Question number nine. Which of the figures below has the greatest area? We look at each figure and we see that according to the grid in which it is drawn it consists of full squares like this or half squares like that. So we will count the number of full squares then add to that the number of half squares which combine into full squares. So for example our counting would go along the following lines. In picture A we have 1, 2, 3, 4, 5, 6, 7 and 8 of these full squares and then 1, 2, 3, 4 half squares so that together gives us 10 because the half squares count two of them for one of the full squares and so here the area of the first figure is 10. So we can keep going like that. In B we have 1, 2, 3, 4, 5, 6, 7, 8 again of the full squares and 1, 2, 3, 4, 5, 6 now of the half squares so that gives us a total of 11 of the full squares. So we have found a greater area. We keep going. In C we have 1, 2, 3, 4, 5, 6, 7, 8 again of the full squares and now 1, 2, 3, 4, 5, 6, 7 and 8 of the half squares so that gives us a total of 12 and in D we have 1, 2, 3, 4, 5, 6, 7, 8 of the full squares and 2 of the half squares so that gives us an area of 9 and in E we have 1, 2, 3, 4, 5, 6, 7, 8 of the full squares and 2 of the half squares again so the area is again 9. And we see that the figure presented in C has the greatest area according to the count we have here executed for each of the figures so the answer is C. Question number 10. A farmer who raises chickens has boxes which hold 6 eggs each and boxes which hold 12 eggs each. What is the least number of boxes he needs to store 66 eggs? Let's make a table to keep track of our information here and we will have a table here for the number of boxes that store 12 eggs and another table for six egg boxes but let's work with this table first so we have on the left column the number of boxes and in the right column we will have the number of eggs so if we have one box then we can fit 12 eggs with two boxes 24 with three boxes 36 with four boxes 48 with five boxes we have 60 eggs and if we have another box if we have six boxes we can store 72 eggs 12 more than 60 but that is more eggs than required so we need 5 12 egg boxes and what's left over is exactly going to fit in one box containing six eggs and one six egg box because after we use up 60 eggs we will have six left and exactly one more box is needed for that so the answer is six we have 5 12 egg boxes and one six egg box needed to store 66 eggs question number 11 each of the students in the class drew his or her pets in the picture to the right we have cats we have dogs and fish we know that two of the students each drew a dog and a fish three of the students each drew a cat and a dog the other students drew one animal each how many students are there in this class so just from reading the problem we know that there are at least five students but those five own multiple pets so we can erase or cross off exactly the number of pets that these five students all together have and then we will know that the remaining number of pets is allotted to one per student so let's do that first we know that two students each drew a dog and a fish so let's erase a dog and a fish that's for one student and let's do that again and so that's all the students that drew a dog and a fish now we know that three students drew a cat and a dog so we have to erase three cats and also three dogs and what we have left over here in the diagram after doing this are some animals that belong to one per child so we have one fish then the dog is second three four five six seven one per student so we have five students with multiple pets plus seven students with exactly one pet okay and if we add them together we have a total of 12 12 students in the class and that is our answer to the problem here we choose b b question number 12 during a party each of two identical cakes was divided into four equal pieces then each of these pieces was divided into three equal pieces after that each of the people at this party got a piece of the cake and three pieces were left over how many people were at this party okay so we can imagine if we were to draw a circular cake like so for example and then divide it into four pieces so that means cut it in half and then cut it in half again we would have four pieces already and then dividing each of our four pieces into two uh or rather three equal pieces we would obtain from each piece maybe a picture that looks something like this and that's one cake but we have two and we can count the number of pieces now so we would have one two three same thing in this big slice over here so three more three more that would give us total of six then three more for this slice a total of nine and three more here for this slice a total of 12 so we have a total of 12 slices per cake and then there are two cakes so we have a total of 24 slices to hand out and we know that not all of the pieces were eaten three are left over so 24 slices minus three slices is 21 slices that were actually consumed one per person and so we know that there have to be 21 people at the party and we choose answer c question 13 the sheet is folded along the thick line so here we have a picture of a sheet of paper with the grid three by five on each side of the red line and then we fold it along the red line the question is which letter will not be covered by a gray square so we have some gray squares on the left some letters on the right and we have to figure out how these squares will line up after the sheet is folded over and the way that will happen is we will have each shaded cell move across the red line from left to right exactly as many squares as it is away from the red line so this square here is exactly adjacent to the red line and it will move one space over after crossing the red line so we will have this square cover up a and likewise this square will cover up here that adjacent square and if we have any gray squares that are two away from the red line or three away from the red line they will be exactly that far away after moving over after