Datasets:

math-eval

/

TAL-SCQ5K

like 49

[ "其它" ]

[ [ { "aoVal": "A", "content": "26, 7 " } ], [ { "aoVal": "B", "content": "25, 8 " } ], [ { "aoVal": "C", "content": "26, 8 " } ], [ { "aoVal": "D", "content": "25, 7 " } ] ]

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

[ "Partition 6 into 1 and 5. 19+1=20, 20+5=25 Partition 8 into 5 and 3. 15-5=10, 10-3=7 " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$81$$ " } ], [ { "aoVal": "B", "content": "$$82$$ " } ], [ { "aoVal": "C", "content": "$$83$$ " } ], [ { "aoVal": "D", "content": "$$84$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Word Problem Modules" ]

[ "$7\\times 12=84$ " ]

[]

[ [ { "aoVal": "A", "content": "$$1$$ " } ], [ { "aoVal": "B", "content": "$$2$$ " } ], [ { "aoVal": "C", "content": "$$3$$ " } ], [ { "aoVal": "D", "content": "$$4$$ " } ], [ { "aoVal": "E", "content": "$$5$$ " } ] ]

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

[ "The total number of children is $$13+19=32$$, To form six teams with the same number of children in each team, we divide $$32$$ by $$6$$, $$32\\div6=5$$$$\\rm R$$$$2$$, The remaining $$2$$ children will need $$4$$ more children to form the sixth team. " ]

[]

[ [ { "aoVal": "A", "content": "$$10\\textbackslash\\%$$ " } ], [ { "aoVal": "B", "content": "$$20\\textbackslash\\%$$ " } ], [ { "aoVal": "C", "content": "$$30\\textbackslash\\%$$ " } ], [ { "aoVal": "D", "content": "$$40\\textbackslash\\%$$ " } ], [ { "aoVal": "E", "content": "$$50\\textbackslash\\%$$ " } ] ]

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

[ "The fraction of people living in Wales who speak Welsh is $$\\frac{600000}{3000000}$$. This can be simplified to $$\\frac{1}{5}$$, and so the percentage is $$20\\textbackslash\\%$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$5$$ " } ], [ { "aoVal": "B", "content": "$$7$$ " } ], [ { "aoVal": "C", "content": "$$14$$ " } ], [ { "aoVal": "D", "content": "$$15$$ " } ], [ { "aoVal": "E", "content": "$$21$$ " } ] ]

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

[ "$$35\\times \\frac {2} {2+3} = 14$$ " ]

[]

[ [ { "aoVal": "A", "content": "a fork " } ], [ { "aoVal": "B", "content": "a knife " } ], [ { "aoVal": "C", "content": "two forks " } ], [ { "aoVal": "D", "content": "two knives " } ], [ { "aoVal": "E", "content": "None of the above " } ] ]

[ "Overseas Competition->Knowledge Point->Word Problem Modules->Applying Addition and Subtraction->Including and Excluding " ]

[ "Each plate needs a knife and a fork. There are six plates here, so $6$$\\times$1$=6$pairs of knives and forks are required, that is, six knives and six forks. But now there are only four forks. $6-4=2$ forks are required. " ]

[]

[ [ { "aoVal": "A", "content": "April $27$\\textsuperscript{th} " } ], [ { "aoVal": "B", "content": "May $15$\\textsuperscript{th} " } ], [ { "aoVal": "C", "content": "April $28$\\textsuperscript{th} " } ], [ { "aoVal": "D", "content": "April $26$\\textsuperscript{th} " } ], [ { "aoVal": "E", "content": "May $16$\\textsuperscript{th} " } ] ]

[ "Overseas Competition->Knowledge Point->Word Problem Modules->Periodic Problems->Periodic Problems of Time->Period of Dates" ]

[ "Nine days before May $6$\\textsuperscript{th} is April $27$\\textsuperscript{th} " ]

[]

[ [ { "aoVal": "A", "content": "$$1$$ " } ], [ { "aoVal": "B", "content": "$$3$$ " } ], [ { "aoVal": "C", "content": "$$5$$ " } ], [ { "aoVal": "D", "content": "$$7$$ " } ], [ { "aoVal": "E", "content": "$$9$$ " } ] ]

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

[ "The numbers of white balls and red balls in total must be a multiple of $7+1=8$, so from $1$ to $15$, only $8$ itself can match the condition. Thus, there are $15-8=7$ black balls in the box. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "\\textbf{$160,000 in economic profits.} " } ], [ { "aoVal": "B", "content": "\\textbf{$100,000 in economic profits.} " } ], [ { "aoVal": "C", "content": "\\textbf{$40,000 in economic profits.} " } ], [ { "aoVal": "D", "content": "\\textbf{neither an economic profit or loss.} " } ], [ { "aoVal": "E", "content": "\\textbf{none of the above.} " } ] ]

[ "Overseas Competition->Knowledge Point->Word Problem Modules" ]

[ "\\textbf{Accounting profit = revenue minus explicit costs. Economic profit = revenue minus both explicit and implicit costs. Accounting profit is always greater than economic profit as there's always an opportunity cost. \\textsuperscript{$}100,000 accounting profit minus the implicit cost of \\textsuperscript{$}60,000 = $40,000 in economic profit.} " ]

[]

[ [ { "aoVal": "A", "content": "$$30$$ " } ], [ { "aoVal": "B", "content": "$$40$$ " } ], [ { "aoVal": "C", "content": "$$50$$ " } ], [ { "aoVal": "D", "content": "$$60$$ " } ], [ { "aoVal": "E", "content": "$$80$$ " } ] ]

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

[ "The ratio by weight of carrot: cabbage:yoghurt$$=2:1:0.5$$ which we can double to give $$4:2:1$$. We can see that here $$2$$ represents the amount of cabbage and we want $$2 \\text{kg}$$ of cabbage, so there will be $$4+2+1=7\\text{kg}$$ of coleslaw altogether. Therefore the number of pots is $$7000\\div 175 =40$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$50$$ " } ], [ { "aoVal": "B", "content": "$$51 \\frac{2}{3}$$ " } ], [ { "aoVal": "C", "content": "$$52 \\frac{1}{2}$$ " } ], [ { "aoVal": "D", "content": "$$53 \\frac{1}{3}$$ " } ], [ { "aoVal": "E", "content": "None of the above " } ] ]

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

[ "Kevin\\textquotesingle s speed $=\\dfrac{70}{1\\dfrac{1}{2}}=\\dfrac{140}{3} \\text{km/h}$ Let John\\textquotesingle s speed be $x \\text{km/h}$. $$\\left (x+ \\frac{140}{3}\\right ) \\times 8=800 \\Rightarrow x=100- \\frac{140}{3}= \\frac{160}{3}=53 \\frac{1}{3}\\text{km/h}$$. " ]

[]

[ [ { "aoVal": "A", "content": "Monday " } ], [ { "aoVal": "B", "content": "Tuesday " } ], [ { "aoVal": "C", "content": "Wednesday " } ], [ { "aoVal": "D", "content": "Thursday " } ] ]

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

[ "$29-8+30=51$, $51\\div7=7\\textbackslash{} \\rm R\\textasciitilde2$,~ it\\textquotesingle s Wednesday " ]

[]

[ [ { "aoVal": "A", "content": "$$380$$ " } ], [ { "aoVal": "B", "content": "$$388$$ " } ], [ { "aoVal": "C", "content": "$$390$$ " } ], [ { "aoVal": "D", "content": "$$394$$ " } ] ]

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

[ "The largest possible sum is $$100 +98+96 + 94 =388$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$12$$ " } ], [ { "aoVal": "B", "content": "$$18$$ " } ], [ { "aoVal": "C", "content": "$$20$$ " } ], [ { "aoVal": "D", "content": "$$24$$ " } ], [ { "aoVal": "E", "content": "$$48$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Word Problem Modules->Inverse Operation Problems->Giving Half of a Whole" ]

[ "$6+6=12$ $12+12=24$ $24+24=48$ " ]

[]

[ [ { "aoVal": "A", "content": "$1$ ounces " } ], [ { "aoVal": "B", "content": "$$\\dfrac{4}{3}$$ ounces " } ], [ { "aoVal": "C", "content": "$4$ ounces " } ], [ { "aoVal": "D", "content": "$5$ ounces " } ] ]

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

[ "Method $$1$$: Suppose $$x$$ ounces of salt should be added to the solution: $$\\dfrac{20\\times20\\textbackslash\\%+x}{20+x}=25\\textbackslash\\%$$ $$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde x=\\dfrac{4}{3}$$. Method $$2$$: $$20\\times(1-20\\textbackslash\\%)\\div(1-25\\textbackslash\\%)-20=\\dfrac{4}{3}$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$28$$ " } ], [ { "aoVal": "B", "content": "$$36$$ " } ], [ { "aoVal": "C", "content": "$$49$$ " } ], [ { "aoVal": "D", "content": "$$56$$ " } ] ]

[ "Overseas In-curriculum->Knowledge Point->Fun Problems in Math->Reasoning", "Overseas Competition->Knowledge Point->Word Problem Modules" ]

[ "$7+6+5+4+3+2+1=28$ " ]

[]

[ [ { "aoVal": "A", "content": "$$5$$ " } ], [ { "aoVal": "B", "content": "$$336$$ " } ], [ { "aoVal": "C", "content": "$$1680$$ " } ], [ { "aoVal": "D", "content": "$$2016$$ " } ] ]

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

[ "$$5$$ out of $$6$$ of the $$2016$$ lightbulbs are not defective. Thus $$2016 \\times \\frac{5}{6} =1680$$ lightbulbs are not defective. " ]

[]

[ [ { "aoVal": "A", "content": "$$0$$ " } ], [ { "aoVal": "B", "content": "$$12$$ " } ], [ { "aoVal": "C", "content": "$$15$$ " } ], [ { "aoVal": "D", "content": "$$18$$ " } ], [ { "aoVal": "E", "content": "$$23$$ " } ] ]

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

[ "The line at the door started with $$8$$ people and $$15$$ more came in, for a total of $$23$$ people. After leaving $$11$$ customers, $$11$$ seats were empty, so the number of people left in line was $$23-11=12$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$12$$ " } ], [ { "aoVal": "B", "content": "$$24$$ " } ], [ { "aoVal": "C", "content": "$$30$$ " } ], [ { "aoVal": "D", "content": "$$36$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Word Problem Modules->Distribution Problems->Basic Distribution Problems" ]

[ "$198+18=216$ $216\\div6=36$ " ]

[]

[ [ { "aoVal": "A", "content": "$$$4$$ " } ], [ { "aoVal": "B", "content": "$$$6$$ " } ], [ { "aoVal": "C", "content": "$$$14$$ " } ], [ { "aoVal": "D", "content": "$$$40$$ " } ] ]

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

[ "$$10-4=6$$ " ]

[]

[ [ { "aoVal": "A", "content": "$$149$$ " } ], [ { "aoVal": "B", "content": "$$150$$ " } ], [ { "aoVal": "C", "content": "$$151$$ " } ], [ { "aoVal": "D", "content": "$$152$$ " } ] ]

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

[ "$2+(50-1)\\times3=149$. " ]

[]

[ [ { "aoVal": "A", "content": "$$300$$ " } ], [ { "aoVal": "B", "content": "$$400$$ " } ], [ { "aoVal": "C", "content": "$$450$$ " } ], [ { "aoVal": "D", "content": "$$600$$ " } ], [ { "aoVal": "E", "content": "$$900$$ " } ] ]

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

[ "Time: $$300\\div20=15$$ hours. So, Factory $$B$$ can assemble $$15\\times30=450$$ televisions. " ]

[]

[ [ { "aoVal": "A", "content": "$21\\textbackslash\\%$ " } ], [ { "aoVal": "B", "content": "$11\\textbackslash\\%$ " } ], [ { "aoVal": "C", "content": "$10\\textbackslash\\%$ " } ], [ { "aoVal": "D", "content": "$9\\textbackslash\\%$ " } ] ]

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

[ "Suppose the interest rate is $$m$$: $$\\begin{eqnarray}10000\\times \\left( 1+m \\right)\\times \\left( 1+m \\right)\\&=\\&10000+2100 \\textbackslash\\textbackslash{} m\\&=\\&0.1 \\end{eqnarray}$$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "Saturday " } ], [ { "aoVal": "B", "content": "Wednesday " } ], [ { "aoVal": "C", "content": "Thursday " } ], [ { "aoVal": "D", "content": "Tuesday " } ], [ { "aoVal": "E", "content": "Monday " } ] ]

[ "Overseas Competition->Knowledge Point->Word Problem Modules->Periodic Problems" ]

[ "$30-4=26$ days later, it will be May $30$\\textsuperscript{th}. $26\\div7=3R5$, which means May $30$\\textsuperscript{th}~is Monday.~ " ]

[]

[ [ { "aoVal": "A", "content": "$$38$$ " } ], [ { "aoVal": "B", "content": "$$34$$ " } ], [ { "aoVal": "C", "content": "$$24$$ " } ], [ { "aoVal": "D", "content": "$$22$$ " } ], [ { "aoVal": "E", "content": "$$18$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Word Problem Modules->Average Problems " ]

[ "$$36\\times (89-81)\\div (101-89)=24$$ " ]

[]

[ [ { "aoVal": "A", "content": "$$50$$ " } ], [ { "aoVal": "B", "content": "$$51 \\frac{2}{3}$$ " } ], [ { "aoVal": "C", "content": "$$52 \\frac{1}{2}$$ " } ], [ { "aoVal": "D", "content": "$$53 \\frac{1}{3}$$ " } ], [ { "aoVal": "E", "content": "None of the above " } ] ]

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

[ "Kevin\\textquotesingle s speed $=\\dfrac{70}{1\\dfrac{1}{2}}=\\dfrac{140}{3} \\text{km/h}$ Let John\\textquotesingle s speed be $x \\text{km/h}$. $$\\left (x+ \\frac{140}{3}\\right ) \\times 8=800 \\Rightarrow x=100- \\frac{140}{3}= \\frac{160}{3}=53 \\frac{1}{3}\\text{km/h}$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$60$$ " } ], [ { "aoVal": "B", "content": "$$90$$ " } ], [ { "aoVal": "C", "content": "$$120$$ " } ], [ { "aoVal": "D", "content": "$$150$$ " } ], [ { "aoVal": "E", "content": "$$180$$ " } ], [ { "aoVal": "F", "content": "None of the above " } ] ]

[ "Overseas Competition->Knowledge Point->Word Problem Modules->Work Word Problems" ]

[ "E " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$10$$ " } ], [ { "aoVal": "B", "content": "$$18$$ " } ], [ { "aoVal": "C", "content": "$$20$$ " } ], [ { "aoVal": "D", "content": "$$22$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Word Problem Modules->Age Problems" ]

[ "Jill is now $$14$$. Two years ago, she was $$12$$. Since Jack is now $$6$$ years older than Jill was $$2$$ years ago, Jack is now $$12 + 6= 18$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$4$$ " } ], [ { "aoVal": "B", "content": "$$6$$ " } ], [ { "aoVal": "C", "content": "$$8$$ " } ], [ { "aoVal": "D", "content": "$$10$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Word Problem Modules" ]

[ "$64\\div 8=8$ " ]

[]

[ [ { "aoVal": "A", "content": "$$2$$ " } ], [ { "aoVal": "B", "content": "$$3$$ " } ], [ { "aoVal": "C", "content": "$$4$$ " } ], [ { "aoVal": "D", "content": "$$5$$ " } ], [ { "aoVal": "E", "content": "$$6$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Word Problem Modules->Average Problems ->Giving and Receiving" ]

[ "Difference: $17-9=8$ Move: Half of $8=4$ " ]

[]

[ [ { "aoVal": "A", "content": "$$5$$ " } ], [ { "aoVal": "B", "content": "$$6$$ " } ], [ { "aoVal": "C", "content": "$$7$$ " } ], [ { "aoVal": "D", "content": "$$8$$ " } ], [ { "aoVal": "E", "content": "$$9$$ " } ] ]

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

[ "Suppose Olivia has $$x$$ correct answers. $$5x-2(10-x)=29$$, $$x=7$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$8$$ " } ], [ { "aoVal": "B", "content": "$$9$$ " } ], [ { "aoVal": "C", "content": "$$10$$ " } ], [ { "aoVal": "D", "content": "$$11$$ " } ], [ { "aoVal": "E", "content": "$$12$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Word Problem Modules->Interval Problems" ]

[ "$10 - 1 = 9$ " ]

[]

[ [ { "aoVal": "A", "content": "$$2$$ " } ], [ { "aoVal": "B", "content": "$$6$$ " } ], [ { "aoVal": "C", "content": "$$3$$ " } ], [ { "aoVal": "D", "content": "$$4$$ " } ], [ { "aoVal": "E", "content": "$$5$$ " } ] ]

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

[ "Jimmy\\textquotesingle s speed is twice as fast as Mary\\textquotesingle s, and the ice cream they eat is the same size, so Jimmy\\textquotesingle s time is only half as long as Mary\\textquotesingle s, that is, $4$$\\div$$2= $$2$ minutes. " ]

[]

[ [ { "aoVal": "A", "content": "$$90\\text{cm}$$ " } ], [ { "aoVal": "B", "content": "$$1\\text{m}$$ " } ], [ { "aoVal": "C", "content": "$$110\\text{cm}$$ " } ], [ { "aoVal": "D", "content": "$$120\\text{cm}$$ " } ] ]

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

[ "$$60 \\text{cm}/\\text{hr} =1 \\text{cm}/\\min = 100 \\text{cm}/100 \\text{mins} = 1 \\text{m}/100 \\text{mins}$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$6$$ " } ], [ { "aoVal": "B", "content": "$$64$$ " } ], [ { "aoVal": "C", "content": "$$24$$ " } ], [ { "aoVal": "D", "content": "$$48$$ " } ], [ { "aoVal": "E", "content": "$$60$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Word Problem Modules->Average Problems " ]

[ "$(78-72)\\times8=48$, $72-48=24$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$28$$ " } ], [ { "aoVal": "B", "content": "$$34$$ " } ], [ { "aoVal": "C", "content": "$$38$$ " } ], [ { "aoVal": "D", "content": "$$42$$ " } ], [ { "aoVal": "E", "content": "$$60$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Word Problem Modules->Average Problems " ]

[ "$4\\times3+6\\times2+18=42$ km. " ]

[]

[ [ { "aoVal": "A", "content": "$$20$$ " } ], [ { "aoVal": "B", "content": "$$30$$ " } ], [ { "aoVal": "C", "content": "$$40$$ " } ], [ { "aoVal": "D", "content": "$$50$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Word Problem Modules->Age Problems->When..., When... Type Age Problems" ]

[ "If $$20$$ years ago Allen was half as old as he is today, then today he is $$40$$. Thus, $$10$$ years ago he was $$30$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$2$$ " } ], [ { "aoVal": "B", "content": "$$3$$ " } ], [ { "aoVal": "C", "content": "$$4$$ " } ], [ { "aoVal": "D", "content": "$$5$$ " } ], [ { "aoVal": "E", "content": "$$6$$ " } ] ]

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

[ "There are 3 families with 3 children = 3 x 3=9. If there are 2 twin boys, 2 x 2 = 4, 0-4=5. " ]

[]

[ [ { "aoVal": "A", "content": "$$20$$ " } ], [ { "aoVal": "B", "content": "$$25$$ " } ], [ { "aoVal": "C", "content": "$$30$$ " } ], [ { "aoVal": "D", "content": "$$35$$ " } ], [ { "aoVal": "E", "content": "$$40$$ " } ] ]

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

[ "£$$20 \\div 50p=40$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$2$$ " } ], [ { "aoVal": "B", "content": "$$4$$ " } ], [ { "aoVal": "C", "content": "$$6$$ " } ], [ { "aoVal": "D", "content": "$$8$$ " } ] ]

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

[ "I run $$2\\text{km}$$ in $$15$$ minutes, or $$8\\text{km}$$ in $$1$$ hour. In $$2$$ hours, I will run $$16\\text{km}$$ and my sister will run $$20\\text{km}$$. She will run $$4\\text{km}$$ farther than I will. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$42$$ " } ], [ { "aoVal": "B", "content": "$$32$$ " } ], [ { "aoVal": "C", "content": "$$18$$ " } ], [ { "aoVal": "D", "content": "$$8$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Word Problem Modules->Chicken-Rabbit Problems" ]

[ "Assume all questions were answered correctly, Total score $=50\\times4=200$ Difference in total score $=200-110=90$ Change in score $=4+1=5$ No. of incorrect answer $=90\\div5=18$ Correct answers $=50-18=32$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$33$$ " } ], [ { "aoVal": "B", "content": "$$44$$ " } ], [ { "aoVal": "C", "content": "$$55$$ " } ], [ { "aoVal": "D", "content": "$$66$$ " } ], [ { "aoVal": "E", "content": "$$99$$ " } ] ]

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

[ "$50 \\textbackslash\\% \\cdot 10=5$ $30 \\textbackslash\\% \\cdot 20=6$ $30 \\textbackslash\\% \\cdot 30=9$ Adding them up gets $5+6+9=20$. The overall percentage correct would be $\\frac{20}{60}=\\frac{1}{3}=0 . \\overline{3} \\approx(\\mathbf{A}) 33$. " ]

[]

[ [ { "aoVal": "A", "content": "＄$$28000$$ " } ], [ { "aoVal": "B", "content": "＄$$32000$$ " } ], [ { "aoVal": "C", "content": "＄$$35000$$ " } ], [ { "aoVal": "D", "content": "＄$$42000$$ " } ], [ { "aoVal": "E", "content": "＄$$56000$$ " } ] ]

[ "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 $$A$$: $$\\left( p+.25 \\right)A=28000p+Ap+2A-28000p-56000$$ $$.25A=2A-56000$$ $$A=32000$$ 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 $$A=32000$$. $$\\boxed{ \\text {B}}$$ " ]

[]

[ [ { "aoVal": "A", "content": "$$10\\textbackslash\\%$$ " } ], [ { "aoVal": "B", "content": "$$15\\textbackslash\\%$$ " } ], [ { "aoVal": "C", "content": "$$20\\textbackslash\\%$$ " } ], [ { "aoVal": "D", "content": "$$25\\textbackslash\\%$$ " } ] ]

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

[ "$$8\\div(8+32)=20\\textbackslash\\%$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$8$ meters " } ], [ { "aoVal": "B", "content": "$$11$$ meters " } ], [ { "aoVal": "C", "content": "$$14$$ meters " } ], [ { "aoVal": "D", "content": "$15$ meters " } ], [ { "aoVal": "E", "content": "$16$ meters " } ] ]

[ "Overseas Competition->Knowledge Point->Word Problem Modules->Interval Problems" ]

[ "There are $7$ intervals between the first student and the last student. Thus, the line is $7 \\times2 = 14$ meters long. " ]

[]

[ [ { "aoVal": "A", "content": "Monday " } ], [ { "aoVal": "B", "content": "Wednesday " } ], [ { "aoVal": "C", "content": "Friday " } ], [ { "aoVal": "D", "content": "Saturday " } ], [ { "aoVal": "E", "content": "Sunday " } ] ]

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

[ "$$30-1+31+21=81$$, $$81\\div7=11$$R$$4$$. $$4$$ days after a Saturday is a Wednesday. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$50$$ " } ], [ { "aoVal": "B", "content": "$$60$$ " } ], [ { "aoVal": "C", "content": "$$180$$ " } ], [ { "aoVal": "D", "content": "$$200$$ " } ], [ { "aoVal": "E", "content": "None of the above " } ] ]

[ "Overseas Competition->Knowledge Point->Word Problem Modules->Unitary Method Problems" ]

[ "x = 1u y = 5u 6u = 300 1u = 50 So, x=50, y=250 Difference 250-50=200 " ]

[]

[ [ { "aoVal": "A", "content": "$$$$Blue " } ], [ { "aoVal": "B", "content": "$$$$Pink " } ], [ { "aoVal": "C", "content": "$$$$Red " } ], [ { "aoVal": "D", "content": "$$$$Black " } ], [ { "aoVal": "E", "content": "$$$$Yellow " } ] ]

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

[ "$19\\div4=4R3$, so the $19$\\textsuperscript{th} flower is yellow. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$x+2$ " } ], [ { "aoVal": "B", "content": "$5x+2$ " } ], [ { "aoVal": "C", "content": "$5x+4$ " } ], [ { "aoVal": "D", "content": "$5x-4$ " } ] ]

[ "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: $$2\\times 2 =4$$ The total boba that John drinks during a week: $$5x+4$$ " ]

[]

[ [ { "aoVal": "A", "content": "$$80$$ " } ], [ { "aoVal": "B", "content": "$$90$$ " } ], [ { "aoVal": "C", "content": "$$100$$ " } ], [ { "aoVal": "D", "content": "$$110$$ " } ], [ { "aoVal": "E", "content": "$$120$$ " } ] ]

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

[ "Of $$60$$ flamingos, $$40$$ stood on $$2$$ legs and $$20$$ stood on $$1$$. All together, these flamingos stood on $$[(40\\times2)+20]$$ legs $$=100$$ legs. " ]

[]

[ [ { "aoVal": "A", "content": "The $$6$$th day " } ], [ { "aoVal": "B", "content": "The $$7$$th day " } ], [ { "aoVal": "C", "content": "The $$8$$th day " } ], [ { "aoVal": "D", "content": "The $$9$$th day " } ] ]

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

[ "omitted " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$16$$ " } ], [ { "aoVal": "B", "content": "$$48$$ " } ], [ { "aoVal": "C", "content": "$$20$$ " } ], [ { "aoVal": "D", "content": "$$12$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Word Problem Modules->Questions Involving Sum, Difference and Multiples->Problems Involving Sum and Difference->Sum and Differences Problems with multiple Variables" ]

[ "Paul has $$(28 + 4) \\div 2 = 16$$ toy cars. " ]

[]

[ [ { "aoVal": "A", "content": "$$5$$ " } ], [ { "aoVal": "B", "content": "$$9$$ " } ], [ { "aoVal": "C", "content": "$$12$$ " } ], [ { "aoVal": "D", "content": "$$18$$ " } ], [ { "aoVal": "E", "content": "$$21$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Word Problem Modules->Average Problems " ]

[ "Total age less than the average: $(34-27)\\times3=21$. Thus, there are $21\\div(35-34)=21$ female teachers. " ]

[]

[ [ { "aoVal": "A", "content": "$$10$$ " } ], [ { "aoVal": "B", "content": "$$15$$ " } ], [ { "aoVal": "C", "content": "$$20$$ " } ], [ { "aoVal": "D", "content": "$$25$$ " } ], [ { "aoVal": "E", "content": "$$30$$ " } ] ]

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

[ "Method1: Cagney can frost one in $$20$$ seconds, and Lacey can frost one in $$30$$ seconds. Working together, they can frost one in $$\\dfrac{20\\cdot30}{20+30}=\\dfrac{600}{50}=12$$ seconds. In $$300$$ seconds ( $$5$$ minutes), they can frost $$\\boxed{(\\text{D})25}$$ cupcakes. Method2: In $$300$$ seconds ($$5$$ minutes), Cagney will frost $$\\dfrac{300}{20}=15$$ cupcakes, and Lacey will frost $$\\dfrac{300}{30}=10$$ ~cupcakes. Therefore, working together they will frost $$15+10=\\boxed{(\\text{D})25}$$ cupcakes. Method3: Since Cagney frosts $$3$$ cupcakes a minute, and Lacey frosts $$2$$ cupcakes a minute, they together frost $$3+2=5$$ cupcakes a minute. Therefore, in $$5$$ minutes, they frost $$5\\times5=25\\Rightarrow\\boxed{(\\text{D})}$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$0$$ " } ], [ { "aoVal": "B", "content": "$$5$$ " } ], [ { "aoVal": "C", "content": "$$10$$ " } ], [ { "aoVal": "D", "content": "$$12$$ " } ], [ { "aoVal": "E", "content": "$$15$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Word Problem Modules" ]

[ "Giving candies to each other won\\textquotesingle t affect the sum of their candies, so there are $$4+4+4=12$$~in total. " ]

[]

[ [ { "aoVal": "A", "content": "$$10$$ " } ], [ { "aoVal": "B", "content": "$$21$$ " } ], [ { "aoVal": "C", "content": "$$20$$ " } ], [ { "aoVal": "D", "content": "$$15$$ " } ] ]

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

[ "$$(31-11)\\div2=10.$$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$24.2m$$ " } ], [ { "aoVal": "B", "content": "$$21m$$ " } ], [ { "aoVal": "C", "content": "$$23m$$ " } ], [ { "aoVal": "D", "content": "$$2.3m$$ " } ], [ { "aoVal": "E", "content": "$$69m$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Word Problem Modules->Periodic Problems->Snail Climbing out of Well Problems->Finding the Height in Snail Climbing out of Wall Problems (completed) " ]

[ "Key: object is thrown upward. When it reaches the highest point, the velocity is 0. The whole journey is divided into 2 parts: upward - highest point - downward to hit the ground. 2 parts have different displacement (and different time). " ]

[]

[ [ { "aoVal": "A", "content": "$$$$Amit " } ], [ { "aoVal": "B", "content": "$$$$Bede " } ], [ { "aoVal": "C", "content": "$$$$Cain " } ], [ { "aoVal": "D", "content": "$$$$Devi " } ], [ { "aoVal": "E", "content": "$$$$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. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$90$$ " } ], [ { "aoVal": "B", "content": "$$95$$ " } ], [ { "aoVal": "C", "content": "$$100$$ " } ], [ { "aoVal": "D", "content": "$$110$$ " } ], [ { "aoVal": "E", "content": "$$120$$ " } ] ]

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

[ "$1.5\\times 0.8= 120\\textbackslash\\%$. " ]

[]

[ [ { "aoVal": "A", "content": "$$$1$$ " } ], [ { "aoVal": "B", "content": "$$$2$$ " } ], [ { "aoVal": "C", "content": "$$$3$$ " } ], [ { "aoVal": "D", "content": "$$$4$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Word Problem Modules->Equation Word Problems->Equivalent Substitution in Equation Word Problems" ]

[ "$$2$$ apples $$+\\textasciitilde3$$ peaches$$\\to $$ ＄$$11$$ $$2$$ apples $$+\\textasciitilde2$$ peaches $$\\to $$ ＄$$8$$ $$1$$ peach $$\\to $$ ＄$$3$$ $$2$$ apples $$+$$ ($$2$$ $$\\times$$ ＄$$3$$) $$\\to $$ ＄$$8$$ $$2$$ apples $$\\to $$ ＄$$8\\textasciitilde-$$ ＄$$6=$$＄$$2$$ $$1$$ apple $$\\to $$ ＄$$2\\div2=$$＄$$1$$ The cost of an apple is ＄$$1$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$2.05 \\rm{a.m.}$$ " } ], [ { "aoVal": "B", "content": "$$2.05 \\rm{p.m.}$$ " } ], [ { "aoVal": "C", "content": "$$1.55 \\rm{a.m.}$$ " } ], [ { "aoVal": "D", "content": "$$1.55 \\rm{p.m.}$$ " } ], [ { "aoVal": "E", "content": "$$2.25 \\rm{p.m.}$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Word Problem Modules" ]

[ "She reached at $$2.25 \\rm{p.m.}$$, thus, $$20$$ minutes before that is $$2.05 \\rm{p.m.}$$ " ]

[]

[ [ { "aoVal": "A", "content": "$$9\\frac{1}{2}$$ " } ], [ { "aoVal": "B", "content": "$$18$$ " } ], [ { "aoVal": "C", "content": "$$19$$ " } ], [ { "aoVal": "D", "content": "$$36$$ " } ], [ { "aoVal": "E", "content": "$$38$$ " } ] ]

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

[ "The three ages have a total of four times Barry\\textquotesingle s age. So Barry is $$76\\div4 = 19$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$25$$ ounces " } ], [ { "aoVal": "B", "content": "$$35$$ ounces " } ], [ { "aoVal": "C", "content": "$$45$$ ounces " } ], [ { "aoVal": "D", "content": "$$59$$ ounces " } ] ]

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

[ "$$47 -- 12 = 35$$ ounces. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$48$ dollars " } ], [ { "aoVal": "B", "content": "$52$ dollars " } ], [ { "aoVal": "C", "content": "$56$ dollars " } ], [ { "aoVal": "D", "content": "$60$ dollars " } ] ]

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

[ "$12\\div (1+2)=4$ $4\\times (1\\times 5+2\\times 4)=4\\times 13=52$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$25$$ " } ], [ { "aoVal": "B", "content": "$$30$$ " } ], [ { "aoVal": "C", "content": "$$33$$ " } ], [ { "aoVal": "D", "content": "$$40$$ " } ], [ { "aoVal": "E", "content": "$$45$$ " } ] ]

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

[ "First, find the amount of money one will pay for three sandals without the discount. We have $\\textbackslash$ 50 \\times 3$ sandals $=\\textbackslash$ 150$. Then, find the amount of money using the discount: $50+0.6 \\times 50+\\frac{1}{2} \\times 50=\\textbackslash$ 105$. Finding the percentage yields $\\frac{105}{150}=70 \\textbackslash\\%$ To find the percent saved, we have $100 \\textbackslash\\%-70 \\textbackslash\\%=(\\text{B}) 30$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$6$$ " } ], [ { "aoVal": "B", "content": "$$14$$ " } ], [ { "aoVal": "C", "content": "$$16$$ " } ], [ { "aoVal": "D", "content": "$$24$$ " } ], [ { "aoVal": "E", "content": "$$26$$ " } ] ]

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

[ "USA won - 104 medals in total\\textquotesingle{} China won 88 medals. Difference = 16. " ]

[]

[ [ { "aoVal": "A", "content": "$$37\\text{m}$$ " } ], [ { "aoVal": "B", "content": "$$41\\text{m}$$ " } ], [ { "aoVal": "C", "content": "$$50\\text{m}$$ " } ], [ { "aoVal": "D", "content": "$$55\\text{m}$$ " } ], [ { "aoVal": "E", "content": "$$60\\text{m}$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Word Problem Modules" ]

[ "Carol finishes $$25$$ metres behind Bridgit, so she travels $$75$$ metres while Bridgit runs $$100$$ metres. Therefore she runs $$3$$ metres for every $$4$$ metres Bridgit runs. When Anna finishes, Bridgit has run $$84$$ metres, so that at that time Carol has run $$\\frac{3}{4}\\times 84$$ metres $$=63$$ metres. Hence Carol finishes $$\\left( 100-63 \\right)$$ metres $$= 37$$ metres behind Anna. " ]

[]

[ [ { "aoVal": "A", "content": "$$3$$miles " } ], [ { "aoVal": "B", "content": "$$4$$miles " } ], [ { "aoVal": "C", "content": "$$6$$miles " } ], [ { "aoVal": "D", "content": "$$10$$miles " } ], [ { "aoVal": "E", "content": "$$20$$miles " } ] ]

[ "Overseas Competition->Knowledge Point->Word Problem Modules" ]

[ "Cycling at $$12$$ mph, Speedy would go $$12$$ miles in an hour. So in $$30$$ minutes he cycles $$12$$ miles $$\\div2 = 6$$ miles. " ]

[]

[ [ { "aoVal": "A", "content": "$$20{\\text{km}}/{\\text{h}}\\textbackslash;$$ " } ], [ { "aoVal": "B", "content": "$$21{\\text{km}}/{\\text{h}}\\textbackslash;$$ " } ], [ { "aoVal": "C", "content": "$$22.5{\\text{km}}/{\\text{h}}\\textbackslash;$$ " } ], [ { "aoVal": "D", "content": "$$24{\\text{km}}/{\\text{h}}\\textbackslash;$$ " } ], [ { "aoVal": "E", "content": "$$25{\\text{km}}/{\\text{h}}\\textbackslash;$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Word Problem Modules" ]

[ "Let $$x\\text{km}$$ be the distance Geraint cycles and let $$t $$ hours be the time his journey should take if he is to be on time. Since $$\\frac{\\text{distance}}{\\text{speed}}\\text{=time}$$, the information in the question tells us that $$\\frac{x}{15}=t+\\frac{1}{6}$$ and that $$\\frac{x}{30}=t-\\frac{1}{6}$$. When we subtract the second equation from the first, we obtain $$\\frac{x}{30}=\\frac{2}{6}$$ and so $$x = 10$$. Hence, from the second equation, $$\\frac{10}{30}=t-\\frac{1}{6}$$ and so $$t=\\frac{1}{3}+\\frac{1}{6}=\\frac{1}{2}$$. Therefore, to arrive on time, Geraint needs to travel $$10\\text{km}$$ in~ $$\\frac{1}{2}$$ hour, which is an average speed of $$20{\\text{km}}/{\\text{h}}\\textbackslash;$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$13$$ " } ], [ { "aoVal": "B", "content": "$$57$$ " } ], [ { "aoVal": "C", "content": "$$62$$ " } ], [ { "aoVal": "D", "content": "$$65$$ " } ], [ { "aoVal": "E", "content": "$$70$$ " } ] ]

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

[ "There are $$6$$ starfish and $$5$$ octopuses. So the number of boxes $$=6 \\times5+5\\times8 =70$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$36$$ days " } ], [ { "aoVal": "B", "content": "$$72$$ days " } ], [ { "aoVal": "C", "content": "$$120$$ days " } ], [ { "aoVal": "D", "content": "$$144$$ days " } ], [ { "aoVal": "E", "content": "$60$ days " } ] ]

[ "Overseas Competition->Knowledge Point->Distance Word Problems->Clock Problems" ]

[ "Clock shows the correct time after a $$12-$$hr loss. With a $$10-$$min. loss daily, it takes $$6$$ days to lose $$1$$ hr \\& $$72$$ days to lose $$12$$ hr. " ]

[]

[ [ { "aoVal": "A", "content": "$$30\\textbackslash\\%$$ " } ], [ { "aoVal": "B", "content": "$$32\\textbackslash\\%$$ " } ], [ { "aoVal": "C", "content": "$$35\\textbackslash\\%$$ " } ], [ { "aoVal": "D", "content": "$$36\\textbackslash\\%$$ " } ] ]

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

[ "Original price: $$\\frac{210}{(1-30\\textbackslash\\%)}=300$$. Selling price after discount: $$210-15=195$$. Discount rate: $$1-\\left( {\\frac{195}{300}} \\right)=1-65\\textbackslash\\%=35\\textbackslash\\%$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$235$$ " } ], [ { "aoVal": "B", "content": "$$230$$ " } ], [ { "aoVal": "C", "content": "$$225$$ " } ], [ { "aoVal": "D", "content": "$$220$$ " } ] ]

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

[ "$241-42+36=235$ " ]

[]

[ [ { "aoVal": "A", "content": "$$110$$ " } ], [ { "aoVal": "B", "content": "$$109$$ " } ], [ { "aoVal": "C", "content": "$$221$$ " } ], [ { "aoVal": "D", "content": "$$222$$ " } ], [ { "aoVal": "E", "content": "$$330$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Word Problem Modules->Page Number Problem->Correspondence between Numbers and Page Numbers->Applying the Total Number of Numbers" ]

[ "From $$1$$ to $$9$$ there are~$1\\times9=9$~digits. From $$10$$ to $$99$$ there are~$2\\times90=180$~digits. From $$100$$ to $$110$$ there are~$3\\times11=33$~digits. In total, there are$9+180+33=222$~digits " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$18$$ " } ], [ { "aoVal": "B", "content": "$$13$$ " } ], [ { "aoVal": "C", "content": "$$9$$ " } ], [ { "aoVal": "D", "content": "$$7$$ " } ], [ { "aoVal": "E", "content": "$$5$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Word Problem Modules->Chicken-Rabbit Problems" ]

[ "$5$ pairs of wings means there are $5$ ducks, so there are $18-5\\times2=8$ legs for dogs, which is $8\\div4=2$. Thus, there are $2+5=7$ animals and they will lift $7$ legs. " ]

[]

[ [ { "aoVal": "A", "content": "$$55\\text{mph}$$ " } ], [ { "aoVal": "B", "content": "$$50\\text{mph}$$ " } ], [ { "aoVal": "C", "content": "$$48\\text{mph}$$ " } ], [ { "aoVal": "D", "content": "$$45\\text{mph}$$ " } ], [ { "aoVal": "E", "content": "Impossible to determine " } ] ]

[ "Overseas Competition->Knowledge Point->Distance Word Problems->Bus Departure Time Word Problems" ]

[ "Let the distance from London to Brighton be $$d$$ miles. Since time $$=$$ distance $$\\div$$ speed, the times Lewis spent on the two parts of his journey are $$\\frac{d}{60}$$ hours and $$\\frac{d}{40}$$ hours. Hence the total time in hours that he travelled is$$\\frac{d}{60}+ \\frac{d}{40}= \\frac{2d+3d}{120}= \\frac{5d}{120}= \\frac{d}{24}$$. Therefore his average speed for the whole journey is $$2d \\div \\left( \\frac{d}{24}\\right)\\text{mph}=48 \\text{mph}$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$90$$ grams " } ], [ { "aoVal": "B", "content": "$$100$$ grams " } ], [ { "aoVal": "C", "content": "$$120$$ grams " } ], [ { "aoVal": "D", "content": "$$150$$ grams " } ] ]

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

[ "$$18\\div15\\textbackslash\\% = 120$$ grams. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$60$$ " } ], [ { "aoVal": "B", "content": "$$70$$ " } ], [ { "aoVal": "C", "content": "$$80$$ " } ], [ { "aoVal": "D", "content": "$$90$$ " } ], [ { "aoVal": "E", "content": "$$100$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Word Problem Modules->Interval Problems" ]

[ "$40 \\div (5 - 1) = 10$ $(10 - 3) \\times 10 = 70$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$0$$ " } ], [ { "aoVal": "B", "content": "$$5$$ " } ], [ { "aoVal": "C", "content": "$$10$$ " } ], [ { "aoVal": "D", "content": "$$12$$ " } ], [ { "aoVal": "E", "content": "$$15$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Word Problem Modules" ]

[ "Giving cookies to each other won\\textquotesingle t affect the sum of their cookies, so there are $$4+4+4=12$$~in total. " ]

[]

[ [ { "aoVal": "A", "content": "$$1:2$$ " } ], [ { "aoVal": "B", "content": "$$1:3$$ " } ], [ { "aoVal": "C", "content": "$$2:3$$ " } ], [ { "aoVal": "D", "content": "$$2:1$$ " } ] ]

[ "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 \\emph{not} wearing jeans is $$6:12=1:2$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$6$$ " } ], [ { "aoVal": "B", "content": "$$7$$ " } ], [ { "aoVal": "C", "content": "$$8$$ " } ], [ { "aoVal": "D", "content": "$$9$$ " } ], [ { "aoVal": "E", "content": "$$10$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Word Problem Modules->Average Problems " ]

[ "$$14\\times(15+25)=560$$ $$560-15\\times24=200$$ $$200\\div25=8$$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$36$$ " } ], [ { "aoVal": "B", "content": "$$40$$ " } ], [ { "aoVal": "C", "content": "$$45$$ " } ], [ { "aoVal": "D", "content": "$$54$$ " } ], [ { "aoVal": "E", "content": "$$63$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Word Problem Modules" ]

[ "$(3 + 2 + 1) \\times (6 +2 +1) = 54$ " ]

[]

[ [ { "aoVal": "A", "content": "$$5$$ " } ], [ { "aoVal": "B", "content": "$$6$$ " } ], [ { "aoVal": "C", "content": "$$7$$ " } ], [ { "aoVal": "D", "content": "$$8$$ " } ], [ { "aoVal": "E", "content": "$$9$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Word Problem Modules->Equation Word Problems->Indefinite Equation Word Problems" ]

[ "Let Billy have $$b$$ soccer balls and $$2b$$ basketballs. Let Milly have $$m$$ basketballs and $$4m$$ soccer balls. Therefore, as they have $$17$$ balls in total, $$3b + 5m= 18$$. The only positive integer solution of this equation is $$b= 1$$, $$m= 3$$. So the number of basketballs is $$2b + m = 2 \\times 1 + 3 = 5$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$10$$ " } ], [ { "aoVal": "B", "content": "$$15$$ " } ], [ { "aoVal": "C", "content": "$$30$$ " } ], [ { "aoVal": "D", "content": "$$45$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Word Problem Modules" ]

[ "$30\\div2=15$ " ]

[]

[ [ { "aoVal": "A", "content": "$$82.5^{}\\circ$$ " } ], [ { "aoVal": "B", "content": "$$90^{}\\circ$$ " } ], [ { "aoVal": "C", "content": "$$97.5^{}\\circ$$ " } ], [ { "aoVal": "D", "content": "$$270^{}\\circ$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Distance Word Problems->Clock Problems" ]

[ "At $$6:15$$, hr. hand has moved $$\\left(1/4\\right)\\times 30^{}\\circ$$ from $$6$$, so $$\\angle =90^{}\\circ+30^{}\\circ/4=97.5^{}\\circ$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$15$$ " } ], [ { "aoVal": "B", "content": "$$20$$ " } ], [ { "aoVal": "C", "content": "$$40$$ " } ], [ { "aoVal": "D", "content": "$$60$$ " } ] ]

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

[ "$$(50-10)\\div 2=20$$ " ]

[]

[ [ { "aoVal": "A", "content": "February " } ], [ { "aoVal": "B", "content": "April " } ], [ { "aoVal": "C", "content": "August " } ], [ { "aoVal": "D", "content": "November " } ] ]

[ "Overseas Competition->Knowledge Point->Word Problem Modules->Periodic Problems->Periodic Problems of Time->Period of Dates" ]

[ "omitted " ]

[]

[ [ { "aoVal": "A", "content": "$$30$$ and $$20$$ " } ], [ { "aoVal": "B", "content": "$$20$$ and $$30$$ " } ], [ { "aoVal": "C", "content": "$$15$$ and $$20$$ " } ], [ { "aoVal": "D", "content": "$$20$$ and $$15$$ " } ] ]

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

[ "According to the ratio $$2:3$$, we could know the total is $$5$$. Then the number of boys is $$50\\times\\frac{2}{5}=20$$. The number of girls is $$50\\times\\frac{3}{5}=30$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "Saturday " } ], [ { "aoVal": "B", "content": "Wednesday " } ], [ { "aoVal": "C", "content": "Thursday " } ], [ { "aoVal": "D", "content": "Tuesday " } ] ]

[ "Overseas Competition->Knowledge Point->Word Problem Modules->Periodic Problems" ]

[ "$29-2=27$ days later, it will be May $29$th. $27\\div7=3R6$, which means May $29$th is Wednesday. " ]

[]

[ [ { "aoVal": "A", "content": "$$39$$ " } ], [ { "aoVal": "B", "content": "$$42$$ " } ], [ { "aoVal": "C", "content": "$$44$$ " } ], [ { "aoVal": "D", "content": "$$47$$ " } ], [ { "aoVal": "E", "content": "None of the above " } ] ]

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

[ "We know that Tim has $$20+2=22$$ toys. The sum of Tom\\textquotesingle s and Tim\\textquotesingle s toys is $$20+22=42$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$46$$ " } ], [ { "aoVal": "B", "content": "$$52$$ " } ], [ { "aoVal": "C", "content": "$$91$$ " } ], [ { "aoVal": "D", "content": "$$104$$ " } ] ]

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

[ "$$143-39=104$$, $$104$$~$\\div$ $$2$$ = $$52$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$90.5$$ " } ], [ { "aoVal": "B", "content": "$$96.5$$ " } ], [ { "aoVal": "C", "content": "$$97.5$$ " } ], [ { "aoVal": "D", "content": "$$98.5$$ " } ], [ { "aoVal": "E", "content": "$$110.5$$ " } ] ]

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

[ "$1.3 \\times 0.75 = 97.5\\textbackslash\\%$. " ]

[]

[ [ { "aoVal": "A", "content": "$$27$$ " } ], [ { "aoVal": "B", "content": "$$28$$ " } ], [ { "aoVal": "C", "content": "$$29$$ " } ], [ { "aoVal": "D", "content": "$$30$$ " } ] ]

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

[ "A month has at most $$31$$ days, Hence, the greatest number of days in that month, after the first, is $$31 -1=30$$. Then, the next day is another $$1\\text{st}$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$1$$ " } ], [ { "aoVal": "B", "content": "$$2$$ " } ], [ { "aoVal": "C", "content": "$$3$$ " } ], [ { "aoVal": "D", "content": "$$4$$ " } ], [ { "aoVal": "E", "content": "$$5$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Word Problem Modules->Work Word Problems" ]

[ "The efficiency of the first faucet is $\\frac1{18}$ and that of the other two is $\\frac29$. Thus it takes $1\\div (\\frac1{18}+\\frac29\\times2)=2$ hours to fill the pool. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "Susan " } ], [ { "aoVal": "B", "content": "Ashley " } ], [ { "aoVal": "C", "content": "Sam " } ], [ { "aoVal": "D", "content": "Max " } ], [ { "aoVal": "E", "content": "Olivia " } ] ]

[ "Overseas Competition->Knowledge Point->Word Problem Modules" ]

[ "Susan: +1+1 Ashley: +1 Sam: +1 Max: +1 Olivia: +1 " ]

[]

[ [ { "aoVal": "A", "content": "$$4$$ " } ], [ { "aoVal": "B", "content": "$$6$$ " } ], [ { "aoVal": "C", "content": "$$24$$ " } ], [ { "aoVal": "D", "content": "$$124$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Word Problem Modules" ]

[ "As they are driving toward each other, the total speed is $$75+80=155$$ miles per hour. Then the time needed is $$620\\div155=4$$ hours. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "28th March " } ], [ { "aoVal": "B", "content": "29th March " } ], [ { "aoVal": "C", "content": "30th March " } ], [ { "aoVal": "D", "content": "31st March " } ], [ { "aoVal": "E", "content": "None of the above. " } ] ]

[ "Overseas Competition->Knowledge Point->Word Problem Modules->Periodic Problems" ]

[ "Draw the calendar out. " ]

[]

[ [ { "aoVal": "A", "content": "$$4$$ " } ], [ { "aoVal": "B", "content": "$$6$$ " } ], [ { "aoVal": "C", "content": "$$8$$ " } ], [ { "aoVal": "D", "content": "$$11$$ " } ], [ { "aoVal": "E", "content": "$$16$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Word Problem Modules->Inverse Operation Problems->Giving Half of a Whole" ]

[ "$2+2=4$ $4+4=8$ " ]

[]

[ [ { "aoVal": "A", "content": "red " } ], [ { "aoVal": "B", "content": "green " } ], [ { "aoVal": "C", "content": "yellow " } ] ]

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

[ "Since $$52\\div(3+2)=10r2$$, the $$52^{\\text{th}}$$ colored lantern is red. " ]

[]

[ [ { "aoVal": "A", "content": "$$36$$ " } ], [ { "aoVal": "B", "content": "$$48$$ " } ], [ { "aoVal": "C", "content": "$$60$$ " } ], [ { "aoVal": "D", "content": "$$72$$ " } ] ]

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

[ "If $$12$$ carrots are left after Jin eats $$\\frac{2}{3}$$ of today\\textquotesingle s carrots, then $$12$$ carrots are $$\\frac{1}{3}$$ of the carrots she started with today. So Jin began today with $$36$$ carrots. Since Jin ate $$\\frac{1}{2}$$ yesterday, she started with $$2\\times36$$ carrots. " ]