math-eval/TAL-SCQ5K huggingface dataset:
Problem: One kind of computer can make $$4\times {{10}^{9}}$$ operations per second.~ How many operations can it make in $$5\times {{10}^{2}}$$ seconds? 
Options:
aoVal: A
content: $$4\times {{10}^{11}}$$ 
aoVal: B
content: $$2\times {{10}^{11}}$$ 
aoVal: C
content: $$2\times {{10}^{12}}$$ 
aoVal: D
content: $$20\times {{10}^{18}}$$ 
aoVal: E
content: $$2\times {{10}^{19}}$$ 
Answer: C
Analysis: $$4\times {{10}^{9}}\times 5\times {{10}^{2}}$$  $$=20\times {{10}^{9}}+2$$  $$=2\times {{10}^{1}}\times {{10}^{11}}$$  $$=2\times {{10}^{1+11}}$$  $$=2\times {{10}^{12}}$$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Operations through Formulas
Problem: $$9+9+9+9 +9+9+9=9\times$$． 
Options:
aoVal: A
content: $$6$$ 
aoVal: B
content: $$7$$ 
aoVal: C
content: $$8$$ 
aoVal: D
content: $$9$$ 
Answer: B
Analysis: $$9+9+9+9 +9+9+9=$$ seven $$9$$\textquotesingle s $$=9\times7$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Multiplication and Division
Problem: $$(3\times 1)+(3\times 2)+(3\times 3)+(3\times 4)=3\times $$． 
Options:
aoVal: A
content: $$5$$ 
aoVal: B
content: $$1+2+3+4$$ 
aoVal: C
content: $$12$$ 
aoVal: D
content: $$1\times 2\times 3\times 4$$ 
Answer: B
Analysis: $$3+6+9+12=30=3\times 10=3\times (1+2+3+4)$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Mixed Operations
Problem: Lucas pays for $4$ pens and $3$ pencils. Bryan spends twice as much paying for $2$ pens and $16$ pencils. A pen is how many times as expensive as a pencil? 
Options:
aoVal: A
content: $\frac{3}{2}$ 
aoVal: B
content: $\frac{5}{3}$ 
aoVal: C
content: $\frac{7}{4}$ 
aoVal: D
content: $$2$$ 
aoVal: E
content: $\frac{13}{4}$ 
Answer: B
Analysis: B 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: Abel and Bella are friends and they always want to sit next to each other. How many ways are there to arrange Abel, Bella and three of their acquaintances to sit in a row, while satisfying the condition? 
Options:
aoVal: A
content: $$12$$ 
aoVal: B
content: $$16$$ 
aoVal: C
content: $$24$$ 
aoVal: D
content: $$32$$ 
aoVal: E
content: $$48$$ 
Answer: E
Analysis: E 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: $$100$$ hundreds $$+10$$ tens $$+1$$ one $$=$$． 
Options:
aoVal: A
content: $$111$$ 
aoVal: B
content: $$1101$$ 
aoVal: C
content: $$1011$$ 
aoVal: D
content: $$10101$$ 
Answer: D
Analysis: $$100\times 100+10\times 10+1\times 1=10000+100+1=10101$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Mixed Operations
Problem: When calculating $$16\times 29$$, which of the following choices is correct? 
Options:
aoVal: A
content: $$144$$ in the blue box represents $$9\times 16$$. 
aoVal: B
content: $$32$$ in the red box represents $$2\times 16$$. 
aoVal: C
content: The product is $$32+144 =176$$. 
aoVal: D
content: There is no regrouping in the calculation. 
Answer: A
Analysis: $$32$$ in the red box represents $$20 \times 16 = 320$$, and the product should be $$320 +144 = 464$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Multiplication and Division
Problem: David measured the length of his garden.  It was $$15$$ metres to the nearest tenth of a metre.  Between what limits was the actual length? 
