math-eval/TAL-SCQ5K huggingface dataset:
Problem: An automobile manufacturer wished to know which of two new paint colors were preferred on its newest line of vehicles. A~large simple random sample was taken from people throughout the USA who had purchased one of their vehicles in the~previous five years. What is the safest generalization of this survey? 
Options:
aoVal: A
content: Only those people who took this particular survey 
aoVal: B
content: All future customers of this automobile manufacturer 
aoVal: C
content: All people who have purchased from that manufacturer in the previous five years 
aoVal: D
content: Only those people who have purchased a vehicle in the previous five years 
aoVal: E
content: All vehicle owners in the USA 
Answer: C
Analysis: A safe generalization will broaden the survey to a population that closely resembles the group of people surveyed. Since the sample was taken from purchasers of a single auto manufacturer over the past five years, the safest generalization is a population with these same qualities. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: 14. Consider the following system of equations: $$\begin{cases} 5 x+7 y=8 \textbackslash\textbackslash{} 2 x+q y=r \end{cases}$$, where $q$ and $r$ are constants. If $q$ and $r$ are chosen such that this system has infinitely many solutions $(x, y)$, find $q+r=$~\uline{~~~~~~~~~~}~. 
Options:
aoVal: A
content: $$6$$ 
aoVal: B
content: $$8$$ 
aoVal: C
content: $$9$$ 
aoVal: D
content: $$12$$ 
Answer: A
Analysis: For the system to have infinitely many solutions, the second equation must be a constant multiple of the first. Comparing the coefficients of $x$, the second equation must be $\frac{2}{5}$ times the first equation. Therefore, $q+r=\left(\frac{2}{5}\right)\times (7)+\left(\frac{2}{5}\right)\times(8)=6$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Indefinite Equations->System of Indefinite Equations
Source: 其它
Problem: For $\triangle ABC$, all of its side lengths are integers. The primeter of $\triangle ABC$ with a side of length $12$ and a side length of $7$ is at least ． 
Options:
aoVal: A
content: $$24$$ 
aoVal: B
content: $$25$$ 
aoVal: C
content: $$26$$ 
aoVal: D
content: $$27$$ 
aoVal: E
content: $$28$$ 
Answer: B
Analysis: We know from the triangle inequality that the last side, $s$, fulfills $s+7\textgreater12$. $P=s+7+12\textgreater12+12$. Therefore, $P\textgreater24+1=25$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Inequalities
Source: 其它
Problem: What is the sum of the numbers that range from $$1$$ to $$100$$ when divide by $$3$$ give you a remainder of $$1$$ ? 
Options:
aoVal: A
content: $$1700$$ 
aoVal: B
content: $$1716$$ 
aoVal: C
content: $$1717$$ 
Answer: C
Analysis: $$\left( 1+100 \right)\times 34\div 2=1717$$． 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Sequences and Number Tables->Arithmetic Sequences
Problem: A cup of boiling water ($212$\textsuperscript{◦}F) is placed to cool in a room whose temper- ature remains constant at $68$\textsuperscript{◦}F. Suppose the difference between the water temperature and the room temperature is halved every $5$ minutes. What is the water temperature, in degrees Fahrenheit, after $15$ minutes? 
