math-eval/TAL-SCQ5K huggingface dataset:
Problem: Alice\textquotesingle s first day in Caterpillar Club was Tuesday. She wants to throw a party on her $40\rm{th}$ day in the club. If Alice attends the club everyday, on which day of the week will the party be? 
Options:
aoVal: A
content: Friday 
aoVal: B
content: Saturday 
aoVal: C
content: Sunday 
aoVal: D
content: Monday 
aoVal: E
content: None of the above 
Answer: B
Analysis: 40th day = 39th day after first day joining.  39 days/7 = 5 weeks R 4 days. 4 days after Tues is Saturday. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Periodic Problems->Periodic Problems of Time
Source: 其它
Problem: Jenny is $5$ years old. She has $2$ sisters, Julie and Jas. Julie is $1$ year older than Jas. In two year\textquotesingle s time, their total age will be $18$ years. How old is Julie now? 
Options:
aoVal: A
content: $$3$$ 
aoVal: B
content: $$4$$ 
aoVal: C
content: $$5$$ 
aoVal: D
content: $$6$$ 
aoVal: E
content: $$7$$ 
Answer: B
Analysis: NA 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Age Problems
Source: 其它
Problem: Julia has eaten one-quarters 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-quarters of the pizza. So three-quarter 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
Problem: A gardener has a box of bulbs to plant in a garden. The garden has three sections.  She plants 1/2~of the bulbs in the first section.  She plants 3 /4~of the remaining bulbs in the second section.  She has 6 bulbs left, which she plants in the third section.  How many bulbs were in the box at the start? 
Options:
aoVal: A
content: $$16$$ 
aoVal: B
content: $$18$$ 
aoVal: C
content: $$24$$ 
aoVal: D
content: $$32$$ 
aoVal: E
content: $$48$$ 
Answer: E
Analysis: 1-1/2-1/2*3/4=1/8，1/8means 6 bulbs，$$so$$ in total there are 6\div 1/8=48 bulbs. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules
Source: 其它
Problem: Will is selling apples at the farmer\textquotesingle s market. After selling $\dfrac{2}{5}$ of them, he has $150$ left. How many apples did Will have to start with? 
Options:
aoVal: A
content: $$250$$ 
aoVal: B
content: $$375$$ 
aoVal: C
content: $$90$$ 
aoVal: D
content: $$60$$ 
Answer: A
Analysis: He has $\dfrac{3}{5}$~remaining after selling $\dfrac{2}{5}$. We can write the equation as:~$150\div\dfrac{3}{5}=250$. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Solving Problems Involving Fractions and Percentages->Finding the Base
Problem: Brandon is engaging sprint race on the track, and he is in the first group. On the track, there are $5$ players in front of him, and there are $8$ players behind him. There are $5$ extra groups signing up for the sprint race, and the number of players of each group are identical. How many sprint players are there in total? 
Options:
aoVal: A
content: $$65$$ 
aoVal: B
content: $$70$$ 
aoVal: C
content: $$78$$ 
aoVal: D
content: $$84$$ 
aoVal: E
content: $$96$$ 
Answer: D
Analysis: There are $5+8+1=14$ players in each group. There are $14\times 6=84$ players in total. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules
Source: 其它
Problem: Niko counted a total of $$60$$ birds perching in three trees. Five minutes later, $$6$$ birds had flown away from the first tree, $$8$$ birds had flown away from the second tree and $$4$$ birds had flown away from the third tree. He noticed that there was now the same number of birds in each tree. How many birds were originally perched in the second tree? 
Options:
aoVal: A
content: $$14$$ 
aoVal: B
content: $$18$$ 
aoVal: C
content: $$20$$ 
aoVal: D
content: $$21$$ 
aoVal: E
content: $$22$$ 
Answer: E
Analysis: Let the number of birds remaining in each tree be $$x$$. Therefore $$x+6+x+8+x+4= 60$$, which has solution $$x=14$$. Hence the number of birds originally perched in the second tree is $$14 + 8 = 22$$. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Equation Word Problems->Word Problems for Linear Equations with One Variable
Problem: A tank filled with $200$ litres of water weighs $350\rm{kg}$. The same tank filled with $150$ litres of water weighs $315\rm{kg}$. What is the weight of the empty tank? 
Options:
aoVal: A
content: $$120\rm{kg}$$ 
aoVal: B
content: $$150\rm{kg}$$ 
aoVal: C
content: $$165\rm{kg}$$ 
aoVal: D
content: $$210\rm{kg}$$ 
aoVal: E
content: None of the above 
Answer: D
Analysis: T + 200L = 350kg $\cdots $ ①  -) T + 150L = 315kg $\cdots $ ②  \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_  (50L = 35kg) x3 = 150L = 105kg  315kg - 105kg = 210kg 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Applying Addition and Subtraction
Source: 其它
Problem: A total of $$42$$ chickens and rabbits are caged together. If there are $$124$$ legs in total, how many chickens and rabbits are in the cage, respectively? 