folding the sheet of paper so this square here will fold over and cover up b this one will fold over and cover up d and then we do one more set of uh transformations like this this square here will come all the way to the end and cover up nothing really and the last one here three away will be three away on the other side and cover up c but we did not find any square opposite of e so here would have to be a gray square if e were to be covered up so that's our remaining letter that does not get covered by gray squares and that is the answer here e and choice e is the solution question number 14 there are 13 coins in john's pocket and each of them is either a five cent coin or a 10 cent coin which of the numbers below cannot be the total value of john's coins before we start thinking about the problem in detail what we can do right away is find the greatest possible value of john's coins and also the least possible value and we do that in the following way the greatest value would be if we pick as many of the 13 coins as we can to be 10 cent coins so we have to pick 12 12 times 10 cents plus we have to use a five cent coin that's stated in the problem plus one times five cents gives us a total of 120 cents plus five cents and that is 125 cents like that so we have a possible answer like that that would be answer e and now let's see what is the smallest possible value so we would obtain the smallest possible value if we assigned now in reverse 12 of john's coins to be five cent coins and one to be a 10 cent coin so let's calculate 12 times five cents plus one times 10 cents that gives us 12 times five is 60 so we have 60 cents here plus 10 cents and that gives us 70 cents and we see that 70 cents is also a possible answer but we do have a smaller value here in a for 60 cents so if we use as many five cent coins as we can and only one 10 cent coin we already obtain a value that is larger than this answer a here so that has to be the impossible amount it is not possible to have john's pockets be filled with coins in such a way that with 13 of them he has 60 cents if the coins are only five and 10 cent coins so the answer is a question number 15 erry chuck darius jack mark and tom were rolling a six-sided die each of the boys rolled the die once and each one got a different number erry got a result four times greater than chuck darius got a result twice as big as jack and three times as big as mark what number did tom roll so we have exactly six boys and each of them rolls a different number we have to keep track of their face values that they have obtained so starting with the first piece of information we see that erry got a result four times greater than chuck so what happens if chuck rolls a one if chuck has a one then erry has a four which is possible if chuck has a two then erry has an eight which is not possible so we stop here and we conclude that chuck has a one and erry has a four so the numbers one and four are already accounted for so moving on we see that darius got a result twice as big as jack and also three times as big as mark so let's start with mark and if mark has the lowest remaining number so a two is the remaining number that's smallest uh and darius has a result three times as large which is a six now if mark rolls a three then darius would have to roll a nine which is also not possible so we assign two to mark and six to darius now we also know that darius got a result twice as big as jack we are sure that his result is six so jack rolls a result half as large as ray as darius is so that's a three and so now we have the numbers one two three four and six accounted for and we don't know what tom rolls but that must be the remaining number so that tells us finally that all together the numbers one two three four and six are assigned so tom rolled a five the number that is left over so that is the answer here to the problem which is d5 question number 16 a squirrel which is misspelled here in the problem went through the maze gathering nuts and so i have a larger picture of the maze here this is where he enters uh each of these is a nut and he can go through these doors in various uh order but he must exit through here and we know that he can only go once through each door between the rooms of the maze so once he passes through a door he cannot go back through the same door the question is what is the greatest number of nuts it could have gathered so we see that most of this space here in the maze is open except here is a solid wall and to get to this room we have to pass through this room and then once we gather here this nut and that nut then we have to return and we cannot because we can only enter each door once and we may not return through it so this area here is kind of off limits because once the squirrel goes in he cannot come back now we look for another area like this and we see a solid wall here and to collect this nut in the lower left hand corner we would have to pass through this room but once we do that we cannot go back so this area here is also off limits and now we can proceed so collecting the first nut here we have one we can go to the right or we can go to the left so either one of those would be the second nut and then we either from above enter this way or from the left we collect that nut that would be nut number three and then we have then we have a choice uh the greatest number of nuts would be collected following a zigzag pattern like this remembering that we can only use one door one door uh ever once so then continuing on uh number four would be here number five would be here then we go down number six then to the left seven eight then down again nine to the right ten and then we have uh eleven so unfortunately there are some off limits areas that we may not go to and then we have to decide between which of these two nuts is collected but the most we can find is 11 without doubling back on ourselves which is not allowed so the answer here would be c 11

mathematical problems

logical solutions

area calculation

egg boxes

class tally

party attendees