Options:
aoVal: A
content: $$14.995\leqslant $$ the actual length $$\leqslant 15.005$$ 
aoVal: B
content: $$14.9\leqslant $$ the actual length $$\textless~15.1$$ 
aoVal: C
content: $$14.95\leqslant $$ the actual length $$\textless{} 15.05$$ 
aoVal: D
content: $$14.99\leqslant $$ the actual length $$\leqslant 15.01$$ 
aoVal: E
content: $$14.5\textless$$ the actual length $$\textless15.5$$ 
Answer: C
Analysis: 14.9500000\ldots{}  15.0499999\ldots{} 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Decimals->Basic Understanding of Decimals->Finding Approximate Values
Problem: Let $$a\square b=ab+a+b$$ for any integers $$a$$ and $$b$$.  The solution of the equation $$3\square 5=2\square x$$ is: $x = $~\uline{~~~~~~~~~~}~ 
Options:
aoVal: A
content: $$3$$ 
aoVal: B
content: $$6$$ 
aoVal: C
content: $$7$$ 
aoVal: D
content: $$10$$ 
aoVal: E
content: $$12$$ 
Answer: C
Analysis: $3\square 5=2\square x \implies 15+3+5=2x+2+x$, hence $x=7$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Operations with New Definition->Finding Unknowns Using the Given Operations
Problem: If $A=1+3+5+7+\cdots +99$, and $B=2+4+6+\cdots +100$, what is the value of $B-A$? 
Options:
aoVal: A
content: $$0$$ 
aoVal: B
content: $$49$$ 
aoVal: C
content: $$50$$ 
aoVal: D
content: $$99$$ 
aoVal: E
content: $$100$$ 
Answer: C
Analysis: $B-A=(2-1)+(4-3)+(6-5)+\cdots +(100-99)=50$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Sequences and Number Tables->Arithmetic Sequences
Source: 其它
Problem: A certain game uses a set of cards containing odd numbers. There are one 1, three 3s, two 5s, one 7, and three 9s. The mean value of this deck is 5.4, with a standard deviation of 2.8. If someone purchased three more decks of these cards and combined~them into single large deck, what would the new mean and standard deviation be? 
Options:
aoVal: A
content: Mean = 5.4; standard deviation = 2.8 
aoVal: B
content: Mean = 5.4; standard deviation = 2.95 
aoVal: C
content: Mean = 16.2; standard deviation = 2.90 
aoVal: D
content: Mean = 16.2; standard deviation = 8.4 
aoVal: E
content: Mean = 21.6standard deviation = 11.2 
Answer: A
Analysis: The mean is a measure of center. While it can be affected by adding new values ,the purchase of second deck would merely add more cards with the same values .As a result ,the mean would not change .Eliminate (C), (D),and (E). The standard deviation is a measure of sprees. As above, unique values will cause the standard deviation to change. However, since the new decks are identic -call to the old, the standard deviation will not change. Eliminate (B). 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: If $A:B=2:3$, $B:C=3:4$, find $A:B:C$=~\uline{~~~~~~~~~~}~. 
Options:
aoVal: A
content: $2:3:4$ 
aoVal: B
content: $8:6:15$ 
aoVal: C
content: $6:3:9$ 
aoVal: D
content: $16:6:15$ 
Answer: B
Analysis: $A:B=4:3$  $B:C=2:5$  $A:B:C=8:6:15$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Ratios and Proportions
Source: 其它
Problem: Boyuan bought $$3$$ types of candies. He bought $$5$$ times as many toffee as lollipop. He bought $$5$$ more chocolate bar than lollipop. If he has $$7$$ lollipop, how many candies did he bought in total? 
Options:
aoVal: A
content: $$17$$ 
aoVal: B
content: $$32$$ 
aoVal: C
content: $$40$$ 
aoVal: D
content: $$54$$ 
Answer: D
Analysis: Lollipop: $$7$$  Toffee: $$7\times5=35$$  Chocolate bar: $$7+5=12$$  Total: $$7+35+12=54$$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers
Problem: Tom and Jerry found $\frac{1}{2}$ of a steak in the kitchen. Tom ate $\frac{1}{2}$ of the leftover steak and Jerry ate $\frac{1}{4}$ of it. They decided to give the rest to their friend Spike. What portion of the original steak would Spike get? 
Options:
aoVal: A
content: $\frac{1}{4}$ 
aoVal: B
content: $\frac{1}{8}$ 
aoVal: C
content: $\frac{1}{16}$ 
aoVal: D
content: $\frac{1}{32}$ 
Answer: B
Analysis: $\frac{1}{2}-\frac{1}{2}\times \frac{1}{2}-\frac{1}{2}\times \frac{1}{4}=\frac{1}{8}$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Fractions
Source: 其它
Problem: Free-response questions on the AP Statistics Exam are graded 4, 3, 2, 1, or~ 0. Question 2 on the exam was of moderate difficulty. The average score on question 2 was 2.05 with a standard deviation of 1. To the nearest tenth, what score was achieved by a student who was at the 90th percentile of all students on the test? You may assume that the scores on the question were approximately normally distributed. 
Options:
aoVal: A
content: $$3.5$$ 
aoVal: B
content: $$3.3$$ 
aoVal: C
content: $$2.9$$ 
aoVal: D
content: $$3.7$$ 
aoVal: E
content: $$3.1$$ 
Answer: B
Analysis: \textbf{Z = 1.28}  \textbf{$$\frac{x-2.05}{1}$$ = 1.28}  \textbf{x= 3.33} 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Fractions
Source: 其它
Problem: Josie and Holly share $$48$$ dollars between them.  Josie has three times as much as Holly.  How much does Holly have? 
Options:
aoVal: A
content: $12$ dollars 
aoVal: B
content: $24$ dollars 
aoVal: C
content: $$36$$ dolllars 
aoVal: D
content: $$40$$ dollars 
aoVal: E
content: $$45$$ dollars 
Answer: A
Analysis: $48 \div (3 + 1) = 12$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: In an arithmetic sequence with $$13$$ terms, if the seventh term is $$20$$, what is the sum of all the terms? 
Options:
aoVal: A
content: $130 
aoVal: B
content: $$260$$ 
aoVal: C
content: $$300$$ 
aoVal: D
content: $$360$$ 
Answer: B
Analysis: Sum $$=$$ Middle Term $$\times$$ Number of Terms, $$20 \times 13 = 260$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Sequences and Number Tables->Arithmetic Sequences
Problem: Find the value of $$123\times (-129)+123\times130$$. 
Options:
aoVal: A
content: $$123$$ 
aoVal: B
content: $$132$$ 
aoVal: C
content: $$129$$ 
aoVal: D
content: $$-123$$ 
Answer: A
Analysis: $$\begin{eqnarray}\&\&123\times (-129+130)\textbackslash\textbackslash{} \&=\&123\times (130-129)\textbackslash\textbackslash{} \&=\&123\times1\textbackslash\textbackslash{} \&=\&123.\end{eqnarray}$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Mixed Operations
Problem: If $A$ ◆$B=(A\times A)-B$ , then $(3$◆$4)$ ◆$5$ ． 
Options:
aoVal: A
content: $$0$$ 
aoVal: B
content: $$6$$ 
aoVal: C
content: $$20$$ 
aoVal: D
content: $$35$$ 
Answer: C
Analysis: $(3$◆$4)=(3\times 3)-4=5$ , and $(3$◆$4)$◆$5=5$◆$5=(5\times5)-5=20$ . 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Operations with New Definition->Operating Directly
Problem: The ratio of the perimeter of a rectangle to the length of one of its sides is~$14:3$. If the area is 27 square inches, how many inches long is one of the longer sides? 
Options:
aoVal: A
content: $$3$$ 
aoVal: B
content: $$4$$ 
aoVal: C
content: $$5$$ 
aoVal: D
content: $$7$$ 
aoVal: E
content: $$6$$ 
Answer: E
Analysis: $\dfrac{2\left( x+y\right)}{x}=\dfrac{14}{3}\textbackslash{} \textbackslash{} \Rightarrow\textbackslash{} \textbackslash{} \dfrac{x+y}{x}$~$=\dfrac{7}{3}\textbackslash{} \textbackslash{} \Rightarrow\textbackslash{} \textbackslash{} 1+\dfrac{y}{x}=1+\dfrac{4}{3}$  $\Rightarrow\textbackslash{} \textbackslash{} \dfrac{y}{x}=\dfrac{4}{3}\textbackslash{} \textbackslash{} \Rightarrow\textbackslash{} \textbackslash{} \dfrac{y\times y}{x\times y}=\dfrac{4}{3}$  The area is~$x\times y=27\textbackslash{} \textbackslash{} \Rightarrow\textbackslash{} \textbackslash{} \dfrac{y\times y}{27}=\dfrac{4}{3}\textbackslash{} \textbackslash{} \Rightarrow\textbackslash{} \textbackslash{} y\times y=36\textbackslash{} \textbackslash{} \Rightarrow\textbackslash{} \textbackslash{} y=6$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Ratios and Proportions
Source: 其它
Problem: The $2021^{st}$ digit at the right of the decimal point in the decimal expression of $\dfrac{6}{7}$ is~\uline{~~~~~~~~~~}~. 
Options:
aoVal: A
content: $$2$$ 
aoVal: B
content: $$4$$ 
aoVal: C
content: $$5$$ 
aoVal: D
content: $$8$$ 
Answer: B
Analysis: $$\frac{6}{7}=0.\overline{857142}$$, it is a decimal which repeats in cycles of $6$ digits. Every $6$$^{th}$ digit goes back to $2$. The $2022$$$^{nd}$$ digit is $2$, so the $2021$$^{st}$ digit is $4$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Decimals->Basic Understanding of Decimals
Problem: Two non-zero real numbers, $x$ and $y$, satisfy $x = 6-y$. Which of the following is a possible value of $$ \frac{{{x}^{2}}}{x-y}+\frac{{{y}^{2}}}{y-x}$$? (Adapted From 2000 AMC 12 Problems, Question \#11) 
Options:
aoVal: A
content: $-1$ 
aoVal: B
content: $-\frac{1}{2}$ 
aoVal: C
content: $\frac{1}{2}$ 
aoVal: D
content: $1$ 
aoVal: E
content: $$2$$ 
Answer: D
Analysis: Note that $x = 6-y \Rightarrow x+y = 6$. Then, $$ \frac{{{x}^{2}}}{x-y}+\frac{{{y}^{2}}}{y-x}=\frac{{{x}^{2}}-{{y}^{2}}}{x-y}=\frac{(x+y)(x-y)}{x-y}=x+y = 6$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: If $ a◆b$ means $(a\times b)+b$, then $2◆3$ has the value of~\uline{~~~~~~~~~~}~. 
Options:
aoVal: A
content: $$12$$ 
aoVal: B
content: $$9$$ 
aoVal: C
content: $$8$$ 
aoVal: D
content: $$6$$ 
Answer: B
Analysis: $2◆3=(2\times3)+3=9$.  So the answer is $\rm B$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Operations with New Definition->Operating Directly
Problem: Each of two boxes contains three chips numbered $$1$$, $$2$$, and $$3$$. A chip is drawn randomly from each box and the numbers on the two chips are multiplied. What is the probability that their product is even? (2015 AMC 8 Problem, Question \#7) 
Options:
aoVal: A
content: $$\frac{1}{9}$$ 
aoVal: B
content: $$\frac{2}{9}$$ 
aoVal: C
content: $$\frac{4}{9}$$ 
aoVal: D
content: $$\frac{1}{2}$$ 
aoVal: E
content: $$\frac{5}{9}$$ 
Answer: E
Analysis: We can instead calculate the probability that their product is odd, and subtract this from $$1$$. In order to get an odd product, we have to draw an odd number from each box. We have a $$\frac23$$ probability of drawing an odd mumber from one box, so there is a $$\left(\frac{2}{3}\right)^{2}= \frac{4}{9}$$ probability of having an odd product. Thus, there is a $$1- \frac{4}{9}= \frac{5}{9}$$ probability of having an even product. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Problem: If $$8$$ watermelons cost as much as $$12$$ pears, then $$24$$ waternelons cost as much as pears. 
Options:
aoVal: A
content: $$6$$ 
aoVal: B
content: $$18$$ 
aoVal: C
content: $$36$$ 
aoVal: D
content: $$48$$ 
Answer: C
Analysis: $$8$$ watermelons = $$12$$ pears  $$8\times3$$ watermelons = $24$ watermelons  $$12\times3$$ pears = $36$ pears 
Knowledge points: Overseas Competition->Knowledge Point->Heuristics Skills-> Equivalent Substitution
Problem: An object of mass m is traveling at constant speed v in acircular path of radius r. How much work is done by the centripetal force during one-half of a revolution? 
Options:
aoVal: A
content: $\pi mv^{2}$ 
aoVal: B
content: $$0$$ 
aoVal: C
content: $\pi mv^{2}r$ 
aoVal: D
content: $2\pi mv^{2}r$ 
Answer: B
Analysis: Since the centripetal force always points along a raduis toward the center of the od the circle, and the velocity of the object is always tangent to the circle (and thus perpendicular to the radius), the work done by the centripetal force is zero. Alternatively, since the object\textquotesingle s speed remains constant, the Work-Energy Theorem tells you that no work is being performed. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Sequences and Number Tables
Source: 其它
Problem: When Xiao Dong did the subtraction equation, he saw the minuend as $$90$$ instead of $$75$$, and the difference he got was $$55$$. The correct result should be~\uline{~~~~~~~~~~}~. 
Options:
aoVal: A
content: $$75$$ 
aoVal: B
content: $$40$$ 
aoVal: C
content: $$55$$ 
aoVal: D
content: $$35$$ 
aoVal: E
content: None of the above 
Answer: B
Analysis: Wrong: $$90 - A = 55$$ -\/-\textgreater{} $$A=90-55=35$$  Correct: $$75 - 35 = 40$$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: The number of papers I have is $$4\times 21$$ more than $6$ dogs. I have~\uline{~ ~?~ ~ ~ ~}~dogs． 
Options:
aoVal: A
content: $$54$$ 
aoVal: B
content: $$78$$ 
aoVal: C
content: $$84$$ 
aoVal: D
content: $$90$$ 
Answer: D
Analysis: $$4\times 21+ 6 = 90$$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Multiplication and Division
Problem: $$4\times 9=$$． 
Options:
aoVal: A
content: $$16\times 2$$ 
aoVal: B
content: $$12\times 3$$ 
aoVal: C
content: $$7\times 5$$ 
aoVal: D
content: $$38\times 1$$ 
Answer: B
Analysis: $$4\times 9=36$$.  $$\text{A}$$: $$16\times 2=32$$;  $$\text{B}$$: $$12\times 3=36$$;  $$\text{C}$$: $$7\times 5=35$$;  $$\text{D}$$: $$38\times 1=38$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Multiplication and Division
Problem: What is the remainder when $$987\textasciitilde654\textasciitilde321$$ is divided by $$100$$? 
Options:
aoVal: A
content: $$21$$ 
aoVal: B
content: $$42$$ 
aoVal: C
content: $$65$$ 
aoVal: D
content: $$98$$ 
Answer: A
Analysis: If a number is divided by $$100$$, the remainder is the number\textquotesingle s last $$2$$ digits. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Multiplication and Division
Problem: $$629+=1000 -174$$． 
Options:
aoVal: A
content: $$197$$ 
aoVal: B
content: $$371$$ 
aoVal: C
content: $$826$$ 
aoVal: D
content: $$1455$$ 
Answer: A
Analysis: $$1000-174=826$$  $$826-629=197$$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Addition and Subtraction
Problem: The distance between A and B is $$350$$km. Kin and Mary drive away from A and B respectively at $8$ a.m. and go towards each other at same time. Kin drives $$40$$km/h, and Mary drives $$50$$km/h. Mary rested for $$2$$ hours on her way and then continues driving. They will meet at~\uline{~~~~~~~~~~}~. 
Options:
aoVal: A
content: $$10$$ a.m. 
aoVal: B
content: $$11$$ a.m. 
aoVal: C
content: $$12$$ p.m. 
aoVal: D
content: $$1$$ p.m. 
Answer: D
Analysis: $$(350-80)$$$\div$$$(40+50)=3$$hr  $$8+2+3=13$$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Fractions
Source: 其它
Problem: A list of $2018$ positive integers has a unique mode, which occurs exactly $10$ times. What is the least number of distinct values that can occur in the list? (2018 AMC 10B Problems, Question \#14) 
Options:
aoVal: A
content: $$202$$ 
aoVal: B
content: $$223$$ 
aoVal: C
content: $$224$$ 
aoVal: D
content: $$225$$ 
aoVal: E
content: $$234$$ 
Answer: D
Analysis: To minimize the number of distinct values, we want to maximize the number of times they appear. So, we could have $223$ numbers appear $9$ times, $1$ number appear once, and the mode appear $10$ times, giving us a total of $223+1+1=$ (D) $225$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: Which of the following is a solution of $$\begin{cases}2x-4=0 \textbackslash\textbackslash{} 4x-y=7 \end{cases}$$? 
Options:
aoVal: A
content: ($x$,$y$)=($2$,$-1$) 
aoVal: B
content: ($x$,$y$)=($2$,$1$) 
aoVal: C
content: ($x$,$y$)=($-2$,$1$) 
aoVal: D
content: ($x$,$y$)=($-2$,$-1$) 
Answer: B
Analysis: $x=2$  $8-y=7$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Linear Equations with Multiple Variables
Source: 其它
Problem: A bakery sells cakes in three different sizes: small ($S$), medium ($M$), and large ($L$). The medium size costs $40\textbackslash\%$ more than the small size and contains $25\textbackslash\%$ less cake than the large size. The large size contains twice as much cake as the small size and costs $60\textbackslash\%$ more than the medium size. Rank the three sizes from best to worst buy in terms of cost per unit of cake. (Adapted From 2005 AMC 8 Problems, Question \#22) 
Options:
aoVal: A
content: $MSL$ 
aoVal: B
content: $SML$ 
aoVal: C
content: $LSM$ 
aoVal: D
content: $SLM$ 
Answer: A
Analysis: Suppose the small size costs $$$1$$ and the large size has $10$ grams of cake. The medium size then costs $$$1.40$$ and has $7.5$ grams of cake. The small size has $5$ grams of cake and the large size costs $$$2.24$$. The small, medium, and large size cost respectively, $0.200$, $0.187$, $0.224$ dollars per gram. The sizes from best to worst buy are $MSL$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: Find the value of the expression $100-98+96-94+92-90+\cdots +8-6+4-2$. 
Options:
aoVal: A
content: $$20$$ 
aoVal: B
content: $$40$$ 
aoVal: C
content: $$50$$ 
aoVal: D
content: $$80$$ 
aoVal: E
content: $$100$$ 
Answer: C
Analysis: $(100-98)+(96-94)+(92-90)+\cdots +(8-6)+(4-2)=2\times25=50$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Addition and Subtraction ->Grouping in Fast Addition and Subtraction of Whole Numbers
Source: 其它
Problem: If~$\dfrac{y}{x-z}=\dfrac{x+y}{z}=\dfrac{x}{y}$~for three positive numbers x, y and z, all different, then what is the value of~$\dfrac{x}{y}$. 
Options:
aoVal: A
content: $$3$$ 
aoVal: B
content: $$5$$ 
aoVal: C
content: $$1$$ 
aoVal: D
content: $$2$$ 
aoVal: E
content: $$6$$ 
Answer: D
Analysis: $\dfrac{x}{y}=\dfrac{y}{x-z}=\dfrac{x+y}{z}\textbackslash{} \textbackslash{} \textbackslash{} \textbackslash{} \Rightarrow\textbackslash{} \textbackslash{} \textbackslash{} \dfrac{x}{y}=\dfrac{x+y+(x+y)}{y+\left( x-z\right)+z}=\dfrac{2x+2y}{x+y}=\dfrac{2\left( x+y\right)}{x+y}=2.$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Ratios and Proportions
Source: 其它
Problem: $$280$$ adults and $$445$$ children visited Universal Studios last week. $330$ visitors did not take the rollercoasters. How many visitors took the roller coaster last week? 
Options:
aoVal: A
content: $$145$$ 
aoVal: B
content: $$245$$ 
aoVal: C
content: $$625$$ 
aoVal: D
content: $$395$$ 
Answer: D
Analysis: $280+445=725$.  $725-330=395$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Addition and Subtraction
Problem: What is the simplest form of $3\dfrac{8}{12}$? 
Options:
aoVal: A
content: $3\dfrac{3}{4}$ 
aoVal: B
content: $\frac{42}{12}$ 
aoVal: C
content: $\frac{10}{3}$ 
aoVal: D
content: $3\dfrac{2}{3}$ 
Answer: D
Analysis: Factors of $8$ are $1, 2, 4, 8$ and factors of $12$ are $1, 2, 3, 4, 6, 12$.  The highest commom factor of $8$ and $12$ is $4$.  Dividing $8$ and $12$ by $4$ respectively, we get  $\frac{8 \div 4}{12 \div 4}$=$\frac{2}{3}$;  Write the whole and the simplified fraction together,  $3\dfrac{2}{3}$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Fractions->Basic Understanding of Fractions->Using Common Factors to Simplify Fractions
Source: 其它
Problem: $$5\textasciitilde\text{m}+5\textasciitilde\text{cm}+5\textasciitilde\text{mm}=$$$$\text{mm}$$. 
Options:
aoVal: A
content: $$5055$$ 
aoVal: B
content: $$5505$$ 
aoVal: C
content: $$5550$$ 
aoVal: D
content: $$55550$$ 
Answer: A
Analysis: $$5\textasciitilde\text{m}+5\textasciitilde\text{cm}+5\textasciitilde\text{mm}=5000\textasciitilde\text{mm}+50\textasciitilde\text{mm}+5\textasciitilde\text{mm}=5055\textasciitilde\text{mm}$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Unit Conversion->Converting between Units of Length