math-eval/TAL-SCQ5K huggingface dataset:
Problem: Felix and Marmalade are two cats. Together they weigh $$10\text{kg}$$.  Felix weighs $$4\text{kg}$$ less than Marmalade. How much does Marmalade weigh? 
Options:
aoVal: A
content: $$3\text{kg}$$ 
aoVal: B
content: $$6\text{kg}$$ 
aoVal: C
content: $$7 \text{kg}$$ 
aoVal: D
content: $$9\text{kg}$$ 
aoVal: E
content: $$14 \text{kg}$$ 
Answer: C
Analysis: $$F=M-4$$  $$m-4+M=10$$  $$M=7$$ 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Questions Involving Sum, Difference and Multiples->Problems Involving Sum and Difference
Problem: $9$ more than $43$ is~\uline{~~~~~~~~~~}~. 
Options:
aoVal: A
content: $$42$$ 
aoVal: B
content: $$52$$ 
aoVal: C
content: $$62$$ 
aoVal: D
content: $$50$$ 
Answer: B
Analysis: $$43+9=52$$ 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Applying Addition and Subtraction
Source: 其它
Problem: My mom\textquotesingle s birthday is on Sunday, and my dad\textquotesingle s birthday is $$55$$ days later. On what day of the week will my dad\textquotesingle s birthday be? ． 
Options:
aoVal: A
content: $$ $$Sunday$$ $$ 
aoVal: B
content: $$ $$Monday$$ $$ 
aoVal: C
content: $$ $$Tuesday$$ $$ 
aoVal: D
content: $$ $$Thursday$$ $$ 
aoVal: E
content: $$ $$Saturday$$ $$ 
Answer: E
Analysis: After $$1$$ day, it is Monday; after $$2$$ days, it is Tuesday; after $$3$$ days, it is Wednesday $$\cdots$$ After $$8$$ days, it is still Monday, and the period is $$7$$ days. $$55\div 7=7$$(weeks) $$\rm R$$ $$6$$ (days). The birthday is Saturday. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Periodic Problems->Periodic Problems of Time
Problem: Sammy\textquotesingle s mom is $$6$$ years older than his dad. Right now, his dad is $$27$$ years old. How old was his mother $$10$$ years ago? 
Options:
aoVal: A
content: $$32$$ 
aoVal: B
content: $$21$$ 
aoVal: C
content: $$15$$ 
aoVal: D
content: $$23$$ 
aoVal: E
content: $$22$$ 
Answer: D
Analysis: Now, his dad is $27$ years old, his mom is $$6$$ years older than his dad, so his mom is $33$ years old. $10$ years ago, his mom was $33-10=23$ years old. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Age Problems->Basic Relationships in Age Problems->Difference of Ages
Problem: Find the exact number of minutes after $3.00 \text{pm}$ when the minute and hour hands are first at $90^{\circ}$ to each other. 
Options:
aoVal: A
content: $$31 \frac{2}{11}$$ 
aoVal: B
content: $$31 \frac{3}{11}$$ 
aoVal: C
content: $$32 \frac{8}{11}$$ 
aoVal: D
content: $$33 \frac{3}{11}$$ 
aoVal: E
content: None of the above 
Answer: C
Analysis: The minute hand rotates $6^{\circ}$ per minute.  The hour hand rotates $\dfrac{1}{60}\times30^{\circ}=\left (\dfrac{1}{2}\right )^{\circ}$ per mimte.  Starting at $3$ O\textquotesingle clock, the minute hand rotates $\left (6t\right )^{\circ}$ and the hour hand rotates ~$$(90+ \frac{1}{2}t)^{ \circ }$$.  When the minute and hour hands are next perpendicular:  $$6t-\left (90+ \frac{1}{2}t\right )=90$$,  $$\frac{11}{2}t=180$$,  $$t= \frac{360}{11}=32 \frac{8}{11}$$. 
Knowledge points: Overseas Competition->Knowledge Point->Distance Word Problems->Clock Problems
Problem: Given that October $$10$$th, $$2018$$ was Wednesday, on what day will October $$10$$th, $$2022$$ fall? 