Options:
aoVal: A
content: $$77$$ 
aoVal: B
content: $$86$$ 
aoVal: C
content: $$92$$ 
aoVal: D
content: $$98$$ 
aoVal: E
content: $$104$$ 
Answer: B
Analysis: NA 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: What is the missing number in the box?  $2:3=\boxed{?}:9$ 
Options:
aoVal: A
content: $$8$$ 
aoVal: B
content: $$7$$ 
aoVal: C
content: $$6$$ 
aoVal: D
content: $$4$$ 
Answer: C
Analysis: $\frac{2}{3}=\frac{6}{9}$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Ratios and Proportions->Proportions
Problem: (the number of $$\rm cm$$ in $$1\rm m$$):(the number of $$\rm m$$ in $$1\rm km$$)$$=$$． 
Options:
aoVal: A
content: $$100:1000$$ 
aoVal: B
content: $$1000:100$$ 
aoVal: C
content: $$1:100$$ 
aoVal: D
content: $$100:1$$ 
Answer: A
Analysis: $$1\rm m=100\rm cm$$ and $$1\rm km=1000m$$; the correct ratio is $$100:1000$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Unit Conversion->Converting between Units of Length
Problem: The house number of a community consists of $5$ digits: $1, 2, 3, 6, 7$. Which number is not used if these $5$ digits are filled in the squares below to make the equation correct? Each number can only be filled in once.  $$\huge\square -\square =\square +\square $$ 
Options:
aoVal: A
content: $$7$$ 
aoVal: B
content: $$6$$ 
aoVal: C
content: $$2$$ 
aoVal: D
content: $$3$$ 
aoVal: E
content: $$1$$ 
Answer: A
Analysis: Pay attention to look at both sides of the formula. One side is addition, and the other side is subtraction. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Addition and Subtraction ->Addition of Whole Numbers->Addition in Horizontal Form
Source: 其它
Problem: Kangaroo came up with a new operation * for positive integers. He gave a few examples as shown below:  $$2$$ * $$3$$$$=\left( 2+1 \right)\times 3=9$$;  $$4$$ * $$2$$$$=\left( 4+3+2+1 \right)\times 2=20$$;  $$3$$ * $$5$$$$=\left( 3+2+1 \right)\times 5=30$$.  What is the value of the expression $$6$$ * $$5$$ ? 
Options:
aoVal: A
content: $$30$$ 
aoVal: B
content: $$90$$ 
aoVal: C
content: $$105$$ 
aoVal: D
content: $$210$$ 
aoVal: E
content: $$315$$ 
Answer: C
Analysis: $$6$$ * $$5 = (6+5+4+3+2+1) \times 5 = 21\times 5 = 105$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Operations with New Definition->Operating Directly->Ordinary Type
Problem: \textbf{A copy machine dealer has data on the number x of copy machines at each of 89 customer locations and the number y of service calls in a month at each location. Summary calculations give $$\bar{}x$$ = 8.4, $$S\_x$$ = 2.1, ȳ = 14.2, $$S\_y$$ = 3.8, and r = 0.86. What is the slope of the least squares regression line of number of service calls on number of copiers?} 
Options:
aoVal: A
content: $$0.86$$ 
aoVal: B
content: $$1.56$$ 
aoVal: C
content: $$0.48$$ 
aoVal: D
content: None of the above 
aoVal: E
content: \textbf{Can't tell from the information given} 
Answer: B
Analysis: $$\beta\_1 =r \frac{S\_y}{S\_x} = 0.86 * \frac{3.8}{2.1} = 1.56$$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: A firm had sales revenue of 1 million dollars last year. It spent 600,000 dollars on labor, 150,000 dollars on capital, and 200,000 dollars on materials. What was the firm\textquotesingle s accounting profit? 
Options:
aoVal: A
content: 0 dollar 
aoVal: B
content: 50,000 dollars 
aoVal: C
content: 400,000 dollars 
aoVal: D
content: 650,000 dollars 
aoVal: E
content: 1,000,000 dollars 
Answer: B
Analysis: Accounting profit = total revenue - explicit cost 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: Nicolas is planning to send a package to his friend Anton, who is a stamp collector. To pay for the postage, Nicolas, would like to cover the package with a large number of stamps. Suppose he has a collection of $5$-cent, $10$-cent, and $25$-cent stamps, with exactly $20$ of each type. What is the greatest number of stamps Nicolas can use to make exactly $\textbackslash$7.10$ in postage? 
