math-eval/TAL-SCQ5K huggingface dataset:
Problem: Grandma made some cheese dumplings and some blueberry dumplings. Altogether, she made $$31$$ dumplings. If she had made $$11$$ more cheese dumplings, then there would be the same number of blueberry dumplings as cheese dumplings. How many cheese dumplings did grandma make?（$$2009$$ Math Kangaroo Problem, Levels $$1-2$$, Question \#$$19$$） 
Options:
aoVal: A
content: $$10$$ 
aoVal: B
content: $$21$$ 
aoVal: C
content: $$20$$ 
aoVal: D
content: $$15$$ 
Answer: A
Analysis: $$(31-11)\div2=10.$$ 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Questions Involving Sum, Difference and Multiples->Problems Involving Sum and Difference
Problem: A worm is staying at the bottom of a well with a depth of $$13$$ metres. If it climbs up $$4$$ metres in the day time and slips down $$1$$ metre at night, which day will it climb up to the ground? 
Options:
aoVal: A
content: The $$4$$th day 
aoVal: B
content: The $$5$$th day 
aoVal: C
content: The $$6$$th day 
aoVal: D
content: The $$7$$th day 
Answer: A
Analysis: It climbs up $$4$$ metres in the day time and slips down $$1$$ metre at night, so it climbs up $$3$$ metres in a whole day. For the first $$3$$ days, it climbs $$9$$ metres. During the day time of the forth day, it climbs $$4$$ metres and reaches the ground. Then it needs four days to climb up to the ground. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Periodic Problems->Snail Climbing out of Well Problems
Problem: The volume of a cuboid is $216$. What is the least sum of the length of all its edges? 
Options:
aoVal: A
content: $$216$$ 
aoVal: B
content: $$32$$ 
aoVal: C
content: $$72$$ 
aoVal: D
content: $$96$$ 
aoVal: E
content: $$120$$ 
Answer: C
Analysis: Given the product of three numbers, the smaller the difference, the smaller the sum. $6\times6\times6=216$, so the answer is $6\times 12=72.$ 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Applying Addition and Subtraction->Problems Involving Extreme Value
Source: 其它
Problem: The state income tax where Kristin lives is levied at the rate of $$p\textbackslash\%$$ of the first ＄$$28000$$ of annual income plus $$ \left( {p+2} \right) \textbackslash\%$$ of any amount above ＄$$28000$$. Kristin noticed that the state income tax she paid amounted to $$ \left( {p+0.25} \right) \textbackslash\%$$ of her annual income. What was her annual income? ($$2001$$ AMC $$12$$ Problem, Question \#$$3$$) 
Options:
aoVal: A
content: ＄$$28000$$ 
aoVal: B
content: ＄$$32000$$ 
aoVal: C
content: ＄$$35000$$ 
aoVal: D
content: ＄$$42000$$ 
aoVal: E
content: ＄$$56000$$ 
Answer: B
Analysis: Method $$1$$：$$Let$$ the income amount be denoted by $$A$$.  We know that $$\frac{A\left( p+.25 \right)}{100}=\frac{28000p}{100}+\frac{\left( p+2 \right)\left( A-28000 \right)}{100}$$  We can now try to solve for $$A$$:  $$\left( p+.25 \right)A=28000p+Ap+2A-28000p-56000$$  $$0.25A=2A-56000$$  $$A=32000$$  so the answer is $$\boxed{ \text {B}}$$.  Method $$2$$: Let $$A$$, $$T$$ be Kristin\textquotesingle s annual income and the income tax total, respectively.  Notice that　  $$T=p\textbackslash\%\cdot 28000+\left( p+2 \right)\textbackslash\%\cdot \left( A-28000 \right)$$  $$\textasciitilde\textasciitilde\textasciitilde\textasciitilde$$$$=\left[ p\textbackslash\%\cdot 28000+p\textbackslash\%\cdot \left( A-28000 \right) \right]+2\textbackslash\%\cdot \left( A-28000 \right)$$  $$\textasciitilde\textasciitilde\textasciitilde\textasciitilde$$$$=p\textbackslash\%\cdot A+2\textbackslash\%\cdot \left( A-28000 \right)$$  We are also given that  $$T=\left( p+0.25 \right)\textbackslash\%\cdot A=p\textbackslash\%\cdot A+0.25\textbackslash\%\cdot A$$  Thus, $$p\textbackslash\%\cdot A+2\textbackslash\%\cdot \left( A-28000 \right)=p\textbackslash\%\cdot A+0.25\textbackslash\%\cdot A$$  $$2\textbackslash\%\cdot \left( A-28000 \right)=0.25\textbackslash\%\cdot A$$  Solve for $$A$$ to obtain $$A=32000$$. $$\boxed{ \text {B}}$$ 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Word Problems in Economics->Complex Money Word Problems
Problem: Kate sold $$10$$ dresses and Tasha sold thrice as many dresses as Kate. How many dresses did Tasha sell? 
Options:
aoVal: A
content: $$30$$ 
aoVal: B
content: $$26$$ 
aoVal: C
content: $$20$$ 
aoVal: D
content: $$42$$ 
Answer: A
Analysis: $3$~$\times$~$10$ = $30$ 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules
Problem: My cat snoozes for $$50$$ minutes in each hour. For how many hours a day does my cat snooze?　 
Options:
aoVal: A
content: $$5$$ 
aoVal: B
content: $$10$$ 
aoVal: C
content: $$15$$ 
aoVal: D
content: $$20$$ 
aoVal: E
content: $$50$$ 
Answer: D
Analysis: Sleeping for $$50$$ minutes every hour through the day is $$\frac{5}{6}$$ of every hour and thus $$\frac{5}{6}$$ of the day. Of $$24$$ hours, $$\frac{5}{6}$$ is $$20$$ hours. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Solving Problems Involving Fractions and Percentages->Knowing the Base
Problem: In a candy shop, the shopkeeper allows the exchange of $$3$$ sweet wrappers for $$1$$ new sweet. Peter has $$81$$ sweets at first. What is the largest possible number of sweets that Peter could have eaten? 
Options:
aoVal: A
content: $$119$$ 
aoVal: B
content: $$120$$ 
aoVal: C
content: $$121$$ 
aoVal: D
content: $$122$$ 
aoVal: E
content: Non of the above 
Answer: C
Analysis: With $$81$$ sweets wrappers. Peter can exchange for $$\frac{81}{3}=27$$ more sweets With the additional $$27$$ sweets wrappers, Peter can exchange for $$\frac{27}{3}=9$$ more sweets.  with the~~additional $$9$$ sweets wrappes. Peter can exchange for $$\frac{9}{3}=3$$ more sweets.  with the additional $$3$$ sweets wrappers. Peter can exchange for $$\frac{3}{3}=1$$ more sweet.  Therefore, the largest possible number of sweets Peter could have  eaten is $$81+27+9+3+1=121$$. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Applying Addition and Subtraction->Simple Addition and Subtraction
Problem: There are $$15$$ roses in a garden, and the number of daisies is $$5$$ more than $6$ times that of roses. How many flowers are there in the garden altogether? 
Options:
aoVal: A
content: $$95$$ 
aoVal: B
content: $$110$$ 
aoVal: C
content: $$120$$ 
aoVal: D
content: $$135$$ 
Answer: B
Analysis: $15\times6=90$  $90+5=95$  $95+15=110$ 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Questions Involving Sum, Difference and Multiples->Basic Computation of the Multiples
Problem: The sum of the ages of four children Anil, Benu, Chas and Dave is $$20$$. Dave is $$5$$ years older than Chas, and $$3$$ years older than Benu. Anil is $$3$$ years younger than Chas. How old is Anil? ． 
Options:
aoVal: A
content: $$1$$ 
aoVal: B
content: $$4$$ 
aoVal: C
content: $$5$$ 
aoVal: D
content: $$6$$ 
aoVal: E
content: $$9$$ 
Answer: A
Analysis: Two of the pieces of information relate to Chas, so we can try to relate the ages of the others to his age. Given Dave is $$5$$ years older than Chas, but $$3$$ years older than Benu, we can tell that Benu is $$2$$ years older than Chas. Given also that Anilis $$3$$ years younger than Chas, we can say that the sum of the ages of the four children is $$\left( -3+2+5=4 \right) $$ more than four times Chas' age. So four times Chas' age plus four is twenty, and so Chas must be four. So Anil is just $$ 1$$ year old. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Age Problems->Basic Relationships in Age Problems
Problem: Allen has a collection of $240$ fossils. Of these, $35$\% are fossilized snail shells. How many fossilized snail shells does Allen have? 
Options:
aoVal: A
content: $$205$$ 
aoVal: B
content: $$156$$ 
aoVal: C
content: $$84$$ 
aoVal: D
content: $$35$$ 
Answer: C
Analysis: $240 \times 35$\%=$84$,  Allen has $84$ fossilized snail shells. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Solving Problems Involving Fractions and Percentages->Percentages Word Problems
Source: 其它
Problem: Daniel had $51$ Lego pieces, he used $42$ pieces to make a mini statue. How many pieces of Lego does Daniel have now? 
Options:
aoVal: A
content: $$42$$ 
aoVal: B
content: $$19$$ 
aoVal: C
content: $$9$$ 
aoVal: D
content: $$0$$ 
Answer: C
Analysis: $51-42=9$ 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules
Source: 其它
Problem: Grandma made some cheese dumplings and some blueberry dumplings. Altogether, she made $$31$$ dumplings. If she had made $$11$$ more cheese dumplings, then there would be the same number of blueberry dumplings as cheese dumplings. How many cheese dumplings did grandma make?（2009 Math Kangaroo Problem, Question \#19） 
Options:
aoVal: A
content: $$10$$ 
aoVal: B
content: $$21$$ 
aoVal: C
content: $$20$$ 
aoVal: D
content: $$15$$ 
Answer: A
Analysis: $$(31-11)\div2=10.$$ 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Questions Involving Sum, Difference and Multiples->Problems Involving Sum and Difference
Problem: Wendy had $$30$$ stickers. First, she gave Aiden $3$ stickers. Then, she gave Terry $7$ stickers. Now, each of the three people has 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 has $30-3-7=20$ stickers. Thus, all of them have $20\times3=60$ stickers, which is equal to the total number of stickers they had at the beginning. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Inverse Operation Problems