7=12$ points.  $(20\times5-4)\div(7+5)=8$, so she got $20-8=12$ problems correct. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Chicken-Rabbit Problems
Source: 其它
Problem: Nathan has $$4$$ red shirts, $$6$$ yellow shirts and $$8$$ white shirts. What fraction of his shirts are white? Give your answer in its simplest form. 
Options:
aoVal: A
content: $$\frac{8}{18}$$ 
aoVal: B
content: $$\frac{3}{9}$$ 
aoVal: C
content: $$\frac{4}{9}$$ 
aoVal: D
content: $$\frac{4}{5}$$ 
Answer: C
Analysis: White shirts $$=8$$  Total shirts $$=4+6+8=18$$  Fraction of shirts that are white $$=\frac{8}{18}=\frac{4}{9}$$ 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Solving Problems Involving Fractions and Percentages->Finding the Rate
Problem: An empty truck weighs $$2000$$ kg. After the truck was loaded, the freight (that is, the load) made up $$80\textbackslash\%$$ of the weight of the loaded truck. At the first stop one fourth of the freight was unloaded. What percent of the loaded truck\textquotesingle s weight did the freight make up after that? ($$2003$$ Math Kangaroo Problem, Level $$7-8$$, Question \#$$17$$) 
Options:
aoVal: A
content: $$20\textbackslash\% $$ 
aoVal: B
content: $$25\textbackslash\%$$ 
aoVal: C
content: $$55\textbackslash\%$$ 
aoVal: D
content: $$ 60\textbackslash\%$$ 
aoVal: E
content: $$75\textbackslash\%$$ 
Answer: E
Analysis: Actually, we don\textquotesingle t need the weight of the empty truck.  Suppose the weight of the loaded truck before the first stop is $5x$ kg, and the weight of the feight is $4x$ kg. After the first stop, $4x \times \frac14=x$ kg of freight is unloaded. Now, the freight weighs $4x-x=3x$ kg, and the loaded truck weighs $5x-x=4x$ kg. Thus, the percent is $\frac{3x}{4x}=\frac34=75\textbackslash\%$. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Solving Problems Involving Fractions and Percentages->Finding the Rate
Problem: Jodie has just begun to read a $$160$$-page book. If she reads $$20$$ pages every day, she will finish the book in ． 
Options:
aoVal: A
content: $$8$$ days 
aoVal: B
content: $$18$$ days 
aoVal: C
content: $$20$$ days 
aoVal: D
content: $$80$$ days 
Answer: A
Analysis: Jodie has just begun to read a $$160$$-page book. If she reads $$20$$ pages every day, she will finish the book in $$160 \div 20 =8$$ days. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Unitary Method Problems->Finding the Total Value using Unitary Method
Problem: Tadek has $$7$$ zloty (unit of money) more than Witek. Witek has $$10$$ zloty less than Karol. Witek and Karol have $$28$$ zloty together. How much money does Tadek have? 
Options:
aoVal: A
content: $$4$$ 
aoVal: B
content: $$5$$ 
aoVal: C
content: $$10$$ 
aoVal: D
content: $$16$$ 
aoVal: E
content: $$9$$ 
Answer: D
Analysis: Witek has $$(28 - 10) \div 2 = 9$$ zloty, so Tadek has $$9 + 7 = 16$$ zloty. 
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
Source: 其它
Problem: Ben and Ken have some books. After Ben sends $4$ books to Ken, he has $1$ book less than Ken. Which of the following is true? 
Options:
aoVal: A
content: Originally, Ben had $3$ books more than Ken. 
aoVal: B
content: Originally, Ben had $4$ books more than Ken. 
aoVal: C
content: Originally, Ben had $1$ books less than Ken. 
aoVal: D
content: Originally, Ben had $9$ books more than Ken. 
aoVal: E
content: Originally, Ben had $7$ books more than Ken. 