Options:
aoVal: A
content: $$45$$ 
aoVal: B
content: $$46$$ 
aoVal: C
content: $$51$$ 
aoVal: D
content: $$54$$ 
aoVal: E
content: $$55$$ 
Answer: A
Analysis: E 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: \textbf{In a statistics course, a linear regression equation was computed to predict the final exam score from the score on the first test. The equation was y = 10 + .9x where y is the final exam score and x is the score on the first test. Carla scored 95 on the first test. What is the predicted value of her score on the final exam?} 
Options:
aoVal: A
content: $$95$$ 
aoVal: B
content: $$85.5$$ 
aoVal: C
content: $$90$$ 
aoVal: D
content: $$95.5$$ 
aoVal: E
content: None of the above 
Answer: D
Analysis: \textbf{$$\hat{y}$$= 10 + .9*95=95.5. The predicted final exam score is 95.5.} 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: A rectangular water tank is filled to a depth of $$70$$cm. It contains $$1050$$ litres of water. Some water is taken out of the tank. The water level drops by $$25$$cm. How much water is left in the tank? 
Options:
aoVal: A
content: $$625\rm L$$ 
aoVal: B
content: $$375\rm L$$ 
aoVal: C
content: $$525\rm L$$ 
aoVal: D
content: $$270\rm L$$ 
aoVal: E
content: $$675\rm L$$ 
Answer: E
Analysis: $$1050\times \frac {70-25}{70} = 675$$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Ratios and Proportions
Source: 其它
Problem: Calculate: $$19.2\div6\times1.1=$$． 
Options:
aoVal: A
content: $35.2$ 
aoVal: B
content: $3.52$ 
aoVal: C
content: $37.4$ 
aoVal: D
content: $3.74$ 
aoVal: E
content: $3.47$ 
Answer: B
Analysis: $$19.2\div6\times1.1$$  $$=3.2\times1.1$$  $$=3.52$$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Decimals->Multiplication and Division of Decimals
Source: 其它
Problem: Pip and Bud buys some £$$10$$ books and £$$15$$ books together. If they spent £$$90$$ on the books, how many books have they bought at most in total? 
Options:
aoVal: A
content: $$14$$ 
aoVal: B
content: $$15$$ 
aoVal: C
content: $$16$$ 
aoVal: D
content: $$7$$ 
aoVal: E
content: $$8$$ 
Answer: E
Analysis: Let $$x$$ be the number of £$$10$$ books and $$y$$ be the number of £$$15$$ books  $$10x+15y=90$$, which has two positive integer solutions only.  $$\begin{cases} x=6 \textbackslash\textbackslash{} y=2 \end{cases}$$, $$\begin{cases} x=3 \textbackslash\textbackslash{} y=4 \end{cases}$$  $$6+2=8$$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Indefinite Equations
Source: 其它
Problem: For how many values of $a$ is it true that the line $y=x+a^{2}-a-3$ passes through the vertex of the parabola $y=x^{2}-6x+a^{2}$? (Adapted From 2005 AMC 12B Problem, Question \#8) 
Options:
aoVal: A
content: $$0$$ 
aoVal: B
content: $$1$$ 
aoVal: C
content: $$2$$ 
aoVal: D
content: $$10$$ 
aoVal: E
content: infinitely many 
Answer: B
Analysis: We see that the vertex of the quadratic function $y=x^{2}-6x+a^{2}$ is $\left(3, a^{2}-9\right)$. If $\left(2, a^{2}-1\right)$ will be on the line $y=x+a^{2}-a-3$, $a^{2} -9=3+a^{2}-a-3$. Solve for $a$, there is one solution, $a=9$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: For how many values of $a$ is it true that the line $y=x+a$ passes through the vertex of the parabola $y=x^{2}+a^{2}$? (2005 AMC 12B Problem, Question \#8) 
Options:
aoVal: A
content: $$0$$ 
aoVal: B
content: $$1$$ 
aoVal: C
content: $$2$$ 
aoVal: D
content: $$10$$ 
aoVal: E
content: infinitely many 
Answer: C
Analysis: We see that the vertex of the quadratic function $y=x^{2}+a^{2}$ is $\left(0, a^{2}\right)$. The $y$-intercept of the line $y=x+a$ is $(0, a)$. We want to find the values (if any) such that $a=a^{2}$. Solving for $a$, the only values that satisfy this are 0 and 1 , so the answer is $(\text{C}) 2$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: $$(100 + 98 + 96 + \cdots +2) -(99 + 97 + 95 + \cdots + 1)=$$． 
Options:
aoVal: A
content: $$25$$ 
aoVal: B
content: $$50$$ 
aoVal: C
content: $$99$$ 
aoVal: D
content: $$100$$ 
Answer: B
Analysis: $$(100-99)+(98-97)+ \cdots +(2-1) = 1+1+ \cdots +1 = 50$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Addition and Subtraction
Problem: For the ratio of $$3:8$$, if the consequent increases by $$24$$, the antecedent should~\uline{~~~~~~~~~~}~. 