Options:
aoVal: A
content: $20$; $12$ 
aoVal: B
content: $22$; $20$ 
aoVal: C
content: $20$; $24$ 
aoVal: D
content: $19$; $21$ 
Answer: B
Analysis: Suppose all the animals in the cage are chickens: there should be $$2\times 42=84$$ legs. However, $$124-84=40$$ legs are missing because we counted $$40\div2=20$$ rabbits. Hence, there are $$20$$ rabbits and $$42-20=22$$ chickens. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Chicken-Rabbit Problems->Solving Chicken-Rabbit Problems by Using Hypothesis
Problem: A string $$12$$ meters long is cut into $$6$$ pieces of equal length.  What is the sum of the lengths of any $$4$$ of these pieces? 
Options:
aoVal: A
content: $$2\text{m}$$ 
aoVal: B
content: $$4\text{m}$$ 
aoVal: C
content: $$6\text{m}$$ 
aoVal: D
content: $$8\text{m}$$ 
Answer: D
Analysis: The string $$12$$ meters long is cut into $$6$$ pieces of length $$2\text{m}$$.  The sum of the lengths of any $$4$$ of these pieces is $$4\times2\text{m}=8\text{m}$$. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Unitary Method Problems->Finding the Unit Rate and then the Total Value
Problem: William plans to take $$16$$ days to drive to New York City from his home. However, since there is heavy traffic, he drives $$20$$ miles less each day and arrives in New York City $$8$$ days later than his plan. What is the distance between William\textquotesingle s home and New York City? 
Options:
aoVal: A
content: $$480$$ miles 
aoVal: B
content: $$1080$$ miles 
aoVal: C
content: $$960$$ miles 
aoVal: D
content: $$720$$ miles 
aoVal: E
content: $$1200$$ miles 
Answer: C
Analysis: We set up an equation in terms of $d$ as the distance and $x$ as the speed in miles per day originally. We have  $d=16x=24(x-20)$  $16x=24x-480$  $480=8x$  $x=60$  Thus, the distance between William\textquotesingle s home and New York City is $60\times16=960$ miles. 
Knowledge points: Overseas Competition->Knowledge Point->Distance Word Problems->Distance Word Problems on Straight Road
Problem: Calculate: $1\times 10+2\times 9+3\times 8+ \cdots + 9\times 2+ 10\times 1$ 
Options:
aoVal: A
content: $$200$$ 
aoVal: B
content: $$140$$ 
aoVal: C
content: $$210$$ 
aoVal: D
content: $$180$$ 
aoVal: E
content: None of the above 
Answer: E
Analysis: E 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Questions Involving Sum, Difference and Multiples
Source: 其它
Problem: The sum of Ben\textquotesingle s age and Ben\textquotesingle s grandma\textquotesingle s age is $$96$$. Ben\textquotesingle s grandma\textquotesingle s age is three times older than Ben\textquotesingle s age. How old is Ben\textquotesingle s grandma? 
Options:
aoVal: A
content: $$24$$ 
aoVal: B
content: $$48$$ 
aoVal: C
content: $$72$$ 
aoVal: D
content: $$96$$ 
Answer: C
Analysis: The sum of Ben\textquotesingle s age and Ben\textquotesingle s grandma\textquotesingle s age is $$96$$.  Since grandma is three times older than Ben, on the number line, $96$ can be divided into $$4$$ portions, and each portion is $$96\div(1+3)=24$$.  Ben is $$24$$ years old, and grandma is three times older.  $$24\times 3=72$$ 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Age Problems->Sums and Multiples in Age Problems
Source: 其它
Problem: Five children divided some cakes equally. What percents of the cakes two of them could get in total? 
Options:
aoVal: A
content: $$5\textbackslash\%$$ 
aoVal: B
content: $$10\textbackslash\%$$ 
aoVal: C
content: $$20\textbackslash\%$$ 
aoVal: D
content: $$40\textbackslash\%$$ 
Answer: D
Analysis: Five children divided some cake equally. Two of the children totally got $$\dfrac{2}{5}$$ of the cake and $$\dfrac{2}{5}=0.4 = 40\textbackslash\%$$. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Solving Problems Involving Fractions and Percentages->Finding the Rate
Problem: This year, March $$19$$ falls on a Thursday. What day of the week will it be in $$30$$ days? 
Options:
aoVal: A
content: Wednesday　 
aoVal: B
content: Thursday　 
aoVal: C
content: Friday　 
aoVal: D
content: Saturday　 
aoVal: E
content: Sunday　 
Answer: D
Analysis: $$30\div4=7 \text{ R } 2$$, two days after a Thursday is a Saturday. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Periodic Problems->Periodic Problems of Time
Problem: A group of students go apple-picking. On average, each boy picks $30$ apples and each girl picks $20$ apples. The average number of apples picked by everyone in this groups is $26$. If there are $15$ boys, then how many girls are there in the group? 
Options:
aoVal: A
content: $$15$$ 
aoVal: B
content: $$10$$ 
aoVal: C
content: $$9$$ 
aoVal: D
content: $$20$$ 
aoVal: E
content: $$23$$ 
Answer: B
Analysis: Every boy picks $$30-26=4$$ more than the average, so boys pick $$4\times 15=60$$ more in total, which is equal to the number that girls picked less than the average in total. Every girl picks $$26-20=6$$ less than the average, so there are $$60\div 6=10$$ girls. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Average Problems
Problem: If $$10^{}\text{th}$$June is Sunday, what day of the week is $$1$$\textsuperscript{st} June? 
Options:
aoVal: A
content: Sunday　 
aoVal: B
content: Monday　 
aoVal: C
content: Wednesday　 
aoVal: D
content: Friday 
aoVal: E
content: Saturday　 
Answer: D
Analysis: We can count backward by days or by weeks. There are 9 days in between, that means 1 week and 2 days. Then count a few days back to find that $$1$$\textsuperscript{st} June is a Friday. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Periodic Problems->Periodic Problems of Time
Problem: The possible grades on an exam are $$5$$, $$4$$, $$3$$, $$2$$, or $$1$$. In a class of $$200$$ students, $$\frac{1}{10}$$ of them got $$5$$s, $$\frac{1}{5}$$ them got $$4$$s, $$25 \textbackslash\% $$ of them got $$3$$s, and $$15 \textbackslash\%$$ of them got $$2$$s. How many students got $$1$$s ? 
Options:
aoVal: A
content: $$40$$ 
aoVal: B
content: $$60$$ 
aoVal: C
content: $$80$$ 
aoVal: D
content: $$100$$ 
Answer: B
Analysis: $$200\times \left( 1-\frac{1}{10}-\frac{1}{5}-25 \textbackslash\% -15 \textbackslash\% ~\right)=60$$. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Solving Problems Involving Fractions and Percentages->Percentages Word Problems
Source: 其它
Problem: In $$2005$$, both John and Mary have the same amount of pocket money per month. In $$2006$$, John had an increase of $$10\textbackslash\%$$ and Mary a decrease of $$10\textbackslash\%$$ in their monthly pocket money. In $$2007$$, John had a decrease of $$10\textbackslash\%$$ and Mary an increase of $$10\textbackslash\%$$ in their monthly pocket money.  $\textasciitilde$  John had a~\uline{~~~~~~~~~~}~$\textbackslash\%$ increase/decrease in his pocket money in $2007$ compared to $2005$.  Mary had a~\uline{~~~~~~~~~~}~$\textbackslash\%$ increase/decrease in her pocket money in $2007$ compared to $2005$.  $\textasciitilde$  Which one of the following statements is correct? 
Options:
aoVal: A
content: Both have the same amount of pocket money now. 
aoVal: B
content: John has more pocket money now. 
aoVal: C
content: Mary has more pocket money now. 
aoVal: D
content: It is impossible to tell who has more pocket money now. 
Answer: A
Analysis: Statement $$\text{A}$$ is correct.  In the amount of pocket money that each has be $$x$$.  In $$2007$$, John has $$1.1\times0.9x=0.99x$$ while Mary has $$0.9\times1.1x=0.99x$$. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Solving Problems Involving Fractions and Percentages->Knowing the Base
Problem: Lena the panda has $$30$$ bamboos. She eats $$3$$ bamboos each day. She ate the $$18th$$ bamboo on Tuesday. On which day did she start eating the bamboos? 