Options:
aoVal: A
content: Monday 
aoVal: B
content: Tuesday 
aoVal: C
content: Wednesday 
aoVal: D
content: Thursday 
Answer: A
Analysis: During this period, $$2020$$ is a leap years. Therefore, October $$10$$, $$2022$$ is $$1+2+1+1=5$$ days after Wednesday, which is Monday. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Periodic Problems->Periodic Problems of Time
Problem: The average cost of a long-distance call in the USA in 1985 was 56 cents per minute, and the average cost of a long-distance call in the USA in 2018 was 2 cents per minute. Find the approximate percent decrease in the cost per minute of a long- distance call. (adapted from 2007 AMC 8, Question \#6) 
Options:
aoVal: A
content: $$90$$ 
aoVal: B
content: $$95$$ 
aoVal: C
content: $$96$$ 
aoVal: D
content: $$97$$ 
aoVal: E
content: $$98$$ 
Answer: C
Analysis: The percent decrease is (the amount of decrease)/(original amount) the amount of decrease is $56-2=54$ so the percent decrease is $\frac{54}{56}$ which is about $ 96 \textbackslash\%$ 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Solving Problems Involving Fractions and Percentages->Percentages Word Problems
Source: 其它
Problem: At which of these times is the angle between the minute hand and the hour hand of a clock equal to $$150^{}\circ$$? 
Options:
aoVal: A
content: $$9\text{pm}$$ 
aoVal: B
content: $$8\text{pm}$$ 
aoVal: C
content: $$6\text{pm}$$ 
aoVal: D
content: $$5\text{pm}$$ 
aoVal: E
content: $$4\text{pm}$$ 
Answer: D
Analysis: At all the times given, the minute hand is pointing to $$12$$. When the minute hand is pointing to $$12$$ and the angle between the hands is $$150^{}\circ$$, the hour hand has turned $$\frac{150}{360}= \frac{5}{12}$$ of a complete turn. Therefore the hour hand will point at $$5$$ and the time will be $$5\text{pm}$$. (There are other times when the angle between the hands is $$150^{}\circ$$ but, of these, only at $$7\text{pm}$$ does the minute hand point to $$12$$ and $$7\text{pm}$$ is not one of the times given.) 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules
Problem: In $$4$$ hours Rob drove $$120 \text{km}$$. He drove at an average rate of $$\text{m/min}$$. 
Options:
aoVal: A
content: $$30$$ 
aoVal: B
content: $$105$$ 
aoVal: C
content: $$120$$ 
aoVal: D
content: $$500$$ 
Answer: D
Analysis: NA 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules
Problem: SASMO 2015 P2 Q10  John is 5 years old this year. His sister is 2 years old. What is the sum of their age after 3 years? 
Options:
aoVal: A
content: $$18$$ 
aoVal: B
content: $$16$$ 
aoVal: C
content: $$14$$ 
aoVal: D
content: $$12$$ 
aoVal: E
content: $$10$$ 
Answer: A
Analysis: $$5+3=8$$  $$7+3=10$$  $$10+8=18$$ 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Age Problems
Source: 其它
Problem: Jack was able to sell $30$\% of his vegetables before noon. If he had 200 pounds of vegetables in the morning, how many pounds of vegetables was he able to sell by noon? 
Options:
aoVal: A
content: $$30$$ 
aoVal: B
content: $$50$$ 
aoVal: C
content: $$60$$ 
aoVal: D
content: $$140$$ 
Answer: C
Analysis: $200 \times 30$\%=$60$,  He was able to sell $60$ pounds of vegetables by noon. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Solving Problems Involving Fractions and Percentages->Percentages Word Problems
Source: 其它
Problem: I am going to travel $$200\text{km}$$ at a rate of $$60\text{km}$$ per hour. How many minutes will my trip take? 
Options:
aoVal: A
content: $$60$$ 
aoVal: B
content: $$200$$ 
aoVal: C
content: $$260$$ 
aoVal: D
content: $$320$$ 
Answer: B
Analysis: I am going to travel $$200\text{km}$$ at $$60\text{km}$$ per hour. Since $$60\text{km}$$ per hour is the same as $$1\text{km}$$ per minute, my trip will take $$200$$ minutes. 
Knowledge points: Overseas Competition->Knowledge Point->Distance Word Problems->Distance Word Problems on Straight Road
Problem: Last year, at the school where Gill teaches Mathematics, $$315$$ out of the $$600$$ pupils were girls. This year, the number of pupils in the school has increased to $$640$$. The proportion of girls is the same as it was last year.  How many girls are there at the school this year? 