Options:
aoVal: A
content: $$+24$$ 
aoVal: B
content: $$\times 4$$ 
aoVal: C
content: $$\times 24$$ 
aoVal: D
content: $$+4$$ 
Answer: B
Analysis: $$8+24=32=8\times 4$$  So, $$3\times 4=12$$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Ratios and Proportions->Ratio
Source: 其它
Problem: What is the value of $$1+3+5+\cdots +2017+2019-2-4-6-\cdots -2016-2018$$? 
Options:
aoVal: A
content: $$-1010$$ 
aoVal: B
content: $$-1009$$ 
aoVal: C
content: $$1008$$ 
aoVal: D
content: $$1009$$ 
aoVal: E
content: $$1010$$ 
Answer: E
Analysis: Solution 1  Rearranging the terms, we get $$(1-2)+(3-4)+(5-6)+\cdots (2017-2018)+2019$$, and our answer is $$-1009+2019=1010$$.  Solution 2  We can rewrite the given expression as  $$1+(3-2)+(5-4)+\cdots +(2017-2016)+(2019-2018)=1+1+1+\cdots +1$$. The number of $$1$$s is the same as the number of terms in $$1$$, $$3$$, $$5$$, $$7\cdots $$, $$2017$$, $$2019$$. Thus the answer is $$1010$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Addition and Subtraction ->Grouping in Fast Addition and Subtraction of Whole Numbers
Problem: Seagram knows that $$1111\times 2222=2468642$$  Which of the following answers should he decide is $$3333\times 4444$$? 
Options:
aoVal: A
content: $$14811850$$ 
aoVal: B
content: $$14811851$$ 
aoVal: C
content: $$14811852$$ 
aoVal: D
content: $$14811853$$ 
aoVal: E
content: $$14811854$$ 
Answer: C
Analysis: $$14811852$$  It should be noted that the units digit of the product of $$3333$$ and $$4444$$ will be the same as the units digit of the product of $$3$$ and $$4$$, namely $$2$$: this succinctly identifies the correct option as $$14811852$$. Alternatively, the calculation $$3333\times 4444$$ will have an answer that is $$3\times 2$$ times greater than $$1111\times 2222$$. Now we can work out $$2468642\times 6$$ exactly, which is indeed $$14811852$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Calculation of Multi-digit Numbers
Problem: For each patient visiting a doctor\textquotesingle s office, the nurse records the patient\textquotesingle s body temperature. The plot above shows the temperatures of 28 patients for one particular day. Which of the following statements is true about the distribution of body temperatures? 
Options:
aoVal: A
content: The distribution is skewed to the left. 
aoVal: B
content: The median temperature could be 98.2$^{\circ}F$ 
aoVal: C
content: The median temperature could be 98.4$^{\circ}F$ 
aoVal: D
content: The minimum temperature is exactly 97.0$^{\circ}F$ 
aoVal: E
content: The distribution is normal. 