Answer: E
Analysis: $4+4-1=7$ 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Average Problems 
Source: 其它
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 $90\textbackslash\%$ of the problems she solved alone, but overall $60\textbackslash\%$ of her answers were correct. Zoe had correct answers to $70\textbackslash\%$ of the problems she solved alone. What was Zoe\textquotesingle s overall percentage of correct answers? (adapted from 2017 AMC 8, Question \#14) 
Options:
aoVal: A
content: $$45$$ 
aoVal: B
content: $$48$$ 
aoVal: C
content: $$50$$ 
aoVal: D
content: $$60$$ 
aoVal: E
content: $$65$$ 
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 $90$ problems correct by herself, and got $120$ problems correct overall.  We also know that Zoe had $70$ problems she did correct alone.  We can see that the total amount of correct problems Chloe and Zoe did was $120-90=30$. Therefore Zoe has $30+70=100$ problems out of $200$ problems correct. Thus $\frac{100}{200} = 50$ percent. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Solving Problems Involving Fractions and Percentages->Percentages Word Problems
Source: 其它
Problem: In the enchanted garden of the Green King, there are apple trees that grow golden apples. Every day, $$5$$ golden apples become ripe on each tree, and at the end of each day they fall from the trees. Today, the Green Gardener has picked up $$20$$ ripe apples that fell under the trees last night. How many enchanted trees are there in the garden?（$$2005$$ Math Kangaroo Problem, Levels $$1-2$$, Question \#$$1$$） 
Options:
aoVal: A
content: $$4$$ 
aoVal: B
content: $$5$$ 
aoVal: C
content: $$6$$ 
aoVal: D
content: $$7$$ 
aoVal: E
content: $$8$$ 
Answer: A
Analysis: $$5$$ apples fall down from each trees. "Each"~is a sign of division. So, the answer is $$20\div5=4$$. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Unitary Method Problems->Applying Multiplication and Division
Problem: Add the number of days in January March, April, May, June, July, August, September, October, November and December. 
Options:
aoVal: A
content: $$334$$ 
aoVal: B
content: $$335$$ 
aoVal: C
content: $$336$$ 
aoVal: D
content: $$337$$ 
Answer: D
Analysis: The only missing month is February. The sum will be either $$365$$ - $$28$$ or (in leap years) $$366$$ - $$29$$. Both are equal to $$337$$． 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Periodic Problems->Periodic Problems of Time
Problem: Some cows live on a grassland. $10$ cows can eat all grass in $9$ days. $12$ cows can eat all grass in $7$ days. The amount of new grass that grows each day is constant. $24$ cows can eat all grass in~\uline{~~~~~~~~~~}~days. 
Options:
aoVal: A
content: $$2$$ 
aoVal: B
content: $$3$$ 
aoVal: C
content: $$4$$ 
aoVal: D
content: $$4.5$$ 
aoVal: E
content: $$5$$ 
Answer: B
Analysis: Suppose a cow can eat $1$ m\textsuperscript{2}~of grass each day.  The amount of new grass that grows each day: $(9\times10-12\times7)\div(9-7)=3$ m\textsuperscript{2}.  The amount of grass originally: $$90-9\times 3=63$$ m\textsuperscript{2}.  $$24$$ cows can eat all grass in $$63\div \left( 24\times1-3 \right)=3$$ days. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Newton's Problem of Cows and Fields
Problem: The distance between Exeter and London is $$175$$ miles. Sam left Exeter at $$10:00$$ on Tuesday for London. Morgan left London for Exeter at $$13:00$$ the same day. They travelled on the same road. Up to the time when they met, Sam\textquotesingle s average speed was $$25$$ miles per hour, and Morgan\textquotesingle s average speed was $$35$$ miles an hour.  At what time did Sam and Morgan meet? 
Options:
aoVal: A
content: $$17:00$$ 
aoVal: B
content: $$15:55$$ 
aoVal: C
content: $$15:30$$ 
aoVal: D
content: $$15:00$$ 
aoVal: E
content: $$14:40$$ 
Answer: E
Analysis: Sam left Exeter three hours before Morgan left London, and travelled $$3\times 25$$ miles $$=75$$ miles in the three hours to $$13:00$$. So at $$13:00$$, the distance between Sam and Morgan was $$ \left( {175-75} \right) $$ miles $$=100$$ miles.  Let the time in hours between $$13:00$$ and the time at which Sam and Morgan met be $$t$$.  Then $$25t+35t=100$$. So $$t=\frac{{100}}{{60}}$$ hours $$=100$$ minutes $$=1$$ hour $$40$$ minutes. So Sam and Morgan met at $$14:40$$. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules