On Monday morning, a snail fell down a well which is $$10$$ meters deep. During the day, it climbs up $$2$$ meters, and during the night it slides down $$1$$ meter. On what day of the week will the snail get out of the well? ($$1998$$ Math kangaroo Problem, Level $$5-6$$, Question \#$$18$$)

$$$$Tuesday

$$$$Thursday

$$$$Saturday

$$$$Sunday

$$$$Monday

Overseas Competition->Knowledge Point->Word Problem Modules->Periodic Problems->Snail Climbing out of Well Problems->Snail Climbing out of Well (completed)

Except the last day, the snail could climb up $2-1=1$ m. Thus, it takes the snail $(10-2)\div1+1=9$ days to get out of the well. $9$ days from Monday is the next Tuesday.


When one ounce of water is added to a mixture of acid and water, the new mixture is $$20\textbackslash\%$$ acid. When one ounce of acid is added to the new mixture, the result is $$33\frac{1}{3}\textbackslash\%$$ acid. The percentage of acid in the original mixture is ($$1973$$ AHSME Problem, Question \#$$33$$)

Overseas Competition->Knowledge Point->Word Problem Modules->Concentration Problems->Basic Concentration Problems

Let $$a$$ be the original number of ounces of acid and $$w$$ be the original number of ounces of water. We can write two equations $$\frac{a}{a+w+1}=\frac{1}{5}$$ and $$\frac{a+1}{a+w+2}=\frac{1}{3}$$. Then cross-multiply to get rid of the fractions. $$5a=a+w+1$$, $$3a+3=a+w+1$$ Solve the system to get $$a=1$$ and $$w=3$$. The percentage of acid in the original mixture is $$\frac{1}{1+3}=25 \textbackslash\% $$.


Brandon is engaging sprint race on the track. He is in the fifth place now, and there are $8$ players behind him. There are $5$ extra groups signing up for sprint race, and the number of players of each group are idential. How many sprint players are there in total?

Overseas Competition->Knowledge Point->Word Problem Modules


A shop advertises everything is "half price in today\textquotesingle s sale." In addition, a coupon gives a $25 \textbackslash\%$ discount on sale prices. Using the coupon, the price today represents what percentage off the original price? (adapted from 2012 AMC 8, Question \#8)

Overseas Competition->Knowledge Point->Word Problem Modules->Solving Problems Involving Fractions and Percentages->Percentages Word Problems

$75 \textbackslash\% \cdot 50\textbackslash\% =37.5\textbackslash\%$  $1 -37.5\textbackslash\% = 62.5\textbackslash\%$


James can frost a cookie every $$12$$ seconds and Lacey can frost a cookie every $$15$$ seconds. Working together, how many cookies can they frost in $$10$$ minutes? (Adapted from $$2012$$ AMC $$10\rm A$$ Problem, Question \#$$11$$)

Overseas Competition->Knowledge Point->Word Problem Modules->Work Word Problems->Collaborative Work Word Problems

In $$600$$ seconds ($$10$$ minutes), James will frost $$\dfrac{600}{12}=50$$ cookies, and Lacey will frost $$\dfrac{600}{15}=40$$ cookies. Therefore, working together they will frost $$50+40=\boxed{(\text{D})90}$$ cookies.


Judy ran for $$16$$ minutes. If her average speed was $$80$$ meters per minute, then how many meters did Judy run?


The length of a rectangle is increased by $10 \textbackslash\%$ and the width is decreased by $10 \textbackslash\%$. What percent of the old area is the new area? (2009 AMC 8, Question \#8)

In a rectangle with dimensions $10 \times 10$, the new rectangle would have dimensions $11 \times 9$. The ratio of the new area to the old area is $99 / 100=(\text{B}) 99$


The price ratio of goods $$A$$ to $$B$$ is $$7:2$$. If their prices were increased by $$\textbackslash$60$$ respectively, the price ratio would be $$5:2$$. What are the prices of these two goods?

$$\textbackslash$315$$, $$\textbackslash$90$$

$$\textbackslash$320$$, $$\textbackslash$100$$

$$\textbackslash$330$$, $$\textbackslash$120$$

$$\textbackslash$350$$, $$\textbackslash$200$$

Overseas Competition->Knowledge Point->Word Problem Modules->Solving Problems Involving Proportions->Complex Ratio Problems

The price difference between the two goods did not change.  Before $\to$ $$A:B:Difference$$ is $$7:2:5$$  After $\to$ $$A:B:Difference$$ is $$5:2:3$$  $\textasciitilde$  Make difference the same number of units in both ratios by finding LCM.  Before $\to$ $$A:B:Difference$$ is $$21:6:15$$  After $\to$ $$A:B:Difference$$ is $$25:10:15$$  $\textasciitilde$  Therefore,  $$4u=\textbackslash$60$$  $$1u=$$ $$\textbackslash$60\div4=\textbackslash$15$$  $\textasciitilde$  The original prices of these two goods were  $$A=21u=21\times15=\textbackslash$315$$ and  $$B=6u=6\times15=\textbackslash$90$$.


Ela and Kasia boarded a super-train. Ela took a seat in the seventeenth car counting from the front of the train, and Kasia was seated in the thirty-fourth car counting from the end. The girls were sitting in the same car. How many cars did the super-train have? (2001 Math Kangaroo Problem, Level 3-4, Question \#13)

Overseas Competition->Knowledge Point->Word Problem Modules->Problems Involving Queuing->Finding the Total Number According to the Position of Two Characters in a Line


James had $$10000$$ dollars, and he deposited ＄$$7000$$ in Bank $$A$$ and ＄$$3000$$ in Bank $$B$$. At the end of the year, James received an interest of＄$$460$$ in total. Given that the sum of the annual interest rates of the two banks is $$10\textbackslash\%$$, find the interest rates of the two banks, respectively.

$$2\textbackslash\%$$; $$8\textbackslash\%$$.

$$3\textbackslash\%$$; $$7\textbackslash\%$$.

$$4\textbackslash\%$$; $$6\textbackslash\%$$.

$$5\textbackslash\%$$; $$5\textbackslash\%$$.

Overseas Competition->Knowledge Point->Word Problem Modules->Word Problems in Economics->Interest Problems

Let $$x$$ be the interest rate of bank $$A$$:  $$7000x+3000(10\textbackslash\%-x)=460$$  $$x=4\textbackslash\%$$.


Peter has a rope, and he ties some knots, each using $10$-cm rope. He ties $25$ knots in total. After he ties the last knot, the knots divide the rope into pieces~ equally, and the length of each piece is $7$ cm. What is the length of the entire rope?

$$425$$ cm

$$432$$ cm

$$456$$ cm

$$478$$ cm

$$480$$ cm

Overseas Competition->Knowledge Point->Word Problem Modules->Interval Problems


The scale of a map is:~$\dfrac{3}{4}$ of an inch $$=10$$ miles. If the distance on the map between two towns is $$12$$ inches, the actual distance between the towns is．

$$90$$ miles

$$120$$ miles

$$150$$ miles

$$160$$ miles

Overseas Competition->Knowledge Point->Word Problem Modules->Solving Problems Involving Proportions->Ratio Word Problems with an Invariant Part

Since $$12 \div \frac{3}{4}=16$$, the actual distance between the towns is $$16\times10 = 160$$ miles.


Gawel lives on the second floor, and Pawel lives in the same building but has to walk up twice as many stairs as Gawel. There are no stairs to the entrance of the building. On which floor does Pawel live? (2000 Math Kangaroo Problem, Level 3-4, Question \#4)

on the $2$\textsuperscript{nd} floor

on the $3$\textsuperscript{rd} floor

on the $4$\textsuperscript{th} floor

on the $5$\textsuperscript{th} floor

on the $6$\textsuperscript{th} floor

Gawel lives on the second floor, so he only needs to walk up one floor.  Pawel has to walk up twice as many stairs as Gawel, so he needs to walk up two floors. So, Pawel lives on the $3$\textsuperscript{rd} floor.


I ate half an apple pie on Saturday and two thirds of the remainder on Sunday. What fraction of the pie was left for Monday?

Overseas Competition->Knowledge Point->Word Problem Modules->Solving Problems Involving Fractions and Percentages->Understanding the Base

After I eat one half, half of the apple pie is left. Eating two thirds of this half leaves one third of one half of the pie, which is one sixth, for Monday.


The length of Amy\textquotesingle s string is $$35\text{cm}$$. The length of David\textquotesingle s string is $$29\text{cm}$$. What is the total length of both strings?

Overseas Competition->Knowledge Point->Word Problem Modules->Applying Addition and Subtraction->Simple Addition and Subtraction


Cici bought $$12$$ pencils and notebooks in total for $$66$$. Each pencil cost $$2$$. Each notebook costs $8$. How many notebooks did Cici buy?

Overseas Competition->Knowledge Point->Word Problem Modules->Chicken-Rabbit Problems->Solving Chicken-Rabbit Problems by Using Hypothesis


There are oranges, apricots and peaches in a big basket. How many fruits are there in the basket if the peaches and the apricots together are 18, the oranges and the apricots together are 28 and 30 fruits are not apricots?

Overseas Competition->Knowledge Point->Word Problem Modules->Unitary Method Problems->Unitary Method with Two Variables

If we sum up the number of A and P, the result is 18.  But if we sum up A and O, result is 28 instead.  Hence, there must be 10 more O than P.  Note that there are a total of 30 I and P.  Therefore, there are (30+10)/2 = 20 Oranges.  Total = 20+18=38.


Suppose $15 \textbackslash\%$ of $x$ equals $20 \textbackslash\%$ of $y$. What percentage of $x$ is $y$ ? (2020 AMC 8, Question 15)

Since $20 \textbackslash\%=\frac{1}{5}$, multiplying the given condition by 5 shows that $y$ is $15 \cdot 5=(\mathbf{C}) 75$ percent of $x$.


$$90$$ miles

$$120$$ miles

$$150$$ miles

$$160$$ miles


Five people took the same test and got an average score of $80$. Four of the people scored $80$, $85$, $82$, and $90$, respectively. What is the score of the remaining person?

Overseas Competition->Knowledge Point->Word Problem Modules->Average Problems

There are $0+5+2+10=17$ points more than the average, so the last score is $80-17=63$.


It is $4:00$ now. How many minutes does it take for the clock\textquotesingle s hour and minute hands to overlap for the first time?

$43\frac 7{11}$ minutes

$10\frac {10}{11}$ minutes

$21\frac 9{11}$ minutes

$22\frac 9{11}$ minutes

$42\frac 7{11}$ minutes

Overseas Competition->Knowledge Point->Distance Word Problems->Clock Problems


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 only $80 \textbackslash\%$ of the problems she solved alone, but overall $88 \textbackslash\%$ of her answers were correct. Zoe had correct answers to $90 \textbackslash\%$ of the problems she solved alone. What was Zoe\textquotesingle s overall percentage of correct answers? (2017 AMC 8, Question \#14)

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 80 problems correct by herself, and got 176 problems correct overall. We also know that Zoe had 90 problems she did correct alone. We can see that the total amount of correct problems Chloe and Zoe did was $176-80=96$. Therefore Zoe has $96+90=186$ problems out of 200 problems correct. This is (C) $93$ percent.


Three boxes under my stairs contain apples or pears or both. Each box contains the same number of pieces of fruit. The first box contains all twelve of the apples and one-ninth of the pears. How many pieces of fruit are there in each box?

Overseas Competition->Knowledge Point->Word Problem Modules->Equation Word Problems->Word Problems for Linear Equations with One Variable


Duckings are walking in a queue when they are following the mother duck. Joey the duckling is the 6th from the front and 8th from the back (including the mother duck). Some of the ducklings at the back moved away from the queue making Joey the middle duckling in the queue. How many ducklings moved away from the queue?

None of the above.

total number of duck and ducklings in the queue is 6+(8-1) = 13.  if 2 ducklings at the back moved away, it would be 13-2=11 in the queue, and the middle ducking would be 6th from the front, which is Joey\textquotesingle s position.


Every week, Robyn swims while Sam sings.  For every $$2$$ of Sam\textquotesingle s songs, Robyn swims $$9$$ lengths.  If Robyn swims $$54$$ lengths, how many songs does Sam sing?

Overseas Competition->Knowledge Point->Word Problem Modules->Unitary Method Problems->Applying Multiplication and Division

Sam sings $$2$$ songs in the time that Robyn swims $$9$$ lengths. The time that she takes for $$54$$ lengths is $$54\div9 =6$$ times longer, so Sam will sing $$2\times6 =12$$ songs.


If today is a Wednesday, then $$22$$ days ago was a．

Tuesday

Wednesday

Thursday

Friday

Overseas Competition->Knowledge Point->Word Problem Modules->Periodic Problems->Periodic Problems of Time

Today is Wednesday. $$21$$ days ago was Wednesday. $$22$$ days ago was Tuesday.  $\textasciitilde$  or  $\textasciitilde$  $$22\div7=3$$ weeks $$1$$ day.  $$1$$ day before Wednesday was a Tuesday.


John drinks $x$ cups of boba everyday on weekdays and $1$ cups of boba everyday on the weekends. Which of the following equations represents how many cups of boba John drinks every week?

Overseas Competition->Knowledge Point->Word Problem Modules->Equation Word Problems

The boba that John drinks during weekdays: $$5x$$  The boba that John drinks during weekends: $$2\times 1 =2$$  The total boba that John drinks during a week: $$5x+2$$


This year Gareth is seven times the age of his son Herbie. In six years\textquotesingle{} time Gareth will be four times as old as Herbie. What is the difference between Gareth\textquotesingle s and Herbie\textquotesingle s ages this year?

$$16$$ years

$$20$$ years

$$24$$ years

$$36$$ years

$$42$$ years

Overseas Competition->Knowledge Point->Word Problem Modules->Age Problems->Basic Relationships in Age Problems

Let Herbie\textquotesingle s age this year be $$h$$ years. Then Gareth\textquotesingle s age this year is $$7h$$ years. In $$6$$ years\textquotesingle{} time, their ages will be $$h+6$$ years and $$7h+6$$ years respectively. We are told that $$7h+6=$$$$4$$$$\left(h+6\right)$$. This gives $$7h+6=4h+24$$ so that $$3h=18$$ and $$h=6$$.  So the present ages are $$6$$ and $$7\times 6=42$$, giving a difference of $$36$$ years.


A piece of pasture can feed $$17$$ cow for $$30$$ days or $$19$$ cows for $$24$$ days. If the grass grows at the same speed every day, how many cows can it feed for $$6$$ days? ．

Overseas Competition->Knowledge Point->Word Problem Modules->Newton's Problem of Cows and Fields


The month of February of a given year has five Fridays. What day of the week is January $$31$$ of that year?

Tuesday

Thursday

Friday

Saturday

There are seven days in a week; so, in order to have five Fridays, a month has at least $$4\times7+1=29$$ days. Since February has at most $$29$$ days, the $$1^{\rm st}$$ and $$29^{\rm th}$$ days of this February must both be Fridays, which means that January $$31$$ is a Thursday.


Several students are singing in a line. Amy is the fourth one from the left. From the right, Amy is the fourth too. How many students are singing? (adapted from $$2019$$ Math kangaroo Problems, Level $$1-2$$, Question \#$$5$$)

Overseas Competition->Knowledge Point->Word Problem Modules->Applying Addition and Subtraction

It comes $4+4=8$ when we count Amy twice.  The number of all singing students is $4+4-1=7$


In a class of pupils, $80\textbackslash\%$ participated in basketball, $85\textbackslash\%$ participated in football, $74\textbackslash\%$ participated in softball and $68\textbackslash\%$ participated in squash. Find the minimum percentage of pupils who participated in all the four sports events.

None of the above

Overseas Competition->Knowledge Point->Word Problem Modules->Solving Problems Involving Fractions and Percentages->Finding the Rate

Least percentage of pupils who participated in both basketball and football  $=80\textbackslash\%+ 85\textbackslash\%-100\textbackslash\% =65\textbackslash\%$,  Least percentage of pupils who participated in basketball, football and softball  $=65\textbackslash\%+74\textbackslash\%-100\textbackslash\% =39\textbackslash\%$,  Least percentage of pupils who participated in basketball, football, softball and squash  $=39\textbackslash\%+ 68\textbackslash\%-100\textbackslash\% = 7\textbackslash\%$.


If six days ago was a Tuesday, then eight days from today is a．

Monday

Tuesday

Wednesday

Thursday

$$6$$ days ago was a Tuesday. Today is Monday. Thus, in $$8$$ days\textquotesingle{} time, it will be a Tuesday.


In the sequence of letters $$\text{KANGAROOKANGAROOKANG}\cdots$$ the word $$\text{KANGAROO}$$ is repeated indefinitely. What is the $$2017\text{th}$$ letter in this sequence?

$$\text{K}$$

$$\text{N}$$

$$\text{G}$$

$$\text{R}$$

$$\text{O}$$

Overseas Competition->Knowledge Point->Word Problem Modules->Periodic Problems->Periodic Problems of Basic Permutation

The sequence $$\text{KANGAROOKANGAROOKANG}\cdots$$ repeats every $$8$$ letters. Since $$2017=8\times252+1$$, the $$2017\text{th}$$ letter in the sequence is the first of the repeating sequence and hence is $$\text{K}$$.


There are $$30$$ students in your class. The ratio of boys to girls is $$3:2$$. How many boys and girls are there?.

$$30$$ and $$20$$

$$20$$ and $$30$$

$$18$$ and $$12$$

$$12$$ and $$18$$

Overseas Competition->Knowledge Point->Word Problem Modules->Solving Problems Involving Proportions->Basic Units

According to the ratio $$3:2$$, we could know the total is $$5$$. Then the number of boys is $$30\times \dfrac{3}{5} =18$$.The number of girls is $$30\times\dfrac{2}{5}=12$$.


A bag contains $$100$$ raffle tickets in red, white and blue. The ratio of red to white tickets is $$5:4$$ and the ratio of white to blue tickets is $$8:7$$. What is the smallest number of tickets that would have to be picked to make sure having selected at least half of \textbf{one} colour?

Overseas Competition->Knowledge Point->Word Problem Modules->Solving Problems Involving Proportions->Complex Ratio Problems

First we must work out the number of tickets of each colour. Let the numbers of red, white and blue tickets be $$r$$, $$w$$ and b respectively; we have $$r : w = 5 : 4$$ and $$w : b = 8 : 7$$. Writing $$r : w$$ as $$10 : 8$$, we can combine the ratios to find $$r : w : b = 10 : 8 : 7$$. Hence for a total of $$100$$ tickets, there must be $$40$$ red, $$32$$ white and $$28$$ blue tickets. Now the largest number of red tickets one can select without having half of them is $$19$$, and likewise $$15$$ and $$13$$ for white and blue tickets. So it would be possible to select $$19 + 15 + 13 = 47$$ tickets and still not have half or more of any particular colour of ticket. However, the very next ticket, the $$48\text{th}$$, would have to be a red, a white or a blue one, and so one of the colours would now have at least half of its number selected.


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?

Friday

Saturday

Sunday

Monday

None of the above

40th day = 39th day after first day joining.  39 days/7 = 5 weeks R 4 days. 4 days after Tues is Saturday.


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?

Overseas Competition->Knowledge Point->Word Problem Modules->Age Problems


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?

$$3:1 $$

$$1:4 $$

Julia has eaten one-quarters of the pizza. So three-quarter of the pizza remain.  Therefore the required ratio is $$\frac{3}{4}:\frac{1}{4}=3:1$$.


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?

Overseas Competition->Knowledge Point->Word Problem Modules

1-1/2-1/2*3/4=1/8，1/8means 6 bulbs，$$so$$ in total there are 6\div 1/8=48 bulbs.


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?

Overseas Competition->Knowledge Point->Word Problem Modules->Solving Problems Involving Fractions and Percentages->Finding the Base

He has $\dfrac{3}{5}$~remaining after selling $\dfrac{2}{5}$. We can write the equation as:~$150\div\dfrac{3}{5}=250$.


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?

There are $5+8+1=14$ players in each group. There are $14\times 6=84$ players in total.


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?

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 $$14 + 8 = 22$$.


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?

None of the above


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?

$20$; $12$

$22$; $20$

$20$; $24$

$19$; $21$

Overseas Competition->Knowledge Point->Word Problem Modules->Chicken-Rabbit Problems->Solving Chicken-Rabbit Problems by Using Hypothesis

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.


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?

Overseas Competition->Knowledge Point->Word Problem Modules->Unitary Method Problems->Finding the Unit Rate and then the Total Value

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 $$4\times2\text{m}=8\text{m}$$.


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?

$$480$$ miles

$$1080$$ miles

$$960$$ miles

$$720$$ miles

$$1200$$ miles

Overseas Competition->Knowledge Point->Distance Word Problems->Distance Word Problems on Straight Road

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.


Calculate: $1\times 10+2\times 9+3\times 8+ \cdots + 9\times 2+ 10\times 1$

None of the above

Overseas Competition->Knowledge Point->Word Problem Modules->Questions Involving Sum, Difference and Multiples


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?

Overseas Competition->Knowledge Point->Word Problem Modules->Age Problems->Sums and Multiples in Age Problems

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.  $$24\times 3=72$$


Five children divided some cakes equally. What percents of the cakes two of them could get in total?

Five children divided some cake equally. Two of the children totally got $$\dfrac{2}{5}$$ of the cake and $$\dfrac{2}{5}=0.4 = 40\textbackslash\%$$.


This year, March $$19$$ falls on a Thursday. What day of the week will it be in $$30$$ days?

Wednesday

Thursday

Friday

Saturday

Sunday

Overseas Competition->Knowledge Point->Word Problem Modules->Periodic Problems->Periodic Problems of Time

$$30\div4=7 \text{ R } 2$$, two days after a Thursday is a Saturday.


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?

Overseas Competition->Knowledge Point->Word Problem Modules->Average Problems

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.


If $$10^{}\text{th}$$June is Sunday, what day of the week is $$1$$\textsuperscript{st} June?

Sunday

Monday

Wednesday

Friday

Saturday

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.


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 ?

Overseas Competition->Knowledge Point->Word Problem Modules->Solving Problems Involving Fractions and Percentages->Percentages Word Problems

$$200\times \left( 1-\frac{1}{10}-\frac{1}{5}-25 \textbackslash\% -15 \textbackslash\% ~\right)=60$$.


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?

$36$ kilograms

$38$ kilograms

$40$ kilograms

$43$ kilograms


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?

Both have the same amount of pocket money now.

John has more pocket money now.

Mary has more pocket money now.

It is impossible to tell who has more pocket money now.

Overseas Competition->Knowledge Point->Word Problem Modules->Solving Problems Involving Fractions and Percentages->Knowing the Base

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 $$0.9\times1.1x=0.99x$$.


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?

Monday

Tuesday

Wednesday

Thursday

Friday

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


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 $2018 \text{kg}$.

None of the above

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$=42 \text{kg}$.  $2018 =48\times6+1= 289$.


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)

Overseas Competition->Knowledge Point->Word Problem Modules->Age Problems->Basic Relationships in Age Problems->Difference of Ages

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.


Sinnce $$7^{2}=49$$, its ones\textquotesingle{} digit is a $$9$$. What is the ones\textquotesingle{} digit of $$2^{50}$$?

Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders->Applying the Properties of Dividing without Remainders->Maximum/Minimum Problems of Division without Remainders


If six days ago was a Tuesday, then eight days from today is a．

Monday

Tuesday

Wednesday

Thursday

$$6$$ days ago was Tues. Today is Mon. In $$8$$ days it will be Tues.


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?

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 $$2004$$.


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?

The $15\text{-ounce}$ bag

The $20\text{-ounce}$ bag


If February is a month that contains Friday the $$13^{}\text{th}$$, what day of the week is February $$1$$?

Sunday

Monday

Wednesday

Thursday

Saturday

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.


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{~~~~~~~~~~}~$$\rm m$$.

Overseas Competition->Knowledge Point->Word Problem Modules->Applying Addition and Subtraction->Simple Addition and Subtraction


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?

Overseas Competition->Knowledge Point->Word Problem Modules->Solving Problems Involving Proportions->Complex Ratio Problems

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 ￡$$6$$.


Adam paid £6 for $$15$$ buns. How many dollars did Tom pay for the same kind of buns if he bought $$5$$?

Overseas Competition->Knowledge Point->Word Problem Modules->Unitary Method Problems->Applying Multiplication and Division


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?

Overseas Competition->Knowledge Point->Word Problem Modules->Applying Addition and Subtraction

14 - 1 (eagle) - 1 (mother hen) = 12 (chicks)  12 - 5 7 chicks


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.

Overseas Competition->Knowledge Point->Word Problem Modules->Unitary Method Problems->Finding the Total Value using Unitary Method

$$7$$ 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.


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)

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.


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?

Overseas Competition->Knowledge Point->Word Problem Modules->Age Problems

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.


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? ．

$$ $$Sunday$$ $$

$$ $$Tuesday$$ $$

$$ $$Thursday$$ $$

$$ $$Saturday$$ $$

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.


My age is $$3$$ years less than $$5$$ times my sister\textquotesingle s age. If I am $$27$$, the sum of my age and my sister\textquotesingle s age is．

Overseas Competition->Knowledge Point->Word Problem Modules->Age Problems->Differences and Multiples in Age Problems

My age, $$27$$, is $$3$$ years less than $$5$$ times my sister\textquotesingle s age, so $$30$$ is $$5$$ times her age. Thus, my sister is $$6$$. The sum of our ages is $$27+6=33$$.


Valeria has $$65$$ grams of a $$20\textbackslash\%$$ sugar solution. How many grams of sugar is in the solution? How many grams of water is in the solution?

$$20$$; $$65$$

$$13$$; $$52$$

$$13$$; $$65$$

$$20$$; $$52$$

Overseas Competition->Knowledge Point->Word Problem Modules->Concentration Problems->Basic Concentration Problems

Sugar: $$65\times 20\textbackslash\% = 13$$ ounces;  Water: $$65 - 13 = 52$$ ounces.


Each watermelon\textquotesingle s full price is $$4$$ dollars. Currently, the watermelons on a discounted price for $$3$$ dollars each. Ms. Lee ~wants to buy $$3$$ of them. She needs to pay~\uline{~~~~~~~~~~}~dollars in total.

Overseas Competition->Knowledge Point->Word Problem Modules->Unitary Method Problems->Finding the Unit Rate and then the Total Value


Mother\textquotesingle s Day in $$2020$$ was May $$10$$th, which was Sunday. Father\textquotesingle s Day in $$2020$$ was June $$21$$st. On what day did Father\textquotesingle s Day fall?

Thursday

Saturday

Sunday

Tuesday

It passed $$31-10 + 21 = 42$$ days in total. Since $$42 \div 7 = 6$$, which means the Father\textquotesingle s Day fell on Sunday as well.


The ages of Tom\textquotesingle s father, Tom\textquotesingle s mother and Tom are $86$ together. Tom\textquotesingle s mother is $24$ years older than Tom and~~$2$ years younger than Tom\textquotesingle s father.How old is Tom?

Overseas Competition->Knowledge Point->Word Problem Modules->Questions Involving Sum, Difference and Multiples->Problems of Sum and Multiple

Tom\textquotesingle s mother is $(86+24-2)\div3=36$ years old.  Tom is $36-24=12$ years old.


How many ways are there to make $$$80$$ using some combination of $$$5$$, $$$10$$ and $$$20$$ notes?

None of the above

Condition on the number of $$$20$$ note you use.  $$4$$ gives $$1$$ possibility,  $$3$$ gives $$3$$ possibilities,  $$2$$ gives $$5$$ possibilities,  $$1$$ gives $$7$$ possibilities,  $$0$$ gives $$9$$ possibilities.


Mark and Karl have $$100$$ dollars in total. The money Mark owns is three times the money Karl owns. How many dollars does Mark have? (Adapted from 1999 Math Kangaroo Problem, Level 3 - 4, Question \#6)

Overseas Competition->Knowledge Point->Word Problem Modules->Questions Involving Sum, Difference and Multiples->Problems of Sum and Multiple->Sum and Multiple of Two Variables


If today is Monday, which day of the week will it be $$22$$ days later? ．

Sunday

Monday

Tuesday

Wednesday

Overseas Competition->Knowledge Point->Word Problem Modules->Periodic Problems->Periodic Problems of Time

Since there are 7 days in a week, the day will repeat again after 7 days.  Thus, if today is Monday, it will be Monday again a week later.  $$22\div 7=3$$ week R $$1$$.  The day after $$22$$ days is the same as the day after $$1$$ day.  $$1$$ day after Monday is Tuesday.  Therefore, it will be \textbf{Tuesday} $$22$$ days later.


In a maths test, the average score of $$40$$ students is $$25$$. The lowest score is $$5$$. At most how many student(s) can get $$100$$ points?

Overseas Competition->Knowledge Point->Word Problem Modules->Average Problems

Total points: $$40\times 25=1000$$.  To have as many $100$ points as possible, we should consider there are only two groups of students: everyone gets $5$ points in a group and everyone gets $100$ in the other group.  $$10\times 100=1000$$, so there must be less than $10$ people getting $100$ points.  If there are nine $100$s: $$9\times 100+31\times 5=1055\textgreater1000$$, which is impossible.  If there are eight $100$s:$$8\times 100+32\times 5=960\textless{}1000$$.  Thus, the answer is $$8$$.


The class organized students to do gymnastics side by side，$$and$$ there are two people in each row. Bob and Tick observed that there were $8$ rows in front of them and $11$ rows in the back. How many people in the class did gymnastics together?~(adapted from 2000 Math Kangaroo Problem, Level 3 - 4, Question \#12)

Overseas Competition->Knowledge Point->Word Problem Modules->Questions Involving Squares->Solid Squares->Basic Solid Square Problems

It can be seen from the question that Bob and Tick have $8$ rows in front and $11$ rows in the back. There are $20$ rows in total, and there are two people in each row, so there are $40$ people in total.


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?

$$7 \text{kg}$$

$$14 \text{kg}$$

Overseas Competition->Knowledge Point->Word Problem Modules->Questions Involving Sum, Difference and Multiples->Problems Involving Sum and Difference

It should be a moment\textquotesingle s work to see that Felix weighs $$3\text{kg}$$ , and Marmalade $$7\text{kg}$$ .


If $6\text{cm}$ represents $50\text{km}$ on a map and the actual distance between two towns is $125\text{km}$, then their distance apart on the map is ．

Overseas Competition->Knowledge Point->Word Problem Modules->Solving Problems Involving Proportions->Basic Units

If $$6\textasciitilde\text{cm}$$ represents $$50\textasciitilde\text{km}$$, $$3\textasciitilde\text{cm}$$ represents $$25\textasciitilde\text{km}$$. So, their distance apart on the map is $$125\div25\times 3=15\textasciitilde\text{cm}$$.


In the yard there is an equal number of pigs, ducks and chickens. Together, they have $$144$$ legs. How many ducks are there in the yard?

Overseas Competition->Knowledge Point->Word Problem Modules->Equation Word Problems->Word Problems for Linear Equations with One Variable

Suppose there are $$x$$ ducks in the yard. Since there is an equal number of pigs, ducks and chickens, $$2x+4x+2x=144$$, $$x=18$$.


How many days are there in exactly $$52$$ weeks?

Each week has $$7$$ days, so $$52$$ weeks has $$52 \times 7 = 364$$ days.


A "combo"~ticket to enter the fair and ride unlimited rides is ＄$$30$$. A``per ride" ticket costs ＄$$12.50$$ to enter and ＄$$5$$ per ride. For a"combo''ticket to cost less than a "per ride" ticket, a person must go on at least  rides.

Overseas Competition->Knowledge Point->Word Problem Modules->Word Problems in Economics->Basic Profit and Loss Concepts

A fair sells a``combo" ticket for ＄$$30$$ entry and a ``per ride" ticket for ＄$$12.50$$ to enter plus ＄$$5$$ per ride. A``per ride" ticket costs ＄$$12.50+$$ ＄$$15 =$$＄$$27.50$$ for $$3$$ rides and ＄$$12.50 +$$ ＄$$20 = $$＄$$32.50$$ for $$4$$ rides.


Niki usually leaves her cell phone on. If her cell phone is on but she is not actually using it, the battery will last for $$24$$ hours. If she is using it constantly, the battery will last for only $$3$$ hours. Since the last recharge, her phone has been on $$9$$ hours, and during that time she has used it for $$60$$ minutes. If she doesn\textquotesingle t use it any more but leaves the phone on, how many more hours will the battery last?

Overseas Competition->Knowledge Point->Word Problem Modules->Work Word Problems->Simple Work Word Problems

When not being used, the cell phone uses up $$\dfrac{1}{24}$$ of its battery per hour. When being used, the cell phone uses up $$\dfrac{1}{3}$$ of its battery per hour. Since Niki\textquotesingle s phone has been on for $$9$$ hours, of those $$8$$ simply on and~~being used to talk, $$8\left(\dfrac{1}{24}\right)+1\left(\dfrac{1}{3}\right)=\dfrac{2}{3}$$ of its battery has been used up. To drain the remaining $$\dfrac{1}{3}$$ the phone can last for $$\dfrac{\dfrac{1}{3}}{\dfrac{1}{24}}=\boxed{(\text{B})8}$$ more hours without being used.


A worm is staying at the bottom of a well with a depth of $$11$$ meters. If it climbs up $$3$$ meters in the day time and slips down $$1$$ meter at night, which day will it climb up to the ground?

Overseas Competition->Knowledge Point->Word Problem Modules->Periodic Problems->Snail Climbing out of Well Problems

It climbs up $$3$$ meters in the day time and slips down $$1$$ meter at night, so it climbs up $$2$$ meters in a whole day. For the first $$4$$ days, it climbs $$8$$ meters. During the day time of the fifth day, it climbs $$3$$ meters and reaches the ground. Then it needs five days to climb up to the ground.


The sum of the ages of Anita and Peter is $$20$$ years. What is the sum of their ages three years ago?

Overseas Competition->Knowledge Point->Word Problem Modules->Age Problems->Basic Relationships in Age Problems->Basic Sum and Differences Problems in Age Problems


James makes a long noodle. He takes some bites of the noodle, each time eating $5$ cm of the noodle from the middle of only one piece. At last, he has $12$ pieces of the noodle with the total length of $24$ cm. How long was the entire noodle in cm at the beginning? (Adapted from 2022 AMC 8 Problem, Question \#11)

Overseas Competition->Knowledge Point->Word Problem Modules->Interval Problems

There are $12$ pieces of noodle left, so he takes $12-1=11$ bites of the noodle. Thus, the length of the noodle he ate is $11\times5=55$ cm. The total length of the entire noodle was $55+24=79$ cm.


A mixture of $25$ liters of paint is $16 \textbackslash\%$ red tint, $24\textbackslash\%$ yellow tint and $60\textbackslash\%$ water. $15$ liters of yellow tint are added to the original mixture. What is the percent of yellow tint in the new mixture? (adapted from 2007 AMC 8, Question \#17 )

Overseas Competition->Knowledge Point->Word Problem Modules->Solving Problems Involving Fractions and Percentages->Percentages Word Problems

The original mixture contains $25\times 24\textbackslash\% = 6$ liters of yellow tint.  $$15$$ liters of yellow tint is added to the mixture, the new mixture now has $6+15=21$ liters of yellow tint.  New percent of yellow =$\frac{21}{25+15} =52.5\textbackslash\% $


Edwin has three flowers, one red, one blue and one yellow. They have $50$ petals in total. The blue flower has $4$ petals less than the red one and $5$ petals more than the yellow one. How many petals does the red flower have?

Overseas Competition->Knowledge Point->Word Problem Modules->Applying Addition and Subtraction->Simple Word Problems Involving Comparing and Ordering


Mike has a bookcase with three layers, and the proportion of books placed on the upper, middle and lower layers is $$5:6:4$$. Given that there are $$240$$ books on the middle level,  books should be removed from middle level and added to the lower level so that the number of books on each level can be exactly the same.

Overseas Competition->Knowledge Point->Word Problem Modules->Solving Problems Involving Proportions->Ratio Word Problems with an Invariant Part

Lower level: $$240\div6\times4=160$$,  The number of books to be added to the lower level: $$(240-160)\div2=40$$.  So, the answer is $$\text{B}$$.


Barb likes to help her father with housework. She dusts every $3$ days, sweeps every $4$ days, and cooks dinner every $5$ days. If she does all $3$ chores on one Sunday, she next does all $3$ on the same day on a ．

Wednesday

Thursday

Friday

Saturday

The least common multiple of $3$, $4$ and $5$ is $60$. We know that $60\div7=8\ldots\ldots4$, then the remainder tells us that it will be on Thursday.


This year, October $$28$$ falls on a Sunday. What day of the week will it be after $$38$$ days? (adapted from 2015 Math Kangaroo Problem, Level 3-4, Question \#8)

Wednesday

Thursday

Tuesday

Friday

Sunday

$$38\div7=5 \text{R} 3$$, three days after Sunday, which is Wednesday.


A family made a $$100\text{km}$$ trip. For half the distance, they drove at $$50\text{km}$$~ per hour; for the other half, they drove at $$25\text{km}$$~ per hour How many hours did this trip take?

Overseas Competition->Knowledge Point->Word Problem Modules

For the first $$50\text{km}$$, they drove at $$50\text{km/hr}$$-that took $$1\text{hr}$$. For the other second $$50\text{km}$$, they drove at $$25\text{km/hr}$$-that took $$2\text{hrs}$$ . The trip took $$3\text{hrs}$$ .


James has $600$ g of a $$8\textbackslash\%$$ salt solution, and then $120$ g of water evaporates. Find the percent concentration of salt solution at this time.

Overseas Competition->Knowledge Point->Word Problem Modules->Concentration Problems


Alex is $8$ years older than her two sisters who are twins. The sum of the ages of all three girls is $32$ years old. How old is Alex? ($2007$ Math Kangaroo Problem, Level $5$-$6$, Question \#$9$)

Overseas Competition->Knowledge Point->Word Problem Modules->Age Problems

The age of each twin is $(32-8) \div 3 =8$ years old. Thus, Alex is $8 +8 =16$ years old.


The rabbit Cici picks mushrooms in the forest. She finds one mushroom every $5$ meters, and she finds $8$ mushrooms in total. How long does Cici walk from where she finds the first mushroom to where she finds the last one?

There are $8-1=7$ spaces. $7\times 5=35$.


Given March $$25$$ of a certain year is Monday, what day of the week would May $$1$$ fall on that year? ．

Tuesday

Wednesday

Thursday

Friday

Overseas Competition->Knowledge Point->Word Problem Modules->Periodic Problems->Periodic Problems of Time

$$1$$ May is $$6+30+1=37$$ days after March $$25$$. Since $$37\div7=5 \text{ R }2$$, $$5$$ weeks after is still Monday and $$2$$ days after that is a Wednesday.


Jane released a homing pigeon at $$5:30\rm$$ a.m.. The pigeon arrived at its destination at $$11:30\rm$$ a.m.. How many miles did the pigeon travel if it flies $$120$$ miles in $$2$$ hours? (adapted from 2007 Math Kangaroo Problem, Level 5-6, Question \#18)

The flying speed is $$120\div2=60$$ miles/hr, and the flying time is $$11:30-5:30=6$$ hours. So, the total distance is $$60\times 6=360$$ miles.


A bag of toffee is $8$ dollars, a bag of cotton candy is $6$ dollars, and a bag of orange candy is $13$ dollars. Now, the candy shop decides to mix $30$ bags of toffee, $50$ bags of cotton candy, and $20$ bags of orange candy for $100$ bags of assorted candy. What should be the price of the assorted candy in dollars to keep the total revenue unchanged?

Overseas Competition->Knowledge Point->Word Problem Modules->Average Problems

Total revenue: $30\times 8+50\times 6+20\times 13=800$ dollars  A bag of assorted candy: $$800\div 100=8$$ dollars


There are some ducks and sheep in a farm. The number of sheep is $10$ more than ducks. The number of ducks is half the number of sheep. In total, how many ducks and sheep are there in the farm?  ~


Pat paints twice as fast as Lee. If it took Lee $6$ hours to paint a room, how long would it have taken if both had painted the room together?

$2$ hours

$3$ hours

$9$ hours

$18$ hours

Overseas Competition->Knowledge Point->Word Problem Modules->Work Word Problems->Collaborative Work Word Problems

Lee paints the room once in $6$ hours. Pat paints the room twice in $6$ hours. Together, they paint the room $3$ times in $6$ hours. So, it takes them $2$ hours to paint it once together.


Ben and Ken have some books. After Ben sends $4$ books to Ken, he has $1$ fewer book than Ken. Which of the following is true?

Originally, Ben had $3$ more books than Ken.

Originally, Ben had $4$ more books than Ken.

Originally, Ben had $1$ fewer book than Ken.

Originally, Ben had $9$ more books than Ken.

Originally, Ben had $7$ more books than Ken.


John bought $10$ pencils and notebooks in total for £$44$. Each pencil cost £$$2$$. Each notebook costs £$$8$$. How many notebooks did John buy?

Overseas Competition->Knowledge Point->Word Problem Modules->Chicken-Rabbit Problems->Solving Chicken-Rabbit Problems by Using Hypothesis

If all were pencils,  $10\times 2= 20$  the total cost would be £$$20$$.  $44- 20=24$  The difference is £$$24$$.  $8-2=6$  $24\div 6=4$  John bought $4$ notebooks.


Lily\textquotesingle s age plus Judy\textquotesingle s age was equal to $$17$$ two years ago. What is the sum of their ages this year? (adapted from 2016 Math Kangaroo Problem, Level 1-2, Question \#16)

Overseas Competition->Knowledge Point->Word Problem Modules->Age Problems->Basic Relationships in Age Problems->Sum of Ages


If I start with $2$, and begin to count by $$3\textquotesingle$$s, my $50^{}\text{th}$ number will be~\uline{~~~~~~~~~~}~.

Overseas Competition->Knowledge Point->Word Problem Modules->Practical Application of Arithmetic Progression


Jucy and Nicole are celebrating their birthdays together. Seven years ago, when Jucy turned $6$ years old, she received a newborn cat as a birthday present. Today the sum of the ages of the two people and the cat is $34$ years. Jucy is~\uline{~~~~~~~~~~}~years younger than Nicole.


Bowen has $$40$$ grams of a $$25\textbackslash\%$$ sugar solution. After~\uline{~~~~~~~~~~}~of water evaporates, the percent concentration of the solution is $$40\textbackslash\%$$. (Evaporate: the water turns from liquid into vapor and is no longer in the solution anymore.)

$$15$$ grams

$$12$$ grams

$$9$$ grams

$$6$$ grams

Overseas Competition->Knowledge Point->Word Problem Modules->Concentration Problems->Finding and Making Use of Invariants


Tom goes to the supermarket to buy egg tarts. There are two kinds of egg tarts. One has $$6$$ tarts in the box and the other has $$10$$ tarts in the box. The unit price of the egg tarts is the same. If Tom wants $$76$$ egg tarts, how many boxes does he need at least?~(adapted from 2011 Math Kangaroo Problem, Level 3 - 4, Question \#10)

Overseas Competition->Knowledge Point->Word Problem Modules->Unitary Method Problems->Unitary Method with One Variable


A gift store buys some gifts at＄$$125$$ each. It wants to earn＄$$25 $$ for each gift after a discount of $$25 \textbackslash\% $$. What is the selling price before the discount for each gift?

$$160$$ dollars

$$180$$ dollars

$$190$$ dollars

$$200$$ dollars

Overseas Competition->Knowledge Point->Word Problem Modules->Word Problems in Economics->Basic Profit and Loss Concepts

$$\left( 125+25 \right)\div \left( 1-25 \textbackslash\% \right)=200$$ dollars.


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?

$36$ kilograms

$38$ kilograms

$40$ kilograms

$43$ kilograms


It takes Mary $$30$$ minutes to walk uphill $$300 $$ m from her home to school, but it takes her only $$10$$ minutes to walk from school to home along the same route. What is her average speed, in m/min, for the round trip? (adapted from 2003 AMC 10 Problem, Question \#4)

Overseas Competition->Knowledge Point->Distance Word Problems->Distance Word Problems on Straight Road

Mary walks a total of $$300+300=600\rm$$ meters in $$40$$ minutes.  Average speed: $600\div40=15$ m/min


If $$6$$ cans of Sweet Stuff Soda all together contain $$96$$ teaspoons of sugar, then there are a total of teaspoons of sugar in $$15$$ cans.

Overseas Competition->Knowledge Point->Word Problem Modules->Unitary Method Problems->Finding the Unit Rate and then the Total Value

If $$6$$ cans contain $$96$$ teaspoons of sugar, $$1$$ can contains $$96\div6 = 16$$ teaspoons of sugar. Thus $$15$$ cans contain $$16\times15 =240$$ teaspoons of sugar.


Twenty-eight students from the fourth grade competed in a math competition. Each student earned a different number of points. The number of students who received more points than Tomek is two times smaller than the number of students who had fewer points than Tomek. In which position did Tomek finish that competition?~ ($2002$~Math Kangaroo Problem, Level $3-4$, Question \#$23$)

Overseas Competition->Knowledge Point->Word Problem Modules->Questions Involving Sum, Difference and Multiples->Problems of Sum and Multiple

Except for Tomek, there are~$28-1=27$~students. The number of students who received more points than Tomek is~$27\div(2+1)=9$, so Tomek is in the~$9+1=10^{th}$~place.


January $$1$$, $$1989$$ was a Sunday. January $$1$$, $$1988$$ (a leap year) was a．

Friday

Saturday

Sunday

Monday

$$366$$ days $$=52$$ weeks $$2$$ days.  Jan $$1$$ $$1988$$ is $$52$$ weeks $$2$$ days before Jan $$1$$ $$1989$$. This date in $$1988$$ is $$2$$ days before Sunday so it is on a Friday.


Algebra books cost $$$12.50$$ each, and geometry books cost $$$14$$ each. How much do $$6$$ algebra books and $$5$$ geometry books cost in total?

Overseas Competition->Knowledge Point->Word Problem Modules->Unitary Method Problems->Finding the Total Value using Unitary Method

Six algebra books cost $$$12.50 \times 6 = $75$$. Five geometry books cost $$$14\times5 = $70$$. All together, they cost a total of $$$145$$.


In $$1975$$ it was estimated that there were $$250000$$ lions in Africa.  Over $$40$$ years this figure has decreased by $$90\textbackslash\%$$.  What is the current estimate for the number of lions in Africa ?

Overseas Competition->Knowledge Point->Word Problem Modules->Solving Problems Involving Fractions and Percentages->Finding the Base

A $$90\textbackslash\%$$ decrease means $$10\textbackslash\%$$ of $$250000 = 250000 \div 10 = 25000$$ of the lions remain.


A test is with a full score of $100$ points. Five students in a group wants to reach an average of $91$ points. If everyone scores a different whole number, the lowest score among them could be~\uline{~~~~~~~~~~}~.


Jack\textquotesingle s sister is $$22$$ years old, which is $$2$$ less than $$4$$ times Jack\textquotesingle s age. Assume Jack is $$x$$ years old, then．

Overseas Competition->Knowledge Point->Word Problem Modules->Equation Word Problems

The equation can represents this situation is $$4x-2=22$$,  so, $$4x=24$$, $$x=6$$.  So, the answer is $$6$$.


Jason and Ross are in line to check tickets. There are $36$ people in front of Jason and $16$ people behind. Ross happens to be in the middle of the entire line. How many people are there between Ross and Jason?~(adapted from 2000 Math Kangaroo Problem, Level 3 - 4, Question \#12)

Overseas Competition->Knowledge Point->Word Problem Modules->Questions Involving Squares->Solid Squares->Basic Solid Square Problems

The total: $36+16+1=53$ , Jack\textquotesingle s location: $$(53-1)\div2=21$$, so there are $15$ people between Jack and Rose.


Amy mixes $$30$$ grams of a $$30\textbackslash\%$$ salt solution and $$20$$ grams of a $$20\textbackslash\%$$ salt solution together. How many grams of water should she add to the mixture to make it a $$10\textbackslash\% $$ solution?

$$30\times30\textbackslash\%+20\times 20\textbackslash\%=9+4=13$$ ounces.  $$13\div10\textbackslash\%-(30+20)=80$$ ounces.


$$12$$ dogs are cqually divided into $$3$$ groups, how many dogs are there in each group.

Overseas Competition->Knowledge Point->Word Problem Modules


My pet frog jumps $$3\text{m}$$ per jump. If it wants to jump from one end of a $$100\text{m}$$ field to another, the least number of jumps it will take is．

Since $$100\div 3=33$$ with remainder $$1$$, it takes my frog $$33$$ jumps to jump $$99\text{m}$$. It needs $$1$$ more jump, for $$34$$ jumps in all, to go at least $$100\text{m}$$.


In $$2010$$, the Math Kangaroo competition is taking place in some schools for the $19$\textsuperscript{th}~time. Alice took part in the $10$\textsuperscript{th}~Math Kangaroo when she was $$9$$ years old. In what year was Alice born? (Adapted from 2008 Math kangaroo Problem, Level 3-4, Question \#9)

Overseas Competition->Knowledge Point->Word Problem Modules->Periodic Problems->Periodic Problems of Time->Period of Time

The $10$\textsuperscript{th}~Math Kangaroo was taking place in $2001$, which means that Alice was $9$ years old in $2001$, so she was born in $1992$.


Half a loaf of bread costs $$6$$ pence more than one-fourth of a loaf of bread. How many pence does a whole loaf of bread cost?

One-fourth of a loaf of bread is ``$$1$$.'' Half a loaf of bread is ``$$2$$.''  One-fourth of a loaf of bread: $$6 \div (2-1) =6$$. A whole loaf of bread: $$6 \times 4 =24$$.


Ben has $$9$$ pairs of shoes, of which $$4$$ shoes are red, $$2$$ shoes are bule, and the rest are green. How many pairs of shoes are green?


Antonette gets $70 \textbackslash\%$ on a 10 -problem test, $80 \textbackslash\%$ on a 20 -problem test and $90 \textbackslash\%$ on a 30 -problem test. If the three tests are combined into one 60 problem test, which percent is closest to her overall score? (2006 AMC 8, Question \#12)

Overseas Competition->Knowledge Point->Word Problem Modules->Solving Problems Involving Fractions and Percentages->Percentages Word Problems

$70 \textbackslash\% \cdot 10=7$ $80 \textbackslash\% \cdot 20=16$ $90 \textbackslash\% \cdot 30=27$ Adding them up gets $7+16+27=50$. The overall percentage correct would be $\frac{50}{60}=\frac{5}{6}=5 \cdot 16 . \overline{6}=83 . \overline{3} \approx(\mathbf{D}) 83$.


Rick and Roy each stands at different ends of a straight road that is $$2021$$m long. They run toward each other. Rick\textquotesingle s speed is $$21\text{m/s}$$ and Roy\textquotesingle s speed is $$22\text{m/s}$$. They will meet in~\uline{~~~~~~~~~~}~seconds.


You have $12$ green cards and the ratio of your red cards to green cards is $3:2$. How many cards do you have?

Overseas Competition->Knowledge Point->Word Problem Modules->Solving Problems Involving Proportions->Basic Units

$12$ is the product of $6$ and $2$.Then multiple $3$ by $6$ is the number of red cards. The total number of cards is $30$.


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 $80\textbackslash\%$ of the problems she solved alone, but overall $90\textbackslash\%$ of her answers were correct. Zoe had correct answers to $75\textbackslash\%$ of the problems she solved alone. What was Zoe\textquotesingle s overall percentage of correct answers? (2017 AMC 8, Question \#14)

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 80 problems correct by herself, and got 180 problems correct overall.  We also know that Zoe had 75 problems she did correct alone.  We can see that the total amount of correct problems Chloe and Zoe did was $180-80=100$. Therefore Zoe has $100+75=175$ problems out of $200$ problems correct. Thus $\frac{175}{200} = 87.5$ percent.


Mr. $$A$$ owns a home worth ＄$$10000$$. He sells it to Mr. $$B$$ at a $$10\textbackslash\%$$ profit based on the worth of the house. Mr. $$B$$ sells the house back to Mr. $$A$$ at a $$10\textbackslash\%$$ loss. Then: ($$1951$$ AHSME Problem, Question \#$$5$$)

$$A$$ comes out even

$$A$$ makes ＄$$1100$$ on the deal

$$A$$ makes ＄$$1000$$ on the deal

$$A$$ loses ＄$$900$$ on the deal

$$A$$ loses ＄$$1000$$ on the deal

Mr. $$ A$$ sells his home for $$(1 + 10\textbackslash\%)\times 10000$$ dollars$$= 1.1\times 10000$$ dollars$$ = 11000$$ dollars to Mr. $$ B$$. Then, Mr. $$ B$$ sells it at a price of $$(1- 10\textbackslash\%)\times 11000$$ dollars $$= 0.9\times 11000$$ dollars $$= 9900$$ dollars, thus $$11000- 9900=\boxed{ (\rm B) \textasciitilde A\textasciitilde makes\textasciitilde1100\textasciitilde on\textasciitilde the\textasciitilde deal}$$.


A $60\textbackslash\%$ alcohol solution contains $120$ grams of water. How many grams of solution are there?

Overseas Competition->Knowledge Point->Word Problem Modules->Concentration Problems


A, B, C, D and E sit around a circular table in that order. A starts by saying "$$100$$", B says "$$101$$", C says~"$$102$$" and so on round the table. Who will eventually say"$$2023$$"?

$$$$A

$$$$B

$$$$Cain

$$$$Devi

$$$$Emily

Overseas Competition->Knowledge Point->Word Problem Modules->Periodic Problems->Periodic Problems of Circular Operations

After Amit says "$$2015$$", there are $$5102-2015=3087$$ numbers remaining. Thus the counting goes round the table $$3087\div6= 514$$ times with a remainder of $$3$$, so the last to count is Devi.


Brown wants to put some board between his books. He has $10$ books in total. If he wants to put one board between every two adjacent books, how many boards does he need to prepare?

Overseas Competition->Knowledge Point->Word Problem Modules->Interval Problems


In a class of $$18$$ students, $$6$$ are wearing jeans. What is the ratio of students wearing jeans to students not wearing jeans?

Overseas Competition->Knowledge Point->Word Problem Modules->Solving Problems Involving Proportions->Dividing Quantities Based on Ratio

If $$6$$ students are wearing jeans, then $$18-6=12$$ are not. The ratio of students wearing jeans to students not wearing jeans is $$6:12=1:2$$.


How many two-digit numbers are divisible by $3$ but not by $6$?

None of the above

Overseas Competition->Knowledge Point->Word Problem Modules->Applying of Multiplication and Division

There are $99$ $4-3 = 33$-multiples of $3$ less than $100$.  $3$ of them are one-digit numbers: $3$, $6$ and $9$.  Hence there are $30$ $2$-digit multiples of $3$.  Similarly, there are $96 + 6 -1 = 16 -1 = 15$ $2$-digit multiples of $6$.  Hence, the answer is 30 - 15 = \textbf{15.}  Answer： \textbf{(B)}


Water from the first faucet fills the swimming pool in $70$ minutes. Water from each of the two other faucets fills the same swimming pool $3$ times faster. In how many minutes will the swimming pool be filled if all three faucets are opened?

Overseas Competition->Knowledge Point->Word Problem Modules->Work Word Problems

The efficiency of the first faucet is $\frac1{70}$ and that of the other two is $\frac3{70}$.  Thus, it takes $1\div (\frac1{70}+\frac3{70}\times2)=10$ minutes to fill the pool.


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 $80\textbackslash\%$ of the problems she solved alone, but overall $90\textbackslash\%$ of her answers were correct. Zoe had correct answers to $75\textbackslash\%$ of the problems she solved alone. What was Zoe\textquotesingle s overall percentage of correct answers? (adapted from 2017 AMC 8, Question \#14)


There is an empty tank which needs $29$ bottles of water to fill with. The workers pour $5$ bottles of water during the day, but uses $1$ bottle of water during the night. In how many days can the tank be filled with water?

Every day: $5-1=4$ bottles.  $29\div4=7R1$  But for the last day, the tank can be filled with water without using during the night.


The selling price of a sofa set is＄$$6500$$ and the profit percentage is $$30\textbackslash\%$$ for each set sold. If the cost of the sofa set is not changed, what is the profit percentage if the profit is ＄$$1200$$?

Overseas Competition->Knowledge Point->Word Problem Modules->Word Problems in Economics->Basic Profit and Loss Concepts


Owen sees an archery right after he entered the carnival. After Owen takes 20 shots, he has made $55 \textbackslash\%$ of his shots. After he takes 5 more shots, he raises his percentage to $56 \textbackslash\%$. How many of the last 5 shots did she make?~ (adapted from 2004 AMC 8, Question\#6)

Owen made $0.55 * 20=11$ shots originally. Letting $x$ be the number of shots she made, we have $\frac{11+x}{25}=0.56$. Solving for $x$ gives us $x=$ (C) 3


I read $1$ page of a book on Nov $1$, $2$ pages on Nov $2$, $3$ pages on Nov $3$, and so on. I followed this pattern for the whole month. Altogether, how many pages did I read in November?

Overseas Competition->Knowledge Point->Word Problem Modules->Page Number Problem

There are in total $30$ days in Nov.  $1+2+3+4+\cdots+30=(1+30)\times30\div2=465$


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 $84 \textbackslash\%$ of her answers were correct. Zoe had correct answers to $80 \textbackslash\%$ of the problems she solved alone. What was Zoe\textquotesingle s overall percentage of correct answers? (adapted from 2017 AMC 8, Question \#14)

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 $168$ problems correct overall. We also know that Zoe had $80$ problems she did correct alone. We can see that the total amount of correct problems Chloe and Zoe did was $168-90=78$. Therefore Zoe has $80+78=158$ problems out of $200$ problems correct. This is (C) $79$ percent.


If $$15^{}\text{th}$$April is Friday, what day of the week is $$1$$\textsuperscript{st~}April ?

Sunday

Monday

Wednesday

Friday

Saturday

Overseas Competition->Knowledge Point->Word Problem Modules->Periodic Problems->Periodic Problems of Time

There are 14 days between $$1$$\textsuperscript{st~}April and $$15^{}\text{th}$$April.  14 days = 2 week  Therefore, $$1$$\textsuperscript{st~}April is Friday.


Grandma made some cheese dumplings and some blueberry dumplings. Altogether, she made $$31$$ dumplings. If she had made $$11$$ more cheese dumplings, then there would be the same number of blueberry dumplings as cheese dumplings. How many cheese dumplings did grandma make?（$$2009$$ Math Kangaroo Problem, Levels $$1-2$$, Question \#$$19$$）

Overseas Competition->Knowledge Point->Word Problem Modules->Questions Involving Sum, Difference and Multiples->Problems Involving Sum and Difference


A worm is staying at the bottom of a well with a depth of $$13$$ metres. If it climbs up $$4$$ metres in the day time and slips down $$1$$ metre at night, which day will it climb up to the ground?

The $$4$$th day

The $$5$$th day

The $$6$$th day

The $$7$$th day

Overseas Competition->Knowledge Point->Word Problem Modules->Periodic Problems->Snail Climbing out of Well Problems

It climbs up $$4$$ metres in the day time and slips down $$1$$ metre at night, so it climbs up $$3$$ metres in a whole day. For the first $$3$$ days, it climbs $$9$$ metres. During the day time of the forth day, it climbs $$4$$ metres and reaches the ground. Then it needs four days to climb up to the ground.


The volume of a cuboid is $216$. What is the least sum of the length of all its edges?

Overseas Competition->Knowledge Point->Word Problem Modules->Applying Addition and Subtraction->Problems Involving Extreme Value

Given the product of three numbers, the smaller the difference, the smaller the sum. $6\times6\times6=216$, so the answer is $6\times 12=72.$


The state income tax where Kristin lives is levied at the rate of $$p\textbackslash\%$$ of the first ＄$$28000$$ of annual income plus $$ \left( {p+2} \right) \textbackslash\%$$ of any amount above ＄$$28000$$. Kristin noticed that the state income tax she paid amounted to $$ \left( {p+0.25} \right) \textbackslash\%$$ of her annual income. What was her annual income? ($$2001$$ AMC $$12$$ Problem, Question \#$$3$$)

Overseas Competition->Knowledge Point->Word Problem Modules->Word Problems in Economics->Complex Money Word Problems

Method $$1$$：$$Let$$ the income amount be denoted by $$A$$.  We know that $$\frac{A\left( p+.25 \right)}{100}=\frac{28000p}{100}+\frac{\left( p+2 \right)\left( A-28000 \right)}{100}$$  We can now try to solve for $$A$$:  $$\left( p+.25 \right)A=28000p+Ap+2A-28000p-56000$$  $$0.25A=2A-56000$$  $$A=32000$$  so the answer is $$\boxed{ \text {B}}$$.  Method $$2$$: Let $$A$$, $$T$$ be Kristin\textquotesingle s annual income and the income tax total, respectively.  Notice that　  $$T=p\textbackslash\%\cdot 28000+\left( p+2 \right)\textbackslash\%\cdot \left( A-28000 \right)$$  $$\textasciitilde\textasciitilde\textasciitilde\textasciitilde$$$$=\left[ p\textbackslash\%\cdot 28000+p\textbackslash\%\cdot \left( A-28000 \right) \right]+2\textbackslash\%\cdot \left( A-28000 \right)$$  $$\textasciitilde\textasciitilde\textasciitilde\textasciitilde$$$$=p\textbackslash\%\cdot A+2\textbackslash\%\cdot \left( A-28000 \right)$$  We are also given that  $$T=\left( p+0.25 \right)\textbackslash\%\cdot A=p\textbackslash\%\cdot A+0.25\textbackslash\%\cdot A$$  Thus, $$p\textbackslash\%\cdot A+2\textbackslash\%\cdot \left( A-28000 \right)=p\textbackslash\%\cdot A+0.25\textbackslash\%\cdot A$$  $$2\textbackslash\%\cdot \left( A-28000 \right)=0.25\textbackslash\%\cdot A$$  Solve for $$A$$ to obtain $$A=32000$$. $$\boxed{ \text {B}}$$


Kate sold $$10$$ dresses and Tasha sold thrice as many dresses as Kate. How many dresses did Tasha sell?

Overseas Competition->Knowledge Point->Word Problem Modules


My cat snoozes for $$50$$ minutes in each hour. For how many hours a day does my cat snooze?

Overseas Competition->Knowledge Point->Word Problem Modules->Solving Problems Involving Fractions and Percentages->Knowing the Base

Sleeping for $$50$$ minutes every hour through the day is $$\frac{5}{6}$$ of every hour and thus $$\frac{5}{6}$$ of the day. Of $$24$$ hours, $$\frac{5}{6}$$ is $$20$$ hours.


In a candy shop, the shopkeeper allows the exchange of $$3$$ sweet wrappers for $$1$$ new sweet. Peter has $$81$$ sweets at first. What is the largest possible number of sweets that Peter could have eaten?

Non of the above

Overseas Competition->Knowledge Point->Word Problem Modules->Applying Addition and Subtraction->Simple Addition and Subtraction

With $$81$$ sweets wrappers. Peter can exchange for $$\frac{81}{3}=27$$ more sweets With the additional $$27$$ sweets wrappers, Peter can exchange for $$\frac{27}{3}=9$$ more sweets.  with the~~additional $$9$$ sweets wrappes. Peter can exchange for $$\frac{9}{3}=3$$ more sweets.  with the additional $$3$$ sweets wrappers. Peter can exchange for $$\frac{3}{3}=1$$ more sweet.  Therefore, the largest possible number of sweets Peter could have  eaten is $$81+27+9+3+1=121$$.


There are $$15$$ roses in a garden, and the number of daisies is $$5$$ more than $6$ times that of roses. How many flowers are there in the garden altogether?

Overseas Competition->Knowledge Point->Word Problem Modules->Questions Involving Sum, Difference and Multiples->Basic Computation of the Multiples


The sum of the ages of four children Anil, Benu, Chas and Dave is $$20$$. Dave is $$5$$ years older than Chas, and $$3$$ years older than Benu. Anil is $$3$$ years younger than Chas. How old is Anil? ．

Overseas Competition->Knowledge Point->Word Problem Modules->Age Problems->Basic Relationships in Age Problems

Two of the pieces of information relate to Chas, so we can try to relate the ages of the others to his age. Given Dave is $$5$$ years older than Chas, but $$3$$ years older than Benu, we can tell that Benu is $$2$$ years older than Chas. Given also that Anilis $$3$$ years younger than Chas, we can say that the sum of the ages of the four children is $$\left( -3+2+5=4 \right) $$ more than four times Chas' age. So four times Chas' age plus four is twenty, and so Chas must be four. So Anil is just $$ 1$$ year old.


Allen has a collection of $240$ fossils. Of these, $35$\% are fossilized snail shells. How many fossilized snail shells does Allen have?

Overseas Competition->Knowledge Point->Word Problem Modules->Solving Problems Involving Fractions and Percentages->Percentages Word Problems

$240 \times 35$\%=$84$,  Allen has $84$ fossilized snail shells.


Daniel had $51$ Lego pieces, he used $42$ pieces to make a mini statue. How many pieces of Lego does Daniel have now?


Grandma made some cheese dumplings and some blueberry dumplings. Altogether, she made $$31$$ dumplings. If she had made $$11$$ more cheese dumplings, then there would be the same number of blueberry dumplings as cheese dumplings. How many cheese dumplings did grandma make?（2009 Math Kangaroo Problem, Question \#19）


Wendy had $$30$$ stickers. First, she gave Aiden $3$ stickers. Then, she gave Terry $7$ stickers. Now, each of the three people has the same number of stickers. At the beginning, how many stickers did they have in total?

Overseas Competition->Knowledge Point->Word Problem Modules->Inverse Operation Problems

Now, Wendy has $30-3-7=20$ stickers. Thus, all of them have $20\times3=60$ stickers, which is equal to the total number of stickers they had at the beginning.