Options:
aoVal: A
content: $$339$$ 
aoVal: B
content: $$338$$ 
aoVal: C
content: $$337$$ 
aoVal: D
content: $$336$$ 
aoVal: E
content: $$335$$ 
Answer: D
Analysis: Last year, the fraction of girls at the school was $$\frac{{315}}{{600}}=\frac{{63}}{{120}}=\frac{{21}}{{40}}$$. This year, there are $$40$$ more pupils at the school, but the proportion of girls has remained the same. So there are $$21$$ more girls at the school this year, making a total of $$315+21=336$$. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Solving Problems Involving Fractions and Percentages->Knowing the Base
Problem: Chloe is buying candies at a grocery store. She can either spend $8$ dollars on a $15$ ounce bag or $12$ dollars on a $20$ ounce bag. Which is a better buy? 
Options:
aoVal: A
content: $15$ ounce bag 
aoVal: B
content: $20$ ounce bag 
Answer: A
Analysis: $\frac{8\text{dollars}}{15\text{ounces}}$  $\frac{12\text{dollars}}{20\text{ounces}}$ 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Unitary Method Problems->Finding the Unit Rate and then the Total Value
Source: 其它
Problem: Jin loves carrots! Yesterday she ate $$\frac{1}{4}$$ of her carrots, and today she ate $$\frac{2}{7}$$ of the number of she ate yesterday. She ate 10 carrots today. Yesterday she have started with  carrots. 
Options:
aoVal: A
content: $$36$$ 
aoVal: B
content: $$48$$ 
aoVal: C
content: $$60$$ 
aoVal: D
content: $$72$$ 
aoVal: E
content: $$140$$ 
Answer: E
Analysis: $10\div\dfrac{2}{7}\div\dfrac{1}{4}=140$ 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Solving Problems Involving Fractions and Percentages->Finding the Base
Problem: Tom scored a basket at the end of the third quarter of a basketball game between the Lakers and the Warriors. At that point, what fraction of the whole game remained to be played? 
Options:
aoVal: A
content: $$\frac{1}{4}$$ 
aoVal: B
content: $$\frac{1}{3}$$ 
aoVal: C
content: $$\frac{1}{2}$$ 
aoVal: D
content: $$\frac{3}{4}$$ 
aoVal: E
content: $$\frac{2}{3}$$ 
Answer: A
Analysis: Basketball games consist of four quarters. Tom\textquotesingle s scoring at the end of the third quarter implies that there is one quarter left, which is equivalent to 1/4 of the entire game. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Solving Problems Involving Fractions and Percentages->Finding the Rate
Problem: Five positive integers (not necessarily all different) are written on five cards. Boris calculates the sum of the numbers on every pair of cards. He obtains only three different totals: 57, 70 and 83. What is the largest integer on any card? 
Options:
aoVal: A
content: $$35$$ 
aoVal: B
content: $$42$$ 
aoVal: C
content: $$48$$ 
aoVal: D
content: $$53$$ 
aoVal: E
content: $$82$$ 
Answer: C
Analysis: Let the five integers be a, b, c, d and e with a ≤ b ≤ c ≤ d ≤ e. The smallest total is 57, which is an odd number so b ≠ a. Similarly, the largest total is 83, which is also an odd number so d ≠ e. Hence we now have a \textless{} b ≤ c ≤ d \textless{} e and a + b = 57 and d + e = 83. Only one possible total remains and so b = c = d with c + d = 70. This gives c = d (= b) = 35 and therefore e, the largest integer, is 83 − 35 = 48 (whilst a = 22). 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules
Source: 其它
Problem: The average cost of a long-distance call in the USA in 1985 was 44 cents per minute, and the average cost of a long-distance call in the USA in 2005 was 11 cents per minute. Find the approximate percent decrease in the cost per minute of a long- distance call. (adapted from 2007 AMC 8, Question \#6) 
Options:
aoVal: A
content: $$25$$ 
aoVal: B
content: $$50$$ 
aoVal: C
content: $$75$$ 
aoVal: D
content: $$100$$ 
aoVal: E
content: $$125$$ 
Answer: C
Analysis: The percent decrease is (the amount of decrease)/(original amount) the amount of decrease is $44-11=33$ so the percent decrease is $\frac{33}{44}$ which is about $(\mathbf{c}) 75 \textbackslash\%$ 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Solving Problems Involving Fractions and Percentages->Percentages Word Problems
Source: 其它
Problem: James lives on the second floor, and Paul lives in the same building but has to walk up five times as many stairs as James. There are no stairs to the entrance of the building. On which floor does Paul live? (Adapted from 2000 Math Kangaroo Problem, Level 3-4, Question \#4) 
Options:
aoVal: A
content: On the $$2^{\rm nd}$$ floor 
aoVal: B
content: On the $$3^{\rm rd}$$ floor 
aoVal: C
content: On the $$4^{\rm th}$$ floor 
aoVal: D
content: On the $$5^{\rm th}$$ floor 
aoVal: E
content: On the $$6^{\rm th}$$ floor 
Answer: E
Analysis: $5+1=6$ 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Interval Problems->Straight Line Type with Objects on Both Sides->Taking the Stairs
Problem: Amy mixes $$10$$ grams of a $$20\textbackslash\%$$ sugar solution and $$40$$ grams of a $$25\textbackslash\%$$ sugar solution together. After~\uline{~~~~~~~~~~}~of water evaporates, the percent concentration of solution is $40\textbackslash\%$. 
Options:
aoVal: A
content: $$15$$ grams 
aoVal: B
content: $$18$$ grams 
aoVal: C
content: $$20$$ grams 
aoVal: D
content: $$25$$ grams 
Answer: C
Analysis: $$10\times20\textbackslash\%+40\times 25\textbackslash\%=12$$ grams.  $$(10+40)-12\div40\textbackslash\%=20$$ grams. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Concentration Problems->Finding and Making Use of Invariants
Problem: Greta was $$110\text{cm}$$ tall $$2$$ years ago, when she was $$10\text{cm}$$ taller than her brother. They both have grown, but now Greta is $$10\text{cm}$$ shorter than her brother. If her brother grew $$40\text{cm}$$ in those two years, Greta is now$$\text{cm}$$ tall. 
Options:
aoVal: A
content: $$120$$ 
aoVal: B
content: $$130$$ 
aoVal: C
content: $$140$$ 
aoVal: D
content: $$150$$ 
Answer: B
Analysis: Greta was $$110\text{cm}$$ tall $$2$$ years ago, when she was $$10\text{cm}$$ taller than her brother. Her brother was $$100\text{cm}$$ then. Now Greta is $$10\text{cm}$$ shorter than her brother. If her brother grew $$40\text{cm}$$, he is now $$140\text{cm}$$. Greta is now $$130\text{cm}$$ tall. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Applying Addition and Subtraction->Simple Addition and Subtraction
Problem: The average weight of $4$ students in Class $A$ is $37$ kilograms. The average weight of $11$ students in Class $B$ is $41$ kilograms. The total weight of $5$ students in Class $C$ is $161$ kilograms. What is the average weight of the students in the $3$ classes? 
Options:
aoVal: A
content: $36$ kilograms 
aoVal: B
content: $38$ kilograms 
aoVal: C
content: $40$ kilograms 
aoVal: D
content: $43$ kilograms 
Answer: B
Analysis: $4 \times 37 + 11 \times 41+ 161 = 760$ kilograms in total.  $760 \div (4 + 11 + 5) = 38$ kg. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Average Problems 
Source: 其它
Problem: Cici and Kiki took the same quiz. They agreed on the following rules: a person got $10$ points for a correct answer and lost $5$ points otherwise (getting a wrong answer or skipping a question). They each answered $10$ questions and together got $95$ points. If Cici got $15$ more points than Kiki, how many questions did Kiki answer correctly? 
Options:
aoVal: A
content: $$6$$ 
aoVal: B
content: $$7$$ 
aoVal: C
content: $$8$$ 
aoVal: D
content: $$9$$ 
Answer: A
Analysis: Kiki got $(95-15)\div2=40$ points. If she answered all the questions correctly, she would have gotten $100$ points. Hence, she skipped or got $(100-40)\div(10+5)=4$ wrong answers and she answered $10-4=6$ questions correctly. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Chicken-Rabbit Problems->Solving Chicken-Rabbit Problems by Using Hypothesis
Problem: Amy and Bella bought some cookies in the ratio of $$3:4$$. Bella gave $$9$$ cookies to Amy. Then the ratio of the number of cookies Amy had to that of Bella was $$1:1$$. How many cookies did they have at first? 
Options:
aoVal: A
content: $$54$$ 
aoVal: B
content: $$63$$ 
aoVal: C
content: $$72$$ 
aoVal: D
content: $$126$$ 
Answer: D
Analysis: Total number of cookies both of them had did not change.  Make total the same number of units by finding LCM.  Before $\to$ $A:B:Total$ $\to$ $3:4:7$ $\to$ $6:8:14$  After $\to$ $A:B:Total$ $\to$ $1:1:2$ $\to$ $7:7:14$  $1u=9$  $14u=14\times9=126$ cookies at first (and also in the end) in total. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Solving Problems Involving Proportions->Ratio Word Problems with Invariant Sums
Problem: Alex, John and Sam went to buy oranges. Alex paid $\textbackslash$20$, John paid $\textbackslash$15$, and Sam only paid $\textbackslash$5$. They bought $120$ oranges altogether. They divided them in proportion to the amount of money each of them had paid. How many oranges did John get? 
Options:
aoVal: A
content: $$15$$ 
aoVal: B
content: $$30$$ 
aoVal: C
content: $$45$$ 
aoVal: D
content: $$60$$ 
aoVal: E
content: None of the above 
Answer: C
Analysis: A: J: S  20:15:5 = 40  120/40 = 3 oranges per person  3*15= 45 oranges 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Applying of Multiplication and Division
Source: 其它
Problem: If $$3^{}\text{th}$$February is Monday, what day of the week is $$27$$\textsuperscript{th~}February? 
Options:
aoVal: A
content: Sunday　 
aoVal: B
content: Monday　 
aoVal: C
content: Wednesday　 
aoVal: D
content: Thursday　 
aoVal: E
content: Saturday　 
Answer: D
Analysis: We can count backward by days or by weeks. Count a few weeks back to find that $$22$$\textsuperscript{nd} February is a Monday. Then count a few days back to find that $$24$$\textsuperscript{th~}February is a Wednesday. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Periodic Problems->Periodic Problems of Time
Problem: The speed of high-speed train is approximately $$350$$ kilometres per hour, while the walking speed of a person is approximately $$5$$ metres per second.  Approximately, how many times is the speed of a high-speed train faster than the walking speed of a person?． 
Options:
aoVal: A
content: $$2 $$ 
aoVal: B
content: $$20 $$ 
aoVal: C
content: $$70 $$ 
aoVal: D
content: $$200 $$ 
aoVal: E
content: $$700$$ 
Answer: B
Analysis: The speed of high-speed train is approximately $$350$$ kilometres per hour, which is approximately $$100$$ metres per second. So its speed is roughly $$20$$ times faster than $$5$$ metres per second. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Questions Involving Sum, Difference and Multiples->Basic Computation of the Multiples
Problem: Martha had ＄$$1.50$$. She bought $$12$$ caramels at $$5$$￠ each. How many chocolate mints at $$10$$￠ each can she buy with the money she has left? 
Options:
aoVal: A
content: $$12$$ 
aoVal: B
content: $$11$$ 
aoVal: C
content: $$10$$ 
aoVal: D
content: $$9$$ 
Answer: D
Analysis: Martha had ＄$$1.50$$. She bought $$12$$ caramels at $$5$$￠ each, spengding  $$60$$￠ and leaving her with $$90$$￠. With $$90$$￠, she can buy $$90 \div 10 =9$$ mints at $$10$$￠ each. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Word Problems in Economics->Basic Profit and Loss Concepts
Problem: Balloons are sold in packets of $$5$$, $$10$$, and $$25$$. Marius buys exactly $$70$$ balloons. What is the smallest number of packets he can buy? (2017 Math Kangaroo Problem, Level 3 - 4, Question \#11) 
Options:
aoVal: A
content: $$3$$ 
aoVal: B
content: $$4$$ 
aoVal: C
content: $$5$$ 
aoVal: D
content: $$6$$ 
aoVal: E
content: $$7$$ 
Answer: B
Analysis: $70 = 25 + 25 + 10 + 10$ 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Unitary Method Problems->Applying Multiplication and Division->Simple Multiplication Applications
Source: 其它
Problem: Three hens and four rabbits are in a cage. How many legs do they have?~(adapted from $$2011$$ Math kangaroo Problems, Level $$1-2$$, Question \#$$4$$) 
Options:
aoVal: A
content: $$12$$ 
aoVal: B
content: $$15$$ 
aoVal: C
content: $$18$$ 
aoVal: D
content: $$22$$ 
aoVal: E
content: $$24$$ 
Answer: D
Analysis: A hen has two legs and a rabbit has four legs, so three hens have six legs and four chickens have sixteen legs.  $6+16=22$ 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Unitary Method Problems->Finding the Total Value using Unitary Method
Problem: A police spots a burglar from $$480$$ m apart. The burglar immdiately runs away at a speed of $$8$$ m/s and the police starts chasing him at $$12$$ m/s at the same time. At this rate, how long will it take the police to catch the burglar? 
Options:
aoVal: A
content: $$24$$ seconds 
aoVal: B
content: $$40$$ seconds 
aoVal: C
content: $$1$$ minutes 
aoVal: D
content: $$2$$ minutes 
Answer: D
Analysis: The distance between them is $$480$$ m, and the difference between their speeds is $$12-8=4$$ m/s. It takes $$480\div4=120$$ seconds to catch the burglar, which is $$2$$ minutes 
Knowledge points: Overseas Competition->Knowledge Point->Distance Word Problems->Distance Word Problems on Straight Road
Problem: At the 2013 Winnebago County Fair a vendor is offering a "fair special" on sandals. If you buy one pair of sandals at the regular price of $\textbackslash$ 50$, you get a second pair at a $40 \textbackslash\%$ discount, and a third pair at half the regular price. Javier took advantage of the "fair special" to buy three pairs of sandals. What percentage of the $\textbackslash$ 150$ regular price did he save? (2013 AMC 8, Question \#12) 
Options:
aoVal: A
content: $$25$$ 
aoVal: B
content: $$30$$ 
aoVal: C
content: $$33$$ 
aoVal: D
content: $$40$$ 
aoVal: E
content: $$45$$ 
Answer: B
Analysis: First, find the amount of money one will pay for three sandals without the discount. We have $\textbackslash$ 50 \times 3$ sandals $=\textbackslash$ 150$. Then, find the amount of money using the discount: $50+0.6 \times 50+\frac{1}{2} \times 50=\textbackslash$ 105$. Finding the percentage yields $\frac{105}{150}=70 \textbackslash\%$ To find the percent saved, we have $100 \textbackslash\%-70 \textbackslash\%=(\text{B}) 30$ 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Solving Problems Involving Fractions and Percentages->Percentages Word Problems
Source: 其它
Problem: There was a $15$-meter long corridor in the school. The teacher asked Jack to put a flowerpot on the corridor every three meters. How many flowerpots did Mike put?~(adapted from $$2008$$ Math kangaroo Problems, Level $$1-2$$, Question \#$$7$$) 
Options:
aoVal: A
content: $$4$$ 
aoVal: B
content: $$5$$ 
aoVal: C
content: $$7$$ 
aoVal: D
content: $$6$$ 
aoVal: E
content: $$8$$ 
Answer: D
Analysis: The corridor is divided into $15\div3=5$ segments. The number of flowerpots is one more than that of the segments, so $5+1=6$. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Interval Problems->Planting no Trees on either Side (straight line type)->Cutting the Ropes
Problem: The original price of a pair of shoes was $$\textbackslash$85$$. Then, it was sold at a discount of $$20\textbackslash\%$$. What was the price of the pair of shoes after the discount? 
Options:
aoVal: A
content: $$$17$$ 
aoVal: B
content: $$$52$$ 
aoVal: C
content: $$$68$$ 
aoVal: D
content: $$$102$$ 
Answer: C
Analysis: After discount, price became $100\textbackslash\%-20\textbackslash\%=80\textbackslash\%$ of original price.  $\textbackslash$85\times80\textbackslash\%=\textbackslash$85\times0.8=\textbackslash$68$ 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Word Problems in Economics->Basic Profit and Loss Concepts
Problem: Julia has eaten one-quarter of the pizza.  What is the ratio of the number of pizza that remain to the number Julia has eaten? 
Options:
aoVal: A
content: $$1:3$$ 
aoVal: B
content: $$3:1 $$ 
aoVal: C
content: $$1:4 $$ 
aoVal: D
content: $$4:1$$ 
aoVal: E
content: $$4:3$$ 
Answer: B
Analysis: Julia has eaten one-quarter of the pizza. So three-quarters of the pizza remain.  Therefore the required ratio is $$\frac{3}{4}:\frac{1}{4}=3:1$$. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Solving Problems Involving Proportions->Basic Units