Options:
aoVal: A
content: Monday 
aoVal: B
content: Tuesday 
aoVal: C
content: Wednesday 
aoVal: D
content: Thursday 
aoVal: E
content: Friday 
Answer: D
Analysis: NA 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Newton's Problem of Cows and Fields->Basic Newton's Problem of Cows and Fields->Finding the Number of Days
Problem: A particular cattle is fed $7 \text{kg}$, $8 \text{kg}$ and $9 \text{kg}$ of special food on Day $1$, $2$ and $3$ respectively by a farmer. From Day $4$ onwards, the amount of food fed to the cattle follows the following rule:  (i) if the total amount of food fed to the cattle in the past $3$ days is $24 \text{kg}$ or more, the cattle will be fed $2 \text{kg}$ less than the day before.  (ⅱ) if the total amount of food fed to the cattle in the past $3$ days is less than $24 \text{kg}$, the cattle will be fed $1 \text{kg}$ more than the day before.  Calculate the minimum number of days for the total amount of food fed to the cattle to exceed $2018 \text{kg}$. 
Options:
aoVal: A
content: $$280$$ 
aoVal: B
content: $$284$$ 
aoVal: C
content: $$288$$ 
aoVal: D
content: $$292$$ 
aoVal: E
content: None of the above 
Answer: E
Analysis: Amount of food fed to the cattle follows the pattern:  $789756$~ ~ $$789756$$~ ~ $789756$~ ~ $789756\cdots\cdots$  Total amount of food fed in the first $6$ days$=42 \text{kg}$.  $2018 =48\times6+1= 289$. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Periodic Problems->Periodic Problems of Basic Permutation
Problem: Emily was $$6$$ years old last year, and her sister was $$6$$ years old six years ago. What is the age difference between Emily and her sister today? (adapted from 2005 Math Kangaroo Problem, Level 1-2, Question \#17) 
Options:
aoVal: A
content: $$4$$ 
aoVal: B
content: $$5$$ 
aoVal: C
content: $$7$$ 
aoVal: D
content: $$13$$ 
aoVal: E
content: $$14$$ 
Answer: B
Analysis: Emily was $$6$$ years old last year , which means Emily is $7$ years old this year. Her sister was~ $$6$$ years old six years ago, which tells us her sister is~ $12$ years old this year, so the age difference is $5$ today. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Age Problems->Basic Relationships in Age Problems->Difference of Ages
Problem: Sinnce $$7^{2}=49$$, its ones\textquotesingle{} digit is a $$9$$. What is the ones\textquotesingle{} digit of $$2^{50}$$? 
Options:
aoVal: A
content: $$0$$ 
aoVal: B
content: $$2$$ 
aoVal: C
content: $$4$$ 
aoVal: D
content: $$8$$ 
Answer: C
Analysis: The ones\textquotesingle{} digits repeat in groups of four: $$2$$, $$4$$, $$8$$, $$6$$, $$2$$, $$4$$, $$8$$, $$6$$, $$\cdots $$. The ones\textquotesingle{} digits of $$2^{48}$$, $$2^{49}$$, $$2\%{50}$$ are $$6$$, $$2$$, $$4$$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders->Applying the Properties of Dividing without Remainders->Maximum/Minimum Problems of Division without Remainders
Problem: If six days ago was a Tuesday, then eight days from today is a． 
Options:
aoVal: A
content: Monday　 
aoVal: B
content: Tuesday　 
aoVal: C
content: Wednesday　 
aoVal: D
content: Thursday　 
Answer: B
Analysis: $$6$$ days ago was Tues. Today is Mon. In $$8$$ days it will be Tues. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Periodic Problems->Periodic Problems of Time
Problem: Granny and the triplets Cara, Cate and Chris all have their birthdays today.  The ages of all four of them total $$120$$ years.  Granny is $$5$$ times as old as each of the triplets.  When were the triplets born? 
Options:
aoVal: A
content: $$2000$$ 
aoVal: B
content: $$2002$$ 
aoVal: C
content: $$2004$$ 
aoVal: D
content: $$2006$$ 
aoVal: E
content: $$2008$$ 
Answer: C
Analysis: Given that Granny is $$5$$ times older than each of the three triplets, we know that the age of each triplet is $$120\div (5+3)= 15$$. So Cara, Cate and Chris were born in $$2004$$. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Age Problems->Sums and Multiples in Age Problems
Problem: Chloe is buying candies at a grocery store. She can either spend $8$ dollars on a $15\text{-ounce}$ bag or $12$ dollars on a $20\text{-ounce}$ bag. Which is a better buy? 
Options:
aoVal: A
content: The $15\text{-ounce}$ bag 
aoVal: B
content: The $20\text{-ounce}$ 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: If February is a month that contains Friday the $$13^{}\text{th}$$, what day of the week is February $$1$$? 