Answer: B
Analysis: The correct answer is (b).Since there are 28 patients, the median will be between the 14th and 15th data points, which puts it in the bar centered at 98.25°$F$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: Points $(\sqrt{\pi}, a)$ and $(\sqrt{\pi}, b)$ are distinct points on the graph of $y^{2}+x^{4}=2 x^{2} y+1$. What is $\textbar a-b\textbar$? 
Options:
aoVal: A
content: $$1$$ 
aoVal: B
content: $\frac{\pi}{2}$ 
aoVal: C
content: $$2$$ 
aoVal: D
content: $\sqrt{1+\pi}$ 
aoVal: E
content: $1+\sqrt{\pi}$ 
Answer: C
Analysis: Since points on the graph make the equation true, substitute $\sqrt{\pi}$ in to the equation and then solve to find $a$ and $b$.  $$ \begin{aligned} \&y^{2}+\sqrt{\pi}^{4}=2 \sqrt{\pi}^{2} y+1 \textbackslash\textbackslash{} \&y^{2}+\pi^{2}=2 \pi y+1 \textbackslash\textbackslash{} \&y^{2}-2 \pi y+\pi^{2}=1 \textbackslash\textbackslash{} \&(y-\pi)^{2}=1 \textbackslash\textbackslash{} \&y-\pi=\pm 1 \textbackslash\textbackslash{} \&y=\pi+1 \textbackslash\textbackslash{} \&y=\pi-1 \end{aligned} $$  There are only two solutions to the equation $(y-\pi)^{2}=1$, so one of them is the value of $a$ and the other is $b$. The order does not matter because of the absolute value sign. $$ \textbar(\pi+1)-(\pi-1)\textbar=2 $$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation
Source: 其它
Problem: Henry the donkey has a very long piece of pasta. He takes a number of bites of pasta, each time eating $3$ inches of pasta from the middle of one piece. In the end, he has $10$ pieces of pasta whose total length is $17$ inches. How long, in inches, was the piece of pasta he started with? 
Options:
aoVal: A
content: $$34$$ 
aoVal: B
content: $$38$$ 
aoVal: C
content: $$41$$ 
aoVal: D
content: $$44$$ 
aoVal: E
content: $$47$$ 
Answer: D
Analysis: NA 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: A new operation $$⊕$$ is defined as $$a⊕b=\frac{2}{a^{2}}+\frac{1}{b}$$:  Which of the equations below is/are correct? 
Options:
aoVal: A
content: $$2⊕4=4⊕2$$ 
aoVal: B
content: $$3⊕6=6⊕3$$ 
aoVal: C
content: $$4⊕8=8⊕4$$ 
aoVal: D
content: $$1008⊕2016=2016⊕1008$$ 
Answer: B
Analysis: method $$1$$:  （$$1$$）$$\frac{2}{2^{2}}+ \frac{1}{4}= \frac{2}{4}+ \frac{1}{4}= \frac{3}{4}$$， $$\frac{2}{4^{2}}+ \frac{1}{2}= \frac{2}{16}+ \frac{1}{2}= \frac{10}{16}= \frac{5}{8}$$. Rejected.  （$$2$$）$$\frac{2}{3^{2}}+ \frac{1}{6}= \frac{2}{9}+ \frac{1}{6}= \frac{4}{18}+ \frac{3}{18}= \frac{7}{18}$$,~ $$\frac{2}{6^{2}}+ \frac{1}{3}= \frac{2}{36}+\frac{1}{3}=\frac{1}{18}+\frac{1}{3}=\frac{1}{18} +\frac{6}{18}=\frac{7}{18}$$.  ($$3$$）$$\frac{2}{4^{2}}+ \frac{1}{8}= \frac{2}{16}+ \frac{1}{8}= \frac{1}{4}$$,  $$\frac{2}{8^{2}}+ \frac{1}{4}= \frac{2}{64}+ \frac{1}{4}= \frac{1}{32}+ \frac{8}{32}= \frac{9}{32} \neq \frac{1}{4}.$$  ($$4$$)$$\frac{2}{1008^{2}}+ \frac{1}{2016}= \frac{1}{252 \times 2016}+ \frac{1}{2016}= \frac{253}{252 \times 2016}.$$  $$\frac{2}{2016^{2}}+ \frac{1}{1008}= \frac{1}{2016 \times 1008}+ \frac{1}{1008}= \frac{2017}{2016 \times 1008}.$$  $$1008⊕2016\neq 2016⊕1008$$.  