Options:
aoVal: A
content: Sunday　 
aoVal: B
content: Monday　 
aoVal: C
content: Wednesday　 
aoVal: D
content: Thursday　 
aoVal: E
content: Saturday　 
Answer: A
Analysis: We can count backward by days or by weeks. Count a few weeks back to find that February $$6$$ is a Friday. Then count a few days back to find that February $$1$$ is a Sunday. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Periodic Problems->Periodic Problems of Time
Problem: Tower $$A$$ is $$\rm 64 m$$ high.The total height of Tower $$A$$ and Tower $$B$$ is $$\rm 112 m$$.What is the difference between the height of the two towers?~\uline{~~~~~~~~~~}~$$\rm m$$. 
Options:
aoVal: A
content: $$14$$ 
aoVal: B
content: $$15$$ 
aoVal: C
content: $$16$$ 
aoVal: D
content: $$17$$ 
aoVal: E
content: $$18$$ 
Answer: C
Analysis: $$112-64=48$$  $$64-48=16$$ 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Applying Addition and Subtraction->Simple Addition and Subtraction
Problem: Clara, Pablo and Miguel share some money in the ratio $$3:1:4$$. Miguel then gave Pablo and Clara each $$25\textbackslash\%$$ of his share, leaving Clara with ￡$$6$$ more than Miguel. How much money does Pablo now have? 
Options:
aoVal: A
content: ￡$$2$$ 
aoVal: B
content: ￡$$3.60$$ 
aoVal: C
content: ￡$$4$$ 
aoVal: D
content: ￡$$6$$ 
aoVal: E
content: ￡$$12$$ 
Answer: D
Analysis: The initial ratio of the children\textquotesingle s money is $$3:1:4$$. When Miguel shares half of his portion, the ratio becomee $$3 + 1:1+ 1:4-2=4:2:2=2:1:1$$. Given that Clara now has ￡$$6$$ more than Miguel, Pablo now has ￡$$6$$. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Solving Problems Involving Proportions->Complex Ratio Problems
Problem: Adam paid £6 for $$15$$ buns. How many dollars did Tom pay for the same kind of buns if he bought $$5$$? 
Options:
aoVal: A
content: $$2$$ 
aoVal: B
content: $$3$$ 
aoVal: C
content: $$4$$ 
aoVal: D
content: $$5$$ 
Answer: A
Analysis: There $$3$$ groups of $$5$$ buns in $$15$$ buns: $$15\div 5=3$$, so $$5$$ buns cost $$6\div 3=$$£$$2$$. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Unitary Method Problems->Applying Multiplication and Division
Problem: SASMO 2014 P2 Q7  There are 14 children playing "The eagle catches the chicks." One of them is the \textquotesingle eagle\textquotesingle{} while another child is the \textquotesingle mother hen\textquotesingle{} whose job is to protect the \textquotesingle chicks\textquotesingle. The rest of the children are the \textquotesingle chicks\textquotesingle. After a while, the \textquotesingle eagle\textquotesingle{} has caught 5 \textquotesingle chicks\textquotesingle. How many \textquotesingle chicks\textquotesingle{} are still around? 
Options:
aoVal: A
content: $$6$$ 
aoVal: B
content: $$7$$ 
aoVal: C
content: $$8$$ 
aoVal: D
content: $$9$$ 
aoVal: E
content: $$10$$ 
Answer: B
Analysis: 14 - 1 (eagle) - 1 (mother hen) = 12 (chicks)  12 - 5 7 chicks 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Applying Addition and Subtraction
Source: 其它
Problem: For every $$7$$ soccer balls Elena bought for the gym, she bought $$4$$ basketballs. If she bought $$35$$ soccer balls, she bought a total ofballs. 
Options:
aoVal: A
content: $$20$$ 
aoVal: B
content: $$35$$ 
aoVal: C
content: $$45$$ 
aoVal: D
content: $$55$$ 
Answer: D
Analysis: $$7$$ out of $$11$$ balls that Elena bought are soccer balls. If there are $$7\times5 =35$$ soccer balls, then there is a total of $$11\times5=55$$ balls. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Unitary Method Problems->Finding the Total Value using Unitary Method
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? (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: When Cici was born, Linda was $11$ years old. The sum of their ages $4$ years later will be $37$. How old will be Linda $3$ years later? 
Options:
aoVal: A
content: $$12$$ 
aoVal: B
content: $$22$$ 
aoVal: C
content: $$24$$ 
aoVal: D
content: $$26$$ 
Answer: C
Analysis: Suppose that Cici will be $x$ years old $4$ years later, Linda will be $$(x+11)$$ years old. $x+(x+11)=37$, so $x=13$. Thus, Linda will be $13+11=24$ years old $3$ years later. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Age Problems
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: $$ $$Tuesday$$ $$ 
aoVal: C
content: $$ $$Thursday$$ $$ 
aoVal: D
content: $$ $$Saturday$$ $$ 
Answer: D
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