method $$2$$:  $$4$$ is twice of $$2$$, $$6$$ is twice of $$3$$, $$8$$ is twice of $$4 $$, $$2016$$ is twice of $$1008$$.  Specifically, the question is asking when is $$k⊕2k=2k⊕k$$?  $$\frac{2}{k^{2}}+ \frac{1}{2k}= \frac{2}{\left(2k\right)^{2}}+ \frac{1}{k} \Rightarrow \frac{2}{k^{2}}+ \frac{1}{2k}= \frac{1}{2k^{2}}+ \frac{1}{k}$$  $$\Rightarrow \frac{4}{2k^{2}}+ \frac{1}{2k}= \frac{1}{2k^{2}}+ \frac{2}{2k} \Rightarrow \frac{3}{2k^{2}}= \frac{1}{2k}$$  $$\Rightarrow \frac{3}{k}=1 \Rightarrow k=3$$  Hence, ($$2$$) is the only possible answer. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Operations with New Definition->Operating Directly
Problem: $$10+8\times6-4\div2=$$． 
Options:
aoVal: A
content: $$18$$ 
aoVal: B
content: $$27$$ 
aoVal: C
content: $$52$$ 
aoVal: D
content: $$56$$ 
Answer: D
Analysis: $$10+8\times6-4\div2=10+48-2=56$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Mixed Operations
Problem: How many even numbers are there?  2, 3, 5, 6, 7, 9, 10. 
Options:
aoVal: A
content: $$3$$ 
aoVal: B
content: $$4$$ 
aoVal: C
content: $$5$$ 
Answer: A
Analysis: $$Omitted.$$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers
Source: 其它
Problem: \textbf{If I toss a fair coin 5000 times} 
Options:
aoVal: A
content: \textbf{and I get anything other than 2500 heads, then something is wrong with the way I flip coins.} 
aoVal: B
content: \textbf{the proportion of heads will be close to 0.5} 
aoVal: C
content: \textbf{a run of 10 heads in a row will increase the probability of getting a run of 10 tails in a row.} 
aoVal: D
content: \textbf{the proportion of heads in these tosses is a parameter} 
aoVal: E
content: \textbf{the proportion of heads will be close to 50.} 
Answer: B
Analysis: \textbf{Since this is a fair coin, the probability to get a head is always 0.5.} 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: True or False: It is impossible for the $25$-th percentile to be equal to the median. 
Options:
aoVal: A
content: True 
aoVal: B
content: False 
Answer: B
Analysis: For example, consider a dataset with the following values: 0, 2, 2, 3, 3. The median of this dataset is $2$, and the $25$-th percentile is also $2$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules
Source: 其它
Problem: $$98 + 99 + 100 + 101 + 102 =$$． 
Options:
aoVal: A
content: $$497$$ 
aoVal: B
content: $$498$$ 
aoVal: C
content: $$499$$ 
aoVal: D
content: $$500$$ 
Answer: D
Analysis: Regrouping,$$ (98+102) + (99+101) + 100 = 500$$. 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Whole Numbers->Whole Numbers Addition and Subtraction
Problem: Alex and Felicia each have cats as pets. Alex buys cat food in cylindrical cans that are 6 cm in diameter and 12 cm high. Felicia buys cat food in cylindrical cans that are 12 cm in diameter and 6 cm high. What is the ratio of the volume of one of Alex\textquotesingle s cans to the volume of one of Felicia\textquotesingle s cans? (2019 AMC 8, 9) 
Options:
aoVal: A
content: $1:4$ 
aoVal: B
content: $1:2$ 
aoVal: C
content: $1:1$ 
aoVal: D
content: $2:1$ 
aoVal: E
content: $4:1$ 
Answer: B
Analysis: Solution 1  Using the formula for the volume of a cylinder, we get Alex,~$\pi108$, and Felicia,~$\pi216$. We can quickly notice that~~cancels out on both sides, and that Alex\textquotesingle s volume is~$\dfrac{1}{2}$~~of Felicia\textquotesingle s leaving~$\dfrac{1}{2}=\boxed{1:2}$~as the answer.  Solution 2  Using the formula for the volume of a cylinder, we get that the volume of Alex\textquotesingle s can is~$3^{2}\cdot12\cdot\pi$, and that the volume of Felicia\textquotesingle s can is~$6^{2}\cdot6\cdot\pi$. Now, we divide the volume of Alex\textquotesingle s can by the volume of Felicia\textquotesingle s can, so we get~$\dfrac{1}{2}$, which is$\boxed{\left( B\right)\textbackslash{} 1:2}.$  Solution 3  The ratio of the numbers is~$\dfrac{1}{2}$. Looking closely at the formula~$r^{2}*h*\pi$, we see that the~$r*h*\pi$~will cancel, meaning that the ratio of them will be~$\dfrac{1\left( 2\right)}{2\left( 2\right)}=\boxed{\left( B\right)\textbackslash{} 1:2}$  Solution 4  The second can is 2 size in each of 2 dimensions, and~$\dfrac{1}{2}$~size in 1 dimension.~$\dfrac{2^{2}}{2}=\boxed{\left( B\right)\textbackslash{} 1:2}$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Ratios and Proportions
Source: 其它
Problem: For certain real numbers $a, b$, and $c$, the polynomial $$ g(x)=x^{3}+a x^{2}+x+10 $$ has three distinct roots, and each root of $g(x)$ is also a root of the polynomial $$ f(x)=x^{4}+x^{3}+b x^{2}+100 x+c . $$ What is $f(1)$? 
Options:
aoVal: A
content: $$-9009$$ 
aoVal: B
content: $$-8008$$ 
aoVal: C
content: $$-7007$$ 
aoVal: D
content: $$-6006$$ 
aoVal: E
content: $$-5005$$ 
Answer: C
Analysis: $f(x)$ must have four roots, three of which are roots of $g(x)$. Using the fact that every polynomial has a unique factorization into its roots, and since the leading coefficient of $f(x)$ and $g(x)$ are the same, we know that  $$ f(x)=g(x)(x-r) $$  where $r \in \mathbb{C}$ is the fourth root of $f(x)$. Substituting $g(x)$ and expanding, we find that  $$ \begin{aligned} f(x) \&=\left(x^{3}+a x^{2}+x+10\right)(x-r) \textbackslash\textbackslash{} \&=x^{4}+(a-r) x^{3}+(1-a r) x^{2}+(10-r) x-10 r \end{aligned} $$  Comparing coefficients with $f(x)$, we see that  $$ \begin{aligned} a-r \&=1 \textbackslash\textbackslash{} 1-a r \&=b \textbackslash\textbackslash{} 10-r \&=100 \textbackslash\textbackslash{} -10 r \&=c . \end{aligned} $$  Let\textquotesingle s solve for $a, b, c$, and $r$. Since $10-r=100, r=-90$, so $c=(-10)(-90)=900$. Since $a-r=1, a=-89$. Then, since $b=1-a r, b=-8009$. Thus, we know that $$ f(x)=x^{4}+x^{3}-8009 x^{2}+100 x+900 . $$ Taking $f(1)$, we find that $$ \begin{aligned} f(1) \&=1^{4}+1^{3}-8009(1)^{2}+100(1)+900 \textbackslash\textbackslash{} \&=1+1-8009+100+900 \textbackslash\textbackslash{} \&=(\mathbf{C})-7007 . \end{aligned} $$ 
Knowledge points: Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Indefinite Equations->System of Indefinite Equations
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